EC-PROJECT WORKS

Know the Basic Concept in Computer Programming

2010-04-17T06:34:00.000-07:00

Computer programming is a problem solving and writing instructions to computers. Different kinds of computer programs does not always have the same programming language use to create them. Different languages have their different strengths and weaknesses, making some kinds of programs easier or more difficult to create. But the basic principles for creating computer programs remains the same. Skilled programmers can switch to a new programming language in a few hourswhile beginners should pick one language at a time.

Three Basic Sizes of Computer Programs:

1 - Trivial programs

* These are programs that a skilled programmer can write in less than two days of coding.

2 - Small Programs

* These are programs that one skilled programmer can write in less than one year of full time work.

3 - Large programs

* These are programs that require more than two to five man-years of labor, normally written by programming teams(which can exceed 1000 skilled workers).

Know the basic Networking & Basic Concepts computer networks

2010-04-17T06:30:00.000-07:00

Computer Network

A communication system for connecting computers/hosts

Why?

Better connectivity

Better communication

Better sharing of resources

Bring people together

Types of Computer Networks

Local Area Network(LAN)

@ Connects hosts within a relatively small geographical area

Wide Area Network(WAN)

@ Hosts may be widely dispersed

LAN and WAN: Comparison

* LAN

@Typical speeds: 10Mbps to 10Gbps

@Typical cost: 1 crore for a hundred node LAN(one-time cost)

* WAN

@Typical speeds: 64Kbps to 8Mbps

@Typical cost: 30 lakhs(recurring cost)

Circuit Switching

* A dedicated communication path is required between two stations.

@ The path follows a fixed sequence of intermediate links.

@ A logical channel gets defined on each physical link.

* In ciruit switching, three steps are required for communication:

1 - Connection establishment

2 - Data transfer

3 - Connection Termination

*Drawbacks:

@ Channel capacity gets dedicated during the entire duration of communication.

@ There is an initial delay.

Packet Switching

* Modern form of long-distance data communication.

@ Network resources are not dedicated.

@ A link can be shared.

* The basic technology has evolved over time.

* Data are transmitted in short packets(~Kbytes).

@ A longer message is broken up into smallerns.

@ The chunks are caleld packets.

@ Every packet contains a header.

*Packet switching is based on store-and-forward concept.

@ Each intermediate network node receives a whole packet.

@ Decides the route.

@ Forwards the packet along the selected route.

*Advantages:

@ Links can be shared; so link utilization is better.

@ Suitable for computer-generated traffic.

@ Buffering and data rate conversion can be performed easily.

@ Some packets may be given priority over others, if desired.

How are the packets transmitted?

* By using two alternative approches:

1 - Virtual circuits

2 - Datagram.

1- Virtual circuit approach

@ Similar in concept to circuit switcing. It's analogy is like a telephone system.

How it works?

1. Route is established a priori.

2. Packet forwarded from one node to the next using store-and-forward scheme.

3. Only the virtual circuit number need to be carried by a packet.

4. No dynamic routing decision is taken by the intermediate nodes.

2- Datagram Approach

@ Basic concept:

- No route is established beforehand.

- Each packet is transmitted as an independent entity.

- Does not maintain any history.

@ Analogy: Postal System.

@ Every intermediate node has to take routing decisions dynamically.

@ Problems:

1. Packets may be delivered out of order.

2. If a node crashes momentarily, all of its queued packets are lost.

3. Duplicate packets may also be generated.

@ Advantages:

1. Faster than virtual circuit for smaller number of packets.

2. More flexible.

3. Packets between two hosts may follow different paths.

7 Layers of the OSI(Open Systems Interconnection) Model and it's Function:

1- Application - Interface point for user applications.

2- Presentation - Provides data independence.

3- Session - Manages sessions.

4- Transport - End-to-end reliable data transfer, with error recovery and flow control.

5- Network - us to establish, maintain, and terminate connections.

6- Datalink - Reliable transfer of frames over a point-to-point link.

7- Physical - Transmit raw bit stream over a physical medium.

3 Internetworking Devices:

1 - Hub

* Extends the span of a single LAN.

2 - Bridge/Layer-2 Switch

* Connects two or more LAN's together.

3 - Router/Layer-3 Switch

* Connects any combination of LAN's and WAN's

* Works at network layer level.

Know About Basics of the Java Programming Language

2010-04-17T06:26:00.000-07:00

Java was developed by Sun Microsystems as an object-oriented language that is used for general purpose bus: programs and interactive world wide web based internet program. Java is a simple, object-oriented, robust, secure, portable, high performance, architecturally neutral, interpreted, multitreated, and dynamic language. It is used to developed applications such as games, office suites, and other various applications. It was originally developed for consumer devices such as TV set, internet appliances, etc. Originally named OAK but was renamed Java.

3 Categories of Java:

1 - Java Standard Edition(J2SE) - for desktop applications

2 - Java Enterprise Edition - for small footprint devices such as personal digital assitants(pda)

3 - Java micro edition - for mobile devices

4 Types of Java Application:

1. Mini Applications - Applets, the first type are essentially mini applications that run inside a java enabled browser such as netscape navigator and internet explorer.

2. Applications - The second type is your typical GUI(graphical user interface) such as the windows notepad, applications which does not require a web browser to execute it.

3. Line Application - A common line application that can be run from ms dos command prompt or a UNIX shell prompt.

4. Libraries - It is not an application rather it is more a collection of classes that belong to one package.

Java Application Development Cycle:

Edit - using a text editor or an Integrated Development Environment(IDE), Java source codes(.java files) are created and stored to a disk.
Compile - using a java compiler, java source codes are compiled into bytecodes and serve to a disk.
Load - a java bytecode are loaded onto memory or library that are used by the java application.
Bytecode-verify - the Java Video Machine(JVM) varifies that the bytecode does not violate any restrictions and compiles with the java applications.
Execute - the java application is then interpreted.

Sample java code:

public class first java program

{

public static void main(String arg [ ] );

{

System.out.println("My First Java Program");

}

Know How Do the Search Engines Work ?

2010-04-17T05:10:00.000-07:00

Search engines are specialized sites that stores the documents and url's of the websites submitted to it's database. Sites like yahoo and google uses different kinds of softwares(called wandere, crawler, robot, worm, spider) which searches the web and stores the documents and files in it's database. When you perform your search, search engines will find the keywords you input that match the files stored in it's database.

The following are the steps how search engines work:

1. Search engines uses softwares which searches the web and stores the documents and files in it's database.

2. Documents and url's are collected to the search engines database.

3. Indexing softwares extract the informtion from the database.

4. When you input the keywords to the search engine, the database is searched for the document or file that matches the keyword.

5. The search engine arranges the list of web pages that contains the file matching the keyword you input, starting from the highest ranked website.

Know the Top-Level Domain Names of domain name system

2010-04-17T05:07:00.001-07:00

The Top-Level Domain Names(TLD) are the domains that have the highest level in the internet. The following are some of the top-level domains:
.aero - air-transport industry
.asia - Asia-Pacific region .
.biz - business
.cat - Catalan
.com - commercial
.coop - cooperatives
.edu - educational
.gov - government
.info - information
.int - international organizations
.jobs - companies
.mil - U.S. military
.mobi - mobile devices
.name - individuals,
.net - network
.org - organization
.pro - professions
.tel - Internet communication services
.travel - travel and tourism industry related sites

What was The Difference Between The Internet and The World wide web

2010-04-17T04:03:00.000-07:00

Many people basically misunderstood that the internet is the same as the world wide web or web in short. Some say that the internet is a collection of web pages, that is not true. The internet is a global network of computers connected usually through telephone lines. ISP's(Internet Service Provider) provides the telephone lines and others stuff so people all over the world can share information with each other. Nobody owns the internet, it was first used by governements and universities to share information. On the other hand the web is one of the protocols that lets you link to web sites all over the world. When we say protocol it is a set of rules which is used by computers to communicate with each other across a network. A protocol is a convention or standard that controls or enables the connection.

Things You Should Know When Going on internet

2010-04-17T03:56:00.000-07:00

Almost over half of the worlds population is connected to the web. And millions of people are online every day, using emails, blogging, instant messaging, online social networking sites, and surfing web sites. But are there drawbacks or bad effects that it do? Yes it has. Below are brief descriptions of Things You Should Know When Going Online:

E-MAILS

What is it? These are messages that are electronically.

What's the appeal? You can send electronic messages faster than traditional mailing.

Whats the drawback? There are people who might send you spams or hoax e-mails, that may ask something about you or contain viruses. Hoax mails are often use to solicite money or your private information.

BLOG

What is it? Online diaries.

What's the appeal? You can share information about yourself to others publicly through the web.

What's the drawback? There are things that should not be publish on your blog like your true identity, because identity can be stolen. Also some employers consult an applicants blog and sometimes the applicant is not hired because of some information or comments on his blog.

INSTANT MESSAGING

What is it? Live text conversations between individuals.

What's the appeal? You can send messages to your friends right away without waiting long.

Whats's the drawback? It can often distract you especially when doing assignments or doing work in the office. This often lead to addiction.

ONLINE SOCIAL NETWORKING SITES

What is it? These are sites that allow people to create their own web pages with their pictures and vital information about them, often use for online friendship.

What's the appeal? You can find and add as many persons or friends as you want.

What's the drawback? Sometimes some people especially youths add those they haven't even seen face to face. Also there are online sexual predators that are waiting to bait someone. Youths and children are most vulnerable to these so parents should guide them.

WEB SITES

What is it? Collection of electronic pages created and maintained by individuals, organizations, groups, and companies.

What's the appeal? Almost all of the information that you need are contained in websites. You don't need to wait long to do research or assignments.

What's the drawback? There are plenty of sites that feature violence and pornography that are easy for people to stumble upon. Youth's and children are the most vulnerable and need guidance from their parents. There are certain softwares that blocks sites that block these thing like K9 WEB PROTECTION.

What is Blogging?

2010-04-17T03:47:00.000-07:00

Blogging is a short term for web log. The purpose of blogging is to share ideas with others through the web. The 2 most popular sites for blogging are blogger.com and wordpress.com, blogger.com is generally for amateurs because it's more easy to use and wordpress.com is generally for professionals. These sites provides some widgets and plugins for those who generally doesn't have any background in web programming. Themes and design can be downloaded if you don't have enough time to make one, just type the keyword on your search engine.

How to check rank of your website? You can check your sites rank inalexa.com. Ranking is very important in certain ways because it gives reliability. You can also check your pagerank in google.

How to web blog?

1. Through social bookmarking like posting your blog articles in digg.com etc...

2. Submit your websites address to search engines especially to google, yahoo, and bing.

3. Let the search engines know your updates, ping your website. One way is to go to pingomatic.com.

4. Post comments on social networking sites like friendster, facebook, and twitter that contains a link address of your site.

5. Create a lot of links.

Way's on How to make money online:

1. PPC - pay per click http://bidvertiser.com/.

2. PPR - pay per review - find advertisers and review their products - http://sworty.com/ *should have a paypal account.
3. PPV - pay per view - ex: youtube partnership program.

4.PPS - pay per signup.

When were Google, Yahoo, and Microsoft Started or founded?

2010-04-17T03:41:00.001-07:00

Google was co-founded by Larry Page and Sergey Brin while they were students at Stanford University and the company was first incorporated as a privately held company on September 7, 1998. Google's initial public offering took place on August 19, 2004, raising $1.67 billion, making it worth $23 billion.

In April 1994, "Jerry and David's Guide to the World Wide Web"

was renamed "Yahoo!", for which the official backronym is "YetAnother Hierarchical OfficiousOracle". The Yahoo! domain was created on January 18, 1995. Yang and Filo realized their website had massive business potential, and on March 1, 1995, Yahoo! was incorporated.

On April 12, 1996, Yahoo! had its initial public offering, raising $33.8 million,

by selling 2.6 million shares at $13 each.

MSN (originally The Microsoft Network) is a collection of internet sites and services provided by microsoft. It was launched on August 24, 1995.

Do you love reading books and other literature?

2010-04-17T03:40:00.000-07:00

Nowadays, computers, cellphones, mp3's, and other devices have become widely popular. Many people have become addicted on using these devices. Yes, there's no harm in using these devices but there should be limitations on how long you should use these. I observed that most people prefer doing other things rather than reading books and doing much important things. The reason they say is that " reading is not fun". Is that how you also view it? Do you know that reading helps reduce stress levels. Also try not to think that reading is boring, reading helps a person develop his skills and learn different things. Reading daily refreshes our mind.

Know how to Maximize your video streaming & watch the videos faster than before

2010-04-17T03:32:00.000-07:00

Having problems when streaming videos online? Try speedbit video accelerator, even if your internet speed slows down you can still stream videos faster without wasting plenty of time. I've tried this and it works great!!!

And offcourse many people watch youtube videos high in number. many people dont have enough speed to stream the videos. this software helps to get a little bit fast download of the videos from youtube and other video hosting sites.

Download it: www.videoaccelerator.com

Know How to keep Your Files Private in your computer or Pc

2010-04-17T03:27:00.000-07:00

Nowadays, almost every people have computers and portable media devices like flash drives and external hard drives. Sometimes when people borrows/uses one of your flash drives or computers, they tend to open your private files and some of us do not like it especially when they do not ask for our permission. SafeHouse Explorer is a free software that password protects your folders and files in your computer and other portable devices. When you simply cannot afford to let your confidential files fall into the wrong hands, this is the software you need. SafeHouse Explorer hides and protects your private documents and files, keeping them safe from intruders and anyone else who doesn’t have your permission to view them. It does this using passwords and super-strong 256-bit encryption. Visit the site:www.safehousesoftware.com/SafeHouseExplorer.aspx

Let us Know What is PIC MicroController?

2010-04-16T11:25:00.000-07:00

The PIC was developed as a peripheral controller:

PIC(Peripheral Interface Controller) is the IC which was developed to control peripheral devices, alleviating the load from the main CPU.
Compared to a human being, the brain is the main CPU and the PIC is equivalent to the autonomic nervous system.
The PIC is the small computer:

The PIC, like the CPU, has calculation functions and memory, and is controlled by the software.

However, the throughput and the memory capacity are low. Depending on the kind of PIC, the maximum clock operating frequency is about 20 MHz and the memory capacity (to write the program) is about 1K to 4K words.

The clock frequency determines the speed at which a program is read and an instruction is executed. The throughput cannot be judged with the clock frequency alone. It changes with the processor architecture. However within the same architecture, the one with the highest clock frequency has the highest throughput.

I use a 14-bit WORD for program memory capacity. An instruction is a word long. Program memory is measured in BYTES, one byte is 8 bits. The bit is the smallest unit, and can have the value of 1 or 0. The instruction word of the PIC16F84A is composed of 14 bits. 1K words is equal to 1 x 1,024 x 14 = 14,336 bits. To convert this to bytes divide it by 8 x 1024, (14,336 / 8 x 1024 = 1.75K bytes).

A memory capacity of 1G bytes = 1,024M bytes, 1M bytes = 1,024K bytes, 1K bytes = 1,024 bytes. 1K bytes is not equal to 1000 bytes. This is because the calculation is in binary (2 to the tenth power = 1,024).

When using the PIC it is possible to make the circuitry compact:

The PIC is convenient for making calculations. The memory, the input/output ports and so on are incorporated into the IC.
The efficiency and the functions are limited, but the PIC can do the job of many IC's with software. So, the circuit can be compact.

A PORTABLE WIRELESS EYE MOVEMENT- CONTROLLED HUMAN-COMPUTER INTERFACE FOR THE DISABLED

2010-01-03T06:04:00.000-08:00

Human-Computer Interface (HCI) has become an important area of research and development for the disabled. A portable wireless eye movementcontrolled Human-Computer Interface which can be used for the disabled who have motor paralysis and who cannot speak in multiple applications (such as communication aid and smart home applications) is described here.

This Interface consists of four major parts: (1) surface electrodes, (2) a twochannel amplifier, (3) a laptop (or a micro-processor), and (4) a ZigBee wireless module. Horizontal and vertical Electro-Oculography (EOG) signals are measured using five surface electrodes placed on the head.

The vertical electrodes are placed about 1.0 cm above the right eyebrow and 2.0 cm below the lower lid of the right eye, the horizontal electrodes are placed 2.0 cm lateral to the each side of outer canthi and the last electrode is placed on user's forehead to serve as a ground. The two-channel amplifier is comprised of instrumentation amplifiers, band-pass filters and shift circuits.

The EOG signals are sampled at the rate of 250Hz and then sent to a laptop or a micro-processor for signal processing which is based on the method of mathematical morphology to recognize the direction of eye movements and voluntary eye blink. The ZigBee wireless communication technology, which is proved to be reliable, low -power and cost-efficient, is used in the portable interface.

The subjects can control the wireless device or move a cursor over a screen by using this interface. The delay of this interface is less than 0.5s and errors are very limited. This interface provides a flexible method for the disabled to improve the life quality.

Problems on Filters and Amplifiers Log-Log Plots and Decibels Passive RC Filters Low-Pass Filter Approximate Integrater High-Pass Filter Approximate

2009-12-29T10:34:00.000-08:00

Problems

In the following circuit, the input signal is , and the components have been chosen such that and . The output is at the terminalsAB.
1. What is the transfer function H( )?
2. Find the current in the circuit.
3. What is the voltage drop across each element of the circuit?
4. Show algebraically that at any instant the potential difference around the circuit is zero.
5. At time , where N=0,2,4,... make a sketch showing the voltage across each element in the complex plane and show that the vector sum of the voltage drops is equal to the voltage supplied.
6. Write an expression for .
7. What is the limit of as goes to zero?
8. What is the limit of as goes to infinity?
9. What is the corner frequency?
10. What is at the corner frequency?
11. Make a sketch showing the characteristics of on a log-log plot. Label the slope of the curve where possible, the corner frequency, and the value of at the corner frequency.
12. Describe the high and low frequency behavior in dB/octave.
1. Draw a passive LCR low-pass filter and write down the transfer function of your four-terminal network.
2. Determine approximations to the transfer function and filter corner frequency(s).
3. Write the resonance frequency, , in terms of the corner frequency(s).
Write down the transfer function, , for the network shown below, and from it find:
1. the corner frequency(s) and
2. the value(s) of at the corner frequency(s).
3. Sketch and the voltage phase-shift as a function of .
4. What type of filter is this?
1. What is the transfer function for the following circuit?
2. Describe the frequency response at low and high frequencies?
3. Sketch the magnitude on a log-log plot. (label slopes, the corner frequency(s), and )
4. What is the signal attenuation for .
1. Show that the transfer function of a single-pole RC filter drops by 3 db at the corner frequency. This is often refereed to as the 3 db down-point.
2. Design a bandpass RC filter with the transfer function shown below. rad/s and rad/s are the 3 db down-points of the RC filter sections. Choose impedances so that the first section is not much affected by the loading of the second section.
3. Are and the 3 db down-points of a bandpass filter?

Amplifier Model,One-, Two- and Three-Pole Amplifier Models,Amplifier with Negative Feedback, problems

2009-12-29T10:32:00.000-08:00

Amplifier Model

Enough of filters. Lets now look at the simple amplifier model in figure 3.14

Figure 3.14: A simple amplifier model.

Notice that is with respect to ground while is a voltage difference. For a typical operational amplifier at . As , due to internal capacitance, ie. the amplifier behaves like a low-pass filter.

where is the transfer function of a low-pass filter.

One-, Two- and Three-Pole Amplifier Models

The simplest amplifier has

where is a real positive number (corner frequency). There is a single pole at on the negative real axis. This means that the impulse response will decay exponentially without ringing.

If we cascade multiple (three) single-pole amplifiers together

There will still be no oscillation to an impulse.

Amplifier with Negative Feedback

Feedback is a widely used technique to improve the characteristics of an imperfect amplifier. A generalized amplifier with negative voltage feedback is shown in figure 3.15.

Figure 3.15: Amplifier with negative voltage feedback.

The overall transfer function (closed-loop gain) can be written as

Realizing that

we may write

If . The transfer function is now independent of the amplifier gain. is the transfer function of a stable resistive network which means that will also be stable.

Sequential RC Filters,Passive RCL Filters,Series RCL Circuit

2009-12-29T10:23:00.000-08:00

Sequential RC Filters

Single-pole filters are rather limited (6 dB/octave slope). For better band-pass and band-reject filters we require more poles and zeros and thus more reactive circuit elements. A simple solution is to connect two or more single-pole RC filters in sequence. If filter draws no current from filter , the transfer function for the combined filter is

One way to do this is to choose a large impedance for . Hence has more poles and zeros than or .

The corner frequency for a high-pass filter is and the transfer function may be written as

For a low-pass filter the corner frequency is and

We may build a two-section low-pass filter by requiring and , as shown in figure 3.7, so that

Figure 3.7: Two-section low-pass filter.

A special case occurs when and we obtain one corner frequency but the slope of the filter is .

The results are similarly for a two-section high-pass filter.

A band-pass filter can be built from one low-pass filter and one high-pass filter, as shown in figure 3.8. The order of the filter sections does not matter as long as the impedance rule is obeyed.

Figure 3.8: Band-pass filter.

If we have only two straight regions and the band-pass frequency range degenerates to zero.

Example: Show that the magnitude of the transfer function

falls off -6 dB/octave at both the low- and high-frequency extremes.

For small (ie. and .

Thus

For large (ie. and .

And thus

Passive RCL Filters
Sequential RC filters always have real poles and hence smooth rounded corners. To improve these filters we introduce an inductor (which is good for high frequencies).
Series RCL Circuit
Consider the RCL circuits as shown in figure 3.9. Each has a low-frequency and high-frequency approximation. Considering the band-reject filter (figure 3.6d) we obtain for the transfer function

Figure 3.9: LCR filters: a) low-pass, b) high-pass, c) band-pass and d) band-reject.

The approximations are:
We notice a zero in the transfer function at . In the low-medium frequency range
for high-medium frequencies
Solving for the corner frequencies we have

Example: Sketch for the LCR circuit shown in figure 3.10 for the two conditions and . In each case, determine the values of at , , and , and label these points on the sketches.

Figure 3.10: LCR circuit with two components across the output.

The transfer function is
For small
For large : .
For the corner frequency: .
For ,
.
For ,
Figure 3.11 is a sketch of the transfer functions.

Figure 3.11: Sketch of the transfer functions for the above circuit.

Example:

Write an expression for the transfer function of the circuit shown in figure 3.12.

Figure 3.12: Circuit with components in parallel at the output.

What phase shift is introduced by this filter at very small and very large frequencies?

For large
For small

On a log-log scale, sketch and the phase shift as a function of .

For the corner frequency .
.

Figure 3.13: Transfer function and phase shift for the above circuit.

Complex Frequencies and the s-Plane,Poles and Zeros of H

2009-12-29T10:22:00.000-08:00

Complex Frequencies and the s-Plane

We will now consider s-plane techniques. Not because we will use them, but more to understand some of the common electronics terminology.

We can enhance the usefulness of the transfer function by transforming to a complex frequency. Define the complex variable such that

where is an inverse time constant.

Our exponential function now becomes

and we have a rich set of cases

So we can not only describe oscillatory behavior but transient responses as well.

Poles and Zeros of H

As before, consider expanding the transfer function as the ratio of two polynomials

If are the roots of and are the roots of we can write

where A is a real constant, are zeros of and are poles (infinities) of . Knowledge of and determines everywhere.

Lets now look at our two filter circuits. For a low-pass filter

and the filter has one pole at -1/(RC). For a high-pass filter

and it has one pole at -1/(RC) and one zero at 0. We refer to these two types of filters as single-pole filters.

There is a general rule that there must be at least as many reactive elements as poles. Based on the location of the poles we are able to deduce the general response properties of the filter. We will not do this here.

Example: If a transfer function has poles at and and a zero at (0,0), as shown in figure 3.5,

Figure 3.5: Poles and zeros in the complex plane.

sketch on the interval .
The transfer function is given by
Plugging in values for gives the table 3.1.

Table 3.1: Numerical values of the transfer function.

Figure 3.6: The transfer function from the table above.
If , what is the approximate value of at its highest point?
If then at is . Therefore

Passive RC Filters,Low-Pass Filter,Approximate Integrater,High-Pass Filter,Approximate Differentiator

2009-12-29T10:20:00.000-08:00

Passive RC Filters

We will now use our passive circuit elements to design some filter circuits. Inductors are not very good devices and hence we will concentrate on the use of resistors and capacitors.

Low-Pass Filter

Figure 3.2 shows one possible low-pass filter. The circuit is essentially a frequency-sensitive voltage divider. At high frequencies the output behaves as if it is shorted while at low frequencies the output appears as an open circuit.

Figure 3.2: RC low-pass filter

Mathematically we have

The approximations are

At the corner

Therefore

is the corner frequency of the filter. At the corner frequency

We say that the output is down by at the corner frequency.

Approximate Integrater

The low-pass filter acts as an approximate integrater at high frequencies. Assume

and integrate to obtain

The DC term is unimportant and may be dropped to obtain

We define

For a low-pass filter at high frequencies and

Thus the low-pass filter integrates at high frequencies but also attenuates the signal by 1/(RC).

High-Pass Filter

Figure 3.3 shows one possible high-pass filter. Mathematically we can write

Figure 3.3: RC high-pass filter.

At low and high frequencies

At the corner frequency we have

and therefore

Approximate Differentiator

A high-pass filter acts as an approximate differentiator at low frequencies. Consider

and differentiate to obtain

We define

Again the filter attenuates the signal by 1/(RC).

Example: Write the transfer function for the network in figure 3.4 and from it find:
the corner frequency,
Treating the circuit like a voltage divider, the transfer function is
For .
For large .
For the corner frequency .
Therefore

Figure 3.4: Four-terminal network without resistance.
the value of at the corner frequency.
At the corner frequency
How many degrees of phase shift are introduced by this network just below and just above the corner frequency?
Since is always real there is no phase shift.

Filter Circuits,Filters and Amplifiers,Log-Log Plots and Decibels

2009-12-29T10:17:00.002-08:00

Filter Circuits

Lets now apply our knowledge of AC circuits to some practical applications. We will first look at some simple passive filters (skipping active filters) and then an amplifier model. Again we will rely on complex variables.

Filters and Amplifiers

Simplistically, filters and amplifiers can be considered as four-terminal networks described by a transfer function as follows:

Figure 3.1 shows some ideal transfer functions. If is a real constant then we call the network an ideal amplifier. If is a heavyside step function we refer to the circuit as an ideal low-pass filter, and if an ideal high-pass filter.

Figure 3.1: a) Ideal amplifier, b) Ideal low-pass filter, c) ideal high-pass filter, d) low-pass filter and e) high-pass filter.

Log-Log Plots and Decibels

A log-log plot of a circuit's transfer function can be a useful qualitative tool to allow us to understand most of the important features of filter and amplifier circuits. The commonly used decibel unit will be defined. Although unappealing to the physicist this unit is still in wide spread use in electronics. Lets start.

If and are two powers, we define the decibel as

where we have used .

The decibel is a property of the network and not the signals. Hence we can make use of any convenient signals in defining decibel. If two constant equal amplitude sources, , are applied to a four-terminal network we may write

Therefore

Using the approximation procedure of a previous lecture

and if and are not too different

By definition an octave interval is when and hence

Likewise for a decade interval and

Four-Terminal Networks,Single-Term Approximations of H

2009-12-29T10:17:00.001-08:00

Four-Terminal Networks

Our previous resonance circuit is an example of a two-terminal network. A source is present but no load. A four-terminal network also has the source removed. The four-terminal network can be described by a transfer function. A generic four-terminal network is shown in figure 2.17. Such a circuit can be analyzed simply by considering it as a voltage divider. In general

Figure 2.17: Generic four-terminal network.

Single-Term Approximations of H

A circuit with a few components quickly leads to a complicated expression for the transfer function. It is often sufficient, and of course, easier to work with approximations to the transfer function.

Let be the ratio of two polynomials

where

If one term dominates in each polynomial

Thus

We define and plot

which is a straight line on a log-log plot with integer slope.

As an example, consider our RCL circuit:

where and .

At low frequencies and

On a log-log plot

which has a slope of +1.

At high frequencies and

On a log-log plot

which has a slope of -1.

Resonance and the Transfer Function,sinusoidal source in our series RCL circuit,Driven series RCL circuit,

2009-12-29T10:14:00.000-08:00

Resonance and the Transfer Function

Lets now consider putting a sinusoidal source in our series RCL circuit and consider the voltage across one of the circuit elements. The resistor for example in figure 2.7

Figure 2.7: Driven series RCL circuit.

Applying Ohm's law across the resistor gives (cf. a voltage divider)

where is known as the transfer function in the frequency domain. We have changed independent variables from to for convenience.

We define

contains all the information needed to characterize the circuit. In exponential form

where

and

has a maximum (resonance) given by . Or is the resonant frequency.

Example: Consider the series LCR circuit (figure 2.8) driven by a voltage phasor .

Figure 2.8: Driven series LCR circuit.
At an angular frequency such that and , write the current phasor in terms of and R.
v(t) is given by v(t)=Zi(t), where . At and
Therefore
At the instant when is exactly real, calculate the three phasors representing the voltage developed across the R, C, and L circuit elements.
v(t) is real at t=0. Thus
And
Also
Algebraically and with a sketch on the complex plane, show that the complex voltage sum around the closed loop is zero.
The three voltage phasors are
Around the closed loop . If this expression is zero at t=0 it will be zero for all time. Therefore .

Figure 2.9: Complex voltage sum around the closed loop of the driven LCR circuit.

Example: Sketch simplified versions of the circuit shown in figure 2.10 that would be valid at:

Figure 2.10: Example LCR circuit.
;
.

Figure 2.14: Example circuit for .
very low frequencies but not ;
When is small C and L are in parallel and
2L and 100R in parallel gives

Figure 2.12: Example circuit for very low frequencies but not .
very high frequencies but not ;
large (note: ).

Figure 2.13: Example circuit for very high frequencies but not .
.
.

Figure: Example circuit for .

Example: For the circuit shown in figure 2.15 plot as a function of frequency over the range rad/s to rad/s.

Figure 2.15: Example circuit with components in parallel.
The equivalent impedance for the three components in parallel is
Plugging in the numerical values gives
A table of values and its plot follows.

Table 2.1: Numerical values for example circuit.

Figure 2.16: Plot of for example circuit.

Sinusoidal Sources and Complex Impedance,Resistive Impedance,Inductive Impedance,Capacitive Impedance,Combined Impedances

2009-12-29T10:10:00.000-08:00

Sinusoidal Sources and Complex Impedance

We now consider current and voltage sources with time average values of zero. We will use periodic signals but the observation time could well be less than one period. Periodic signals are also useful in the sense that arbitrary signals can usually be expanded in terms of a Fourier series of periodic signals. Lets start with

Notice that I have now switched to lowercase symbols. Lowercase is generally used for AC quantities while uppercase is reserved for DC values.

Now is the time to get into complex notation since it will make our discussion easier and is encountered often in electronics. The above voltage and current signals can be written

To be cleaver we will define one EMF in the circuit to have . In other words, we will pick t=0 to be at the peak of one signal. The vector notation is used to remind us that complex numbers can be considered as vectors in the complex plane. Although not so common in physics, in electronics we refer to these vectors as phasors. Hence you should now review complex notation.

The presence of sinusoidal or in circuits will result in an inhomogeneous differential equation with a time-dependent source term. The solution will contain sinusoidal terms with the source frequency.

The extension of Ohm's law to AC circuits can be written as

where is the source frequency. is a generalized resistance referred to as the impedance.

We can cancel out the common time dependent factors to obtain

and hence you see the power of the complex notation. For a physically quantity we take the amplitude of the real signal

We will now examine each circuit element in turn with a voltage source to deduce its impedance.

Resistive Impedance

Kirchoff's voltage law for a voltage source and resistor is

Trying the solutions

leads to

The impedance is equal to the resistance, as expected.

Capacitive Impedance

Kirchoff's voltage law for a voltage source and capacitor is

Solving this equation gives

For DC circuits and hence . The capacitor acts like an open circuit (infinite resistance) in a DC circuit.

Inductive Impedance

Kirchoff's voltage law for a voltage source and inductor is

Solving this equation gives

For DC circuits and hence . There is no voltage drop across an inductor in DC (zero resistance).

Combined Impedances

We now know the impedance for each of our passive circuit elements:

The equivalent impedance of a circuit can be obtained by using the following rules for combining impedances.

In series

In parallel

Appealing to the complex notation we can write

where R is the resistance and X is called the reactance (always a function of ).

For a series combination of R, L and C

gives a special frequency, .

Example: An inductor and capacitor in parallel form the tank circuit shown in figure 2.5.

Figure 2.5: Tank circuit with inductor and capacitor.
Determine an expression for the impedance of this circuit.
The impedance of an inductor and a capacitor are
Combining the impedances in parallel gives
What is the impedance when ?
Substituting this value for into the above result gives

Example: The tank circuit schematic shown in figure 2.6 results from the use of a real inductor.

Figure 2.6: Tank circuit with real inductor
Find an expression for the impedance of this circuit.
The impedance of an inductor, capacitor and resistor are
The resistor and inductor are in series and this combination of impedance is in parallel with the capacitor. Combining the impedances gives
If L=1H, , and F, what is the impedance when ?
Substituting this value for into the above equation gives
Substituting the numerical values for the inductance, resistance and capacitance gives
What is the impedance when is very small?
What is the phase angle between the voltage and at resonance and at rad/s?
Rationalizing the denominator of the impedance gives
Taking the real and imaginary components gives
The inverse tangent of the ratio of the imaginary to real parts is
There is a resonance at
and hence
At rad/s.

Circuit Equations,RC Circuit,RL Circuit,LC Circuit,RCL Circuit

2009-12-29T10:03:00.000-08:00

Circuit Equations

Recall that voltage V is related to current I, via the passive DC circuit element resistance R, by Ohm's law V=IR. Analogously, the change in voltage and change in current are related to the current and voltage, via the passive AC circuit elements C and L, by

Applying the above three equations, along with Kirchoff's loop rule, to AC circuits results in a set of differential equations. These differential equations are linear with constant coefficients and can easily be solved for Q(t), I(t), and V(t). In general the solutions will consist of a transient response and a steady-state response. The transient response describes the return to equilibrium after the EMFs change suddenly. The steady-state response describes the long term behaviour when the circuit is driven by a sinusoidal source.

We will first consider the transient response. This will be one of the few times we consider non-oscillating AC behaviour. Since Ohm's law and Kirchoff's laws are linear we can use complex exponential signals and take real or imaginary parts in the end. This is not true for power, since it is non-linear (product of signals).

RC Circuit

Consider the resistor R and capacitance C in the circuit loop in figure 2.1. Notice that there is no source.

Figure 2.1: RC circuit.

We start with a differential version of Kirchoff's voltage law.

When applied to our circuit

where is the voltage drop across the capacitor and is the voltage drop across the resistor.

The change in the voltage drop across the capacitor is given by our previous expression,

The change in the voltage drop across the resistor can be obtained from Ohm's law

Substituting these changes in voltage into Kirchoff's equation gives

where the current due to the flow of charge on or off the capacitor is the same as through the resistor.

Now we need some initial conditions. Notice that although the capacitor behaves as an open circuit to DC, current must flow to charge or discharge the capacitor. Lets take the case where the capacitor is initially charged and then the circuit is closed and the charge is allowed to drain off the capacitor (eg. closing a switch). The resulting current will flow through the resistor.

Solving for the current we obtain

where is the initial current given by Ohm's law

Using a time dependent version of Ohm's law we can solve for the voltage across the resistor

where is the initial voltage across the capacitor and is the commonly defined time constant of the decay. You should also be able to solve for the voltage across the capacitor and charge on the capacitor.

For the case of an applying voltage being suddenly placed into the circuit (inserting a battery) the capacitor is initially not charged and the voltage across the capacitor is

In the first case, current and voltage exponentially decay away with time constant when the switch is closed. The charge flows off the capacitor and through the resistor. The energy initially stored in the capacitor is dissipated in the resistor.

In the second case the capacitor charges to a voltage until no current flows and hence the voltage drop across the resistor is zero. Energy from the battery is stored in the capacitor.

In both cases the characteristic RC time constant occurs. In general this is true of all resistor-capacitor combinations and will be important throughout the course.

RL Circuit

The response of the RL circuit, shown in figure 2.2, is similar to that of the RC circuit. There are however some significant differences.

Figure 2.2: RL circuit.

If a battery is inserted into the circuit the current raises quickly from zero to some finite value. The EMF generated in the inductor impedes the current flow until it is constant.

The expression for the current in the RL circuit is

where the time constant is now

The voltage across the resistor is an increasing exponential unlike the RC circuit in which the voltage across the resistor decreased exponentially. Likewise, the voltage across the inductor decreases with time while in the RC circuit the voltage across the capacitor increased with time.

There are other initial conditions we could work with in this circuit but these can now be worked out by the student.

LC Circuit

Lets now consider the LC circuit in figure 2.3 which has no resistive element.

Figure 2.3: LC circuit.

Kirchoff's voltage law applied to the loop is

Substituting our previous expressions for and gives

Using I = dQ/dt gives

The circuit equation is second-order in Q and one possible solution is

where is the initial charge on the capacitor and is an arbitrary phase constant. Considering the cases of , gives . The angular frequency is totally determined by the other parameters of the circuit

and is the natural or resonance frequency of the circuit.

We can also solve for the current and voltage across the capacitor

Notice that unlike the transient current and voltage responses of the RC and RL circuits, the LC circuit oscillates. The energy in the circuit is shared back and forth between the inductor and capacitor.

RCL Circuit

Lets now consider the case of all three passive circuit elements in series, as in figure 2.4.

Figure 2.4: RCL circuit.

Applying Kirchoff's law around the loop and using I=dQ/dt gives

The solution will not only depend on the initial conditions but also the relative values of R,C and L.

There are three possible solutions:

under damped ( ): ,
over damped ( ): , and
critically damped ( ): .

RCL circuits have a variety of properties, especially when driven by sinusoidal sources, which will not be investigated here. My aim is simply to expose you to the area and get on to more interesting topics. Driven oscillating systems also appear in other areas of physics and hopefully you will encounter them there. The detailed considerations lead to discussions on resonance and quality-factor Q.

Alternating Current Circuits,Inductance:The fundamental property of a capacitor,Faraday's law applied to an inductor

2009-12-29T10:01:00.000-08:00

Alternating Current Circuits

We now consider circuits where the currents and voltages may vary with time (V=V(t), I=I(t) (also Q=Q(t))). These lectures will concentrate on the special case in which the signals are periodic, with time average values of zero ( ). Circuits with these signals are referred to as alternating current (AC) circuits. In general signals will have both DC and AC properties ( ). We will concentrate only on the AC components and assume that the DC properties can be treated separately using the methods of the previous lectures.

The algebraic equations representing Kirchoff's laws for DC circuits will take the form of differential equations for AC circuits. So now is a good time to review your differential equations and complex number theory because we will use it.

Capacitance

The fundamental property of a capacitor is that it can store charge and hence electric field energy. The capacitance C between two appropriate surfaces is defined by

where V is the potential difference between the surfaces and Q is the magnitude of the charge distributed on either surface.

In terms of current, I = dQ/dt implies

In electronics we take (displacement current). In other words, the current flowing from or to the capacitor is taken to be equal to the displacement current through the capacitor. You should be able to show that capacitors add linearly when placed in parallel.

There are four principle functions of a capacitor in a circuit.

Since Q and can be stored a capacitor can be used as a (non-ideal) source of I and V.
Since a capacitor passes AC current but not DC current it can be used to connect parts of a circuit that must operate at different DC voltage levels.
A capacitor and resistor in series will limit current and hence smooth sharp edges in voltage signals.
Charging or discharging a capacitor with a constant current results in the capacitor having a voltage signal with a constant slope, ie. dV/dt = I/C = constant if Iis a constant.

Some capacitors (electrolytic) are asymmetric devices with a polarity that must be hooked-up in a definite way. You will learn this in the lab. The SI unit for capacitance is farad (F). The capacitance in a circuit is typically measured in F or pF. Non-ideal circuits will have stray capacitance, leakage currents and inductive coupling at high frequency. Although important in real circuit design we will slip over these nasties at this point.Capacitors can be obtained in various tolerance ratings from % to %. Because of dimensional changes, capacitors have a high temperature dependence of capacitance. A capacitor does not hold a charge indefinitely because the dielectric is never a perfect insulator. Capacitors are rated for leakage, the conduction through the dielectric, by the leakage resistance-capacitance product in . High temperature increases leakage.

Inductance

Faraday's law applied to an inductor states that a changing current induces a back EMF that opposes the change. Or

Where V is the voltage across the inductor and L is the inductance measured in henry (H). The more common units encountered in circuits are H and mH.

The inductance will tend to smooth sudden changes in current just as the capacitance smoothes sudden changes in voltage. Of course, if the current is constant there will be no induced EMF. So unlike the capacitor which behaves like an open-circuit in DC circuits, an inductor behaves like a short-circuit in DC circuits.

Applications using inductors are less common than those using capacitors, but inductors are very common in high frequency circuits. We will again skip over the unpleasantness - that non-ideal inductors have some resistance and some capacitance.

Inductors are never pure inductances because there is always some resistance in and some capacitance between the coil windings. When choosing an inductor (occasionally called a choke) for a specific application, it is necessary to consider the value of the inductance, the DC resistance of the coil, the current-carrying capacity of the coil windings, the breakdown voltage between the coil and the frame, and the frequency range in which the coil is designed to operate. To obtain a very high inductance it is necessary to have a coil of many turns. The inductance can be further increased by winding the coil on a closed-loop iron or ferrite core. To obtain as pure an inductance as possible, the DC resistance of the windings should be reduced to a minimum. This can be done by increasing the wire size, which of course, increases the size of the choke. The size of the wire also determines the current-handling capacity of the choke since the work done in forcing a current through a resistance is converted to heat in the resistance. Magnetic losses in an iron core also account for some heating, and this heating restricts any choke to a certain safe operating current. The windings of the coil must be insulated from the frame as well as from each other. Heavier insulation, which necessarily makes the choke more bulky, is used in applications where there will be a high voltage between the frame and the winding. The losses sustained in the iron core increases as the frequency increases. Large inductors, rated in henries, are used principally in power applications. The frequency in these circuits is relatively low, generally 60 Hz or low multiples thereof. In high-frequency circuits, such as those found in FM radios and television sets, very small inductors (of the order of microhenries) are frequently used.

Problems on Kirchoff's Laws Series and Parallel Combinations of Resistors Voltage Divider Current Divider Branch Current Method Loop Current Method

2009-12-29T09:59:00.000-08:00

Problems

Find the current in each resister in the circuit shown below. V, V, and . Hint: writing down the loop-current equations in terms of the symbols will give you most of the marks.
Determine the Thevenin equivalent circuit of the circuit shown below. Hint: determine and .
Consider the following circuit:
1. What is the Thevenin equivalent voltage?
2. What is the Thevenin equivalent resistance?
3. For a variable load resistance placed externally between the terminals A and B, plot the current through the load as a function of . Label the intercepts on both axes.
Sketch the current through a load resistance as a function of for the circuit shown below. Label both intercepts and the slope.
Find the voltage, , across the 3 load resistor for the circuit below by replacing the remaining circuit by its Thevenin equivalent. Hint: You can check your answer by direct analysis of the entire circuit.

Branch Current Method,Loop Current Method,Equivalent Circuits,Thevenin's and Norton's Theorem,Determination of Thevenin and Norton Circuit Elements

2009-12-29T09:56:00.001-08:00

Branch Current Method

Use both of Kirchoff's laws. But be aware that an arbitrary application of Kirchoff's two equations will not always yield an independent set of equations. But the following approach will probably work.

Label the current in each branch (do not worry about the direction of the actual current).
Use only interior loops and all but one node.
Solve the system of algebraic equations.
Loop Current Method
This method is also referred to as the mesh loop method. The independent current variables are taken to be the circulating current in each of the interior loops.
1. Label interior loop currents on a diagram.
2. Obtain expressions for the voltage changes around each interior loop.
3. Solve the system of algebraic equations.
Depending on the problem, it may ultimately be necessary to algebraically sum two loop currents in order to obtain the needed interior branch current for the final answer.
Lets consider the example of the Wheatstone bridge circuit shown in figure 1.6. We wish to calculate the currents around the loops. The three currents are identified as: the clockwise current around the large interior loop which includes the EMF, the clockwise current around the top equilateral triangle, and the clockwise current around the bottom equilateral triangle. The voltage loop expressions for the three current loops are

Figure 1.6: Loop method for the Wheatstone bridge circuit.
Collecting terms containing the same current gives
If the values for the parameters shown in the diagram are used, the current values can be found by solving the set of simultaneous equations to give
Moreover, if we number the individual currents through each resistor using the same scheme as we have for each component (current through is , has , etc.) and identify as the current out of the battery, then
These are the same currents that would be found using only Kirchoff's equations; however, here we had to handle only three simultaneous equations instead of six.
Example: Use the loop current method to determine the voltage developed across the terminals AB in the circuit shown in figure 1.7.

Figure 1.7: Example circuit for analysis using the loop current method.
Consider the clockwise current loop through the two resistors and the two potentials. Similarly consider the clockwise current around the other internal loop consisting of the three resistors and . Kirchoff's law gives
Solving the above two equations for the unknown loop currents and gives
The voltage across AB is given simply by
Equivalent Circuits
Equivalent circuits is often the hardest concept and most numerically intensive in the course. Learning them well could make a difference on your midterm exam. Look in several books until you find the explanation you understand best.
Since Ohm's law and Kirchoff's equations are linear, we can replace any DC circuit by a simplified circuit. Just like a combination of resistors and Ohm's law could give an equivalent resistor, a combination of circuit elements and Kirchoff's laws can give an equivalent circuit. Two possibilities are shown in figure
Thevenin's and Norton's Theorems
A Thevenin equivalent circuit contains an equivalent voltage source in series with an equivalent resistor . A Norton equivalent circuit contains an equivalent current source in parallel with an equivalent resistor .

Figure 1.8: Thevenin and Norton equivalent circuits.
Determination of Thevenin and Norton Circuit Elements
One approach to determine the equivalent circuits is:
Thevenin - calculate the open-circuit voltage .
Norton - calculate the short-circuit current between A and B; .
.
An alternative to step 2 is to short all voltage sources, open all current sources, and calculate the equivalent resistance remaining between A and B. We will use the latter approach whenever manageable. To see if you understand equivalent circuits so far, convince yourself that .
Solution: From Thevenin's theorem
According to Notron's theorem
Therefore
Lets now return to our Wheatstone bridge example shown in figure 1.6. We will calculate the current through by replacing the rest of the circuit by its Thevenin equivalent.
is removed and the open terminals are labeled . The polarity assigned is arbitrary as will be verified in the calculations.
The evaluation of is performed using Kirchoff's laws:
The result is V. The minus sign means only that the arbitrary choice of polarity was incorrect.

Figure 1.9: Thevenin's theorem applied to the Wheatstone bridge circuit.
The voltage source is shorted out and is calculated (figure 1.9):
Note that when the source is shorted out, the resistors that were in series ( and ; and ) become parallel combinations.
The network is assembled in series as shown in figure 1.9 and the current through is calculated.
Note that the numerical value of the current is the same as that in the preceding calculations, but the sign is opposite. This is simply due to the incorrect choice of polarity of for this calculation. In fact, the current flow is in the same direction in both examples, as would be expected.
Example: Find the Thevenin equivalent components and for the circuit in figure 1.10.

Figure 1.10: Example circuit for analysis using a Thevenin equivalent circuit.
Shorting the V's to find gives two resistors in parallel, which are in series with a third resistor:
The open circuit voltage gives . For the open circuit no current flows from the node joining the two resistors to A. A is thus at -V relative to this node. Around the interior loop (cf. voltage divider).
Therefore

Kirchoff's Laws,Conservation of energy,Conservation of charge,Series and Parallel Combinations of Resistors,Voltage Divider,Current Divider

2009-12-29T09:49:00.000-08:00

Kirchoff's Laws

The conservation of energy and conservation of charge when applied to electrical circuits are known as Kirchoff's laws.

Conservation of energy - zero algebraic sum of the voltage drops around a closed circuit loop (imaginary loop)

Conservation of charge - zero algebraic sum of the currents flowing into a point (total charge in, equals total charge out)

When applying these laws to solve for circuit unknowns we will find the following definitions useful:

an element is an impedance (resistance) or EMF (ideal voltage source or ideal current source),
a node is a point where three or more current-carrying elements are connected,
a branch is one element or several in series connecting two adjacent nodes, and
an interior loop is a circuit loop not subdivided by a branch.

Using these definitions we can apply Kirchoff's laws to a circuit to solve for the unknown quantities. The general procedure is:

define the currents and voltages on a diagram,
apply Kirchoff's laws to loops and nodes,
write down a set of linear algebraic equations, and
solve for the unknowns.

But before we look at general circuits lets consider how simple resistors add.

Series and Parallel Combinations of Resistors

Circuit elements are connected in series when a common current passes through each element. The equivalent resistance of a combination of resistors connected in series is given by summing the voltage drops across each resistor.

If , where are all the other resistors than ; the largest resistor wins.

Circuit elements are connected in parallel when a common voltage is applied across each element. The equivalent resistance of a combination of resistors connected in parallel is given by summing the current through each resistor

If , where are all the other resistors than ; the smallest resistor wins.

The following ``divider'' circuits are useful combinations of resistors. Believe it or not, they are a super useful concept that will often be used in one form or another; learn it.

Voltage Divider

Figure 1.3: Divider circuits: a) voltage divider and b) current divider.

Consider the voltage divider shown in figure 1.3a. The voltage across the input source is and the voltage across the output between terminals A and Bis . The output voltage from the voltage divider in thus

Example: Determine an expression for the voltage on the voltage divider in figure 1.4.

Figure 1.4: Example voltage divider.
We take the bottom line in figure 1.4 to be at ground and define the current flowing between and ground to be I. Ohm's law gives
Applying Kirchoff's voltage law for the input source gives
Combining the above two results and solving for leads to
Current Divider
Consider the current divider shown in figure 1.3b. The source current is divided between the two resistors and is given by . The voltage at the output is . The output current from the current divider is thus

Example: Determine an expression for the current through the resistor in the circuit shown in figure 1.5.

Figure 1.5: Example current divider.

The current I is divided amongst the three resistors and hence we use our expression for resistors in parallel
where is the common voltage across the three parallel resistors. The current through is thus
Now lets consider some general approaches to solving for unknowns in circuits.

Electromotive Force (EMF),EMF (battery, power supply, signal generator, etc.). We will deal with two types of EMFs,ideal voltage source,ideal current

2009-12-29T09:47:00.001-08:00

Electromotive Force (EMF)

Charge can flow in a material under the influence of an external electric field. Eventually the internal field due to the repositioned charge cancels the external electric field resulting in zero current flow. To maintain a potential drop (and flow of charge) requires an external energy source, ie. EMF (battery, power supply, signal generator, etc.). We will deal with two types of EMFs:

The ideal voltage source is able to maintain a constant voltage regardless of the current it must put out ( is possible).

The ideal current source is able to maintain a constant current regardless of the voltage needed ( is possible).

Because a battery cannot produce an infinite amount of current, a model for the behavior of a battery is to put an internal resistance in series with an ideal voltage source (zero resistance). Real-life EMFs can always be approximated with ideal EMFs and appropriate combinations of other circuit elements.

Ground

A voltage must always be measured relative to some reference point. It is proper to speak of the voltage across an electrical component but we often speak of voltage at a point. It is then assumed that the reference voltage point is ground.

Under strict definition, ground is the body of the earth. It is an infinite electrical sink. It can accept or supply any reasonable amount of charge without changing its electrical characteristics.

It is common, but not always necessary, to connect some part of the circuit to earth or ground, which is taken, for convenience and by convention, to be at zero volts. Frequently, a common (or reference) connection of the metal chassis of the instrument suffices. Sometimes there is a common reference voltage that is not at 0 V. Figure 1.2 show some common ways of depicting grounds on a circuit diagram.

Figure 1.2: Some grounding circuit diagram symbols: a) earth ground, b) chassis ground and c) common.

When neither a ground nor any other voltage reference is shown explicitly on a schematic diagram, it is useful for purposes of discussion to adopt the convention that the bottom line on a circuit is at zero potential.

The schematic diagram consists of idealized circuit elements,A two-terminal network is a circuit that has only two points of interest, say A and B

2009-12-29T09:46:00.000-08:00

The Schematic Diagram

The schematic diagram consists of idealized circuit elements each of which represents some property of the actual circuit. Figure 1.1 shows some common circuit elements encountered in DC circuits. A two-terminal network is a circuit that has only two points of interest, say A and B.

Figure 1.1: Common circuit elements encountered in DC circuits: a) ideal voltage source, b) ideal current source and c) resistor.

Direct current (DC) circuit analysis,Current,Potential Difference,Resistance and Ohm's Law,

2009-12-29T09:41:00.001-08:00

Basic Concepts

Direct current (DC) circuit analysis deals with constant currents and voltages, while alternating current (AC) circuit analysis deals with time-varying voltage and current signals whose time average values are zero. Circuits with time-average values of non-zero are also important and will be mentioned briefly in the section on filters. The DC circuit components considered in this course are the constant voltage source, constant current source, and resistor. Electronics also deals with charge Q, electric and magnetic fields, as well as, potential V. We will not be concerned with a detailed description of these quantities but will use approximation methods when dealing with them. Hence electronics can be considered as a more practical approach to these subjects. For the details look at your classical physics and quantum mechanics courses.

Current

The fundamental quantity in electronics is charge and at its basic level is due to the charge properties of the fundamental particles of matter. For all intensive purposes it is the electron (or lack of electrons) that matter. The role of the proton charge is negligible.

The aggregate motion of charges is called current I

where dq is the amount of positive charge crossing a specified surface in a time dt. Be aware that the charges in motion are actually negative electrons. Thus the electrons move in the opposite direction to the current flow.

The SI unit for current is the ampere (A). For most electronic circuits the ampere is a rather large unit so the mA unit is more common.

Potential Difference

It is often more convenient to consider the electrostatic potential V rather than electric field as the motivating influence for the flow of electric charge. The generalized vector properties of are usually unimportant. The change in potential dV across a distance in an electric field is

A positive charge will move from a higher to a lower potential. The potential is also referred to as the potential difference or, incorrectly, as just voltage:

Remember that current flowing in a conductor is due to a potential difference between its ends. Electrons move from a point of less positive potential to more positive potential and the current flows in the opposite direction.

The SI unit of potential difference is the volt (V).

Resistance and Ohm's Law

For most materials

where is the voltage across the object, I is the current through the object, and R is a proportionality constant called the resistance of the object. Resistance is a function of the material and shape of the object, and has SI units of ohms ( ). It is more common to find units of k and M . The inverse of resistivity is conductivity.

Resistor tolerances can be as bad as % for general-purpose resistors to % for ultra-precision resistors. Only wire-wound resistors are capable of ultra-precision applications.

The concept of current through and potential across are key to the understanding of and sounding intelligent about electronics.

Now comes the most useful visual tool of this course.

Positive feedback

2009-07-11T06:43:00.001-07:00

As we've seen, negative feedback is an incredibly useful principle when applied to operational amplifiers. It is what allows us to create all these practical circuits, being able to precisely set gains, rates, and other significant parameters with just a few changes of resistor values. Negative feedback makes all these circuits stable and self-correcting.

The basic principle of negative feedback is that the output tends to drive in a direction that creates a condition of equilibrium (balance). In an op-amp circuit with no feedback, there is no corrective mechanism, and the output voltage will saturate with the tiniest amount of differential voltage applied between the inputs. The result is a comparator:

With negative feedback (the output voltage "fed back" somehow to the inverting input), the circuit tends to prevent itself from driving the output to full saturation. Rather, the output voltage drives only as high or as low as needed to balance the two inputs' voltages:

Whether the output is directly fed back to the inverting (-) input or coupled through a set of components, the effect is the same: the extremely high differential voltage gain of the op-amp will be "tamed" and the circuit will respond according to the dictates of the feedback "loop" connecting output to inverting input.

Another type of feedback, namely positive feedback, also finds application in op-amp circuits. Unlike negative feedback, where the output voltage is "fed back" to the inverting (-) input, with positive feedback the output voltage is somehow routed back to the noninverting (+) input. In its simplest form, we could connect a straight piece of wire from output to noninverting input and see what happens:

The inverting input remains disconnected from the feedback loop, and is free to receive an external voltage. Let's see what happens if we ground the inverting input:

With the inverting input grounded (maintained at zero volts), the output voltage will be dictated by the magnitude and polarity of the voltage at the noninverting input. If that voltage happens to be positive, the op-amp will drive its output positive as well, feeding that positive voltage back to the noninverting input, which will result in full positive output saturation. On the other hand, if the voltage on the noninverting input happens to start out negative, the op-amp's output will drive in the negative direction, feeding back to the noninverting input and resulting in full negative saturation.

What we have here is a circuit whose output is bistable: stable in one of two states (saturated positive or saturated negative). Once it has reached one of those saturated states, it will tend to remain in that state, unchanging. What is necessary to get it to switch states is a voltage placed upon the inverting (-) input of the same polarity, but of a slightly greater magnitude. For example, if our circuit is saturated at an output voltage of +12 volts, it will take an input voltage at the inverting input of at least +12 volts to get the output to change. When it changes, it will saturate fully negative.

So, an op-amp with positive feedback tends to stay in whatever output state its already in. It "latches" between one of two states, saturated positive or saturated negative. Technically, this is known as hysteresis.

Hysteresis can be a useful property for a comparator circuit to have. As we've seen before, comparators can be used to produce a square wave from any sort of ramping waveform (sine wave, triangle wave, sawtooth wave, etc.) input. If the incoming AC waveform is noise-free (that is, a "pure" waveform), a simple comparator will work just fine.

However, if there exist any anomalies in the waveform such as harmonics or "spikes" which cause the voltage to rise and fall significantly within the timespan of a single cycle, a comparator's output might switch states unexpectedly:

Any time there is a transition through the reference voltage level, no matter how tiny that transition may be, the output of the comparator will switch states, producing a square wave with "glitches."

If we add a little positive feedback to the comparator circuit, we will introduce hysteresis into the output. This hysteresis will cause the output to remain in its current state unless the AC input voltage undergoes a major change in magnitude.

What this feedback resistor creates is a dual-reference for the comparator circuit. The voltage applied to the noninverting (+) input as a reference which to compare with the incoming AC voltage changes depending on the value of the op-amp's output voltage. When the op-amp output is saturated positive, the reference voltage at the noninverting input will be more positive than before. Conversely, when the op-amp output is saturated negative, the reference voltage at the noninverting input will be more negative than before. The result is easier to understand on a graph:

When the op-amp output is saturated positive, the upper reference voltage is in effect, and the output won't drop to a negative saturation level unless the AC input rises above that upper reference level. Conversely, when the op-amp output is saturated negative, the lower reference voltage is in effect, and the output won't rise to a positive saturation level unless the AC input drops below that lower reference level. The result is a clean square-wave output again, despite significant amounts of distortion in the AC input signal. In order for a "glitch" to cause the comparator to switch from one state to another, it would have to be at least as big (tall) as the difference between the upper and lower reference voltage levels, and at the right point in time to cross both those levels.

Another application of positive feedback in op-amp circuits is in the construction of oscillator circuits. An oscillator is a device that produces an alternating (AC), or at least pulsing, output voltage. Technically, it is known as an astable device: having no stable output state (no equilibrium whatsoever). Oscillators are very useful devices, and they are easily made with just an op-amp and a few external components.

When the output is saturated positive, the V_ref will be positive, and the capacitor will charge up in a positive direction. When V_ramp exceeds V_ref by the tiniest margin, the output will saturate negative, and the capacitor will charge in the opposite direction (polarity). Oscillation occurs because the positive feedback is instantaneous and the negative feedback is delayed (by means of an RC time constant). The frequency of this oscillator may be adjusted by varying the size of any component.

REVIEW:
Negative feedback creates a condition of equilibrium (balance). Positive feedback creates a condition of hysteresis (the tendency to "latch" in one of two extreme states).
An oscillator is a device producing an alternating or pulsing output voltage.

Practical considerations

2009-07-11T06:42:00.001-07:00

Real operational have some imperfections compared to an “ideal” model. A real device deviates from a perfect difference amplifier. One minus one may not be zero. It may have have an offset like an analog meter which is not zeroed. The inputs may draw current. The characteristics may drift with age and temperature. Gain may be reduced at high frequencies, and phase may shift from input to output. These imperfection may cause no noticable errors in some applications, unacceptable errors in others. In some cases these errors may be compensated for. Sometimes a higher quality, higher cost device is required.

Common-mode gain

As stated before, an ideal differential amplifier only amplifies the voltage difference between its two inputs. If the two inputs of a differential amplifier were to be shorted together (thus ensuring zero potential difference between them), there should be no change in output voltage for any amount of voltage applied between those two shorted inputs and ground:

Voltage that is common between either of the inputs and ground, as "V_common-mode" is in this case, is called common-mode voltage. As we vary this common voltage, the perfect differential amplifier's output voltage should hold absolutely steady (no change in output for any arbitrary change in common-mode input). This translates to a common-mode voltage gain of zero.

The operational amplifier, being a differential amplifier with high differential gain, would ideally have zero common-mode gain as well. In real life, however, this is not easily attained. Thus, common-mode voltages will invariably have some effect on the op-amp's output voltage.

The performance of a real op-amp in this regard is most commonly measured in terms of its differential voltage gain (how much it amplifies the difference between two input voltages) versus its common-mode voltage gain (how much it amplifies a common-mode voltage). The ratio of the former to the latter is called the common-mode rejection ratio, abbreviated as CMRR:

An ideal op-amp, with zero common-mode gain would have an infinite CMRR. Real op-amps have high CMRRs, the ubiquitous 741 having something around 70 dB, which works out to a little over 3,000 in terms of a ratio.

Because the common mode rejection ratio in a typical op-amp is so high, common-mode gain is usually not a great concern in circuits where the op-amp is being used with negative feedback. If the common-mode input voltage of an amplifier circuit were to suddenly change, thus producing a corresponding change in the output due to common-mode gain, that change in output would be quickly corrected as negative feedback and differential gain (being much greater than common-mode gain) worked to bring the system back to equilibrium. Sure enough, a change might be seen at the output, but it would be a lot smaller than what you might expect.

A consideration to keep in mind, though, is common-mode gain in differential op-amp circuits such as instrumentation amplifiers. Outside of the op-amp's sealed package and extremely high differential gain, we may find common-mode gain introduced by an imbalance of resistor values. To demonstrate this, we'll run a SPICE analysis on an instrumentation amplifier with inputs shorted together (no differential voltage), imposing a common-mode voltage to see what happens. First, we'll run the analysis showing the output voltage of a perfectly balanced circuit. We should expect to see no change in output voltage as the common-mode voltage changes:

instrumentation amplifier
v1 1 0 
rin1 1 0 9e12  
rjump 1 4 1e-12
rin2 4 0 9e12  
e1 3 0 1 2 999k
e2 6 0 4 5 999k
e3 9 0 8 7 999k
rload 9 0 10k  
r1 2 3 10k     
rgain 2 5 10k  
r2 5 6 10k     
r3 3 7 10k     
r4 7 9 10k     
r5 6 8 10k     
r6 8 0 10k     
.dc v1 0 10 1  
.print dc v(9) 
.end

v1            v(9)           
0.000E+00     0.000E+00
1.000E+00     1.355E-16
2.000E+00     2.710E-16
3.000E+00     0.000E+00   As you can see, the output voltage v(9)
4.000E+00     5.421E-16   hardly changes at all for a common-mode
5.000E+00     0.000E+00   input voltage (v1) that sweeps from 0
6.000E+00     0.000E+00   to 10 volts.
7.000E+00     0.000E+00
8.000E+00     1.084E-15
9.000E+00    -1.084E-15
1.000E+01     0.000E+00

Aside from very small deviations (actually due to quirks of SPICE rather than real behavior of the circuit), the output remains stable where it should be: at 0 volts, with zero input voltage differential. However, let's introduce a resistor imbalance in the circuit, increasing the value of R₅ from 10,000 Ω to 10,500 Ω, and see what happens (the netlist has been omitted for brevity -- the only thing altered is the value of R₅):

v1           v(9)           
0.000E+00     0.000E+00
1.000E+00    -2.439E-02
2.000E+00    -4.878E-02
3.000E+00    -7.317E-02   This time we see a significant variation
4.000E+00    -9.756E-02   (from 0 to 0.2439 volts) in output voltage
5.000E+00    -1.220E-01   as the common-mode input voltage sweeps
6.000E+00    -1.463E-01   from 0 to 10 volts as it did before.
7.000E+00    -1.707E-01
8.000E+00    -1.951E-01
9.000E+00    -2.195E-01
1.000E+01    -2.439E-01

Our input voltage differential is still zero volts, yet the output voltage changes significantly as the common-mode voltage is changed. This is indicative of a common-mode gain, something we're trying to avoid. More than that, its a common-mode gain of our own making, having nothing to do with imperfections in the op-amps themselves. With a much-tempered differential gain (actually equal to 3 in this particular circuit) and no negative feedback outside the circuit, this common-mode gain will go unchecked in an instrument signal application.

There is only one way to correct this common-mode gain, and that is to balance all the resistor values. When designing an instrumentation amplifier from discrete components (rather than purchasing one in an integrated package), it is wise to provide some means of making fine adjustments to at least one of the four resistors connected to the final op-amp to be able to "trim away" any such common-mode gain. Providing the means to "trim" the resistor network has additional benefits as well. Suppose that all resistor values are exactly as they should be, but a common-mode gain exists due to an imperfection in one of the op-amps. With the adjustment provision, the resistance could be trimmed to compensate for this unwanted gain.

One quirk of some op-amp models is that of output latch-up, usually caused by the common-mode input voltage exceeding allowable limits. If the common-mode voltage falls outside of the manufacturer's specified limits, the output may suddenly "latch" in the high mode (saturate at full output voltage). In JFET-input operational amplifiers, latch-up may occur if the common-mode input voltage approaches too closely to the negative power supply rail voltage. On the TL082 op-amp, for example, this occurs when the common-mode input voltage comes within about 0.7 volts of the negative power supply rail voltage. Such a situation may easily occur in a single-supply circuit, where the negative power supply rail is ground (0 volts), and the input signal is free to swing to 0 volts.

Latch-up may also be triggered by the common-mode input voltage exceeding power supply rail voltages, negative or positive. As a rule, you should never allow either input voltage to rise above the positive power supply rail voltage, or sink below the negative power supply rail voltage, even if the op-amp in question is protected against latch-up (as are the 741 and 1458 op-amp models). At the very least, the op-amp's behavior may become unpredictable. At worst, the kind of latch-up triggered by input voltages exceeding power supply voltages may be destructive to the op-amp.

While this problem may seem easy to avoid, its possibility is more likely than you might think. Consider the case of an operational amplifier circuit during power-up. If the circuit receives full input signal voltage before its own power supply has had time enough to charge the filter capacitors, the common-mode input voltage may easily exceed the power supply rail voltages for a short time. If the op-amp receives signal voltage from a circuit supplied by a different power source, and its own power source fails, the signal voltage(s) may exceed the power supply rail voltages for an indefinite amount of time!

Offset voltage

Another practical concern for op-amp performance is voltage offset. That is, effect of having the output voltage something other than zero volts when the two input terminals are shorted together. Remember that operational amplifiers are differential amplifiers above all: they're supposed to amplify the difference in voltage between the two input connections and nothing more. When that input voltage difference is exactly zero volts, we would (ideally) expect to have exactly zero volts present on the output. However, in the real world this rarely happens. Even if the op-amp in question has zero common-mode gain (infinite CMRR), the output voltage may not be at zero when both inputs are shorted together. This deviation from zero is called offset.

A perfect op-amp would output exactly zero volts with both its inputs shorted together and grounded. However, most op-amps off the shelf will drive their outputs to a saturated level, either negative or positive. In the example shown above, the output voltage is saturated at a value of positive 14.7 volts, just a bit less than +V (+15 volts) due to the positive saturation limit of this particular op-amp. Because the offset in this op-amp is driving the output to a completely saturated point, there's no way of telling how much voltage offset is present at the output. If the +V/-V split power supply was of a high enough voltage, who knows, maybe the output would be several hundred volts one way or the other due to the effects of offset!

For this reason, offset voltage is usually expressed in terms of the equivalent amount of input voltage differential producing this effect. In other words, we imagine that the op-amp is perfect (no offset whatsoever), and a small voltage is being applied in series with one of the inputs to force the output voltage one way or the other away from zero. Being that op-amp differential gains are so high, the figure for "input offset voltage" doesn't have to be much to account for what we see with shorted inputs:

Offset voltage will tend to introduce slight errors in any op-amp circuit. So how do we compensate for it? Unlike common-mode gain, there are usually provisions made by the manufacturer to trim the offset of a packaged op-amp. Usually, two extra terminals on the op-amp package are reserved for connecting an external "trim" potentiometer. These connection points are labeled offset null and are used in this general way:

On single op-amps such as the 741 and 3130, the offset null connection points are pins 1 and 5 on the 8-pin DIP package. Other models of op-amp may have the offset null connections located on different pins, and/or require a slightly difference configuration of trim potentiometer connection. Some op-amps don't provide offset null pins at all! Consult the manufacturer's specifications for details.

Bias current

Inputs on an op-amp have extremely high input impedances. That is, the input currents entering or exiting an op-amp's two input signal connections are extremely small. For most purposes of op-amp circuit analysis, we treat them as though they don't exist at all. We analyze the circuit as though there was absolutely zero current entering or exiting the input connections.

This idyllic picture, however, is not entirely true. Op-amps, especially those op-amps with bipolar transistor inputs, have to have some amount of current through their input connections in order for their internal circuits to be properly biased. These currents, logically, are called bias currents. Under certain conditions, op-amp bias currents may be problematic. The following circuit illustrates one of those problem conditions:

At first glance, we see no apparent problems with this circuit. A thermocouple, generating a small voltage proportional to temperature (actually, a voltage proportional to the difference in temperature between the measurement junction and the "reference" junction formed when the alloy thermocouple wires connect with the copper wires leading to the op-amp) drives the op-amp either positive or negative. In other words, this is a kind of comparator circuit, comparing the temperature between the end thermocouple junction and the reference junction (near the op-amp). The problem is this: the wire loop formed by the thermocouple does not provide a path for both input bias currents, because both bias currents are trying to go the same way (either into the op-amp or out of it).

In order for this circuit to work properly, we must ground one of the input wires, thus providing a path to (or from) ground for both currents:

Not necessarily an obvious problem, but a very real one!

Another way input bias currents may cause trouble is by dropping unwanted voltages across circuit resistances. Take this circuit for example:

We expect a voltage follower circuit such as the one above to reproduce the input voltage precisely at the output. But what about the resistance in series with the input voltage source? If there is any bias current through the noninverting (+) input at all, it will drop some voltage across R_in, thus making the voltage at the noninverting input unequal to the actual V_in value. Bias currents are usually in the microamp range, so the voltage drop across R_in won't be very much, unless R_in is very large. One example of an application where the input resistance (R_in) would be very large is that of pH probe electrodes, where one electrode contains an ion-permeable glass barrier (a very poor conductor, with millions of Ω of resistance).

If we were actually building an op-amp circuit for pH electrode voltage measurement, we'd probably want to use a FET or MOSFET (IGFET) input op-amp instead of one built with bipolar transistors (for less input bias current). But even then, what slight bias currents may remain can cause measurement errors to occur, so we have to find some way to mitigate them through good design.

One way to do so is based on the assumption that the two input bias currents will be the same. In reality, they are often close to being the same, the difference between them referred to as the input offset current. If they are the same, then we should be able to cancel out the effects of input resistance voltage drop by inserting an equal amount of resistance in series with the other input, like this:

With the additional resistance added to the circuit, the output voltage will be closer to V_in than before, even if there is some offset between the two input currents.

For both inverting and noninverting amplifier circuits, the bias current compensating resistor is placed in series with the noninverting (+) input to compensate for bias current voltage drops in the divider network:

In either case, the compensating resistor value is determined by calculating the parallel resistance value of R₁ and R₂. Why is the value equal to the parallel equivalent of R₁ and R₂? When using the Superposition Theorem to figure how much voltage drop will be produced by the inverting (-) input's bias current, we treat the bias current as though it were coming from a current source inside the op-amp and short-circuit all voltage sources (V_in and V_out). This gives two parallel paths for bias current (through R₁ and through R₂, both to ground). We want to duplicate the bias current's effect on the noninverting (+) input, so the resistor value we choose to insert in series with that input needs to be equal to R₁ in parallel with R₂.

A related problem, occasionally experienced by students just learning to build operational amplifier circuits, is caused by a lack of a common ground connection to the power supply. It is imperative to proper op-amp function that some terminal of the DC power supply be common to the "ground" connection of the input signal(s). This provides a complete path for the bias currents, feedback current(s), and for the load (output) current. Take this circuit illustration, for instance, showing a properly grounded power supply:

Here, arrows denote the path of electron flow through the power supply batteries, both for powering the op-amp's internal circuitry (the "potentiometer" inside of it that controls output voltage), and for powering the feedback loop of resistors R₁ and R₂. Suppose, however, that the ground connection for this "split" DC power supply were to be removed. The effect of doing this is profound:

No electrons may flow in or out of the op-amp's output terminal, because the pathway to the power supply is a "dead end." Thus, no electrons flow through the ground connection to the left of R₁, neither through the feedback loop. This effectively renders the op-amp useless: it can neither sustain current through the feedback loop, nor through a grounded load, since there is no connection from any point of the power supply to ground.

The bias currents are also stopped, because they rely on a path to the power supply and back to the input source through ground. The following diagram shows the bias currents (only), as they go through the input terminals of the op-amp, through the base terminals of the input transistors, and eventually through the power supply terminal(s) and back to ground.

Without a ground reference on the power supply, the bias currents will have no complete path for a circuit, and they will halt. Since bipolar junction transistors are current-controlled devices, this renders the input stage of the op-amp useless as well, as both input transistors will be forced into cutoff by the complete lack of base current.

REVIEW:
Op-amp inputs usually conduct very small currents, called bias currents, needed to properly bias the first transistor amplifier stage internal to the op-amps' circuitry. Bias currents are small (in the microamp range), but large enough to cause problems in some applications.
Bias currents in both inputs must have paths to flow to either one of the power supply "rails" or to ground. It is not enough to just have a conductive path from one input to the other.
To cancel any offset voltages caused by bias current flowing through resistances, just add an equivalent resistance in series with the other op-amp input (called a compensating resistor). This corrective measure is based on the assumption that the two input bias currents will be equal.
Any inequality between bias currents in an op-amp constitutes what is called an input offset current.
It is essential for proper op-amp operation that there be a ground reference on some terminal of the power supply, to form complete paths for bias currents, feedback current(s), and load current.

Drift

Being semiconductor devices, op-amps are subject to slight changes in behavior with changes in operating temperature. Any changes in op-amp performance with temperature fall under the category of op-amp drift. Drift parameters can be specified for bias currents, offset voltage, and the like. Consult the manufacturer's data sheet for specifics on any particular op-amp.

To minimize op-amp drift, we can select an op-amp made to have minimum drift, and/or we can do our best to keep the operating temperature as stable as possible. The latter action may involve providing some form of temperature control for the inside of the equipment housing the op-amp(s). This is not as strange as it may first seem. Laboratory-standard precision voltage reference generators, for example, are sometimes known to employ "ovens" for keeping their sensitive components (such as zener diodes) at constant temperatures. If extremely high accuracy is desired over the usual factors of cost and flexibility, this may be an option worth looking at.

REVIEW:
Op-amps, being semiconductor devices, are susceptible to variations in temperature. Any variations in amplifier performance resulting from changes in temperature is known as drift. Drift is best minimized with environmental temperature control.

Frequency response

With their incredibly high differential voltage gains, op-amps are prime candidates for a phenomenon known as feedback oscillation. You've probably heard the equivalent audio effect when the volume (gain) on a public-address or other microphone amplifier system is turned too high: that high pitched squeal resulting from the sound waveform "feeding back" through the microphone to be amplified again. An op-amp circuit can manifest this same effect, with the feedback happening electrically rather than audibly.

A case example of this is seen in the 3130 op-amp, if it is connected as a voltage follower with the bare minimum of wiring connections (the two inputs, output, and the power supply connections). The output of this op-amp will self-oscillate due to its high gain, no matter what the input voltage. To combat this, a small compensation capacitor must be connected to two specially-provided terminals on the op-amp. The capacitor provides a high-impedance path for negative feedback to occur within the op-amp's circuitry, thus decreasing the AC gain and inhibiting unwanted oscillations. If the op-amp is being used to amplify high-frequency signals, this compensation capacitor may not be needed, but it is absolutely essential for DC or low-frequency AC signal operation.

Some op-amps, such as the model 741, have a compensation capacitor built in to minimize the need for external components. This improved simplicity is not without a cost: due to that capacitor's presence inside the op-amp, the negative feedback tends to get stronger as the operating frequency increases (that capacitor's reactance decreases with higher frequencies). As a result, the op-amp's differential voltage gain decreases as frequency goes up: it becomes a less effective amplifier at higher frequencies.

Op-amp manufacturers will publish the frequency response curves for their products. Since a sufficiently high differential gain is absolutely essential to good feedback operation in op-amp circuits, the gain/frequency response of an op-amp effectively limits its "bandwidth" of operation. The circuit designer must take this into account if good performance is to be maintained over the required range of signal frequencies.

REVIEW:
Due to capacitances within op-amps, their differential voltage gain tends to decrease as the input frequency increases. Frequency response curves for op-amps are available from the manufacturer.

Input to output phase shift

In order to illustrate the phase shift from input to output of an operational amplifier (op-amp), the OPA227 was tested in our lab. The OPA227 was constructed in a typical non-inverting configuration (Figure below).

OPA227 Non-inverting stage

The circuit configuration calls for a signal gain of ≅34 V/V or ≅50 dB. The input excitation at Vsrc was set to 10 mVp, and three frequencies of interest: 2.2 kHz, 22 kHz, and 220 MHz. The OPA227's open loop gain and phase curve vs. frequency is shown in Figure below.

A_V and Φ vs. Frequency plot

To help predict the closed loop phase shift from input to output, we can use the open loop gain and phase curve. Since the circuit configuration calls for a closed loop gain, or 1/β, of ≅50 dB, the closed loop gain curve intersects the open loop gain curve at approximately 22 kHz. After this intersection, the closed loop gain curve rolls off at the typical 20 dB/decade for voltage feedback amplifiers, and follows the open loop gain curve.

What is actually at work here is the negative feedback from the closed loop modifies the open loop response. Closing the loop with negative feedback establishes a closed loop pole at 22 kHz. Much like the dominant pole in the open loop phase curve, we will expect phase shift in the closed loop response. How much phase shift will we see?

Since the new pole is now at 22 kHz, this is also the -3 dB point as the pole starts to roll off the closed loop again at 20 dB per decade as stated earlier. As with any pole in basic control theory, phase shift starts to occur one decade in frequency before the pole, and ends at 90^o of phase shift one decade in frequency after the pole. So what does this predict for the closed loop response in our circuit?

This will predict phase shift starting at 2.2 kHz, with 45^o of phase shift at the -3 dB point of 22 kHz, and finally ending with 90^o of phase shift at 220 kHz. The three Figures shown below are oscilloscope captures at the frequencies of interest for our OPA227 circuit. Figure below is set for 2.2 kHz, and no noticeable phase shift is present. Figure below is set for 220 kHz, and ≅45^o of phase shift is recorded. Finally, Figure below is set for 220 MHz, and the expected ≅90^o of phase shift is recorded. The scope plots were captured using a LeCroy 44x Wavesurfer. The final scope plot used a x1 probe with the trigger set to HF reject.

OPA227 Av=50dB @ 2.2 kHz

OPA227 Av=50dB @ 22 kHz

OPA227 Av=50dB @ 220 kHz

Operational amplifier models

2009-07-11T06:40:00.000-07:00

While mention of operational amplifiers typically provokes visions of semiconductor devices built as integrated circuits on a miniature silicon chip, the first op-amps were actually vacuum tube circuits. The first commercial, general purpose operational amplifier was manufactured by the George A. Philbrick Researches, Incorporated, in 1952. Designated the K2-W, it was built around two twin-triode tubes mounted in an assembly with an octal (8-pin) socket for easy installation and servicing in electronic equipment chassis of that era. The assembly looked something like this:

The schematic diagram shows the two tubes, along with ten resistors and two capacitors, a fairly simple circuit design even by 1952 standards:

In case you're unfamiliar with the operation of vacuum tubes, they operate similarly to N-channel depletion-type IGFET transistors: that is, they conduct more current when the control grid (the dashed line) is made more positive with respect to the cathode (the bent line near the bottom of the tube symbol), and conduct less current when the control grid is made less positive (or more negative) than the cathode. The twin triode tube on the left functions as a differential pair, converting the differential inputs (inverting and noninverting input voltage signals) into a single, amplified voltage signal which is then fed to the control grid of the left triode of the second triode pair through a voltage divider (1 MΩ -- 2.2 MΩ). That triode amplifies and inverts the output of the differential pair for a larger voltage gain, then the amplified signal is coupled to the second triode of the same dual-triode tube in a noninverting amplifier configuration for a larger current gain. The two neon "glow tubes" act as voltage regulators, similar to the behavior of semiconductor zener diodes, to provide a bias voltage in the coupling between the two single-ended amplifier triodes.

With a dual-supply voltage of +300/-300 volts, this op-amp could only swing its output +/- 50 volts, which is very poor by today's standards. It had an open-loop voltage gain of 15,000 to 20,000, a slew rate of +/- 12 volts/µsecond, a maximum output current of 1 mA, a quiescent power consumption of over 3 watts (not including power for the tubes' filaments!), and cost about $24 in 1952 dollars. Better performance could have been attained using a more sophisticated circuit design, but only at the expense of greater power consumption, greater cost, and decreased reliability.

With the advent of solid-state transistors, op-amps with far less quiescent power consumption and increased reliability became feasible, but many of the other performance parameters remained about the same. Take for instance Philbrick's model P55A, a general-purpose solid-state op-amp circa 1966. The P55A sported an open-loop gain of 40,000, a slew rate of 1.5 volt/µsecond and an output swing of +/- 11 volts (at a power supply voltage of +/- 15 volts), a maximum output current of 2.2 mA, and a cost of $49 (or about $21 for the "utility grade" version). The P55A, as well as other op-amps in Philbrick's lineup of the time, was of discrete-component construction, its constituent transistors, resistors, and capacitors housed in a solid "brick" resembling a large integrated circuit package.

It isn't very difficult to build a crude operational amplifier using discrete components. A schematic of one such circuit is shown in Figure below.

A simple operational amplifier made from discrete components.

While its performance is rather dismal by modern standards, it demonstrates that complexity is not necessary to create a minimally functional op-amp. Transistors Q₃ and Q₄ form the heart of another differential pair circuit, the semiconductor equivalent of the first triode tube in the K2-W schematic. As it was in the vacuum tube circuit, the purpose of a differential pair is to amplify and convert a differential voltage between the two input terminals to a single-ended output voltage.

With the advent of integrated-circuit (IC) technology, op-amp designs experienced a dramatic increase in performance, reliability, density, and economy. Between the years of 1964 and 1968, the Fairchild corporation introduced three models of IC op-amps: the 702, 709, and the still-popular 741. While the 741 is now considered outdated in terms of performance, it is still a favorite among hobbyists for its simplicity and fault tolerance (short-circuit protection on the output, for instance). Personal experience abusing many 741 op-amps has led me to the conclusion that it is a hard chip to kill . . .

The internal schematic diagram for a model 741 op-amp is shown in Figure below.

Schematic diagram of a model 741 op-amp.

By integrated circuit standards, the 741 is a very simple device: an example of small-scale integration, or SSI technology. It would be no small matter to build this circuit using discrete components, so you can see the advantages of even the most primitive integrated circuit technology over discrete components where high parts counts are involved.

For the hobbyist, student, or engineer desiring greater performance, there are literally hundreds of op-amp models to choose from. Many sell for less than a dollar apiece, even retail! Special-purpose instrumentation and radio-frequency (RF) op-amps may be quite a bit more expensive. In this section I will showcase several popular and affordable op-amps, comparing and contrasting their performance specifications. The venerable 741 is included as a "benchmark" for comparison, although it is, as I said before, considered an obsolete design.

Widely used operational amplifiers

Model	Devices/ package	Power supply	Bandwidth	Bias current	Slew rate	Output current
number	(count)	(V)	(MHz)	(nA)	(V/µS)	(mA)
TL082	2	12 / 36	4	8	13	17
LM301A	1	10 / 36	1	250	0.5	25
LM318	1	10 / 40	15	500	70	20
LM324	4	3 / 32	1	45	0.25	20
LF353	2	12 / 36	4	8	13	20
LF356	1	10 / 36	5	8	12	25
LF411	1	10 / 36	4	20	15	25
741C	1	10 / 36	1	500	0.5	25
LM833	2	10 / 36	15	1050	7	40
LM1458	2	6 / 36	1	800	10	45
CA3130	1	5 / 16	15	0.05	10	20

Listed in Table above are but a few of the low-cost operational amplifier models widely available from electronics suppliers. Most of them are available through retail supply stores such as Radio Shack. All are under $1.00 cost direct from the manufacturer (year 2001 prices). As you can see, there is substantial variation in performance between some of these units. Take for instance the parameter of input bias current: the CA3130 wins the prize for lowest, at 0.05 nA (or 50 pA), and the LM833 has the highest at slightly over 1 µA. The model CA3130 achieves its incredibly low bias current through the use of MOSFET transistors in its input stage. One manufacturer advertises the 3130's input impedance as 1.5 tera-ohms, or 1.5 x 10¹² Ω! Other op-amps shown here with low bias current figures use JFET input transistors, while the high bias current models use bipolar input transistors.

While the 741 is specified in many electronic project schematics and showcased in many textbooks, its performance has long been surpassed by other designs in every measure. Even some designs originally based on the 741 have been improved over the years to far surpass original design specifications. One such example is the model 1458, two op-amps in an 8-pin DIP package, which at one time had the exact same performance specifications as the single 741. In its latest incarnation it boasts a wider power supply voltage range, a slew rate 50 times as great, and almost twice the output current capability of a 741, while still retaining the output short-circuit protection feature of the 741. Op-amps with JFET and MOSFET input transistors far exceed the 741's performance in terms of bias current, and generally manage to beat the 741 in terms of bandwidth and slew rate as well.

My own personal recommendations for op-amps are as such: when low bias current is a priority (such as in low-speed integrator circuits), choose the 3130. For general-purpose DC amplifier work, the 1458 offers good performance (and you get two op-amps in the space of one package). For an upgrade in performance, choose the model 353, as it is a pin-compatible replacement for the 1458. The 353 is designed with JFET input circuitry for very low bias current, and has a bandwidth 4 times are great as the 1458, although its output current limit is lower (but still short-circuit protected). It may be more difficult to find on the shelf of your local electronics supply house, but it is just as reasonably priced as the 1458.

If low power supply voltage is a requirement, I recommend the model 324, as it functions on as low as 3 volts DC. Its input bias current requirements are also low, and it provides four op-amps in a single 14-pin chip. Its major weakness is speed, limited to 1 MHz bandwidth and an output slew rate of only 0.25 volts per µs. For high-frequency AC amplifier circuits, the 318 is a very good "general purpose" model.

Special-purpose op-amps are available for modest cost which provide better performance specifications. Many of these are tailored for a specific type of performance advantage, such as maximum bandwidth or minimum bias current. Take for instance the op-amps, both designed for high bandwidth in Table below.

High bandwidth operational amplifiers

Model	Devices/ package	Power supply	Bandwidth	Bias current	Slew rate	Output current
number	(count)	(V)	(MHz)	(nA)	(V/µS)	(mA)
CLC404	1	10 / 14	232	44,000	2600	70
CLC425	1	5 / 14	1900	40,000	350	90

The CLC404 lists at $21.80 (almost as much as George Philbrick's first commercial op-amp, albeit without correction for inflation), while the CLC425 is quite a bit less expensive at $3.23 per unit. In both cases high speed is achieved at the expense of high bias currents and restrictive power supply voltage ranges. Some op-amps, designed for high power output are listed in Table below.

High current operational amplifiers

Model	Devices/ package	Power supply	Bandwidth	Bias current	Slew rate	Output current
number	(count)	(V)	(MHz)	(nA)	(V/µS)	(mA)
LM12CL	1	15 / 80	0.7	1000	9	13,000
LM7171	1	5.5 / 36	200	12,000	4100	100

Yes, the LM12CL actually has an output current rating of 13 amps (13,000 milliamps)! It lists at $14.40, which is not a lot of money, considering the raw power of the device. The LM7171, on the other hand, trades high current output ability for fast voltage output ability (a high slew rate). It lists at $1.19, about as low as some "general purpose" op-amps.

Amplifier packages may also be purchased as complete application circuits as opposed to bare operational amplifiers. The Burr-Brown and Analog Devices corporations, for example, both long known for their precision amplifier product lines, offer instrumentation amplifiers in pre-designed packages as well as other specialized amplifier devices. In designs where high precision and repeatability after repair is important, it might be advantageous for the circuit designer to choose such a pre-engineered amplifier "block" rather than build the circuit from individual op-amps. Of course, these units typically cost quite a bit more than individual op-amps.

ANALOG INTEGRATED CIRCUITS

2009-06-30T04:42:00.000-07:00

Analog circuits are circuits dealing with signals free to vary from zero to full power supply voltage. This stands in contrast to digital circuits, which almost exclusively employ "all or nothing" signals: voltages restricted to values of zero and full supply voltage, with no valid state in between those extreme limits. Analog circuits are often referred to as linear circuits to emphasize the valid continuity of signal range forbidden in digital circuits, but this label is unfortunately misleading. Just because a voltage or current signal is allowed to vary smoothly between the extremes of zero and full power supply limits does not necessarily mean that all mathematical relationships between these signals are linear in the "straight-line" or "proportional" sense of the word. As you will see in this chapter, many so-called "linear" circuits are quite nonlinear in their behavior, either by necessity of physics or by design.

The circuits in this chapter make use of IC, or integrated circuit, components. Such components are actually networks of interconnected components manufactured on a single wafer of semiconducting material. Integrated circuits providing a multitude of pre-engineered functions are available at very low cost, benefitting students, hobbyists and professional circuit designers alike. Most integrated circuits provide the same functionality as "discrete" semiconductor circuits at higher levels of reliability and at a fraction of the cost. Usually, discrete-component circuit construction is favored only when power dissipation levels are too high for integrated circuits to handle.

Perhaps the most versatile and important analog integrated circuit for the student to master is the operational amplifier, or op-amp. Essentially nothing more than a differential amplifier with very high voltage gain, op-amps are the workhorse of the analog design world. By cleverly applying feedback from the output of an op-amp to one or more of its inputs, a wide variety of behaviors may be obtained from this single device. Many different models of op-amp are available at low cost, but circuits described in this chapter will incorporate only commonly available op-amp models.

Voltage comparator Experiment

2009-06-30T04:41:00.000-07:00

PARTS AND MATERIALS

Operational amplifier, model 1458 or 353 recommended (Radio Shack catalog # 276-038 and 900-6298, respectively)
Three 6 volt batteries
Two 10 kΩ potentiometers, linear taper (Radio Shack catalog # 271-1715)
One light-emitting diode (Radio Shack catalog # 276-026 or equivalent)
One 330 Ω resistor
One 470 Ω resistor

This experiment only requires a single operational amplifier. The model 1458 and 353 are both "dual" op-amp units, with two complete amplifier circuits housed in the same 8-pin DIP package. I recommend that you purchase and use "dual" op-amps over "single" op-amps even if a project only requires one, because they are more versatile (the same op-amp unit can function in projects requiring only one amplifier as well as in projects requiring two). In the interest of purchasing and stocking the least number of components for your home laboratory, this makes sense.

CROSS-REFERENCES

Lessons In Electric Circuits, Volume 3, chapter 8: "Operational Amplifiers"

LEARNING OBJECTIVES

How to use an op-amp as a comparator

SCHEMATIC DIAGRAM

ILLUSTRATION

INSTRUCTIONS

A comparator circuit compares two voltage signals and determines which one is greater. The result of this comparison is indicated by the output voltage: if the op-amp's output is saturated in the positive direction, the noninverting input (+) is a greater, or more positive, voltage than the inverting input (-), all voltages measured with respect to ground. If the op-amp's voltage is near the negative supply voltage (in this case, 0 volts, or ground potential), it means the inverting input (-) has a greater voltage applied to it than the noninverting input (+).

This behavior is much easier understood by experimenting with a comparator circuit than it is by reading someone's verbal description of it. In this experiment, two potentiometers supply variable voltages to be compared by the op-amp. The output status of the op-amp is indicated visually by the LED. By adjusting the two potentiometers and observing the LED, one can easily comprehend the function of a comparator circuit.

For greater insight into this circuit's operation, you might want to connect a pair of voltmeters to the op-amp input terminals (both voltmeters referenced to ground) so that both input voltages may be numerically compared with each other, these meter indications compared to the LED status:

Comparator circuits are widely used to compare physical measurements, provided those physical variables can be translated into voltage signals. For instance, if a small generator were attached to an anemometer wheel to produce a voltage proportional to wind speed, that wind speed signal could be compared with a "set-point" voltage and compared by an op-amp to drive a high wind speed alarm:

Precision voltage follower Experiment

2009-06-30T04:40:00.000-07:00

PARTS AND MATERIALS

Operational amplifier, model 1458 or 353 recommended (Radio Shack catalog # 276-038 and 900-6298, respectively)
Three 6 volt batteries
One 10 kΩ potentiometer, linear taper (Radio Shack catalog # 271-1715)

CROSS-REFERENCES

Lessons In Electric Circuits, Volume 3, chapter 8: "Operational Amplifiers"

LEARNING OBJECTIVES

How to use an op-amp as a voltage follower
Purpose of negative feedback
Troubleshooting strategy

SCHEMATIC DIAGRAM

ILLUSTRATION

INSTRUCTIONS

In the previous op-amp experiment, the amplifier was used in "open-loop" mode; that is, without any feedback from output to input. As such, the full voltage gain of the operational amplifier was available, resulting in the output voltage saturating for virtually any amount of differential voltage applied between the two input terminals. This is good if we desire comparator operation, but if we want the op-amp to behave as a true amplifier, we need it to exhibit a manageable voltage gain.

Since we do not have the luxury of disassembling the integrated circuitry of the op-amp and changing resistor values to give a lesser voltage gain, we are limited to external connections and componentry. Actually, this is not a disadvantage as one might think, because the combination of extremely high open-loop voltage gain coupled with feedback allows us to use the op-amp for a much wider variety of purposes, much easier than if we were to exercise the option of modifying its internal circuitry.

If we connect the output of an op-amp to its inverting (-) input, the output voltage will seek whatever level is necessary to balance the inverting input's voltage with that applied to the noninverting (+) input. If this feedback connection is direct, as in a straight piece of wire, the output voltage will precisely "follow" the noninverting input's voltage. Unlike the voltage follower circuit made from a single transistor (see chapter 5: Discrete Semiconductor Circuits), which approximated the input voltage to within several tenths of a volt, this voltage follower circuit will output a voltage accurate to within mere microvolts of the input voltage!

Measure the input voltage of this circuit with a voltmeter connected between the op-amp's noninverting (+) input terminal and circuit ground (the negative side of the power supply), and the output voltage between the op-amp's output terminal and circuit ground. Watch the op-amp's output voltage follow the input voltage as you adjust the potentiometer through its range.

You may directly measure the difference, or error, between output and input voltages by connecting the voltmeter between the op-amp's two input terminals. Throughout most of the potentiometer's range, this error voltage should be almost zero.

Try moving the potentiometer to one of its extreme positions, far clockwise or far counterclockwise. Measure error voltage, or compare output voltage against input voltage. Do you notice anything unusual? If you are using the model 1458 or model 353 op-amp for this experiment, you should measure a substantial error voltage, or difference between output and input. Many op-amps, the specified models included, cannot "swing" their output voltage exactly to full power supply ("rail") voltage levels. In this case, the "rail" voltages are +18 volts and 0 volts, respectively. Due to limitations in the 1458's internal circuitry, its output voltage is unable to exactly reach these high and low limits. You may find that it can only go within a volt or two of the power supply "rails." This is a very important limitation to understand when designing circuits using operational amplifiers. If full "rail-to-rail" output voltage swing is required in a circuit design, other op-amp models may be selected which offer this capability. The model 3130 is one such op-amp.

Precision voltage follower circuits are useful if the voltage signal to be amplified cannot tolerate "loading;" that is, if it has a high source impedance. Since a voltage follower by definition has a voltage gain of 1, its purpose has nothing to do with amplifying voltage, but rather with amplifying a signal's capacity to deliver current to a load.

Voltage follower circuits have another important use for circuit builders: they allow for simple linear testing of an op-amp. One of the troubleshooting techniques I recommend is to simplify and rebuild. Suppose that you are building a circuit using one or more op-amps to perform some advanced function. If one of those op-amps seems to be causing a problem and you suspect it may be faulty, try re-connecting it as a simple voltage follower and see if it functions in that capacity. An op-amp that fails to work as a voltage follower certainly won't work as anything more complex!

COMPUTER SIMULATION

Schematic with SPICE node numbers:

Netlist (make a text file containing the following text, verbatim):

Voltage follower
vinput 1 0
rbogus 1 0 1meg
e1 2 0 1 2 999meg
rload 2 0 10k
.dc vinput 5 5 1
.print dc v(1,0) v(2,0) v(1,2)
.end

An ideal operational amplifier may be simulated in SPICE using a dependent voltage source (e1 in the netlist). The output nodes are specified first (2 0), then the two input nodes, non-inverting input first (1 2). Open-loop gain is specified last (999meg) in the dependent voltage source line.

Because SPICE views the input impedance of a dependent source as infinite, some finite amount of resistance must be included to avoid an analysis error. This is the purpose of R_bogus: to provide DC path to ground for the V_input voltage source. Such "bogus" resistances should be arbitrarily large. In this simulation I chose 1 MΩ for an R_bogus value.

A load resistor is included in the circuit for much the same reason: to provide a DC path for current at the output of the dependent voltage source. As you can see, SPICE doesn't like open circuits!

Non inverting amplifier Experiment

2009-06-30T04:39:00.000-07:00

PARTS AND MATERIALS

Operational amplifier, model 1458 or 353 recommended (Radio Shack catalog # 276-038 and 900-6298, respectively)
Three 6 volt batteries
Two 10 kΩ potentiometers, linear taper (Radio Shack catalog # 271-1715)

CROSS-REFERENCES

Lessons In Electric Circuits, Volume 3, chapter 8: "Operational Amplifiers"

LEARNING OBJECTIVES

How to use an op-amp as a single-ended amplifier
Using divided, negative feedback

SCHEMATIC DIAGRAM

ILLUSTRATION

INSTRUCTIONS

This circuit differs from the voltage follower in only one respect: output voltage is "fed back" to the inverting (-) input through a voltage-dividing potentiometer rather than being directly connected. With only a fraction of the output voltage fed back to the inverting input, the op-amp will output a corresponding multiple of the voltage sensed at the noninverting (+) input in keeping the input differential voltage near zero. In other words, the op-amp will now function as an amplifier with a controllable voltage gain, that gain being established by the position of the feedback potentiometer (R₂).

Set R₂ to approximately mid-position. This should give a voltage gain of about 2. Measure both input and output voltage for several positions of the input potentiometer R₁. Move R₂ to a different position and re-take voltage measurements for several positions of R₁. For any given R₂ position, the ratio between output and input voltage should be the same.

You will also notice that the input and output voltages are always positive with respect to ground. Because the output voltage increases in a positive direction for a positive increase of the input voltage, this amplifier is referred to as noninverting. If the output and input voltages were related to one another in an inverse fashion (i.e. positive increasing input voltage results in positive decreasing or negative increasing output), then the amplifier would be known as an inverting type.

The ability to leverage an op-amp in this fashion to create an amplifier with controllable voltage gain makes this circuit an extremely useful one. It would take quite a bit more design and troubleshooting effort to produce a similar circuit using discrete transistors.

Try adjusting R₂ for maximum and minimum voltage gain. What is the lowest voltage gain attainable with this amplifier configuration? Why do you think this is?

COMPUTER SIMULATION

Schematic with SPICE node numbers:

Netlist (make a text file containing the following text, verbatim):

Noninverting amplifier
vinput 1 0
r2 3 2 5k
r1 2 0 5k
rbogus 1 0 1meg
e1 3 0 1 2 999meg
rload 3 0 10k
.dc vinput 5 5 1
.print dc v(1,0) v(3,0)
.end

With R₁ and R₂ set equally to 5 kΩ in the simulation, it mimics the feedback potentiometer of the real circuit at mid-position (50%). To simulate the potentiometer at the 75% position, set R₂ to 7.5 kΩ and R₁ to 2.5 kΩ.

High-impedance voltmeter Experiment

2009-06-30T04:38:00.000-07:00

PARTS AND MATERIALS

Operational amplifier, model TL082 recommended (Radio Shack catalog # 276-1715)
Operational amplifier, model LM1458 recommended (Radio Shack catalog # 276-038)
Four 6 volt batteries
One meter movement, 1 mA full-scale deflection (Radio Shack catalog #22-410)
15 kΩ precision resistor
Four 1 MΩ resistors

The 1 mA meter movement sold by Radio Shack is advertised as a 0-15 VDC meter, but is actually a 1 mA movement sold with a 15 kΩ +/- 1% tolerance multiplier resistor. If you get this Radio Shack meter movement, you can use the included 15 kΩ resistor for the resistor specified in the parts list.

This meter experiment is based on a JFET-input op-amp such as the TL082. The other op-amp (model 1458) is used in this experiment to demonstrate the absence of latch-up: a problem inherent to the TL082.

You don't need 1 MΩ resistors, exactly. Any very high resistance resistors will suffice.

CROSS-REFERENCES

Lessons In Electric Circuits, Volume 3, chapter 8: "Operational Amplifiers"

LEARNING OBJECTIVES

Voltmeter loading: its causes and its solution
How to make a high-impedance voltmeter using an op-amp
What op-amp "latch-up" is and how to avoid it

SCHEMATIC DIAGRAM

ILLUSTRATION

INSTRUCTIONS

An ideal voltmeter has infinite input impedance, meaning that it draws zero current from the circuit under test. This way, there will be no "impact" on the circuit as the voltage is being measured. The more current a voltmeter draws from the circuit under test, the more the measured voltage will "sag" under the loading effect of the meter, like a tire-pressure gauge releasing air out of the tire being measured: the more air released from the tire, the more the tire's pressure will be impacted in the act of measurement. This loading is more pronounced on circuits of high resistance, like the voltage divider made of 1 MΩ resistors, shown in the schematic diagram.

If you were to build a simple 0-15 volt range voltmeter by connecting the 1 mA meter movement in series with the 15 kΩ precision resistor, and try to use this voltmeter to measure the voltages at TP1, TP2, or TP3 (with respect to ground), you'd encounter severe measurement errors induced by meter "impact:"

Try using the meter movement and 15 kΩ resistor as shown to measure these three voltages. Does the meter read falsely high or falsely low? Why do you think this is?

If we were to increase the meter's input impedance, we would diminish its current draw or "load" on the circuit under test and consequently improve its measurement accuracy. An op-amp with high-impedance inputs (using a JFET transistor input stage rather than a BJT input stage) works well for this application.

Note that the meter movement is part of the op-amp's feedback loop from output to inverting input. This circuit drives the meter movement with a current proportional to the voltage impressed at the noninverting (+) input, the requisite current supplied directly from the batteries through the op-amp's power supply pins, not from the circuit under test through the test probe. The meter's range is set by the resistor connecting the inverting (-) input to ground.

Build the op-amp meter circuit as shown and re-take voltage measurements at TP1, TP2, and TP3. You should enjoy far better success this time, with the meter movement accurately measuring these voltages (approximately 3, 6, and 9 volts, respectively).

You may witness the extreme sensitivity of this voltmeter by touching the test probe with one hand and the most positive battery terminal with the other. Notice how you can drive the needle upward on the scale simply by measuring battery voltage through your body resistance: an impossible feat with the original, unamplified voltmeter circuit. If you touch the test probe to ground, the meter should read exactly 0 volts.

After you've proven this circuit to work, modify it by changing the power supply from dual to split. This entails removing the center-tap ground connection between the 2nd and 3rd batteries, and grounding the far negative battery terminal instead:

This alteration in the power supply increases the voltages at TP1, TP2, and TP3 to 6, 12, and 18 volts, respectively. With a 15 kΩ range resistor and a 1 mA meter movement, measuring 18 volts will gently "peg" the meter, but you should be able to measure the 6 and 12 volt test points just fine.

Try touching the meter's test probe to ground. This should drive the meter needle to exactly 0 volts as before, but it will not! What is happening here is an op-amp phenomenon called latch-up: where the op-amp output drives to a positive voltage when the input common-mode voltage exceeds the allowable limit. In this case, as with many JFET-input op-amps, neither input should be allowed to come close to either power supply rail voltage. With a single supply, the op-amp's negative power rail is at ground potential (0 volts), so grounding the test probe brings the noninverting (+) input exactly to that rail voltage. This is bad for a JFET op-amp, and drives the output strongly positive, even though it doesn't seem like it should, based on how op-amps are supposed to function.

When the op-amp ran on a "dual" supply (+12/-12 volts, rather than a "single" +24 volt supply), the negative power supply rail was 12 volts away from ground (0 volts), so grounding the test probe didn't violate the op-amp's common-mode voltage limit. However, with the "single" +24 volt supply, we have a problem. Note that some op-amps do not "latch-up" the way the model TL082 does. You may replace the TL082 with an LM1458 op-amp, which is pin-for-pin compatible (no breadboard wiring changes needed). The model 1458 will not "latch-up" when the test probe is grounded, although you may still get incorrect meter readings with the measured voltage exactly equal to the negative power supply rail. As a general rule, you should always be sure the op-amp's power supply rail voltages exceed the expected input voltages.

Integrator Experiment

2009-06-30T04:37:00.002-07:00

PARTS AND MATERIALS

Four 6 volt batteries
Operational amplifier, model 1458 recommended (Radio Shack catalog # 276-038)
One 10 kΩ potentiometer, linear taper (Radio Shack catalog # 271-1715)
Two capacitors, 0.1 µF each, non-polarized (Radio Shack catalog # 272-135)
Two 100 kΩ resistors
Three 1 MΩ resistors

Just about any operational amplifier model will work fine for this integrator experiment, but I'm specifying the model 1458 over the 353 because the 1458 has much higher input bias currents. Normally, high input bias current is a bad characteristic for an op-amp to have in a precision DC amplifier circuit (and especially an integrator circuit!). However, I want the bias current to be high in order that its bad effects may be exaggerated, and so that you will learn one method of counteracting its effects.

CROSS-REFERENCES

Lessons In Electric Circuits, Volume 3, chapter 8: "Operational Amplifiers"

LEARNING OBJECTIVES

Method for limiting the span of a potentiometer
Purpose of an integrator circuit
How to compensate for op-amp bias current

SCHEMATIC DIAGRAM

ILLUSTRATION

INSTRUCTIONS

As you can see from the schematic diagram, the potentiometer is connected to the "rails" of the power source through 100 kΩ resistors, one on each end. This is to limit the span of the potentiometer, so that full movement produces a fairly small range of input voltages for the op-amp to operate on. At one extreme of the potentiometer's motion, a voltage of about 0.5 volt (with respect the the ground point in the middle of the series battery string) will be produced at the potentiometer wiper. At the other extreme of motion, a voltage of about -0.5 volt will be produced. When the potentiometer is positioned dead-center, the wiper voltage should measure zero volts.

Connect a voltmeter between the op-amp's output terminal and the circuit ground point. Slowly move the potentiometer control while monitoring the output voltage. The output voltage should be changing at a rate established by the potentiometer's deviation from zero (center) position. To use calculus terms, we would say that the output voltage represents the integral (with respect to time) of the input voltage function. That is, the input voltage level establishes the output voltage rate of change over time. This is precisely the opposite of differentiation, where the derivative of a signal or function is its instantaneous rate of change.

If you have two voltmeters, you may readily see this relationship between input voltage and output voltage rate of change by measuring the wiper voltage (between the potentiometer wiper and ground) with one meter and the output voltage (between the op-amp output terminal and ground) with the other. Adjusting the potentiometer to give zero volts should result in the slowest output voltage rate-of-change. Conversely, the more voltage input to this circuit, the faster its output voltage will change, or "ramp."

Try connecting the second 0.1 µF capacitor in parallel with the first. This will double the amount of capacitance in the op-amp's feedback loop. What affect does this have on the circuit's integration rate for any given potentiometer position?

Try connecting another 1 MΩ resistor in parallel with the input resistor (the resistor connecting the potentiometer wiper to the inverting terminal of the op-amp). This will halve the integrator's input resistance. What affect does this have on the circuit's integration rate?

Integrator circuits are one of the fundamental "building-block" functions of an analog computer. By connecting integrator circuits with amplifiers, summers, and potentiometers (dividers), almost any differential equation could be modeled, and solutions obtained by measuring voltages produced at various points in the network of circuits. Because differential equations describe so many physical processes, analog computers are useful as simulators. Before the advent of modern digital computers, engineers used analog computers to simulate such processes as machinery vibration, rocket trajectory, and control system response. Even though analog computers are considered obsolete by modern standards, their constituent components still work well as learning tools for calculus concepts.

Move the potentiometer until the op-amp's output voltage is as close to zero as you can get it, and moving as slowly as you can make it. Disconnect the integrator input from the potentiometer wiper terminal and connect it instead to ground, like this:

Applying exactly zero voltage to the input of an integrator circuit should, ideally, cause the output voltage rate-of-change to be zero. When you make this change to the circuit, you should notice the output voltage remaining at a constant level or changing very slowly.

With the integrator input still shorted to ground, short past the 1 MΩ resistor connecting the op-amp's noninverting (+) input to ground. There should be no need for this resistor in an ideal op-amp circuit, so by shorting past it we will see what function it provides in this very real op-amp circuit:

As soon as the "grounding" resistor is shorted with a jumper wire, the op-amp's output voltage will start to change, or drift. Ideally, this should not happen, because the integrator circuit still has an input signal of zero volts. However, real operational amplifiers have a very small amount of current entering each input terminal called the bias current. These bias currents will drop voltage across any resistance in their path. Since the 1 MΩ input resistor conducts some amount of bias current regardless of input signal magnitude, it will drop voltage across its terminals due to bias current, thus "offsetting" the amount of signal voltage seen at the inverting terminal of the op-amp. If the other (noninverting) input is connected directly to ground as we have done here, this "offset" voltage incurred by voltage drop generated by bias current will cause the integrator circuit to slowly "integrate" as though it were receiving a very small input signal.

The "grounding" resistor is better known as a compensating resistor, because it acts to compensate for voltage errors created by bias current. Since the bias currents through each op-amp input terminal are approximately equal to each other, an equal amount of resistance placed in the path of each bias current will produce approximately the same voltage drop. Equal voltage drops seen at the complementary inputs of an op-amp cancel each other out, thus nulling the error otherwise induced by bias current.

Remove the jumper wire shorting past the compensating resistor and notice how the op-amp output returns to a relatively stable state. It may still drift some, most likely due to bias voltage error in the op-amp itself, but that is another subject altogether!

COMPUTER SIMULATION

Schematic with SPICE node numbers:

Netlist (make a text file containing the following text, verbatim):

DC integrator
vinput 1 0 dc 0.05
r1 1 2 1meg
c1 2 3 0.1u ic=0
e1 3 0 0 2 999k
.tran 1 30 uic
.plot tran v(1,0) v(3,0)
.end

555 audio oscillator Experiment

2009-06-30T04:37:00.001-07:00

PARTS AND MATERIALS

Two 6 volt batteries
One capacitor, 0.1 µF, non-polarized (Radio Shack catalog # 272-135)
One 555 timer IC (Radio Shack catalog # 276-1723)
Two light-emitting diodes (Radio Shack catalog # 276-026 or equivalent)
One 1 MΩ resistor
One 100 kΩ resistor
Two 510 Ω resistors
Audio detector with headphones
Oscilloscope (recommended, but not necessary)

A oscilloscope would be useful in analyzing the waveforms produced by this circuit, but it is not essential. An audio detector is a very useful piece of test equipment for this experiment, especially if you don't have an oscilloscope.

CROSS-REFERENCES

Lessons In Electric Circuits, Volume 4, chapter 10: "Multivibrators"

LEARNING OBJECTIVES

How to use the 555 timer as an astable multivibrator
Working knowledge of duty cycle

SCHEMATIC DIAGRAM

ILLUSTRATION

INSTRUCTIONS

The "555" integrated circuit is a general-purpose timer useful for a variety of functions. In this experiment, we explore its use as an astable multivibrator, or oscillator. Connected to a capacitor and two resistors as shown, it will oscillate freely, driving the LEDs on and off with a square-wave output voltage.

This circuit works on the principle of alternately charging and discharging a capacitor. The 555 begins to discharge the capacitor by grounding the Disch terminal when the voltage detected by the Thresh terminal exceeds 2/3 the power supply voltage (V_cc). It stops discharging the capacitor when the voltage detected by the Trig terminal falls below 1/3 the power supply voltage. Thus, when both Thresh and Trig terminals are connected to the capacitor's positive terminal, the capacitor voltage will cycle between 1/3 and 2/3 power supply voltage in a "sawtooth" pattern.

During the charging cycle, the capacitor receives charging current through the series combination of the 1 MΩ and 100 kΩ resistors. As soon as the Disch terminal on the 555 timer goes to ground potential (a transistor inside the 555 connected between that terminal and ground turns on), the capacitor's discharging current only has to go through the 100 kΩ resistor. The result is an RC time constant that is much longer for charging than for discharging, resulting in a charging time greatly exceeding the discharging time.

The 555's Out terminal produces a square-wave voltage signal that is "high" (nearly V_cc) when the capacitor is charging, and "low" (nearly 0 volts) when the capacitor is discharging. This alternating high/low voltage signal drives the two LEDs in opposite modes: when one is on, the other will be off. Because the capacitor's charging and discharging times are unequal, the "high" and "low" times of the output's square-wave waveform will be unequal as well. This can be seen in the relative brightness of the two LEDs: one will be much brighter than the other, because it is on for a longer period of time during each cycle.

The equality or inequality between "high" and "low" times of a square wave is expressed as that wave's duty cycle. A square wave with a 50% duty cycle is perfectly symmetrical: its "high" time is precisely equal to its "low" time. A square wave that is "high" 10% of the time and "low" 90% of the time is said to have a 10% duty cycle. In this circuit, the output waveform has a "high" time exceeding the "low" time, resulting in a duty cycle greater than 50%.

Use the audio detector (or an oscilloscope) to investigate the different voltage waveforms produced by this circuit. Try different resistor values and/or capacitor values to see what effects they have on output frequency or charge/discharge times.

555 ramp generator Experiment

2009-06-30T04:34:00.000-07:00

PARTS AND MATERIALS

Two 6 volt batteries
One capacitor, 470 µF electrolytic, 35 WVDC (Radio Shack catalog # 272-1030 or equivalent)
One capacitor, 0.1 µF, non-polarized (Radio Shack catalog # 272-135)
One 555 timer IC (Radio Shack catalog # 276-1723)
Two PNP transistors -- models 2N2907 or 2N3906 recommended (Radio Shack catalog # 276-1604 is a package of fifteen PNP transistors ideal for this and other experiments)
Two light-emitting diodes (Radio Shack catalog # 276-026 or equivalent)
One 100 kΩ resistor
One 47 kΩ resistor
Two 510 Ω resistors
Audio detector with headphones

The voltage rating on the 470 µF capacitor is not critical, so long as it generously exceeds the maximum power supply voltage. In this particular circuit, that maximum voltage is 12 volts. Be sure you connect this capacitor in the circuit properly, respecting polarity!

CROSS-REFERENCES

Lessons In Electric Circuits, Volume 1, chapter 13: "Capacitors"

Lessons In Electric Circuits, Volume 4, chapter 10: "Multivibrators"

LEARNING OBJECTIVES

How to use the 555 timer as an astable multivibrator
A practical use for a current mirror circuit
Understanding the relationship between capacitor current and capacitor voltage rate-of-change

SCHEMATIC DIAGRAM

ILLUSTRATION

INSTRUCTIONS

Again, we are using a 555 timer IC as an astable multivibrator, or oscillator. This time, however, we will compare its operation in two different capacitor-charging modes: traditional RC and constant-current.

Connecting test point #1 (TP1) to test point #3 (TP3) using a jumper wire. This allows the capacitor to charge through a 47 kΩ resistor. When the capacitor has reached 2/3 supply voltage, the 555 timer switches to "discharge" mode and discharges the capacitor to a level of 1/3 supply voltage almost immediately. The charging cycle begins again at this point. Measure voltage directly across the capacitor with a voltmeter (a digital voltmeter is preferred), and note the rate of capacitor charging over time. It should rise quickly at first, then taper off as it builds up to 2/3 supply voltage, just as you would expect from an RC charging circuit.

Remove the jumper wire from TP3, and re-connect it to TP2. This allows the capacitor to be charged through the controlled-current leg of a current mirror circuit formed by the two PNP transistors. Measure voltage directly across the capacitor again, noting the difference in charging rate over time as compared to the last circuit configuration.

By connecting TP1 to TP2, the capacitor receives a nearly constant charging current. Constant capacitor charging current yields a voltage curve that is linear, as described by the equation i = C(de/dt). If the capacitor's current is constant, so will be its rate-of-change of voltage over time. The result is a "ramp" waveform rather than a "sawtooth" waveform:

The capacitor's charging current may be directly measured by substituting an ammeter in place of the jumper wire. The ammeter will need to be set to measure a current in the range of hundreds of microamps (tenths of a milliamp). Connected between TP1 and TP3, you should see a current that starts at a relatively high value at the beginning of the charging cycle, and tapers off toward the end. Connected between TP1 and TP2, however, the current will be much more stable.

It is an interesting experiment at this point to change the temperature of either current mirror transistor by touching it with your finger. As the transistor warms, it will conduct more collector current for the same base-emitter voltage. If the controlling transistor (the one connected to the 100 kΩ resistor) is touched, the current decreases. If the controlled transistor is touched, the current increases. For the most stable current mirror operation, the two transistors should be cemented together so that their temperatures never differ by any substantial amount.

This circuit works just as well at high frequencies as it does at low frequencies. Replace the 470 µF capacitor with a 0.1 µF capacitor, and use an audio detector to sense the voltage waveform at the 555's output terminal. The detector should produce an audio tone that is easy to hear. The capacitor's voltage will now be changing much too fast to view with a voltmeter in the DC mode, but we can still measure capacitor current with an ammeter.

With the ammeter connected between TP1 and TP3 (RC mode), measure both DC microamps and AC microamps. Record these current figures on paper. Now, connect the ammeter between TP1 and TP2 (constant-current mode). Measure both DC microamps and AC microamps, noting any differences in current readings between this circuit configuration and the last one. Measuring AC current in addition to DC current is an easy way to determine which circuit configuration gives the most stable charging current. If the current mirror circuit were perfect -- the capacitor charging current absolutely constant -- there would be zero AC current measured by the meter.

PWM power controller Experiment

2009-06-30T04:29:00.000-07:00

PARTS AND MATERIALS

Four 6 volt batteries
One capacitor, 100 µF electrolytic, 35 WVDC (Radio Shack catalog # 272-1028 or equivalent)
One capacitor, 0.1 µF, non-polarized (Radio Shack catalog # 272-135)
One 555 timer IC (Radio Shack catalog # 276-1723)
Dual operational amplifier, model 1458 recommended (Radio Shack catalog # 276-038)
One NPN power transistor -- (Radio Shack catalog # 276-2041 or equivalent)
Three 1N4001 rectifying diodes (Radio Shack catalog # 276-1101)
One 10 kΩ potentiometer, linear taper (Radio Shack catalog # 271-1715)
One 33 kΩ resistor
12 volt automotive tail-light lamp
Audio detector with headphones

CROSS-REFERENCES

Lessons In Electric Circuits, Volume 3, chapter 8: "Operational Amplifiers"

Lessons In Electric Circuits, Volume 2, chapter 7: "Mixed-Frequency AC Signals"

LEARNING OBJECTIVES

How to use the 555 timer as an astable multivibrator
How to use an op-amp as a comparator
How to use diodes to drop unwanted DC voltage
How to control power to a load by pulse-width modulation

SCHEMATIC DIAGRAM

ILLUSTRATION

INSTRUCTIONS

This circuit uses a 555 timer to generate a sawtooth voltage waveform across a capacitor, then compares that signal against a steady voltage provided by a potentiometer, using an op-amp as a comparator. The comparison of these two voltage signals produces a square-wave output from the op-amp, varying in duty cycle according to the potentiometer's position. This variable duty cycle signal then drives the base of a power transistor, switching current on and off through the load. The 555's oscillation frequency is much higher than the lamp filament's ability to thermally cycle (heat and cool), so any variation in duty cycle, or pulse width, has the effect of controlling the total power dissipated by the load over time.

Controlling electrical power through a load by means of quickly switching it on and off, and varying the "on" time, is known as pulse-width modulation, or PWM. It is a very efficient means of controlling electrical power because the controlling element (the power transistor) dissipates comparatively little power in switching on and off, especially if compared to the wasted power dissipated of a rheostat in a similar situation. When the transistor is in cutoff, its power dissipation is zero because there is no current through it. When the transistor is saturated, its dissipation is very low because there is little voltage dropped between collector and emitter while it is conducting current.

PWM is a concept easier understood through experimentation than reading. It would be nice to view the capacitor voltage, potentiometer voltage, and op-amp output waveforms all on one (triple-trace) oscilloscope to see how they relate to one another, and to the load power. However, most of us have no access to a triple-trace oscilloscope, much less any oscilloscope at all, so an alternative method is to slow the 555 oscillator down enough that the three voltages may be compared with a simple DC voltmeter. Replace the 0.1 µF capacitor with one that is 100 µF or larger. This will slow the oscillation frequency down by a factor of at least a thousand, enabling you to measure the capacitor voltage slowly rise over time, and the op-amp output transition from "high" to "low" when the capacitor voltage becomes greater than the potentiometer voltage. With such a slow oscillation frequency, the load power will not be proportioned as before. Rather, the lamp will turn on and off at regular intervals. Feel free to experiment with other capacitor or resistor values to speed up the oscillations enough so the lamp never fully turns on or off, but is "throttled" by quick on-and-off pulsing of the transistor.

When you examine the schematic, you will notice two operational amplifiers connected in parallel. This is done to provide maximum current output to the base terminal of the power transistor. A single op-amp (one-half of a 1458 IC) may not be able to provide sufficient output current to drive the transistor into saturation, so two op-amps are used in tandem. This should only be done if the op-amps in question are overload-protected, which the 1458 series of op-amps are. Otherwise, it is possible (though unlikely) that one op-amp could turn on before the other, and damage result from the two outputs short-circuiting each other (one driving "high" and the other driving "low" simultaneously). The inherent short-circuit protection offered by the 1458 allows for direct driving of the power transistor base without any need for a current-limiting resistor.

The three diodes in series connecting the op-amps' outputs to the transistor's base are there to drop voltage and ensure the transistor falls into cutoff when the op-amp outputs go "low." Because the 1458 op-amp cannot swing its output voltage all the way down to ground potential, but only to within about 2 volts of ground, a direct connection from the op-amp to the transistor would mean the transistor would never fully turn off. Adding three silicon diodes in series drops approximately 2.1 volts (0.7 volts times 3) to ensure there is minimal voltage at the transistor's base when the op-amp outputs go "low."

It is interesting to listen to the op-amp output signal through an audio detector as the potentiometer is adjusted through its full range of motion. Adjusting the potentiometer has no effect on signal frequency, but it greatly affects duty cycle. Note the difference in tone quality, or timbre, as the potentiometer varies the duty cycle from 0% to 50% to 100%. Varying the duty cycle has the effect of changing the harmonic content of the waveform, which makes the tone sound different.

You might notice a particular uniqueness to the sound heard through the detector headphones when the potentiometer is in center position (50% duty cycle -- 50% load power), versus a kind of similarity in sound just above or below 50% duty cycle. This is due to the absence or presence of even-numbered harmonics. Any waveform that is symmetrical above and below its centerline, such as a square wave with a 50% duty cycle, contains no even-numbered harmonics, only odd-numbered. If the duty cycle is below or above 50%, the waveform will not exhibit this symmetry, and there will be even-numbered harmonics. The presence of these even-numbered harmonic frequencies can be detected by the human ear, as some of them correspond to octaves of the fundamental frequency and thus "fit" more naturally into the tone scheme.