ECE 3355 Electronics Lecture Notes Set 4 -- Version 21

ECE 3355 Electronics Lecture Notes Set 4 -- Version 21
Operational Amplifiers Dr. Dave Shattuck Dept. of ECE, Univ. of Houston

Amplifier Frequency Response
We will cover material from Sections 2.1 through 2.6 from the 5th Edition of the Sedra and Smith text. It may also be useful for you to consult the Nilsson and Riedel Electric Circuits text. The material is from Chapter 5 of the 6th or 7th Editions.

Overview of this Part Fundamentals of Operational Amplifiers
In this part, we will cover the following topics: Basic Operational Amplifier Requirements Equivalent Circuit for Operational Amplifiers Negative Feedback and What it Does You can click on the blue text to jump to the subject that you want to learn about now.

Operational Amplifiers (Op Amps)
Operational Amplifiers (op amps) are devices that amplify voltages. Because of the way the op amps are built, they facilitate the application of negative feedback, which in turn allows easy design of special applications, and for the op amps to behave very ideally. Operational amplifiers are most useful because of their ability to process signals easily. One example: An audio mixer provides the functions of providing variable gain (amplification), and adding different signals together with separate gains for each signal. Op amps can perform these functions, and do so in an easy to use and design form.

Op Amps: A Structural Definition
An op amp is a differential input, single-ended output, amplifier. The schematic symbol for the op amp has three terminals that are always shown: Inverting input (which should not be called the negative input) Noninverting input (which should not be called the positive input) Output (which can be called the output) Schematic Symbol for the Op Amp

This means that at the input, the difference in voltage between the input terminals is used. An op amp is a differential input, single-ended output, amplifier. Schematic Symbol for the Op Amp

This means that at the output, the voltage is obtained with respect to a reference, usually called ground. An op amp is a differential input, single-ended output, amplifier. Schematic Symbol for the Op Amp

There are actually five terminals which are always present in an op amp. These are shown in the more complete schematic shown below. The dc power supplies must be connected for the op amp to work. They may not be shown, since they do not affect the signal behavior in many cases. However, the connections must be present, whether they are shown in the schematic or not. Complete Schematic Symbol for the Op Amp

It is important to note that, in the positive and negative dc power supplies, the “positive” and “negative” here are relative. The voltages for these terminals may have any relationship to ground. For example, the voltages could be: +15[V] and ground, ground and -15[V], +15[V] and -15[V], or +15[V] and +5[V]. Complete Schematic Symbol for the Op Amp

Actually, most op amps have at least two more terminals for use in correcting for some of the non-ideal characteristics of the op amp. However, for this course, we will assume that our op amps are ideal, and will not use these additional terminals. Complete Schematic Symbol for the Op Amp

Op Amps: A Functional Definition
Op amps take the difference between the voltages at the two inputs, and amplify it by a large amount, and provide that voltage at the output with respect to ground. This can be shown with an equivalent circuit. Equivalent Circuit for the Op Amp

The effective resistance between the input terminals, and the resistances between the input terminals and ground, are typically large compared to other resistances in the op-amp circuits, and can often be large enough to be considered effectively infinite. The output resistance is small enough to be ignored. Equivalent Circuit for the Op Amp Ignoring all these resistances gives us the relatively simple equivalent circuit at right. In some situations, these resistances can be estimated and included in the equivalent circuit to make it more accurate. For the purposes of this course, this model will be accurate enough for all the problems we consider.

Note that the output is a function of only the difference between the inputs. This means that if v- = v+ = 500,000[V], the output should be zero volts. Equivalent Circuit for the Op Amp This is hard to obtain in practice, and is called common mode rejection. The part of the two voltages, v- and v+, that are common, is called the common mode.

The coefficient in the dependent source is the gain. Equivalent Circuit for the Op Amp The gain A is called the differential gain, and also called the open loop gain, for reasons that will become obvious soon.

Conceptually, the gain A is a function of frequency. Equivalent Circuit for the Op Amp The response is good for all frequencies, even at dc, in the ideal case. The value for the gain, A, is very large, in general.

The output voltage is limited. The output voltage cannot be higher than the positive dc power supply voltage (+VDC), and cannot be lower than the negative dc power supply voltage (-VDC). Equivalent Circuit for the Op Amp, for Region Marked in Red

Solving Op-Amp Circuits
We will use two assumptions for analysis and design of op amp circuits where the op amp can be considered to be ideal. The Two Assumptions 1) i- = i+ = 0. These currents are small due to the high input resistances. 2) If there is negative feedback, then v- = v+. If there is no negative feedback, the op amp output will saturate. If vi is positive, it saturates at +VDC, and if vi is negative, it saturates at –VDC.

Solving Op-Amp Circuits
We use these two assumptions for the analysis and design of op amp circuits where the op amp can be considered to be ideal. While the equivalent circuit may seem to be easier, we will see that these assumptions make solving op amp circuits much easier. The Two Assumptions 1) i- = i+ = 0. 2) If there is negative feedback, then v- = v+. If not, the output saturates.

First Assumption The Two Assumptions 1) i- = i+ = 0.
The first assumption results from having large resistances at the input, larger than the resistance values typically connected to them. This assumption is not conditional; it happens whether negative feedback is present or not. The Two Assumptions 1) i- = i+ = 0. 2) If there is negative feedback, then v- = v+. If not, the output saturates.

Second Assumption The second assumption results from negative feedback and the very large gain of the op amp. This is called the virtual short, or the summing-point constraint. The two input voltages are constrained to be equal by the presence of negative feedback. Without negative feedback, even a small input will saturate the output. The Two Assumptions 1) i- = i+ = 0. 2) If there is negative feedback, then v- = v+. If not, the output saturates.

Is This Reasonable? Many students who are seeing this for the first time have little trouble accepting the first assumption. It seems reasonable to be able to have large input resistances. However, the notion that the input voltage vi will be forced to zero by something called negative feedback is harder to accept. Some of these students are troubled by the notion that the input would be zero, which is then multiplied by a very large number to get a finite, nonzero output. The Two Assumptions 1) i- = i+ = 0. 2) If there is negative feedback, then v- = v+. If not, the output saturates.

Is This Reasonable? Yes! The notion that the input voltage vi is forced to zero by something called negative feedback is an approximation. It is actually forced to be very small, because the gain is so large. Thus, a very small input, which is almost zero, is then multiplied by a very large number, to get a finite, nonzero output. To understand this better, we need to understand negative feedback better. The Two Assumptions 1) i- = i+ = 0. 2) If there is negative feedback, then v- = v+. If not, the output saturates.

Negative Feedback – Signal Flow Diagrams
Engineers have developed a way of looking at signals called the signal flow diagram. This is not a schematic, and does not represent wire and specific components. A line represents a path that a signal might follow. The signals can be voltages or currents. Therefore, we will label the signals with the symbol x. In the signal flow diagram shown below, there is an input signal, xi. This signal flows into an amplifier with gain A, which is shown with a triangle. This produces an output signal xo. The input is multiplied by the gain, to give the output. Signal Flow Diagram

Negative Feedback – Signal Flow Diagrams
Now, let’s add negative feedback to our signal flow diagram. In the signal flow diagram shown below, we add another amplifier. This amplifier has a gain which is conventionally called b. This amplifier amplifies the output signal, to produce a feedback signal, xf. Finally, this feedback signal is subtracted from the input signal. The symbol for this action is called a summing point or a summing junction. The signs at the junction indicate the signs for the summation.

Negative Feedback – Definition
At this point, we can define negative feedback. Negative feedback is when a portion of the output is taken, returned to the input, and subtracted from this input. If we were to add it to the input, we would call it positive feedback.

Negative Feedback – Notes
The feedback amplifier, with a gain of b, is typically not an amplifier per se, but rather is a resistive network. In any case, the key is that the feedback signal xf is proportional to the output signal, with a multiplier equal to b. The gain A is called the open loop gain, because this would be the gain if the loop were to be opened, that is, if the feedback were removed.

Gain with Negative Feedback
Now, let’s solve for the gain with negative feedback, which is xo/xs. We start by writing an equation for the summing junction, taking into account the signs, to get where the second equation comes by using the definition of the feedback gain b. Next, we use a similar definition for the feed-forward gain, A, to write We then substitute the first equation into the second to get We can combine terms, then we can divide through by xs, and then by (1+Ab), to get

This is the gain with negative feedback If we take the case where A is very large, and it usually is, we can get a special situation. Specifically, take the case where Ab >> 1. Then, and we can use this approximation to simplify the gain with feedback, which we call Af, to

Thus, the gain with negative feedback, Af, is The only requirement is that Ab >> 1. Thus, the gain is not a function of A at all!?! This is a seemingly bizarre, but wondrous result, which is fundamental to the power of negative feedback. The gain of the op amp, which changes from time to time, and from op amp to op amp, does not affect the overall gain with feedback.

Thus, the gain with negative feedback, Af, is Thus, the gain is not a function of A at all!?! The gain of the op amp does not affect the overall gain with feedback. The overall gain, Af, is determined by the way feedback is applied. Feedback is used to allow gain to be traded off for a variety of desirable results. When we use op amps, we have a relatively simple way to determine the presence of negative feedback: If there is a signal path between the output of the op amp, and the inverting input, there will be negative feedback.

Thus, the gain with negative feedback, Af, is With this result, we can look again at the signal flow diagram. The input to the op amp, vi, is the output divided by the gain, vo/A. If A is large, then vi will be much less than vo, and can usually be neglected. This is what we call the virtual short.

How do we use this? This is a good question.
We will use the two assumptions to solve op amp circuits more quickly. We will show how to do this in the next part. The point to recognize here is that negative feedback can be very useful, and makes op amps circuits much easier to analyze, and therefore much easier to design with. Many circuits that we look at will have negative feedback. However, we will always check for negative feedback when we solve op amp problems.

Solving Op Amp Circuits
We have seen that we can solve op amp circuits by using two assumptions: The Two Assumptions 1) i- = i+ = 0. 2) If there is negative feedback, then v- = v+. If not, the output saturates. The key to using these assumptions is being able to determine whether the op amp has negative feedback. Remember that we have negative feedback when a portion of the output is returned to the input, and subtracted from it.

Negative Feedback Identification
We have seen that we can solve op amp circuits by using two assumptions: For ideal op amps, we can assume that the op amp has negative feedback if there is a signal path from the output to the inverting input of the op amp. The Two Assumptions 1) i- = i+ = 0. 2) If there is negative feedback, then v- = v+. If not, the output saturates.

Most of the time, this feedback path is provided by using a resistor. For ideal op amps, we can assume that the op amp has negative feedback if there is a signal path from the output to the inverting input of the op amp.

In general the rule is this: If, when the output voltage increases, the voltage at the inverting input also increases immediately, then we have negative feedback. For ideal op amps, we can assume that the op amp has negative feedback if there is a signal path from the output to the inverting input of the op amp.

In general the rule is this: If, when the output voltage increases, the voltage at the inverting input also increases immediately, then we have negative feedback. For ideal op amps, we can assume that the op amp has negative feedback if there is a signal path from the output to the inverting input of the op amp. These are two different ways of saying the same thing. However, for most students this becomes clearer once we see some examples. We will look at one example in detail next, and then more examples after that.

Inverting Configuration of the Op Amp
One of the simplest op amp amplifiers is called the inverting configuration of the op amp.

The inverting configuration is distinguished by the feedback resistor, Rf, between the output and the inverting input, and the input resistor, Ri, between the input voltage and the inverting input. The noninverting input is grounded.

Note that the feedback resistor, Rf, between the output and the inverting input, means that we have negative feedback.

Note that the feedback resistor, Rf, between the output and the inverting input, means that we have negative feedback. Thus, we will have a virtual short at the input of the op amp,

Gain for the Inverting Configuration
Let’s find the voltage gain, which is the ratio of the output voltage vo to the input voltage vi. To get this, let’s define two currents, ii and if.

Next, since we know that the voltage v- is zero, we can write that the current ii is

Following a similar approach, since we know that the voltage v- is zero, we can write that the current if is

Next, by applying KCL at the inverting input terminal, we can write

Finally, we solve for vo/vi, by dividing both sides by vi, and then by multiplying both sides by -Rf, and we get

This is the result that we were looking for. As implied by our analysis of negative feedback, the gain is not a function of the op amp gain at all. The gain is the ratio of two resistor values,

Input Resistance for the Inverting Configuration
Let’s find the input resistance of this amplifier, which is defined as the Thevenin resistance seen by the source. The source is not shown here, but is assumed to be at the input. Here, we will take the source as the terminals connected to vi.

Input Resistance for the Inverting Configuration
The Thevenin resistance seen by the source will be the ratio of vi/ii. We have already solved for ii, and found that

Output Resistance for the Inverting Configuration
Let’s find the output resistance of this amplifier, which is defined as the Thevenin resistance seen by the load. The load is not shown here, but is assumed to be at the output. Here, we will take the load as the terminals connected to vo.

The Thevenin resistance seen by the load can be found by setting all independent sources equal to zero, and then applying a test source at the output. We do this here, applying a current source.

Now, we solve for vo/it, which is the output resistance, Rout. We know that v- = 0, due to the presence of negative feedback.

Now, we solve for vo/it, which is the output resistance, Rout. We know that v- = 0, due to the presence of negative feedback. Thus,

Now, we solve for vo/it, which is the output resistance, Rout. We know that i- = 0, due our first assumption. Thus,

Now, we solve for vo/it, which is the output resistance, Rout. Next, we write KVL around the loop marked with a dashed green line. Thus,

Now, we solve for vo/it, which is the output resistance, Rout. Since vo = 0, we have

Testing the Virtual Short Assumption
Let’s test the results we have obtained, so test the virtual short assumption. We found the gain, input resistance, and output resistance for this configuration. Let’s check our approach, by going back to the original equivalent circuit for the op amp. That is, we replace the op amp with a dependent source.

Testing the Virtual Short Assumption
Solving this circuit for the gain, vo/vi, we get If we take the limit as A goes to infinity, we get the same answer we had before.

Is This Assumption Really Valid?
This is a good question. You can check this by performing the solutions with actual values for real op amps. Try an open loop gain A of 106, and see how close your answers are. You can also check this by building an op amp circuit, and measuring the actual gain, and other parameters. You might be surprised by how accurate this assumption is. Go back to Overview slide.

ECE 3355 Electronics Lecture Notes Set 4 -- Version 21

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ECE 3355 Electronics Lecture Notes Set 4 -- Version 21

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