So, why are these points important? Well, let's go over a little history. Up until the invention of op-amps, engineers were limited to the use of transistors in amplification circuits. The problem with transistors is being 'current driven' devices, they always affect the signal of the circuit that the designer wants to amplify. Due to manufacturing tolerances, the gain of the circuits would vary significantly. Designing an amplifier circuit was a tedious process that required much trial and error. What engineers wanted was a simple device that they could attach to a signal that could multiply the value by any desired amount. The device should be easy to use and require very few external components. To paraphrase, operation of this amplifier should be a 'piece of cake'. Hence, the operational amplifier or op-amp.

To begin with, let's take a look at the special case I mentioned in the previous discussion. First, return to the previous block diagram and add a feedback loop.

Figure 1

You will see that I have represented the forward, or open loop gain with the value G, and the feedback gain with the value H. (This diagram should look very familiar to those of you who have had training in control theory.) First, you see that the output is tied to the negative input. This is called negative feedback. What good is negative feedback? Let's try an experiment. Take your hand and hold it an inch over your desk, keep it there. You are experiencing negative feedback right now. You are observing via sight and feel the distance from your hand to the desk. If your hand moves, you respond with a movement in the opposite direction. That is negative feedback. You invert the signal you receive via your senses and send it back to your arm. The same thing occurs when a negative feedback is applied to an op-amp. The output signal is sent back to the negative input. A signal change in one direction at the output causes a Vsum to change in the opposite direction.

You should get an intuitive grasp of this negative feedback configuration. Look at the diagram above, and assume a value of 50,000 for 'G' and a value of 1 for 'H'. Now start by applying a 1 to the positive input. Assume the negative input is at 0 to begin with. That puts a value of 1 at the input of the gain block 'G' and the output will start heading for the positive rail. But what happens as the output approaches 1? The negative input also approaches 1. The output of the summing block is getting smaller and smaller. If the negative input goes higher that 1, the input to the gain block 'G' will go negative as well forcing the output to go in the negative direction. Of course that will cause a positive error to appear at the input of the gain block 'G' starting the whole process over again. Where will this all stop? It will stop when the negative input is equal to the positive input. In this case since 'H' is 1 the output will be 1 also.

You have learned this in control theory. Look at the basic control equation in reference to the diagram above.

Vo = Vi * (G/(1+G*H).

What happens when G is very large? The 1 in the denominator becomes insignificant and the equation becomes Vo = approximately Vi*(1/H). Well, H in this case is 1 so it follows that Vo = approximately Vi* (1/1) or,

Vo = Vi

This is the special case where you can assume that the inputs of the op-amp are equal. Apply it ONLY when there is negative feedback. Now when feedback gain is one this also demonstrates another neat op-amp circuit: the voltage follower. Whatever voltage is put on the positive, input will appear at the output.

Figure 2

Take a look at the drawing above. This is an op-amp in the negative feedback configuration. When you look at this, you should see a summer and an amplifier just like in the previous drawing. In this configuration, you can make the assumption that the positive and negative inputs are equal.

Negative feedback is the case that is drilled into you in school, the one that often causes confusion. It is a special case, a very widely used special case. None the less, if you do not have negative feedback and the inputs and output are within operational limits, you must NOT assume the inputs of the op-amp are equal.

So why is this negative feedback configuration used so much? Well remember the reason op-amps were invented? Amplifiers were tough to make. There had to be an easier way. Take a look at the control equation again.

Vo = Vi*(G/(1+G*H))

I have already shown that for large values of G, the equation approximates.

Vo = Vi*(1/H)

You will see that the amplification of Vi depends on the value of H. for example, if we can make H equal 1/10 then Vo = Vi*(1/(1/10)) or,

Vo = Vi * 10.

So how do we go about doing that? Do you remember the voltage divider circuit? That would be very useful here as we would like H to be the equivalent of dividing by 10. Let us insert the voltage divider circuit in place of H. (note, Vi will be connected to V+)

Figure 3

Notice that the input to the voltage divider is coming from the output of the op-amp, Vo. The output of the voltage divider is going to the negative input of the op-amp V-. Will the op-amp input V- affect the voltage divider circuit? NO! It has high impedance. It will not affect the divider. (If you didn't get that, go back and read part 1 till you do!) Now since the input to the divider is hooked to a voltage source, and the output is not affect by the circuit, we can calculate the gain from Vo to V- very easily with the voltage divider rule.

V-/Vo = Ri/(Ri+Rf) = H

Thus is follows that,

1/H = (Ri+Rf)/Ri or with a little algebra, 1/H = (Ri/Ri) + Rf/Ri = (Rf/Ri) + 1

1/H=(Rf/Ri)+1

There you have it, the gain of this op-amp circuit. Let's look at it another way. Go back to the previous equation.

V-/Vo = Ri/(Ri+Rf)

Now we learned that in this special case of negative feedback we can assume that V+ = V-. This is because the negative feedback loop is pushing the output around trying to reach this state. So let's assume that Vi = V+ which is where the input to our amplifier will be hooked up. Now we can replace V- with Vi, and the equation looks like this.

Vi/Vo = Ri/(Ri+Rf)

What we really want to know is what does the circuit do to Vi to get Vo. So lets do a little math to come up with this equation.

Vo = Vi*((Ri+Rf)/Ri) = Vi*(Rf/Ri +1) or Vo/Vi = Rf/Ri + 1

Please note that this is equal to 1/H. The gain of this circuit is controlled by 2 simple resistors. And believe me, that was a whole lot easier to understand and calculate than a transistor amplification circuit. The operation of this amplifier was pretty easy to understand.

Now for a recap. The important points to negative feedback in an op-amp circuit.

1. The negative feedback configuration is the only time you can assume that V- = V+.
2. The high impedance inputs and the low impedance output make it easy to calculate the effects simple resistor networks can have in a feedback loop.
3. The high open loop gain of the op-amp is what makes the output gain of this special case equal to approximately 1/H.
4. All this adds up to a simple and easy way to make an amplification circuit.

In the next installment, I will discuss the virtues of positive feedback.

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