Now we will do a little intuitive analysis. Don't forget the important points we learned in parts 1 and 2. Remember these basic rules still apply.

1. The inputs are high impedance. They have negligible effects on the circuit that they are hooked to.
2. The inputs can have different voltages applied to them. The do NOT have to be equal.
3. The open loop gain of an op-amp is VERY high.
4. Due to the high open loop gain and the output limitations of the op-amp, if one input is higher that the other the output will 'rail' to its maximum or minimum value.
5. The negative feedback configuration is the only time you can assume that V- = V+.
6. 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.
7. The high open loop gain of the op-amp is what makes the output gain of the negative feedback case equal to approximately 1/H.

First, apply 0 volts to Vin. In this case the input is connected to V-. You also see that the output is connected via a resistor to a reference voltage, Vref. What is the voltage at V+? Does the voltage at V+ equal the voltage at V-? NO! (Apply rules 2 and 5) So what is the voltage at V+? Well, that depends on two things. The voltage at Vref and the output voltage of the amplifier, Vo. Does the V+ input load the circuit at all? No, it does not. (Apply rules 1 and 5) To begin the analysis, let Vref = 2.5 V, and assume the output is equal to 0V. Now what is the voltage at V+? Since Vo is equal to 0, we have a basic voltage divider again. Assume Rref = 10K and Rh = 100K

V+ = Vref*(Rh/(Rh+Rref)) = 2.5*100K/110K = 2.275V

So now there is 2.275V at V+ and 0V at V-. What will the op-amp do? Refer to the basic rule number 4 and remember the block diagram of the op-amp. If you memorize this diagram and always apply it when you analyze an op-amp circuit you will come to understand them intuitively. Imagine the signals applied to the positive and negative inputs then determine the output.

Figure 2

So what do we have? Well Vsum is equal to V+ - V- or, in this case, Vsum = 2.275V. Vo is equal to Vsum*G. The output will obviously go to the positive rail. (If this is not obvious to you, you need to review part 1 more carefully.) Now we have Vo at the positive rail. Let's assume that it is 4V for this particular op-amp. (Remember, the output rails depend on the op-amp used, and you should always refer to the data sheets for that information. 4V used in this case is typical for an LM324 with a 0 to 5V supply)

The output is at 4V, V- is at 0V, but what about V+? Well, it has changed. We must go back and analyze it again. (Do you feel like you are going in circles? You should. That is what feedback is all about, outputs affect inputs which affects the outputs, and so on, and so on.) The analysis this time has changed slightly. It is no longer possible to use just the voltage divider rule to calculate V+. We must also use superposition.

In superposition, you set one voltage source to 0 and analyze the results, then you set the other source to 0 and analyze the results. Then you add the two results together to get the complete equation. So let's do that now. We already know the result due to Vref from above. Here is the positive feedback diagram again for reference.

Figure 3

Here is the result due to Vref using the voltage divider rule.

V+ due to Vref = Vref*Rh/(Rh+Rref)

Here is the result due to Vo using the voltage divider rule.

V+ due to Vo = Vo*Rref/(Rref+Rh)

The result due to both is thus.

V+ = (V+ due to Vref) + (V+ due to Vo) or

V+= (Vref*Rh/(Rh+Rref)) + (Vo*Rref/(Rref+Rh))

Now insert all the current values and we have.

V+ = (2.5*100K/110K)+(4*10K/110K) = 2.64V

Is this circuit stable now? Yes it is. We have 0V at V-, and 2.64V at V+ this results in a positive error, which when amplified by the open loop gain of the op-amp causes the output to go to the positive rail. This is 4V, which is the state that we just analyzed.

So let's change something. Slowly ramp up the voltage at V-. At what point will the op-amp output change? Right after the voltage at V- exceeds the voltage at V+. This results in a negative error, causing the output to swing to the negative rail. And what happens to V+? Well, it changes back to 2.275V as we calculated above. So how do we get the output to go positive again? We adjust the input to less that 2.275V. The positive feedback RE-ENFORCES the change in the output. Making it necessary to move the input farther in the opposite direction to affect another change in the output.

The effect that I have just described is called hysteresis. It is an effect very commonly created using a positive feedback loop with an op-amp. What is hysteresis good for you ask? Well, heating your house for one thing. It is hysteresis that keeps your furnace from clicking on and off every few seconds. Your oven and refrigerator use this principle as well. In fact, the disk drive on the computer you are using to read this uses magnetic hysteresis to store information.

One important item to note: The size of the hysteresis window depends on the ratio of the 2 resistors Rref and Rh. In most typical applications Rh is much larger that Rref. If the signal at Vi is smaller than the window, it is possible to create a circuit that latches hi or low and never changes. This is usually not desired and can be avoided be performing the analysis above and comparing the calculated limits to the input signal range.

Now that we have covered these 3 basic configurations of an op-amp, let's put together a simple circuit that uses them. Here we have a voltage follower, hooked to a comparator using hysteresis, with an LED as an indicator.

Figure 4

You should build in your lab gain an intuitive understanding of what has been discussed. Experiment with feedback changes in all parts of the circuit. Note that you can change the input potentiometers from 5K to 100K with out affecting the voltage at which the comparator switches.

There you have it, the basics of op-amp circuits. With this information, you can analyze most op-amp circuits you come across and build some really neat circuits yourself. What about filters you say! Well, isn't a filter nothing more than an amplifier that changes gain depending on the frequency? Simply replace the resistors with an impedance and thus add a frequency component to the circuit. What about oscillators you say? These are feedback circuits where timing of the signals is important. They still follow the rules above. I believe that grasping the basics of any discipline is the most important thing you can do. If you understand the basics, you can always build on that foundation to obtain higher knowledge, but if you do not 'get the basics' you will flounder in your chosen field.

One last question. Is this circuit depicting positive or negative feedback? If you can answer it, you know your op-amps!

Figure 5

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