BASIC FEEDBACK THEORY QUESTION

[walid1]

I read this:
"The ideal parameters are never fully realized but they present a very convenient method for the preliminary analysis of circuitry. So important are these ideal definitions that they are repeated here. The ideal amplifier possesses
1. Infinite gain
2. Infinite input impedance
3. Infinite bandwidth
4. Zero output impedance
From these definitions two important theorems are developed.
1. No current flows into or out of the input terminals.
2. When negative feedback is applied, the differential input voltage is reduced to zero."
My question is about the last line, that is what the meaning of the differential input voltage?

 

[audioguru2]

Hi Walid,
The differential input voltage is the voltage between inputs, and if the amplifier has an infinite gain would be zero when negative feedback is applied. ;D

 

[Kevin Weddle]

I think the real purpose of having differential inputs into an amplifier is so that you can apply two signals on a pn junction and have the resulting voltage be gained. This is what you do if you want to combine two signals but in a way that is different from using resistors. You could use resistors to get a voltage addition and then send this to an amplifier. Or you can do it in the same breathe.

 

[audioguru2]

No Kevin,
Differential inputs are used so that they can be exactly the same, for very low input offset voltage and drift. Also the differential inputs allow negative feedback to be applied, resulting in cancellation of any non-linearity and distortion between the amplifier's input and output.

 

[Kevin Weddle]

The ideal parameters of an opamp are building blocks. I have seen zero difference between the inputs at low gain, but not at higher gain. Imagine an amplifier meant to amplify so much that you use it for very low gain. Maybe even a gain of only one, a buffer you know. Anyways, I like to see and measure things. Not sure what's going on with that low gain. Also, you might want to go with lower value resistors to see the difference. It seems that a sufficient output current from the opamp helps to establish the difference.

In the end, you should have input signal that is gained by whatever your gain is. You can actually take the DC difference and multiply by the gain to get your output.

 

[audioguru2]

you might want to go with lower value resistors to see the difference. It seems that a sufficient output current from the opamp helps to establish the difference.

Opamps can't supply much output current for lower value resistors. The input impedance of opamps is so high the lower value resistors are not necessary. Most opamps have a minimum of 2k ohms load spec'd. In order for the load to receive most of the output then the feedback resistors must be much higher than 2K.

 

[Kevin Weddle]

Yes, I know. Maybe I would go with an Rf of about 10K. Of course the opamp input current is well below the feedback current. Remember that article about using a 10K output resistor to the negative supply. That worked great to get a difference. I think that this should work fairly well too. But yes I have read about the signal being too small to measure. I think there is something going on with low gain. It's like the input floats.

I think I have it. If you have equal RF and RI the divider will set the input at zero. But why a gain of one with no difference on the input? It takes high gain to send the small input to a larger output. The answer is that the gain through the opamp is still high because RF being 100k and RI being 100K puts the impedance seen by the output at 200K. That RI being 100K might as well be 10K to the opamp because from the simplified drawing of an opamp, it is in parallel with RE. Let me sketch what I'm talking about.

View attachment 37341

 

[audioguru2]

Kevin,
The ridiculous article about using a 10K output resistor to the negative supply was to operate its output NPN emitter-follower in class-A, to avoid the 0.00008% crossover distortion from its complimentary output PNP emitter-follower.

You must remember that an opamp with negative feedback uses its voltage gain of from 200,000 to millions to keep the voltage at its inputs the same.

An inverting opamp circuit with equal values for its Ri and Rf has a voltage gain exactly 1, because the opamp senses the current in the input resistor by its voltage drop and negative feedback forces the current in the feedback resistor to be the same and the input voltages to be the same.

Example #1: An opamp with +15V and -15V supplies. The non-inverting input is connected to 0V. Ri and Rf are 10k equally.
1) Feed +10V to the input of Ri.
2) Ri is connected to the inverting input of the opamp and it begins to rise.
3) With the non-inverting input at 0V and the inverting input rising, the output of the opamp goes to exactly -10V to keep the current in the resistors and its input voltages the same. Therefore the inverting voltage gain is exactly 1.

Example #2: As above but Rf is 100k for an inverting gain of 10.
1) Feed +1V to the input of 10k Ri.
2) Ri is connected to the inverting input of the opamp and it begins to rise.
3) The output of the opamp goes negative to exactly -10V to keep the current in the resistors and its input voltages the same. Therefore the inverting voltage gain is exactly 10.

Example #3: As above but Rf is 1M for an inverting voltage gain of 100.
1) Feed +0.1V to the input of 10k Ri. Then figure out what happens at the output yourself. ;D

 

[Kevin Weddle]

Everything you say seems right to me. But can't we explore it a little further. I have read all that stuff and it gets very repetative. It seems like that old list of ideal opamp parameters is always popping up. The circuit isn't that bad. I know that the output resistor isn't a collector resistor and that the input resistor is not an emitter resistor. I am not confused when it comes to opamps.

 

[audioguru2]

Kevin,
I didn't see your sketch before. It is a differential pair of transistors, definitely not an opamp.
Its output resistance is high and its input resistance is fairly low, the exact opposite of an opamp. Therefore the negative feedback voltage divider is messed-up and is difficult to calculate.

I recall testing a differential pair like yours long ago. Its output was distorted like mad if it exceeded only a few hundred millivolts. Unless you keep its input very low like opamps, its gain is impossible to measure.