Hoping you can open the attached asc file. I would like to know why the voltage gain from V1 to RL (from 1 volt to 10 volts) can be calculated from the value of RF2 plus 1k (10k) divided by the RF1 value (1k).

 

Could you clip the schematic and post it for those of us on devices that can’t run ltspice?

 

Gain = (9k+1K)/1k

 

Q1 and Q1 form a differential input with one input being the input signal and the other input being negative feedback from the output as determined by RF2 and RF1.
If we assume the amplifier open-loop gain is very high (the signal difference between the two differential inputs is then very small), then RF1 and RF2 provide attenuation in the negative feedback network that largely determines the amplifier gain , and the gain is as shown by ron.

 

Hoping you can open the attached asc file. I would like to know why the voltage gain from V1 to RL (from 1 volt to 10 volts) can be calculated from the value of RF2 plus 1k (10k) divided by the RF1 value (1k).

Unfortunately, the excerpt from the whole circuit (as prented by ronsimpson) does include neither V1 nor RL - nevertheles, my general comment is as follows:
You are asking " why the voltage gain from V1 to RL.... can be calculated from the value of RF2 ...divided by the RF1. "
This implies that the gain expression would be correct.
What is the source of this knowledge? Your own calculation or simulation or any textbook?
And - is the gain inverting or not?
The ratio of two resistors reminds me - of course - on opamps where the gain either is "-Ra/Rb" (inverting) or (1+Ra/Rb) which is non-invertig.

 

Q1 and Q1 form a differential input with one input being the input signal and the other input being negative feedback from the output as determined by RF2 and RF1.
If we assume the amplifier open-loop gain is very high (the signal difference between the two differential inputs is then very small), then RF1 and RF2 provide attenuation in the negative feedback network that largely determines the amplifier gain , and the gain is as shown by ron.

Unfortunately, the excerpt from the whole circuit (as prented by ronsimpson) does include neither V1 nor RL - nevertheles, my general comment is as follows:
You are asking " why the voltage gain from V1 to RL.... can be calculated from the value of RF2 ...divided by the RF1. "
This implies that the gain expression would be correct.
What is the source of this knowledge? Your own calculation or simulation or any textbook?
And - is the gain inverting or not?
The ratio of two resistors reminds me - of course - on opamps where the gain either is "-Ra/Rb" (inverting) or (1+Ra/Rb) which is non-invertig.

The equation holds true for different values of the two resistances.. I cant unfortunately remember where I got it from but I agree it looks like the op amp equation you mention except that the value added to the numerator is 1k not 1.

 

The equation holds true for different values of the two resistances.. I cant unfortunately remember where I got it from

It's the inverse value of the resistor attenuation.

 

..... except that the value added to the numerator is 1k not 1.

My writing was not correct: I wrote (1+Ra/Rb) and it should be [1+(Ra/Rb)]

 

There are two different equations because there are two different types of amplifiers.

 

This is a sim of the non-inverting opamp circuit for the given amp feedback network, giving the calculated AC gain of 10:

 

This is a sim of the non-inverting opamp circuit for the given amp feedback network, giving the calculated AC gain of 10:

However, the question remains: What is the relation to the circuit as given in post#1 ?
Is the circuit comparable to an ideal non-invering opamp-based amplifier?

 

Think of it this way. The divider in the feedback loop is placing 1/10 of the output on the inverting input. The circuit is stable when the two inputs are at the same voltage. So the output must be 10 times the input.

 

I would like to know why the voltage gain from V1 to RL (from 1 volt to 10 volts) can be calculated from the value of RF2 plus 1k (10k) divided by the RF1 value (1k).

Because that is the math of a non-inverting amplifier, which is derived from the math of a 2-resistor voltage divider.

https://en.wikipedia.org/wiki/Voltage_divider

https://en.wikipedia.org/wiki/Operational_amplifier#Negative-feedback_applications

https://en.wikipedia.org/wiki/Negative-feedback_amplifier

ak

 

Is the circuit comparable to an ideal non-invering opamp-based amplifier?

I think the open-loop gain of the posted audio amp circuit is high enough to give a reasonable approximation of the op amp circuit.
That differential stage with the current-mirror differential-to-single-ended conversion has high voltage gain.

 

So does Q3, although Ccc seems a bit large.

ak

 

I think the open-loop gain of the posted audio amp circuit is high enough to give a reasonable approximation of the op amp circuit.
That differential stage with the current-mirror differential-to-single-ended conversion has high voltage gain.

Yes - most probably.
My only problem is that I cannot identify any input or output signal which - normally - is reqired to say something about gain.

 

I never remember the formulae for op amp configurations. If you assume that they have very high open loop gain then any feedback arrangement will try to make the differential inputs have equal voltage. Typically it’s just a potential divider. Without feedback the output will swing to one rail or the other in a futile attempt to balance the inputs

 

I never remember the formulae for op amp configurations. If you assume that they have very high open loop gain then any feedback arrangement will try to make the differential inputs have equal voltage. Typically it’s just a potential divider. Without feedback the output will swing to one rail or the other in a futile attempt to balance the inputs

If you set RF2 to 1k, the same as RF1, the voltage output becomes 2k. The formula for the op amp (RF2 + 1)/RF! does not work, the formula (RF2 + 1k)/RF1 does, why is this?

 

If you set RF2 to 1k, the same as RF1, the voltage output becomes 2k.

Really ? How do you know?

The formula for the op amp (RF2 + 1)/RF! does not work, the formula (RF2 + 1k)/RF1 does, why is this?

Again - how do you know? (It`s wrong).
As I have already pointed out in my post#5 - it is unusual and problematic to make a false claim and then ask why it is correct.

 

Really ? How do you know?


Again - how do you know? (It`s wrong).
As I have already pointed out in my post#5 - it is unusual and problematic to make a false claim and then ask why it is correct.

Hust by the LTSPICE simulation