# Differential Op amp gain

Discussion in 'Homework Help' started by Electromech man, May 2, 2009.

1. ### Electromech man Thread Starter New Member

May 2, 2009
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Hey folks,

Haven't done any electronics for a couple of years so a bit rusty. I have been staring at this question all day, and although I have five text books on hand still cannot crack it. I have a standard differential op amp circuit, and am given only the value of R2(150kohms). The first part asks if the gain of the amplifier is to be 50, what should be the value of R1.
In the diff op amp equations does the gain represent "Vout" or "V2-V1"? I am thinking that you would make "Vout" 50 and "V2-V1" 1, then just solve for R1. Am I on the right track?

Part b asks if the gain is 50, and V1=2+sin(120t)-0.04sin(500t), and V2=3+sin(120t)+0.02sin(500t) what is the output voltage. To me this is not in polar form or rectangular form, how do I solve it?

As I said I'm a bit rusty...

I have attached a photo of the circuit.

Any assistance would be much appreciated.

Cheers.

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2. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
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It is very difficult to make out the circuit notations - its just obvious it's a diff amp. I'm not sure which is supposed to be R2.

In any case you are on the right track - a diff amp multiplies the difference between V1 and V2 by the overall gain. Since this is an inverting amplifier

Vo = -K*(v1-v2) where K is the gain.

For part b

v1-v2 = -1 - 0.06sin(500t)

So

Vo = -50*(-1-.06sin(500t)) = 50+3sin(500t)

A 50V DC offset in a real diff op amp would be unrealistic for typical designs. However, it's just a question.

3. ### Electromech man Thread Starter New Member

May 2, 2009
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Thanks. The R2 is what I understand to be the feedback resistor, and the resistor between op amp and ground.

Cheers.

4. ### PRS Well-Known Member

Aug 24, 2008
989
35
The part a of the question, about the gain, is, I think, about 30kohms. This is because the configuration is that of a non-inverting op amp which has a gain of (Rf/Rs)+1. Rf is the feedback resistor, Rs is the sampling resistor and I think -- you have a blurry diagram -- you have 150k as the feedback resistor and by the formula 30k = Rs for approximately 50V/V gain.

5. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
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Well spotted PRS.

If the feedback is from the output to the positive terminal - which the schematic might (?) be showing - then the circuit won't even work as a diff amp.

I believe if the feedback was from the output to the negative terminal then it would work. The other input resistor(s) would need to be 3k to give a gain of 50x with the 150k feedback and 150k from positive terminal to ground.

6. ### Electromech man Thread Starter New Member

May 2, 2009
8
0
I have since learned that the +ve and -ve inputs are incorrect in the shown circuit, so upon correction this would make it a standard differential op amp.

Thus gain(A) = -(R2/R1), and with A = 50 and R2 = 150kohms R1 = -3kohms. This makes no sense to me as you can't have a -ve resistance!

What am I doing wrong?

Also in the second part with the subtraction of the sinusoidal voltage format I cannot find examples of this in any of my old notes or the text books. Can you tell me where I can find some good readings/examples of this?

Regards.

7. ### Ron H AAC Fanatic!

Apr 14, 2005
7,050
657
Vout=(V2-V1)*(R2/R1)

8. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
783
Hi Electromech Man,

Ron H has given a nice way of recasting the equation so that you won't be confused about what appears to be negative resistance issue.

While this helps in the diff amp case it won't help you when you come to think about the simpler, single-ended inverting op-amp case.

There you will be again be confronted with the amplifier gain being something like ...

Av=-R2/R1=-50 (say)

What this relationship is actually telling you is that the output signal is an amplified inversion of the input signal. What does this mean?

If the input signal is currently increasing in a positive direction then the output will be decreasing in a negative direction at a proportionate rate. This is often referred to as a phase inversion - common in many voltage amplifier circuit designs.

By way of numerical example for the diff amp case, suppose firstly that

V1=4V and V2=3.9V.

Then
Vo = -50*(4-3.9) = -5V.

If now V1=3.9V and V2=4.0V

Then
Vo = -50*(3.9-4.0) = +5V

Mercifully - one can design a non-inverting op amp amplifier stage.