Expression for closed loop gain of differential amplifier

[nothing909]

Homework Statement

The question is

Derive an expression for the closed loop gain of the differential amplifier.

I have solutions for these questions but the solution for this question is quite vague.

Here is the solution:

I understand part 1 and part 3, but part 2 I don't.

So looking at part 2, it starts by doing a voltage divider to find the voltage at V* by doing V1 x R4/R3xR4

The rest of the step in part 2, I don't understand.

How does it go from from V* = V1 x R4/R3xR4 then to = V1/(R1 + R3/R4)
then to = (V1 x R4/R3)/(1 + R4/R3)

Is there some steps that should be in between to show more clearly what is happening because I can't follow it. Can someone explain how to get from each step to the next?

 

[gneill]

Unless there's some special relationships between the resistor values that we haven't been told about I don't see how the derivation makes sense. certainly the final result is rather dubious, depending as it does upon only two of the four resistor values.

 

[willem2]

\frac {V_1} {R_1 + \frac {R_3} {R_4} } is a mistake, it should be
\frac {V_1} {1+ \frac {R_3} {R_4} }

For the rest, isn't there some assumption like R1/R2 = R3/R4 ?

 

[nothing909]

Yes, R1/R2 = R3/R4, it says that in the solutions.

and yes, that is a mistake.

i'm confused about where the 1 comes from, why does it just appear?

 

[willem2]

i'm confused about where the 1 comes from, why does it just appear?

\frac{ R_4} {R_3+R_4} divide the numerator and the denominator by R4

 

[nothing909]

ok, thanks. can you explain how it goes from

V1/(1 + R3/R4) then to = (V1 x R4/R3)/(1 + R4/R3)

 

[willem2]

ok, thanks. can you explain how it goes from

V1/(1 + R3/R4) then to = (V1 x R4/R3)/(1 + R4/R3)

I also get that from the leftmost expression by dividing numerator and denominator by R3. (So the middle expression seems to have no point)

 

[nothing909]

yea, i see that. i understand all the steps now, thanks for your help