Expression for closed loop gain of differential amplifier

In summary, the conversation is about deriving an expression for the closed loop gain of a differential amplifier. The solution provided is unclear for part 2 and there is confusion about the steps involved. It is later discovered that there is a mistake in the solution and the correct expression is (V1 x R4/R3)/(1 + R4/R3). After clarifying the steps, the confusion is resolved and the conversation ends with a thank you for the help.
  • #1
nothing909
170
4

Homework Statement


The question is

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

IMG_0042.JPG
I have solutions for these questions but the solution for this question is quite vague.

Here is the solution:

IMG_0041.JPG
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?
 

Attachments

  • IMG_0042.JPG
    IMG_0042.JPG
    53.7 KB · Views: 1,243
  • IMG_0041.JPG
    IMG_0041.JPG
    54.2 KB · Views: 823
Physics news on Phys.org
  • #2
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.
 
  • #3
[tex] \frac {V_1} {R_1 + \frac {R_3} {R_4} } [/tex] is a mistake, it should be
[tex] \frac {V_1} {1+ \frac {R_3} {R_4} } [/tex]

For the rest, isn't there some assumption like R1/R2 = R3/R4 ?
 
  • #4
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?
 
  • #5
nothing909 said:
i'm confused about where the 1 comes from, why does it just appear?

[tex] \frac{ R_4} {R_3+R_4} [/tex] divide the numerator and the denominator by R4
 
  • #6
ok, thanks. can you explain how it goes from

V1/(1 + R3/R4) then to = (V1 x R4/R3)/(1 + R4/R3)
 
  • #7
nothing909 said:
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)
 
  • #8
yea, i see that. i understand all the steps now, thanks for your help
 

FAQ: Expression for closed loop gain of differential amplifier

What is a differential amplifier?

A differential amplifier is a type of electronic amplifier that amplifies the difference between two input signals. It is commonly used in electronic circuits to amplify small signals and reject common-mode noise.

How is the closed loop gain of a differential amplifier calculated?

The closed loop gain of a differential amplifier is calculated by dividing the output voltage by the input voltage difference. It can be expressed by the equation A = -(Rf/R1), where Rf is the feedback resistor and R1 is one of the input resistors.

What is the advantage of using a differential amplifier over a single-ended amplifier?

A differential amplifier offers better noise rejection and higher common-mode rejection ratio compared to a single-ended amplifier. This makes it more suitable for amplifying low-level signals in noisy environments.

What factors can affect the closed loop gain of a differential amplifier?

The closed loop gain of a differential amplifier can be affected by the values of the input resistors, the feedback resistor, and the open loop gain of the amplifier. It can also be influenced by temperature, power supply voltage, and component tolerances.

How can the accuracy of the closed loop gain be improved in a differential amplifier?

The accuracy of the closed loop gain can be improved by using precision resistors with low tolerance values, minimizing temperature variations, and carefully selecting components with matching characteristics. Additionally, using a feedback capacitor can help reduce errors caused by the amplifier's open loop gain.

Similar threads

Replies
1
Views
2K
Replies
6
Views
1K
Replies
12
Views
1K
Replies
5
Views
1K
Replies
5
Views
3K
Replies
4
Views
2K
Replies
13
Views
2K
Back
Top