Differential OP AMP Circuit Question

The circuit is already pretty simple, so it's not too hard to find the ac gain.If you accidentally labeled the 300 ohm resistors so it should have been 200 ohms, then you should come up with a computation as gneill asks for. The circuit is already pretty simple, so it's not too hard to find the ac gain.In summary, the given circuit has a 200 ohm resistor in the feedback path and a 300 ohm resistor in the input path. This results in a large common-mode gain, making it not a true differential amplifier. To determine the AC gain, we need to find the voltage at Vout and divide it by the voltage at Vin.
  • #1
joey5981
1
0

Homework Statement


Find Vin/Vout

Homework Equations


Vout = Rf/R1(Vb-Va) ?

The Attempt at a Solution


I think the answer is the first one.

Answer choices:
Vin/Vout = -2, Rin = 100

Vin/Vout = -2, Rin = 200

Vin/Vout = -2.5, Rin = 100

Vin/Vout = -2.5, Rin = 200

Vin/Vout = -3, Rin = 100

Vin/Vout = 2, Rin = 100

Vin/Vout = 2, Rin = 200

Vin/Vout = 2.5, Rin = 100

Vin/Vout = 2.5, Rin = 200

None of the above
 

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  • #2
Can you show us how you arrived at your conclusion? We won't confirm a guess...
 
  • #3
Hi joey5981. http://img96.imageshack.us/img96/5725/red5e5etimes5e5e45e5e25.gif

So you believe the 300 ohms resistor (and its 100 ohm companion) do nothing here? How can you prove this?

Vin/Vout = -2, ...
The gain was given as this? Or did you mis-type what the book showed as Vout/Vin?
 
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  • #4
Due to the unequal 200 and 300 ohm resistors this is not a differential amplifier in the accepted sense. In other words, it has a huge common-mode gain.

If you accidentally labeled the 300 ohm resistors so it should have been 200 ohms, then you should come up with a computation as gneill asks for.
 
  • #5


Assuming the circuit is an inverting operational amplifier, the correct answer would be Vin/Vout = -2, Rin = 100. This is because the inverting amplifier equation is Vout = -Rf/Rin * Vin, and in this case, Vb = 0 (since the negative input is grounded) and Va = Vin. Therefore, substituting these values into the equation given in the homework statement, we get Vout = -Rf/R1 * Vin, which matches the inverting amplifier equation. The value of Rf/R1 in this circuit is -2, and since Rin is equal to R1, the value of Rin is 100.
 

1. What is a differential OP AMP circuit?

A differential OP AMP circuit is a type of operational amplifier (OP AMP) circuit that has two inputs and one output. It is used to amplify the difference between the two input voltages, while rejecting any common-mode signals that are present on both inputs.

2. How does a differential OP AMP circuit work?

A differential OP AMP circuit works by comparing the voltage difference between its two inputs and amplifying that difference to produce an output voltage. The circuit is designed to have very high gain, which means even small differences between the inputs can be amplified significantly.

3. What are the advantages of using a differential OP AMP circuit?

There are several advantages to using a differential OP AMP circuit, including high gain, high input impedance, low output impedance, and the ability to reject common-mode signals. This makes it useful in applications where precise amplification and rejection of noise is necessary, such as in instrumentation and control systems.

4. How is a differential OP AMP circuit different from a regular OP AMP circuit?

A regular OP AMP circuit has only one input and one output, while a differential OP AMP circuit has two inputs and one output. Additionally, a differential OP AMP circuit is designed to amplify the difference between its inputs, while a regular OP AMP circuit amplifies the voltage at its single input.

5. What are some common applications of differential OP AMP circuits?

Differential OP AMP circuits are commonly used in instrumentation and control systems, audio amplifiers, and signal processing circuits. They are also used in communication systems, such as in differential amplifiers for balanced audio signals or in differential receivers for high-speed data transmission.

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