Calculating variable gain for difference amplifier

In summary, the conversation discusses the approach to solving a problem involving a fixed gain, resistors, and an op-amp. The suggested method involves assigning unknown voltages to various nodes and using KCL to solve for the equations. The final result is an instrumentation amplifier where the output voltage is equal to the difference between the two input voltages.
  • #1
Marshillboy
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0

Homework Statement



http://img706.imageshack.us/img706/263/hwstatement.png

Homework Equations



KCL/Node voltage method
Ideal op-amp rules

The Attempt at a Solution



I solved a previous, similar problem where the gain was fixed and the resistor mass going into the inverting and noninverting inputs of the opamp was merely a single resistor from each lead of Vi.

The way I've been attempting to solve this problem is by assigning unknown voltages (say V1, V2) to each of the leads of Vi, assigning two more unknowns to the top and bottom nodes of Rg (say, Rg1, Rg2), and another unknown voltage to the inverting/noninverting potential (say, Vb). Then, I find the node equations using KCL and attempt to solve. The resulting equations are extremely unwieldy and I feel like there must be a better way to approach this problem.
 

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  • #2
Your two input voltages (to each input of Vi) are not unknown. Call them V1 and V2. Then you have 4 independent (unknown) nodes (the two on either end of Rg and the op amp input and output nodes.

Let V2 = 0 and compute output voltage Vo1 for input V1. Then let V1 = 0 and compute Vo2 for input V2. I would use nodal equations so there are 4 equations for the 4 unknown voltages. Software can easily do the algebra for you and it doesn't make mistakes!
Then add the effect of the two input voltages to get the total output Vo = V01 + V02 (superposition principle).

If you do everything right you will note to your amazement that V0 = k(V1 - V2). In other words, if you apply the same voltage V (not saturating the inputs or output) to both inputs you get V0 = 0. This configuration is sometimes called an instrumentation amplifier.
 

1. How do you calculate the gain for a difference amplifier?

The gain for a difference amplifier can be calculated by dividing the output voltage by the input voltage. The formula for gain is (Rf/R1) + 1, where Rf is the feedback resistor and R1 is the input resistor.

2. What is the purpose of calculating variable gain for a difference amplifier?

The purpose of calculating variable gain for a difference amplifier is to control the amplification of the difference between two input signals. This allows for a customized output signal that is based on the difference between the two inputs.

3. How do you choose the values for the feedback resistor and input resistor in a difference amplifier?

The values for the feedback resistor and input resistor should be chosen based on the desired gain and the input signals. A larger feedback resistor will result in a higher gain, while a smaller input resistor will reduce the impact of the input signals on the output.

4. Can the gain for a difference amplifier be negative?

Yes, the gain for a difference amplifier can be negative. This means that the output signal will be inverted compared to the input signals. The gain can be positive, negative, or zero depending on the values of the resistors and the input signals.

5. Are there any limitations to calculating variable gain for a difference amplifier?

One limitation of calculating variable gain for a difference amplifier is that the gain cannot be infinitely large. There is a maximum gain that can be achieved based on the values of the resistors and the input voltage range. Additionally, the difference amplifier may introduce noise or distortion to the output signal if the gain is too high.

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