Gain of Electric Circuit: Vo/Vs and Resistance R

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Discussion Overview

This discussion revolves around the gain of an electric circuit, specifically the relationship between output voltage (Vo) and input voltage (Vs) in terms of resistance (R). The context includes theoretical analysis and application of circuit laws such as Kirchhoff's Current Law (KCL) and Kirchhoff's Voltage Law (KVL) as well as operational amplifier (op-amp) characteristics.

Discussion Character

  • Homework-related
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant calculates the gain as Vo=4Vs based on KCL but finds this result odd since it does not include resistance R, questioning the provided answer of -4R/(R+80).
  • Another participant points out that the ground connection provides an additional current path, which affects the current assumptions in the circuit analysis.
  • A participant questions whether removing the ground connection would change the gain, suggesting that the gain should remain consistent regardless of the ground's presence due to the application of KCL and KVL.
  • Another participant suggests using known formulas for an inverting op-amp adder instead of starting from basic KCL and KVL rules, indicating a different approach to the problem.
  • There is a discussion about the implications of removing the ground connection and how op-amps still provide current paths to ground, which may influence the gain calculation.

Areas of Agreement / Disagreement

Participants express differing views on the impact of the ground connection on gain calculations and the assumptions made regarding current paths in the circuit. There is no consensus on the correct approach or final gain expression.

Contextual Notes

Participants highlight potential limitations in their assumptions regarding current flow and the role of ground in the circuit, indicating that the analysis may depend on specific circuit configurations and definitions.

TwoEG
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Homework Statement



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So this problem is on Introduction to Electric Circuits 9th Edition (Richard C.Dorf)
This asks me to express the gain of this circuit Vo/Vs by terms of the resistance of R.

Homework Equations



KCL, KVL, some ideal OP AMP characteristics...

The Attempt at a Solution



Look at the undermost (horizontal) wire. Leftmost (perpendicular) wire has escaping current Vs/20(A). Middle (perpendicular) wire has no current (connected to OP AMP). Rightmost (perpendicular) wire has entering current Vo/80(A). Apply supernode KCL. Vs/20=Vo/80. Vo=4Vs. So gain is 4. Huh, this is weird. There is no variable R in this gain. Actually, provided answer sheet says gain is -4R/(R+80). But I cannot understand this answer. This gain has minus sign, so Vo and Vs must have different sign. This means undermost wire can only have entering current or escaping current. What's the problem with my logic?
 
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The ground connection on the bottom rail is another current path, as is the output of the first op-amp. So you can't assume that the current that flows through Vs must be the same as the current that flows through the 80k resistor.
 
gneill said:
The ground connection on the bottom rail is another current path, as is the output of the first op-amp. So you can't assume that the current that flows through Vs must be the same as the current that flows through the 80k resistor.

Ahh... I forgot about that. But one more question. Is this mean that if I remove ground in this circuit, than the gain of circuit will become 4? I think gain of circuit must be same whether ground exists or not, since calculation using KCL and KVL on other parts of the circuit will be same.
 
Are you required to start the gain calculation with the basic KCL and KVL rules?
Why not using the known formulas for an inverting opamp adder?
That`s what the 1st stage does: It simply calculates the weighted sum of two input signals (Vs and Vo): V1=-[Vs(100k/20k)+Vo(R/20k)
 
TwoEG said:
Ahh... I forgot about that. But one more question. Is this mean that if I remove ground in this circuit, than the gain of circuit will become 4? I think gain of circuit must be same whether ground exists or not, since calculation using KCL and KVL on other parts of the circuit will be same.

The op-amps will still provide "hidden" current paths to ground, even if the bottom rail is made into an essential node by removing its ground connection. Think of them as controlled voltage sources with one "leg" of the source connected to ground.

My suggestion is to write KCL at the first op-amp's negative input and recognize how V1 is related to Vo.
 
Thanks for your replies! Helped me a lot :)
 

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