Learn How to Calculate g(x) with an Op-Amp Circuit | Helpful Tips and Tricks

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    Circuit Op-amp
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Discussion Overview

The discussion revolves around calculating the function g(x) in an op-amp circuit, specifically in the context of current versus resistance. Participants explore the circuit's configuration, equations, and the expected graph, while addressing uncertainties about the circuit's behavior and design.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant presents a circuit diagram and equations but expresses uncertainty about their correctness and how to derive g(x).
  • Another participant inquires about the input voltage and the meaning of variables in the context of the circuit.
  • A participant clarifies that the input voltage is represented by a specific symbol and identifies resistance as a variable in the equations.
  • Concerns are raised about the clarity of the original post, with one participant requesting a clearer explanation of the circuit's purpose and derivations.
  • One participant describes their intention to use the circuit in parallel with capacitors and an inductor to achieve negative resistance.
  • Another participant questions the necessity of having two identical blocks in the circuit design, suggesting that one may suffice.
  • A later reply suggests that the circuit may not produce the expected g(x) and emphasizes the need for non-linearity to achieve predictable oscillations.
  • One participant proposes a solution involving a one op-amp negative resistance with an additional resistor to ensure positive resistance and suggests testing its frequency response.
  • Another participant provides a mathematical expression related to the circuit's behavior, indicating a relationship between voltage and resistance.
  • A participant claims to have resolved the issue, attributing the behavior of g(x) to the limitations of the op-amp's output voltage and the necessity of using two op-amps in parallel.

Areas of Agreement / Disagreement

Participants express various uncertainties and disagreements regarding the circuit's design, the correctness of equations, and the expected behavior of g(x). No consensus is reached on the optimal configuration or the interpretation of the circuit's function.

Contextual Notes

Limitations include unclear definitions of variables, assumptions about circuit behavior, and unresolved mathematical derivations. The discussion reflects a range of interpretations and approaches to the problem.

Lasha
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So this is the circuit(with equations which were written by me, so I don’t know if they’re correct)
2akefy8.jpg

and this is the graph I should get, but I don’t know how.( g(x) is the current vs resistance)
fjnneb.jpg

I assume that those two equations collide somehow and I get that g(x), but I’m not sure.
So my question is: how do I get that g(x) with that circuit?
thanks
 
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Lasha said:
So this is the circuit(with equations which were written by me, so I don’t know if they’re correct)
2akefy8.jpg

and this is the graph I should get, but I don’t know how.( g(x) is the current vs resistance)
fjnneb.jpg

I assume that those two equations collide somehow and I get that g(x), but I’m not sure.
So my question is: how do I get that g(x) with that circuit?
thanks

How are you setting the voltage ø in your circuit? I see a black line at the top of your diagram -- is that where you are putting in the input voltage ø?

And in your plot, what is x?
 
Yes.Input voltage is that sign.X is resistance.Oh and I forgot to mention R1=R2 and R4=R5
 
I'm sorry, but I'm having a really hard time tracking your post and whatever questions you have. Much of what you have posted is nonsense, IMO.

"X is resistance" -- *What* resistance? What are you trying to do with this circuit?

And your simplifications for I1 and I4 look wrong to me. Can you post your math in those derivations? But *only* after you clarify what you are trying to do with this circuit please.
 
I'm trying to put this circuit in parallel with two capacitors and the inductor so it should work like a negative resistance but with function like that
 
Lasha said:
I'm trying to put this circuit in parallel with two capacitors and the inductor so it should work like a negative resistance but with function like that

Still makes no sense to me. Please post your full circuit diagram with your circuit analysis...
 
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md2vt5.png
 
Lasha said:
So this is the circuit(with equations which were written by me, so I don’t know if they’re correct)
It is not clear to me why the circuit needs two identical blocks. Either one alone seems to be all that is needed. Do you know why it is constructed in duplicate?

and this is the graph I should get, but I don’t know how.( g(x) is the current vs resistance)
fjnneb.jpg

I assume that those two equations collide somehow and I get that g(x), but I’m not sure.
So my question is: how do I get that g(x) with that circuit?
You don't get that g(x). What you obtain with your circuit is a straight line without the bends in it. This means that, while your circuit is capable of oscillating when connected to an LC circuit, you have no control over the amplitude of the oscillations. You will have to build in some non-linearity if you want the oscillations to be predictable and sinusoidal. The g(x) in your figure shows such a non-linearity.

I suggest that you construct a one OP-AMP negative resistance, add a resistor in series with it so that overall it has a positive resistance, then run some tests to see how good its frequency response is, producing a plot of resistance vs. frequency.
 
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i'll use e instead of phi.

it's a safe bet
e = e1 X R3/(R2+R3) = e2 X R6/(R5+R6)

and algebra should take you to I1 and I4
 
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  • #10
I've solved this.It turns out that g(x) ''breaks'' that way because of the fact that output voltage on the op-amp can't exceed supply voltage so the whole function changes.We need two op-amps in parallel so when whenever one of the output voltages reach their limits whole function doesn't become flat.Thanks anyways and sorry for not supplying enough details for it to make sense.
 
  • #11
glad you got it solved.

Indeed, the other safe bet is e1 and e2 won't exceed Vsupply.

Everybody learns that one the hard way.
 

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