How Do You Calculate Thevenin Equivalent Circuits?

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Discussion Overview

The discussion revolves around calculating the Thevenin equivalent circuit with respect to specific terminals in a given circuit. Participants explore the relationships between short circuit current (Isc), open circuit voltage (Voc), and the impact of feedback on these parameters. The scope includes homework-related problem-solving and technical reasoning.

Discussion Character

  • Homework-related
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant calculated the short circuit current (Isc) and expressed confusion about deriving the open circuit voltage (Voc) from the given circuit parameters.
  • Another participant clarified that a fraction of the voltage (v2) is fed back into the circuit, which influences the calculations.
  • A question was raised regarding the relationship between Isc and Voc, specifically why Voc is not simply Isc divided by 50kΩ.
  • It was noted that the Thevenin resistance is not 50kΩ, and that feedback alters the output and input impedance, as well as the gain.
  • Participants discussed whether the voltage being calculated (Vth or Voc) is simply across the dependent current source and confirmed that the 50kΩ resistor remains part of the circuit model.

Areas of Agreement / Disagreement

Participants express uncertainty regarding the calculations of Voc and Isc, with some agreeing on the influence of feedback on circuit parameters, while others seek clarification on the relationships between these values. The discussion remains unresolved as participants explore different aspects of the problem.

Contextual Notes

There are limitations in the assumptions made about the circuit, particularly regarding the impact of feedback on impedance and the definitions of Voc and Isc. The discussion does not resolve these complexities.

orangeincup
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Homework Statement


Determine the Thevenin equivalent with respect to the terminals a,b

Homework Equations


Rth=Voc/Isc

The Attempt at a Solution


I'm doing this practice problem and I calculated the the short circuit current below:

500uA*100ohm/(100ohm+1310ohm)= 35.460 uA

since 80*ib=current in v2, 35.460 A * 80=2836.87 uA

So Isc=2836.87

Now for the open circuit voltage..
80ib* 50kohm= 40.0*10^5 ib
Now to solve for ib is where I'm lost.
In the solution they have 4*10^-5v2=-160ib, but I have no idea where it came from?
 

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A fraction of v2 is fed back to influence the input. That fraction is given as 4x10-5v2, so substitute in this your value for v2 (which you have in terms of ib).
 
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Thank you. I have a question about calcuating Isc vs. Voc, why isn't Voc simply Isc/50kohms?

I know the circuit is suppose to be "open" but isn't it still the same thing?
 
orangeincup said:
Thank you. I have a question about calcuating Isc vs. Voc, why isn't Voc simply Isc/50kohms?
It would be if the Thèvenin resistance were 50kΩ, but it isn't. That's what you are trying to find! The feedback changes the output impedance. And it changes the input impedance, and it changes the gain.
 
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So is the Vth or Voc I'm calculating simply the voltage across the 80ib dependent current source?

In the open circuit voltage I thought the 50kohm resistor would still be part of the circuit, and the voltage across that would equal the voltage across a-b?
 
orangeincup said:
So is the Vth or Voc I'm calculating simply the voltage across the 80ib dependent current source?
Yes, but the 50k remains in place, it's part of the model. (Without some resistive load, the voltage across a current source would be infinite.)

In the open circuit voltage I thought the 50kohm resistor would still be part of the circuit, and the voltage across that would equal the voltage across a-b?
Yes, and yes.
 
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