What is the Thevenin equivalent circuit?

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Discussion Overview

The discussion revolves around the concept of the Thevenin equivalent circuit, specifically how to derive it from a given circuit with three nodes. Participants explore the calculation of Thevenin resistance and voltage, addressing the confusion regarding the configuration of resistors and the measurement points for voltage.

Discussion Character

  • Homework-related
  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • One participant expresses confusion about obtaining the Thevenin equivalent circuit from a circuit with three nodes, questioning the parallel configuration of resistors.
  • Another participant explains how to find the open circuit voltage using a voltage divider, arriving at a value of 3.33V.
  • A later reply confirms understanding of the Thevenin resistance and proposes an equivalent circuit diagram.
  • Participants discuss the expression for Thevenin voltage and resistance, with one providing a formula that includes the parallel resistance of R1 and R2.
  • Questions arise about the correct measurement points for the open-circuit voltage, specifically whether it is between V1 and Vb or V2 and Vb.
  • Clarifications are made regarding the measurement of voltage, with one participant stating that it is measured between Vb and V2, as V2 is considered ground.
  • Another participant emphasizes the relevance of the voltage across the base-emitter junction in the context of a transistor.

Areas of Agreement / Disagreement

Participants express differing views on the measurement points for open-circuit voltage and the configuration of the circuit, indicating that the discussion remains unresolved with multiple competing perspectives.

Contextual Notes

There are unresolved questions regarding the assumptions made about the circuit configuration and the definitions of measurement points, which may affect the calculations presented.

Defennder
Homework Helper
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Homework Statement


I really don't see how the Thevenin equivalent circuit is obtained from the one on the left. The way learned to find Thevenin equivalent circuit considers the circuit network in between 2 nodes. This one appears to have 3 nodes. And why is the thevenin resistance given as though the resistance were connected in parallel? I don't see why they are in parallel.


2. Homework Equations /pictures
Thevenin equivalence.
http://img380.imageshack.us/img380/5082/theveninequivalentvm4.jpg
 
Last edited by a moderator:
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To obtain the open circuit voltage you use a voltage divider with the 10V source and the two resistors, getting 3.33V.

For the equivalent resistance you short circuit the source, so the two resistors are in parallel (the upper terminal of R1 is connected to ground).
 
Changed image

Yeah, thanks, I think I can see it for the Thevenin resistance. I assume that the above gray box is also equivalent to this:

http://img508.imageshack.us/img508/8425/thevenincircuitwi3.jpg

But the open circuit voltage is between V1 and Vb or V2 and Vb?
 
Last edited by a moderator:
for this circuit there is an expression to calculate v(th)& R(th) which is:
R(th)=R1//R2
v(th)=R(th)[(v1/R1)+(v2/R2)]
where v1=10v,v2=0v
// means parallel
 
Where did you get that expression for v(th) from? And more importantly is the open-circuit voltage between V1 and Vb or V2 and Vb? That's all I need to be able to figure out the thevenin voltage.
 
v(th) comes from the Voltage divider rule.

v(th) is measured between vb and v2.
 
But why isn't it measured between V1 and Vb?
 
Defennder said:
But why isn't it measured between V1 and Vb?

Because V2 is ground and you want to polarize the transistor, so you are really interested in the voltage between base and emitter: Vbe = Vb - Ve.
 
Oh yeah you're right. It's the voltage across the Vbe that's relevant here. Thanks a lot!
 

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