Solving for (β+1) in Thevenin Equivalent Homework

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Discussion Overview

The discussion revolves around understanding the derivation of the term (β+1) in the context of Thevenin equivalent circuits, specifically related to a common emitter BJT amplifier. Participants are exploring the relationships between collector current, emitter current, and the Thevenin voltage (vth) in their calculations.

Discussion Character

  • Homework-related
  • Technical explanation
  • Conceptual clarification

Main Points Raised

  • Some participants express confusion about the origin of the (β+1) term in their equations, specifically in relation to the Thevenin voltage (vth).
  • One participant notes that for a common emitter BJT amplifier, the collector current (Ic) is related to the base current (Ib) by Ic = β * Ib, suggesting a connection to the (β+1) term.
  • Another participant attempts to clarify that the (β+1) term arises when factoring the equation (β + 1)i + is = 0, indicating a step in the derivation process.
  • There is a concern raised about the substitution of variables in the equation for vth, with one participant questioning the validity of substituting an expression that still contains the variable i.

Areas of Agreement / Disagreement

Participants do not appear to reach a consensus on the derivation of the (β+1) term, with multiple interpretations and approaches presented. The discussion remains unresolved regarding the clarity of the derivation process.

Contextual Notes

Some participants reference specific values for beta and resistances, but the discussion lacks clarity on how these values affect the derivation of the equations. There is also an indication that the original circuit diagram is not provided, which may limit the understanding of the problem.

nunez2005
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Homework Statement


I'm trying to understand where the (β+1) comes from.


2. Homework Equations
vth=-βiR2
i+βi+is=0

vth=-β(-is/(β+1))R2


3. The Attempt at a Solution

vth=-βiR2
i+βi+is=0

vth=-β(-βi-is)R2
 

Attachments

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nunez2005 said:

Homework Statement


I'm trying to understand where the (β+1) comes from.


2. Homework Equations
vth=-βiR2
i+βi+is=0

vth=-β(-is/(β+1))R2


3. The Attempt at a Solution

vth=-βiR2
i+βi+is=0

vth=-β(-βi-is)R2


Welcome to the PF.

Without seeing the original circuit that this is modeling, it's hard to be sure. But for a CE BJT amplifier, your collector current Ic = β * Ib. So your emitter current is...
 
i've attached the solution that i have in the book, the value of beta is 150 and R1=100k and R2=39k, the only problem I have is that how do you get to the vth while substituting beta and the resistance in both sides to be equal to -Beta (-is/beta+1), the beta + 1 has to come out from simplying and combining like terms but I just can't see it.
 
nunez2005 said:

Homework Statement


I'm trying to understand where the (β+1) comes from.


2. Homework Equations
vth=-βiR2
i+βi+is=0

vth=-β(-is/(β+1))R2


3. The Attempt at a Solution

vth=-βiR2
i+βi+is=0

So factor out the i in that equation above:

(β + 1)i + is = 0

There's your (β + 1) term.

vth=-β(-βi-is)R2

That doesn't look right. You want to substitute for i in your equation for Vth from above, but what you've stuck in there also contains i.
 

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