# Thevenin equivalent

## 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

• problem.png
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berkeman
Mentor

## 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.

gneill
Mentor

## 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.