- #1

- 2

- 0

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter nunez2005
- Start date

- #1

- 2

- 0

- #2

berkeman

Mentor

- 59,634

- 9,771

## 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.....

- #3

- 2

- 0

- #4

gneill

Mentor

- 20,925

- 2,866

## 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 + i

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.

Share: