Continue to Site

# Source-absorption theorem

Status
Not open for further replies.

#### peterpops

##### Junior Member level 3
Evening!

I'm looking at an example (see figure 1).

They want to know an expression of Rin.

Solution:
"From the figure we see that the voltage vΠ will be equal to -ve. Thus looking between E and ground we see a resistance rΠ in parallel with a current source drawing a current gmve away from terminal E. The latter souce can be replaced by a resistance (1/gm), resulting in the input resistance Rin given by Rin = rΠ // (1/gm)"

I understand the solution, except what the bolded sentence has to do with it. Later there is an exercise with the following figure

Same question, an expression of Rin.

My solution:
Rin = (RB + rΠ) // (1/gm)

However that's not correct...
I see that vΠ is not equal to -ve anymore but how should I solve this one?

Applying KVL and KCL we have the following equations:

VE=-VΠ-RB*IR
IE=-IR-gm*VΠ
VΠ=IR*RΠ ==> IR=VΠ/RΠ

substituting this last in the previous two:

VE=-VΠ*(1+RB/RΠ)
IE=-VΠ*(1/RΠ+gm)

that is:

Rin = VE/IE = (RΠ+RB)/(1+gm*RΠ)

reyhane12

Points: 2