Small-signal BJT analysis

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corsair

Joined Mar 6, 2010
51
Hi guys,

I attached the answer for the problem. I am having trouble understanding one part of the solution in finding Rinb.

My original solution was:
Rin = (10k + 10k) || (.0145 + 2k)

When they solve for Rinb, they multiply the resistors in the emitter by beta + 1 (where beta = 100). Why is this?
 

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t_n_k

Joined Mar 6, 2009
5,455
The small signal emitter current ie is (1+β)ib.

"Looking into" the Base one sees an effective small signal resistance of

rB=vb/ib=ie(re+RE)/ib=(1+β)ib(re+RE)/ib=(1+β)(re+RE)
 
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