
#1
Jan2112, 01:11 PM

P: 596

1. The problem statement, all variables and given/known data
Find input resistance of the ckt (see attached) 2. Relevant equations 3. The attempt at a solution Q2 is diode connected, so I replaced Q2 with VBE resistance (rbe or r∏). So, Rin is rbe1+(β+1)rbe2 But the answer is somewhat different, it's rbe1+ (β+1) (rbe21/gm2) Where did 1/gm2 come from? 



#2
Jan2112, 06:26 PM

Sci Advisor
P: 4,003

Q2 is also a transistor and it draws collector current as well as base current.
So, this affects the total resistance of Q2 in this circuit. 



#3
Jan2212, 09:19 AM

P: 596

I tried to draw an equivalent diagram and solve. see attachment.
Basically, rbe and gm*vbe are both shorted (because of collector base short of Q2). I attached a test source Vx to determine the input impedance of just Q2. So impedance will be Vx/ix. After solving, I got 1/gm (assuming β>>1). From the equivalent ckt, Vx=Vbe ix = Vbe/rbe +gmVbe ix=Vbe (1/rbe+gm) ix = Vx (1/rbe+gm) Rin = Vx/ix = 1/(1/rbe+gm) rbe = β/gm Rin = 1/(gm/β +gm) Rin = 1/gm((1+β)/β)) if β>>1, then (1+β)/β = 1 Rin = 1/gm 


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