|Jan21-12, 01:11 PM||#1|
Input impedance of BJT amplifier
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) (rbe2||1/gm2)
Where did 1/gm2 come from?
|Jan21-12, 06:26 PM||#2|
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.
|Jan22-12, 09:19 AM||#3|
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,
ix = Vbe/rbe +gmVbe
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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