# Input impedance of BJT amplifier

#### likephysics

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?

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#### vk6kro

Science Advisor
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.

#### likephysics

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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