The first transistor has one V_{BE} drop and second transistor has second V_{BE} drop. The voltage divider produces V_{TH} to the input base. The dc emitter current of the second stage is

I_{E2} = (V_{TH} – 2 v_{BE} ) / (R_{E} )

The dc emitter current of the first stage that is the base current of second stage is given by

I_{E1} » I_{E2} / b_{2}

If r'_{e(2)} is neglected then input impedance of second stage is

Z_{in (2)} = b_{2} R_{E}

This is the impedance seen by the first transistor. If r'_{e(1)} is also neglected then the input impedance of 1 becomes.

Z_{in (1)} = b_{1} b_{2} R_{E}

which is extremely high because of the products of two betas, so the approximate input impedance of Darlington amplifier is

Z_{in} = R_{1} || R_{2}

The Thevenin impedance at the input is given by

R_{TH} = R_{S} || R_{1} || R_{2}

Similar to single stage common collector amplifier, the output impedance of the two stages z_{out(1)} and z_{out(2)} are given by.

Therefore, t he output impedance of the amplifier is very small.
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