The first transistor has one VBE drop and second transistor has second VBE drop. The voltage divider produces VTH to the input base. The dc emitter current of the second stage is
IE2 = (VTH – 2 vBE ) / (RE )
The dc emitter current of the first stage that is the base current of second stage is given by
IE1 » IE2 / b2
If r'e(2) is neglected then input impedance of second stage is
Zin (2) = b2 RE
This is the impedance seen by the first transistor. If r'e(1) is also neglected then the input impedance of 1 becomes.
Zin (1) = b1 b2 RE
which is extremely high because of the products of two betas, so the approximate input impedance of Darlington amplifier is
Zin = R1 || R2
The Thevenin impedance at the input is given by
RTH = RS || R1 || R2
Similar to single stage common collector amplifier, the output impedance of the two stages zout(1) and zout(2) are given by.
Therefore, t he output impedance of the amplifier is very small.
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