Zout = - dVout / dIout
The minus sign in the derivative comes from the fact the output impedance has the effect of decreasing Vout. The output current IE, which is related to the base current.
Vin = IBRS + VBE +VE [Vout = VE]
Vout = VE = Vin – IBRS – VBE [Iout = IE and IB = IE/(β + 1)]
Vout = Vin – {IE/(β + 1)}RS – VBE = Vin – {Iout/( β + 1)}RS – VBE
= - Iout{RS/(β + 1)} + (Vin – VBE)
So, Zout = - dVout/dIout = -d/dIout [ - Iout{RS/(β + 1)} + (Vin – VBE) ]
Zout = RS/(β + 1)
Thus we obtain the result that the impedance of the source, as viewed by the load, is reduced by the factor ~ 1/ β.
Zout = {RS/( β + 1) ≈ RS/β |
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