We can easily find Zin from what we know already of the behavior of the transistor. We know that the sum of VBE and the IR drop across RE must equal Vin.
Vin = VB = VBE +VE = VBE + IERE [IR = IC + IB = βIB + IB = (β + 1)IB]
Vin =VBE + IERE = IB(β + 1)RE [IB = Iin]
Vin = VBE + Iin (β + 1) RE
Taking the derivative of Vin with respect to Iin, remembering that VBE is a constant, we get the result:
Zin = dVin/dIin = d/dIin(VBE + Iin (β + 1)RE) = (β + 1)RE
Zin = (β + 1)RE ≈ βRE |
Because IE = IB (β + 1). The IR drop across RE is greater then it would be for IB alone. The amplification of the base current causes RE to appear larger to a source looking into the input by (β + 1).
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