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In a source follower like this:enter image description here

When AC source is at its positive half cycle, conventional current passes from base terminal to R_E and R_L and then straight to the ground. The resistance it sees is the input impedance Z_in.

But when AC source is at its negative half cycle, it passes from ground to R_E and R_L and then to the emitter terminal. Why does the resistance that this conventional current sees is still the input impedance Z_in and not the output impedance Z_out?

Zout is the resistance that the load R_L sees in its own perspective, that's why I though that if current passes from the ground to the R_E and R_L, it will also face the resistance of Z_out and not Z_in.

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  • \$\begingroup\$ Draw the small-signal AC equivalent of the circuit with using, for example, transistor's hybrid model, and you'll find the answer. \$\endgroup\$ Commented Jun 25, 2020 at 5:55
  • \$\begingroup\$ Hybrid pi model and t model only shows input impedance and r'e, but it does not show output impedance from the perspective of the emitter load, right? \$\endgroup\$
    – hontou_
    Commented Jun 25, 2020 at 6:24
  • \$\begingroup\$ Take a look here electronics.stackexchange.com/questions/310471/… \$\endgroup\$
    – G36
    Commented Jun 25, 2020 at 12:44

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The input capacitor is making it a little difficult to see what's happening so let me say one thing; when the external input signal is below 0 volts, the DC level at the base is still a positive due to the bias resistors R1 and R2 AND, the signal superimposed on the base DC bias never gets down too low i.e. it always keeps the base emitter junction forward biased. The input capacitor ensures that you can pass a supposed AC current into the base node whilst preventing that node going negative or below around 1 volt in normal operation.

Clearly, if you overdrive the input then serious distortion occurs and the base emitter region gets fully reverse biased - when this happens the base collector region is still reverse biased (as per normal operation) and the input impedance would be see to rise slightly.

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