# Output impedance of current series feedback BJT amplifier?

So according to what I've read, in a common emitter BJT amplifier, an unbypassed emitter resistor acts as a current-series feedback path. The gain without feedback is: $$A=\frac{I_o}{V_i}=\frac{-h_{fe}}{h_{ie}+R_E}$$ and the feedback factor is: $$\beta=-R_E$$ The output impedance should then be: $$Z_{of}=Z_o(1+A\beta)=Z_o(1+\frac{R_Eh_{fe}}{h_{ie}+R_E})$$ This is a different result compared to finding the Thevinin equivalent resistance looking back into the output which just gives the output impedance as $R_C$ whether or not an emitter resistor is bypassed or not.

Any insight on this seeming inconsistency would be helpful. Thanks.

The "gain" formula as given by you is the transconductance of a transistor in common-emitter configuration WITH feedback.

1.) Without feedback (RE=0) this transconductance is simply $$gm = \frac{hfe}{hie}$$

2.) With feedback we have $$A= \frac{gm}{1+gm*R_E}= \frac{hfe}{hie+R_E}$$

From this, we can derive the loop gain Aloop=-gm*Re. Note that gm=hfe/hie.

3.) Based on the loop gain expression (and knowing that the output resistance will be increased using feedback) we can expect (approximation):

$$Z_{out}= Z_o*(1+gm*R_E)=r_{ce}*(1+gm*R_E)$$

4.) Comment: The above expression for Zout is an approximation because it is derived under the assumption of IDEAL voltage feedback. However, the feedback voltage is provided across RE (not an ideal voltage source). More than that, also the influence of the input circuitry was neglected. However, we should not forget, that during practical operation of this circuit the large value of Zout is paralleled with the collector resitor Rc (which is much smaller that Zout). Hence, the approximation seems to be acceptable.

• sherrellbc - thank you for editing. Just now, I have replaced Rce by rce (because it is not a static but a dynamic resistance) – LvW Jul 27 '14 at 13:49