From the wikipedia page on Common Collector Amplifiers there is paragraph describing how the circuit acts as a negative feedback amplifier

The circuit can be explained by viewing the transistor as being under the control of negative feedback. From this viewpoint, a common collector stage (Fig. 1) is an amplifier with full series negative feedback. In this configuration (Fig. 2 with β = 1), the entire output voltage VOUT is placed contrary and in series with the input voltage VIN. Thus the two voltages are subtracted according to Kirchhoff's voltage law (KVL) (the subtractor from the function block diagram is implemented just by the input loop) and their extraordinary difference Vdiff = VIN - VOUT is applied to the base-emitter junction. The transistor continuously monitors Vdiff and adjusts its emitter voltage almost equal (less VBEO) to the input voltage by passing the according collector current through the emitter resistor RE. As a result, the output voltage follows the input voltage variations from VBEO up to V+; hence the name, emitter follower.

Common Collector / Emitter Follower Circuit Negative Feedback Amplifier


The negative feedback amplifier can be expressed with this equation:

\$Out = (In - Out)*Aol \quad if \quad (\beta = 1)\$

How can the emitter follower be expressed in this way?

My Thoughts So Far

\$ In = V_B \quad Out = V_E \$

As mentioned in the wiki article the summation point IN - OUT is applied to Vbe:

\$ V_{BE} = V_B - V_E \$


\$ V_E = R_E(I_C + I_B) \$

\$ I_C = \beta I_B \$

\$ V_E = R_E(\beta I_B + I_B) \$

\$ V_E = R_EI_B(\beta + 1) \$

At this point I have an equation relating output to base current (proportional to input voltage) multiplied by a gain but I cant see the step to get from this to the negative feedback amplifier equation


This isnt a question about how the feedback mechanism works. I understand the gain of the transistor causes base current to be multiplied increasing output voltage across \$R_E\$ which in turn decreases base current. This question is only about modelling the circuit as a negative feedback amplifier


1 Answer 1


With Vin=Vb, Aol=gmRE (gm: transconductance) and beta=1 your block diagram results in the classical common collector transfer function (output VE)

Acl=VE/Vin=gmRE/(1+gmRE). The loop gain is Aloop=gmRE

  • \$\begingroup\$ Please can you expand on how you found gmRE to be the open loop gain \$\endgroup\$
    – andowt
    Feb 2, 2021 at 18:11
  • \$\begingroup\$ I understand now, \$ I_C = g_m V_{BE} \$ then using the approximation \$ I_E \sim I_C \$ output voltage is given by \$ V_E = I_E * R_E = g_m V_{BE} R_E \$ if we expand \$ V_{BE} \$ we get the negative feedback amplifier equation \$ V_E = (V_B - V_E) g_m R_E \$ \$\endgroup\$
    – andowt
    Feb 2, 2021 at 19:46
  • 1
    \$\begingroup\$ Yes - OK. At the input summing junction we have, of course, VBE=VB-VE.. Evaluating the gain of the block diagram, we see that the result meets the classical well-known gain expression. Hence, the block diagram mirrors the reality and we can derive Aol and the loop gain expression from it. \$\endgroup\$
    – LvW
    Feb 3, 2021 at 9:01

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