(I posted a similar question that shows a better understanding of the circuit here.) I don't know if my reputation would suffer if I deleted this question.

I am learning the four negative feedback topologies. I would like to reach a point where I will be able work my way from expressions like \$A_f={A_o\over(1+A_oB)}\$ through algebra to more concrete expressions using component values (resistors, e.g.). My first attempt was with voltage-series negative feedback circuits and a simple common collector. I think I was quite successful. I was able to find good expressions for \$A_o, A_f\$ and \$v_{out}, v_{in} \$. From the expressions I derived I could see why we make approximations in analysis, such as \$v_{out}=v_s\$.

I arrived at the open-loop gain \$A_o={(\beta+1)g_mR_E\over\beta}\approx g_mR_E \$ and the closed-loop gain \$A_f={\frac{(\beta+1)}{\beta}g_mR_E\over1+\frac{(\beta+1)}{\beta}g_mR_E}\approx {g_mR_E\over1+g_mR_E}\approx 1\$.

I was satisfied that my expressions for \$v_{out}\$ and \$v_{in}\$ were accurate enough when I was able to show the tiny amount that \$A_f\$ falls short of unity gain. For example: \$v_{in}={v_s\over 1+\frac{(\beta+1)}{\beta}g_mR_E} \$ gave a good prediction of what appeared across \$v_{be}\$ in simulation and \$v_{out}=\frac{(\beta+1)}{\beta}g_mv_{in}R_E\$ gave a value of 9.95v for a 10v input signal at \$v_s\$.


simulate this circuit – Schematic created using CircuitLab

Then, I attempted to repeat the same process for current-series negative feedback, using a common emitter (with degeneration, not bypassed) as an example. Here's what I came up with for open-loop gain:

\$A_o = {v_{out}\over v_{in}} = {g_mv_{in}R_C\over v_{in}} = g_mR_C\$

For closed-loop gain:

\$A_f = {v_{out}\over v_{s}} = {g_mv_{in}R_C\over v_{in}+v_f} = {g_mv_{in}R_C\over v_{in}-g_mv_{in}R_E}\$

\$A_f(v_{in}-g_mv_{in}R_E) = g_mv_{in}R_C\$


\$A_f = {g_mR_C\over 1-g_mR_E} \$

And for the feedback factor:

\$v_f = Bv_{out}\$


\$B = {g_mv_{in}R_E\over g_mv_{in}R_C} = {R_E \over R_C}\$

Everything looked good to me, and all my equations proved out in a simulator. But then, disaster struck as I read more about the current-series negative feedback. Must my work instead be based on this set of equations \$A_o = {i_o\over v_{in}}\$, \$A_f = {i_o\over v_{s}}\$, and \$B = {v_f\over i_{o}}\$? I can see how using these equations as a starting point would make clear that I am looking at a transconductance amplifier, not a voltage amplifier, but if my equations work, does it matter which I use? Is my understanding of current-series negative feedback sufficient?

  • \$\begingroup\$ Should also mention that I used B for the feedback network, not \$\beta\$, as both symbols in the same set of equations would be too confusing for me! \$\endgroup\$
    – nuggethead
    Commented Nov 30, 2020 at 2:19
  • \$\begingroup\$ Can you add a circuit diagram for folks? \$\endgroup\$ Commented Nov 30, 2020 at 2:20
  • \$\begingroup\$ Yep. Added circuit diagram. \$\endgroup\$
    – nuggethead
    Commented Nov 30, 2020 at 2:28
  • \$\begingroup\$ I posted a new question, showing more analysis and (I think) a better understanding of the transconductance amplifier here. electronics.stackexchange.com/questions/536335/… \$\endgroup\$
    – nuggethead
    Commented Dec 12, 2020 at 2:27

1 Answer 1


Perhaps the shown block diagram can help to clarify the situation? (The last - most right - block should contain a minus sign).

In the diagram I have set Ie=Ic. Of course, we could include the base current into the diagram - however, this would not influence the feedback loop. Note the loop gain is Aloop=-gm*RE.

The circuit establishes current-controlled voltage feedback. (I do not like expressions like "series-parallel" or "parallel-parallel" or....because some authors are counting input-output and some others output-input. So it is somewhat confusing, I think).

enter image description here

  • \$\begingroup\$ Serh schöne illustration LvW, too bad the common-emitter structure wasn't taught to me this way while in college 35 years ago: +1 : ) \$\endgroup\$ Commented Nov 30, 2020 at 9:05
  • \$\begingroup\$ Yes - it is nothing else than a graphical illustration of the well-known transistor equations. \$\endgroup\$
    – LvW
    Commented Nov 30, 2020 at 11:20
  • \$\begingroup\$ Thank you for sharing the illustration, but this doesn't answer my question. Are my expressions correct? \$\endgroup\$
    – nuggethead
    Commented Nov 30, 2020 at 17:30
  • \$\begingroup\$ I was of the opinion that now you can find the answer by yourself.....No - the sign in the denominator for Af is wrong. More than that, what is meaning of vf and B ? What is your definition? \$\endgroup\$
    – LvW
    Commented Dec 1, 2020 at 8:36
  • \$\begingroup\$ @LvW You wrote: "(The last - most right - block should contain a minus sign).". Do you mean to say that the box labelled RC should instead say -RC ? \$\endgroup\$
    – nuggethead
    Commented Dec 2, 2020 at 1:40

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