# Analyzing current-series negative feedback using voltage gain confusion

(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(1-g_mR_E)=g_mR_C\$$

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

And for the feedback factor:

$$\v_f = Bv_{out}\$$

$$\g_mv_{in}R_E=Bg_mv_{in}R_C\$$

$$\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?

• 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! – nuggethead Nov 30 '20 at 2:19
• Can you add a circuit diagram for folks? – relayman357 Nov 30 '20 at 2:20
• Yep. Added circuit diagram. – nuggethead Nov 30 '20 at 2:28
• I posted a new question, showing more analysis and (I think) a better understanding of the transconductance amplifier here. electronics.stackexchange.com/questions/536335/… – nuggethead Dec 12 '20 at 2:27