(This question is related to this one here, where I attempted to understand how the discrete component values in a negative feedback amplifier are represented by the closed-loop gain and feedback factor.)

Today, I'm hoping to understand voltage-controlled voltage feedback a little better. My plan is to begin with a schematic of a common-collector amplifier and use it to define loop gain and feedback factor. Is this analysis correct?

I'll begin with a block diagram showing the internal amplifier and feedback network:

enter image description here

As I understand it, the signal \$v_s\$ will be significantly reduced by the feedback network \$K\$ enough that there will be a minuscule voltage presented to the amplifier \$A_v\$. The voltage output will be nearly the same as the signal, but with a lower output impedance and higher current. Below is the circuit I'm working with:

enter image description here

1. Determining \$v_{in}\$

If I start with a KVL loop on the input side, I get...


Replacing \$v_{f}\$ with an expression including \$R_E\$...

\$v_s=v_{in}+i_ER_E\$, and then replacing \$i_E\$ with a known component value...

\$v_s=v_{in}+(\frac{\beta+1}{\beta})i_bR_E\$, and then replacing \$i_b\$ with \$g_mv_{in}\$...



2. Determining \$v_{out}\$

\$v_{out}\$ is a simpler matter; it is simply dependent on \$R_E\$.




3. Determining open-loop gain

Open-loop gain is simply the action of the internal amplifier, expressed in terms of \$v_{in}\$ and \$v_{out}\$.

\$A_v = \frac{v_{out}}{v_{in}}\$

\$A_v = \frac{(\frac{\beta+1}{\beta})g_mv_{in}R_E}{v_{in}}\$

\$A_v = \frac{(\beta+1)g_mR_E}{\beta}\approx g_mR_E\$

So, the open-loop gain is fairly simple.

4. Determining closed-loop gain

Closed-loop gain will use the signal voltage instead of \$v_{in}\$, so... \$A_f=\frac{v_{out}}{v_s}\$

\$A_f = \frac{(\frac{\beta+1}{\beta})g_mv_{in}R_E}{v_{in}[1+(\frac{\beta+1}{\beta})g_mR_E]}\$

cancelling out \$v_{in}\$...

\$A_f = \frac{(\frac{\beta+1}{\beta})g_mR_E}{[1+(\frac{\beta+1}{\beta})g_mR_E]}\approx 1\$

We can approximate this to 1, as the \$\beta\$ expressions and the 1 in the denominator become negligible.

5. Checking against the circuit

Given the circuit values above, we can verify our results. DC analysis gives a quiescent current of \$3mA\$. That gives us:

\$g_m=\frac{I_{CQ}}{V_T} = \frac{3mA}{25mV} = 0.12\$

\$A_v = \frac{(\beta+1)g_mR_E}{\beta}\approx g_mR_E\$

\$A_v = \frac{(250+1)0.12*4000}{250}=481.92\$

\$A_f = \frac{(\frac{\beta+1}{\beta})g_mR_E}{[1+(\frac{\beta+1}{\beta})g_mR_E]}\approx 1\$

\$A_f = \frac{(\frac{250+1}{250})0.12*4000}{[1+(\frac{250+1}{250})0.12*4000]}=0.998\$

And what is \$K\$, the feedback factor? Well, if

\$A_f = \frac{A_v}{1+A_vK}\$, then \$K = \frac{482.92}{481.92}=1.002\$

So, the circuit should change the voltage by a factor of 0.998, a barely-detectable change. What are the exact voltages \$v_{in}\$ and \$v_f\$?

Assuming a \$v_s=10v\$ input, we would expect:

\$v_{in} = \frac{v_s}{[1+(\frac{\beta+1}{\beta})g_mR_E]}\$ \$v_{in} =\frac{10}{[1+(\frac{250+1}{250})0.12*4000]}=20.7mV\$

And for \$v_{out}\$...



6. To summarize

The output voltage is nearly the input voltage, missing by only \$20mV\$. All of this checks out in simulation, so I ask you ... have I analyzed this correctly? Thanks!

  • \$\begingroup\$ The Vout and gain are incorrect. Av is gmRc(common emitter) but not for a common collector or voltage follower as you have shown . Also for DC Q point base load = Zb= hFE * Re which affects Vb divider then estimate Vbe=0.65 V to get Vout for DC \$\endgroup\$ Dec 12, 2020 at 4:35
  • \$\begingroup\$ your 20mV attenuation depends on a ratio just below 1.000 on the undefined assumed input you used I get Av=0.955 and Vo=12.1, which may be ideal for a 24V supply \$\endgroup\$ Dec 12, 2020 at 4:48
  • \$\begingroup\$ @Tony can you explain more what expression is correct for gain? I neglected the role of \$r_e\$, is that the problem? \$\endgroup\$
    – nuggethead
    Dec 12, 2020 at 12:17
  • \$\begingroup\$ Also, @Tony, I don't see that I used gmRc anywhere (there isn't a collector resistor). Could you please elaborate? \$\endgroup\$
    – nuggethead
    Dec 12, 2020 at 21:21
  • \$\begingroup\$ @Tony Stewart Sunnyskyguy my test input was 10v, how did you get 12.1v output? Could you please clarify? \$\endgroup\$
    – nuggethead
    Dec 13, 2020 at 14:32


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