(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:
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:
1. Determining \$v_{in}\$
If I start with a KVL loop on the input side, I get...
\$v_s=v_{in}+v_f\$
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}\$...
\$v_s=v_{in}+(\frac{\beta+1}{\beta})g_mv_{in}R_E\$
\$v_s=\frac{v_s}{[1+(\frac{\beta+1}{\beta})g_mR_E]}\$
2. Determining \$v_{out}\$
\$v_{out}\$ is a simpler matter; it is simply dependent on \$R_E\$.
\$v_{out}=i_eR_E\$
\$v_{out}=(\frac{\beta+1}{\beta})i_bR_E\$
\$v_{out}=(\frac{\beta+1}{\beta})g_mv_{in}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}\$...
\$v_{out}=(\frac{\beta+1}{\beta})g_mv_{in}R_E\$
\$v_{out}=(\frac{250+1}{250})0.12*0.0207*4000=9.98v\$
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!
