# Voltage-voltage Feedback and common-collector analysis?

(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!

• 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 – Tony Stewart EE75 Dec 12 '20 at 4:35
• 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 – Tony Stewart EE75 Dec 12 '20 at 4:48
• @Tony can you explain more what expression is correct for gain? I neglected the role of $r_e$, is that the problem? – nuggethead Dec 12 '20 at 12:17
• Also, @Tony, I don't see that I used gmRc anywhere (there isn't a collector resistor). Could you please elaborate? – nuggethead Dec 12 '20 at 21:21
• @Tony Stewart Sunnyskyguy my test input was 10v, how did you get 12.1v output? Could you please clarify? – nuggethead Dec 13 '20 at 14:32