Negative feedback from block diagram to circuit components

Question

I can read and understand the block diagrams which show the internal amplifier and the feedback network and I can read and understand the schematics of several single-BJT amplifier circuits, but I have a very difficult time going back-and-forth between the two. I have been unable to find resources that show how to define loop gain and feedback factor when viewing the schematic, or how to predict component values by looking at the block diagram. I tried it on my own here hoping that users can help me understand what I'm missing. My objective was to begin with a common emitter circuit schematic (below right) and use it to derive the familiar equations for the block diagram (below left). In specific, I'd like my analysis to give me the closed-loop gain, feedback factor (\$K\$), and values for \$v_{in}\$, \$v_f\$, and \$i_{out}\$. (I use \$K\$ instead of the more common \$\beta\$ because the latter is already used to describe the transistor.)

Is my analysis correct?

NB: I edited this question to correct an earlier error where I neglected \$r_e\$ when calculating voltage gain.

For the circuit above, I applied a test voltage of \$v_s=100mV\$ and the simulation showed an output current of \$342\mu A\$. If all goes well, I should be able to predict this result using the expressions I find for closed-loop gain.

1. Transconductance
I understand that to properly analyze this circuit I will need to visualize it as a transconductance amplifier, so next I examined the closed-loop gain \$A_f=\frac{i_{out}}{v_s}\$. To get the closed-loop gain, first I'll need some expressions for \$i_{out}\$. I began with the output voltage, which is easily found with the resistors \$R_C\$ and \$R_E\$.

\$i_{out}=\frac{v_{s}A_V}{R_C}\$ where \$A_V=\frac{R_C}{(R_E+r_e)}\$.

\$i_{out}=\frac{v_s({\frac{R_C}{(R_E+r_e)}})}{R_C}=\frac{v_s}{(R_E+r_e)}\$

\$i_{out}=\frac{v_s}{(R_E+r_e)}\$

2. Feedback Voltage

Next, I wanted to understand how the feedback network \$K\$ works. I found a value for \$K\$ by working backwards from \$i_{out}\$. I know the effect of the internal amplifier \$A\$, so I know that \$i_{out}\$ will be \$g_m\$ times \$v_{in}\$.

\$v_{in}g_m=i_{out}=\frac{v_s}{(R_E+r_e)}\$

\$v_{in}=\frac{v_s}{(R_E+r_e)g_m}=\frac{v_s}{R_Eg_m+r_eg_m}\$

Because \$r_e\$ and \$g_m\$ are related through inversion,\$r_eg_m=1\$. So...

\$v_{in}=\frac{v_s}{R_eg_m+1}\$

Because \$v_{in}=v_s+v_f\$, I know that...

\$v_s-\frac{v_s}{R_Eg_m+1}=v_f\$

Finally, I can say that

\$v_f=v_s[1-\frac{1}{(R_Eg_m+1)}]\$

3. Defining K, the feedback factor

I know that \$K\$ will have a value in ohms, or \$(\frac{V}{I})\$. Its effect on the input, the voltage \$v_f\$, will be \$v_f=Ki_{out}\$. If I use my expressions for \$v_f\$ and \$i_{out}\$ I can find an expression for \$K\$.

\$v_f=Ki_{out}=\frac{v_s(1-\frac{1}{R_Eg_m+1})}{\frac{v_s}{R_E+r_e}}\$

\$K=v_s[1-\frac{1}{R_Eg_m+1}]*(\frac{R_E+r_e}{v_s})\$

I can cancel out \$v_s\$, giving...

\$K=[1-\frac{1}{R_Eg_m+1}]*(R_E+r_e)\$

Reworking, I get

\$K=(R_E+r_e) - \frac{R_E+r_e}{(R_Eg_m+1)}\$

This yields a mess of \$R_E\$ variables:

\$K = R_E+r_e-\frac{(R_E+r_e)}{(\frac{R_E}{r_e}+1)}\$

Which eventually boils down to

\$K=R_E\$.

4. Finding the closed-loop gain
Finally, I plugged in my \$K=R_E\$ expression into the standard closed-loop gain \$A_f=\frac{g_m}{1+Kg_m}\$ equation:

\$A_f=\frac{g_m}{1+Kg_m}\$ \$A_f=\frac{g_m}{1+R_Eg_m}\$

5. To Double-Check

I went back to my original schematic and checked my work.

if \$A_f=\frac{g_m}{1+R_Eg_m}\$, then my \$A_f=\frac{0.12}{1+(284*0.12)}=0.00342\$

If \$A_f=0.00342\$, my output current should be \$i_out=A_fv_s=0.00342*0.1v=0.000342A=342\mu A\$

If \$K=R_E=284\Omega\$, then \$v_f=Ki_{out}=284*0.000342=97mV\$

If \$v_f=97mV\$, then \$v_{in}=v_s-v_f=100mV-97mV=3mV\$

So, the negative feedback \$K\$ prevents the majority (97%) of the signal voltage from reaching the transistor. The closed loop gain is defined as

\$A_f=\frac{g_m}{1+Kg_m}\$ and the feedback factor \$K=R_E\$

Everything checks out in simulation and it all makes conceptual sense... so, is my analysis correct?

Answer 1

The resistor RE provides (output)-current-controlled voltage feedback. Therefore, for analyzing the feedback mechanism, it is correct to consider the BJT as a transconductance device with voltage-in and current-out.

From this consideration we already can conclude that the feedback factor will be a resistive quantity.

Applying the transconductance view it is a simple matter to write down the output-to-input ratio ic/vbe with vb=vs and ve=icRE. (Note: All the variables are differential values; for simplicity I have assumed ic=ie with ib=0).

From this, you can find the closed-loop "gain" ic/vs (closed-loop transconductance). Now - analyzing the denominator of this ratio, you can find the loop gain expression and the feedback "factor" (which has the unit "ohm"). Of course, the numerator is identical to the open-loop "gain" (open-loop transconductance).

Answer 2

Ok first things, the block diagram used in control theory need not always give you the same 'answer' (of the form A/1+AK) in an actual circuit due to the following reasons (here A is the open loop gain and K the feedback factor as shown in your diagram):

(a) The control block doesnt show the impact of loading. A and K will have input and output impedances

(b) In the control block, the signal flow is one way. In an actual circuit the signal flows both ways.

(c) Not all circuits can be neatly represented in that control block manner. A good example is a common emitter or common source amplifier with a degenerate resistor as you have shown.

If you want accurate answers in a circuit - use KCL at all nodes. Having said this - this is how you calculate open loop gain, loop gain and closed loop gain 'quickly, but maybe inaccurately':

Step 1: open the feedback path i.e. disconnect the feedback from the output and ground that feedback route. i.e. the input to K is 0. Calculate the small signal gain. The answer is your open loop gain (OLG).

Step 2: Ground your input. Apply a test voltage to your disconnected feedback line i.e. at the input to K. Now calculate the small signal gain. This answer is the loop gain (LG). If you see a negative sign, it is negative feedback.

Step3: Now blindly write the closed loop gain = OLG/(1+LG) Note that this form assumes negative feedback, so dont carry the negative sign from the calculated LG.

This process will allow you to calculate CLG quickly at the cost of some errors.