# How to find feedback coefficient $\beta$ of this current-voltage feedback circuit?

Here is the circuit from YouTube:

In order to determine the feedback, we have to break the loop at the gate of M1.

Assume A is $$\\infty\$$, so $$\V_{GS}\$$ of M1 is 0. As we know the $$\V_S\$$ of M1 is grounded, we have $$\V_S = V_G = 0\$$. And that's why the gate of M1 is grounded.

$$\frac{V_{out}} {I_{in}} = R_2 \parallel R_3$$

The following is how I break the loop and add test voltage.

$$Loop \hspace{1mm} Gain = L = \frac{V_F} {V_t} = g_{m1} \bigr(R_3 \parallel (R_1 + R_2)\bigr) \frac{R_1} {R_1 + R_2}$$

• Why do you call it a current-voltage feedback circuit? It's a potential divider and the attenuation it provides is the feedback factor. Commented Mar 24 at 10:05
• @Andyaka What's the $\beta$ according to you?
– kile
Commented Mar 24 at 10:38
• $\dfrac{R_1}{R_1+R_2}$ and, you don't need to break any loops to see this. Commented Mar 24 at 10:43
• @Andyaka Nope, check youtu.be/…
– kile
Commented Mar 24 at 11:18
• Sorry but youtube is not a technical authority that can be trusted. There are many sources that agree with me: google.com/… Commented Mar 24 at 13:14

The feedback network includes R1, R2, and the transistor M1. The feedback factor $$\\beta\$$ is calculated as follows:

$$\\beta = \frac{i_{fb}}{V_{out}} = g_{m1}\cdot\frac{R1}{R1 + R2}\$$

• I think the trickiest thing is how to find the feedback network. Can you explain how you find this feedback network?
– kile
Commented Mar 25 at 6:53
• @kile - the above consideration is correct (see my detailed answer).
– LvW
Commented Mar 25 at 8:59
• @kile If you compare the circuit with this diagram here, you can identify different parts of the feedback network. Also, check the LvW answer. Commented Mar 25 at 16:21
• @internet I draw a a schematic on how I break the loop. Can you see my updates on the question? Do I break the loop correctly?
– kile
Commented Mar 25 at 22:24
• @kile Yes, you can break the loop like that but you make a mistake in your calculate of LG. Commented Mar 26 at 2:47

Here comes my explanation of the shown circuit:

1.) The input of the amplifier (common gate configuration with a current input) is the source node of M2.

2.) Therefore, the feedback signal (in form of a current) is superimposed at this node with the signal current. That means: Transistor M1 is part of the feedback path. It is its purpose to convert the feedback voltage (at the gate of M1) into a current.

3.) Opening the feedback loop at the gate of M1 and injecting a test voltage Vt,in, we get the loop gain expression:

Loop gain LG=Vt,out/Vt,in with

id1=gm1*Vt,in and

Vt,out=[R1/(R1+R2)] * id2 * R3||(R1+R2)

Loop gain LG=Vt,out/Vt,in=[R1/(R1+R2)] * (id2/id1) * gm1 * R3||(R1+R2)

Because of id2=id1 we can simplify:

LG=[R1/(R1+R2)] * gm1 * R3||(R1+R2)

The gain of the open loop is

Aol=Vout/i,in=R3||(R1+R2)**

Therefore the feedback "factor" is beta=LG/Aol:

beta=[R1/(R1+R2)] * gm1.

• I draw a a schematic on how I break the loop. Can you see my updates on the question?
– kile
Commented Mar 25 at 22:23
• @kile - Yes, and what is the question?
– LvW
Commented Mar 26 at 8:56
• is the schematic the same as your description?
– kile
Commented Mar 26 at 9:59