# Electronic – Art of Electronics Ex.2.20 Series negative feedback amplifier

amplifiercommon-emitternegative feedback

Here is the exercise followed by a schematic:

Applying the techniques we used earlier, one can show that G(open
loop) ≈ 200, loop gain ≈ 20, Zout(open loop) ≈10k, Zout(closed loop) ≈
500Ω, and G(closed loop) ≈9.5. Exercise 2.22. Go for it!

Here is my analysis:

First lets calculate open loop gain cutting the feedback (bottom leg of R5 and grounding it). The amplitude of the gain of the first stage (common emitter amplifier formed by Q1) is

R3/R4 = 0.62

( I am ignoring re1=25 Ohm here, since R4>>re1).

For the second stage the OL Gain is

R5/re2 = 10kOhm/25ohm = 400.

So the total open-loop gain is: A = 400*0.62 = 250.

Feedback network is formed by R5,R4 voltage divider and returns signal voltage to the input with the ratio of B = 1/10.
So the

loop-gain = A*B = 25.

Now total closed loop gain (from previous chapters of the book) is:

A/(1+AB)=250/26 = 9.6.

This is the first problem. Why am I getting 250 for the open loop gain, insted of 200 as it is given in the book?
With the A = 200 all the following values are also exactly correct. Is there a mistake in my reasoning?

Then we calculate Zout (open loop) performing same feedback removal as before. Then our output impedance is simply

R5||(closed collector-base pn junction resistance).

Considering that the later should be in the range of MOhms I can neglect it and Zout (open loop) becomes

Zout(open loop) ≈ R5 = 10kOhm.

Now for Zout(closed loop). Looking into the output:

Zout(closed loop) = R5 + R4||[re+ (R1||R2||Rs)/ß] ≈ R5+50 Ohms ≈ R5,

where Rs is the source impedance (not shown) and not considered (Rs=infinity), and ß is the intrinsic transistor current gain, which I consider to be ß = 100, as was done earlier in the book.
This is the second problem, since in the solution it should come out to be 500.
This is what one gets if we simply say: "since it is a negative feedback, then" :

Zout(closed loop) =Zout(open loop)/(1+AB) ≈ 500 Ohms if we use A = 200

≈ 400 Ohms if we use A = 250.

But why brute force analysis did not work at all (as it was done for the previous similar example in the book)?

I will try to describe the method used by the book.

First, we need to find the open-loop gain.

And I will use this circuit diagram

The Q2 voltage gain is

$$A_{V2} \approx \frac{R_5+R_4}{re2} \approx \frac{11\textrm{k}\Omega}{25\Omega} \approx 440 \: V/V$$

The first stage gain is:

$$A_{V1} \approx \frac{R_3||(\beta +1)re2}{R_4||R_5} \approx \frac{500\Omega}{1\textrm{k} \Omega} \approx 0.5 \: V/V$$

Therefore the open-loop gain is:

$$\ A_{OL} = A_{V1} * A_{V2} = 220 \:V/V \$$

From the diagram we see that:

$$Z_{out(OL)} = R_4+R_5 = 11\textrm{k}\Omega$$

And

$$Z_{out(CL)} = \frac{Z_{out(OL)}}{1+ A_{OL}\times\frac{R_4}{R_4+R_5} } = \frac{11k \Omega}{1 + 220 \times \frac{1}{11}} \approx 520\Omega$$

And now the closed-loop gain:

$$A_{CL} = \frac{A_{OL}}{1+ A_{OL}\times\frac{R_4}{R_4+R_5} } \approx 10.47 \:V/V$$

The LTspice is showing $$\Z_{out(CL)} = 529\Omega\$$ and $$\A_{CL} = 10.47\: V/V\$$ quite good.