# Art of Electronics Ex.2.20 Series negative feedback amplifier

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)?

• Is there anything in parallel with R3? Feb 14, 2020 at 9:29
• yes, thanks inclusion of R3||re2 fixes the gains. But I do not see how it will affect Zout(closed loop). Could you comment on that please?
– kek
Feb 14, 2020 at 10:03

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.

• why do you have the equivalent schematic like this? I just cut the line connecting R5 and R4 and grounded R5, when I wanted to remove feedback. Is it the same?
– kek
Feb 14, 2020 at 17:27
• I do this on purpose to include the feedback network loading effect. I found resistance seen from the emitter into the feedback network resistor (with output shorted because of voltage feedback). Nax I do the same for an output node. Watch this youtu.be/NRYMrheAAJI?t=1746
– G36
Feb 14, 2020 at 19:22
• Oh man... whole new world :). I guess, I will just go continue studying, satisfied with what I got. Thanks
– kek
Feb 14, 2020 at 20:03
• Have you tried to simulate that open loop circuit? When I put that into simulation (Falstad simulator), the output is locked at Vcc, nowhere near the quiescent output voltage of the closed loop circuit (12V). Oct 31, 2020 at 4:15
• @QelopinUacel You cannot break a loop in this circu because in the circuit the negative feedback look also help the set the DC bias point.
– G36
Oct 31, 2020 at 16:59

You have forgotten that R3 is in parallel with Rpi/Hie of Q2. If you include this you will get the answers given in the book. As an aside, I would expect a two transistor amp to have an Aol of 2000.

• @EinarA...I think you have made an error: the ratio rpi/hie (small signal quantities - small letters!) is identical to 1/gm=re.
– LvW
Feb 14, 2020 at 9:51
• okay. So now (open loop) 1st stage gain is R3||ßre2/R4 = 0.5. For the 2nd stage it is: R5/re2||R3 = 416. Which gives total open loop gain of 208, which is close to 200 and also this gives all other gains correct. Am i right?
– kek
Feb 14, 2020 at 10:00
• But the question with Zout(closed loop) still remains, as I dont see how knowing that now R3||re2 would change that.
– kek
Feb 14, 2020 at 10:01
• Serhii...of course, there is an influence. Any test signal at the output (collector of Q2) will influence the emitter of Q1 (now working as a common base stage) and, hence, the collector of Q1 (identical to the base of Q2). And this signal works back to the collector of Q2 (input for the test signal). This is the closed-loop of the circuit.
– LvW
Feb 14, 2020 at 11:13
• Serhii..of course, you could follow the "brute force" approach. In this case, you have to introduce a test voltage Vtest at the output node and find the corresponding current Itest into this node. However, do not forget that this voltage (due to the feedback loop) will also cause a current at the collector node of Q2 - thereby increasing the total current Itest - equivalent to reducing the ratio Vtest/Itest=Zout.
– LvW
Feb 14, 2020 at 17:09