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In Albert Malvino's "Electronic Principles", he said that in a well-designed two-stage feedback amplifier for small signal operation(picture below) , the voltage gain equals $$A_v=\frac{r_f} {r_e}+1$$

Why does the voltage gain equals to this? I read before that the overall voltage gain of a two stage amplifier(without the feedback) is equals to the product of the voltage gain of the each stage. But when feedback resistor is added, it now only depends on this feedback resistor. enter image description here

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    \$\begingroup\$ What do you need to help you? How does this circuit first look to you? What's its open loop gain? Are you familiar with the "generic" idea of closed loop gain with feedback? (In short: \$\frac{A}{1+A\cdot B}\$.)? Please let us know how you "look at" this circuit? How far can you get? (The answer you have is approximate, but close enough in this case for most intents and purposes. It misses the loss due to \$R_L\$, though. Also, the 1st order equation is slightly more complicated. And the first stage should probably be "bootstrapped," as well.) \$\endgroup\$
    – jonk
    May 7 '20 at 5:25
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    \$\begingroup\$ It's a good question, though. So +1 for that. \$\endgroup\$
    – jonk
    May 7 '20 at 5:33
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    \$\begingroup\$ It comes from basic feedback control theory and a mathematical simplification. @jonk is right, you should be familiar with closed loop vs open loop gain. \$\endgroup\$ May 7 '20 at 5:57
  • \$\begingroup\$ Do you understand why a non-inverting op-amp has the same gain formula? \$\endgroup\$
    – Andy aka
    May 7 '20 at 8:30
  • \$\begingroup\$ No, just the basics of amplifier and feedbacks \$\endgroup\$
    – hontou_
    May 7 '20 at 9:02
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You didn't respond to my thoughts. But I'll propose some simplifying ideas that will work okay, I think.

First off, this two-stage amplifier applies some global NFB. It does so by taking the output (which is roughly "in-phase" with the input) and applying it to the emitter of the first stage. How this achieves NFB isn't too difficult to see. If the input rises, the output of the second stage also rises. This output rise is fed back to the emitter of the first stage, causing the emitter to rise a little bit in response. So, when the base rises, the emitter rises a little in response. This acts to maintain the \$V_\text{BE}\$ (keep it from changing as much as it might, otherwise) and therefore acts as NFB (as the collector current is a function of \$V_\text{BE}\$.)

So you need to understand that the NFB here follows the basic idea of NFB illustrated by this wiki page. First, you will want to estimate the open-loop gain of the system (you find that by breaking the NFB) and then you want to estimate the magnitude of the NFB and apply that into the formula. I won't derive it here, because you can go to the above page for that, or just google the development elsewhere. (It's everywhere.)

The equation is: $$A_{\text{closed loop}}=\frac{A_{\text{open loop}}}{1+A_{\text{open loop}}\cdot B_\text{NFB}}$$

You need to find the open loop gain and find the negative feedback proportion in order to compute values from the above equation.

So let's first examine the circuit:

schematic

simulate this circuit – Schematic created using CircuitLab

The first thing you need to do is to estimate the open loop gain. You do this by removing \$C_6\$ and \$R_f\$ in the above circuit, which comprises the NFB, and then working out the combined voltage gain for the two stages.

The DC operating point is found, using \$R_\text{TH}=R_1\mid\mid R_2\$ and \$V_\text{TH}=V_\text{CC}\frac{R_2}{R_1+R_2}\$, by:

$$V_\text{E}=I_\text{B}\cdot\left(\beta+1\right)\cdot\left(R_\text{e}+R_\text{E1}\right)=\frac{\left(V_\text{TH}-V_\text{BE}\right)\cdot\left(\beta+1\right)\cdot\left(R_\text{e}+R_\text{E1}\right)}{R_\text{TH}+\left(\beta+1\right)\cdot\left(R_\text{e}+R_\text{E1}\right)}\approx 1.1\:\text{V}$$

This suggests \$I_\text{Q}\approx 1.1\:\text{mA}\$ and a dynamic resistance of \$r_e\approx 24\:\Omega\$ for stage 1. So, for stage 1, \$A_v=\frac{R_{\text{C}_1}}{R_\text{E}+r_\text{e}}\approx 17.65\$. If you do similar calculations, you'll find that the 2nd stage also has \$I_\text{Q}\approx 1.3\:\text{mA}\$ and that here \$A_v=\frac{R_{\text{C}_1}}{r_\text{e}}\approx 180\$. Combined, this results in \$A_v\approx 3150\$. However, the input of the 2nd stage loads the output of the 1st stage. So, this attenuates by about \$\frac{R_4\mid\mid R_5\mid\mid \left(\beta+1\right)r_e}{R_\text{C1}+R_4\mid\mid R_5\mid\mid \left(\beta+1\right)r_e}\approx 0.257\$. So the total open-loop voltage gain is about \$A_v\approx 810\$.

For calculation purposes I'd call this \$A_v\approx 800\$ for the open-loop voltage gain for both stages.

Now you need to work out the NFB factor, \$B_\text{NFB}\$. For this purpose, it's possible to see that the fraction is \$\frac{R_{\text{e}}}{R_{\text{e}}+R_\text{f}}\$, for AC purposes, or \$B_\text{NFB}=\frac{R_{\text{e}}}{R_{\text{e}}+R_\text{f}}\$. For example, assume \$R_\text{f}=18\:\text{k}\Omega\$. Then we'd find that the closed-loop gain is about \$\frac{800}{1+800\cdot \frac{180\:\Omega}{180\:\Omega+18\:\text{k}\Omega}}\approx 90\$.

From simulation, I find that LTspice provides the closed-loop gain of about \$A_v\approx 88\$. This is very close the above-calculated value.

Suppose we change \$R_\text{f}=180\:\text{k}\Omega\$. That's a one-magnitude change in value! In this case, we find a calculated estimate of \$A_v\approx 440\$ and LTspice reports \$A_v=440\$, as well. Suppose we now change \$R_\text{f}=4.7\:\text{k}\Omega\$. Here, we find a calculated estimate of \$A_v\approx 25.2\$ and LTspice reports \$A_v=25.6\$.

So this works pretty well.

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  • \$\begingroup\$ @jonk...I fully support your calculation. I only like to point to the fact (in order to avoid misunderstandings) that in your calculation you have used re=1/gm=24 Ohms - in contrast to the formula and the value of re as given in the task desription (external re=180 Ohms), which seems to be not correct. \$\endgroup\$
    – LvW
    May 8 '20 at 7:50
  • \$\begingroup\$ @LvW I think the \$r_e\$ in the OP's diagram must be external. \$\endgroup\$
    – jonk
    May 8 '20 at 7:54
  • \$\begingroup\$ @jonk...yes, but in this case, the OP forgot the input resistance 1/gm at the emitter node which is - as you have shown - much smaller. \$\endgroup\$
    – LvW
    May 8 '20 at 8:05
  • \$\begingroup\$ Sorry, it seems there still a lot I need to study first to understand this. The book I am reading does not yet delve deeper into this topic. Maybe after I finish Malvino's book I will be able to understand this circuits and terms \$\endgroup\$
    – hontou_
    May 10 '20 at 3:24
  • \$\begingroup\$ @IwataniNaofumi Is there something hard to follow in what I wrote? \$\endgroup\$
    – jonk
    May 10 '20 at 3:28
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The phase from input (base) to output is ---- Q1 inverts and Q2 inverts, so result is INPHASE.

The phase from feedback node (emitter of Q1) to output is ---- Q1emitter_to_collector is non-inverting while Q2 still inverts., so result is OUT_OF_PHASE.

So, yes, you can use the same gain-set formula as non-inverting opamp circuit.

What is the open loop gain? Q1 has gain of 20x, Q2 gain ~~ 100x. total ~~ 2,000x

If your feedback ratio is 1,000X, you'll have poor gain stability and wandering frequency response. Don't do this, if you want Left and Right RIAA vinyl cartridge channel matching.

If your feedback ratio is 100x (OK 99x), your surplus gain is 2,000/100 and you'll have about 5% accuracy. This tracking of Left/Right channels may be adequate for good sound_stage imaging.

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As JONK suggests, the interface of Q1 and Q2 results in less than ideal gain; the Rin of Q2 if beta=100 will be ~~~ 20 * 100 = 2Kohms, a severe load on that first gain stage.

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  • \$\begingroup\$ The open loop gain won't be close to 2000. There's a horrible loss between the two stages. I'd estimate closer to about 750-800. (And that's assuming an unloaded output.) \$\endgroup\$
    – jonk
    May 8 '20 at 0:16

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