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I'm currently studying the three modes of a BJT differential amplifier, and am struggling with some hypotheses the textbook makes.

When examining single input and double input, the author uses a model like the one below, and goes on to an AC analysis of the circuit.

schematic

simulate this circuit – Schematic created using CircuitLab

While applying KVL from V1 to V2, they attempt to simplify analysis by stating that 'we choose a very large emitter resistor, almost infinite' without further explanation, and they end up with the formula for the single input DA.

$$ A_s = \frac{V_o}{V_i} = \frac{R_C}{2r_e} $$

However, when doing the AC analysis of the common-mode DA, the emitter resistance shows up in the formula

$$ A_c = \frac{V_o}{V_i} = \frac{βR_C}{r_i + 2(β+1)R_E} $$

The author omits further diagrams, by saying we 'apply a similar logic'. What I struggle to understand is, why is \$R_e\$ suddenly not infinite, and we take it into account in common-mode?

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  • \$\begingroup\$ You are looking at the front end of a typical Operational Amplifier. Try this link for a complete explanation even using your circuit diagram and explaining it. electronics-tutorials.ws/opamp/opamp_1.html \$\endgroup\$
    – Gil
    Jan 27, 2021 at 20:17
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    \$\begingroup\$ Did you draw the small signal equivalent circuit? Maybe the reason why "a very large emitter resistor" as used is to mimic a current source in the "tail" instead of a resistor. I think that's not a good teaching method. The analysis is simpler when there's a current source in the place of Re so start with that. After that we can repeat that with a resistor Re. I suggest that you watch: youtube.com/watch?v=Lq9gX05Dqio I didn't watch the whole video so no guarantees that the analysis is fully correct. \$\endgroup\$ Jan 27, 2021 at 20:31
  • \$\begingroup\$ What is the meaning of ri in your last equation? What happens for RE very large? \$\endgroup\$
    – LvW
    Jan 28, 2021 at 8:49
  • \$\begingroup\$ @LvW I think the OP is wondering about what is currently inexplicable to the OP: 'we choose a very large emitter resistor, almost infinite'. It wasn't explained to the OP. Just claimed, by fiat. And the OP doesn't know why it's okay to just make that claim in order to simplify the analysis. And then, why when it is not treated as infinite that the resistor value winds up in a more complex-looking denominator. Perhaps looking for an intuitive approach? Not sure. I doubt the OP even understands why a grounded-base Q2 = 'common base load' or what a common base load means to analysis. \$\endgroup\$
    – jonk
    Jan 28, 2021 at 22:31

2 Answers 2

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I admit I'm not entirely sure of how well your words have communicated to me. And I'm very sure that I do not understand exactly what you need to hear from me. LvW has already provided a standard segue into it, although I've no idea how well that falls upon what you are struggling for. My only addition in that vein is to suggest that you carefully read this page. They cover your questions with all the necessary math and reasoning. If you work at that page, I think you can recover much of what you want.

But I often struggle for two things, simultaneously. One is how to develop quantitative models that predict well enough to be of practical use. The other is how to develop basic intuition -- a more qualitative approach that aids me somewhat when I skim over schematics, identifying the more important aspects of each functional grouping as well as I may, and then move on more quickly to search out what I'm looking for or stop at places that catch my interest and perhaps become the target for more thought.

So in that vein, let me start with what I think you are discussing in your question:

  1. Gaining an idea about a first order approximation of the differential mode voltage gain given a single-ended source, with the other base tied to a voltage source that is at the same voltage as the reference point for the single-ended input source; and,
  2. Gaining an idea about a first order approximation of the voltage gain for a common-mode change at both bases.

I'll start by adding a few details to your schematic:

schematic

simulate this circuit – Schematic created using CircuitLab

I've included the dynamic resistance (just the thermal voltage of the PN junction divided by the current through it) right at the tip of the BJT emitter. This is NOT part of your schematic. I'm just inserting it there to make it visible and hard to ignore. (Do keep in mind that this is only for AC analysis and not for DC operating point purposes, of course.)

Case 1 -- Applying 'emitter-degenerated CE amplifier' model

Here \$Q_2\$'s base (\$V_2\$) is tightly nailed down to a voltage source. So it cannot move. This also means that \$V_{\text{TIP}_2}\$ is similarly and equally nailed down, as well. (The idea of \$r_e\$ is to capture the effect of what happens at the emitter when a tiny change takes place across the base-emitter PN junction. So it is perfectly proper to consider \$V_{\text{TIP}_2}\$ as fixed and equally treatable as an ideal voltage.)

If \$R_\text{E}\$ were discounted (treated as open), then the analysis is pretty easy. You have \$R_\text{C}\$ in the collector and you have two \$r_e\$ values in the emitter leg. If you have done prior work on basic emitter-degenerated CE amplifier stages, then you know that the voltage gain magnitude is just the collector resistance divided by the emitter resistance. Here, and considering \$V_{\text{O}_1}\$ that would clearly mean, \$A_{v_d}\approx -\frac{R_\text{C}}{2\,r_e}\$.

Notice that the very important simplification made was to consider \$R_\text{E}\$ as an open (infinite resistance.)

Since we are talking only about the tiniest of changes, we can reasonably make the statement that \$V_\text{E}\$ doesn't change for these purposes. And if that doesn't change, then the current magnitude in \$R_\text{E}\$ also doesn't change. So we just assume it exactly doesn't change. But that assumption is the same as if \$R_\text{E}\$ were of infinite magnitude. (Think about this.)

This assumption isn't true. But it simplifies analysis. Putting its value back into analysis would result in a more complicated equation. But for many purposes, not a much better one.

The result here depends on dragging over an idea from the emitter-degenerated CE amplifier. But that's cheating because I just hauled over the equation and used it as if it was right to use.

But what if that's wrong? The following will avoid cheating.

Case 1 -- An analytic method

Let's use KCL and get the value for \$V_\text{E}\$:

$$\begin{align*} \frac{V_\text{E}}{r_e}+\frac{V_\text{E}}{r_e} &= I_{\text{R}_\text{E}}+\frac{V_{\text{TIP}_1}}{r_e}+\frac{V_{\text{TIP}_2}}{r_e}\\\\&\therefore\\\\V_\text{E}&=\frac{I_{\text{R}_\text{E}}\cdot r_e + V_{\text{TIP}_1}+V_{\text{TIP}_2}}{2} \end{align*}$$

Now, we want to know how this value changes when \$V_1\$ changes. We can assume that the only part of the above equation that changes is \$V_{\text{TIP}_1}\$. So this means that: \$\Delta \,V_\text{E}\approx \frac12\,\Delta\,V_{\text{TIP}_1}\$.

(Keep in mind that \$V_{\text{TIP}_1}\$ is a fixed distance away from \$V_1\$. So this is the same thing as saying: \$\Delta \,V_\text{E}\approx \frac12\,\Delta\,V_{1}\$. Just FYI.)

We are allowed to shrink \$\Delta\,V_{1}\$ to an arbitrarily small value. In fact, so small that while its magnitude is larger than zero it is also smaller than the smallest finite value nearest to zero.

So, the change in \$Q_1\$'s emitter current will be:

$$\Delta \,I_{\text{E}_1} = \frac{\Delta\,V_{\text{TIP}_1}-\Delta \,V_\text{E}}{r_e}=\frac{\Delta\,V_{\text{TIP}_1}-\frac12\,\Delta\,V_{\text{TIP}_1}}{r_e}=\frac12\,\frac{\Delta\,V_{\text{TIP}_1}}{r_e}$$

Neglecting base currents, such that \$\Delta \,I_{\text{C}_1}=\Delta \,I_{\text{E}_1}\$, that tiny change then also causes a change in the voltage drop across \$Q_1\$'s \$R_\text{C}\$:

$$\Delta \,V_{\text{OUT}_1} = -\Delta \,I_{\text{C}_1}\cdot R_\text{C}=\frac12\,\frac{\Delta\,V_{\text{TIP}_1}}{r_e}\cdot R_\text{C}= \frac{R_\text{C}}{2\,r_e}\cdot \Delta\,V_{\text{TIP}_1}$$

Dividing both sides by \$\Delta\,V_{\text{TIP}_1}\$, find the differential mode gain as:

$$A_{v_d}=\frac{\Delta \,V_{\text{OUT}_1}}{\Delta\,V_{1}}=\frac{\Delta \,V_{\text{OUT}_1}}{\Delta\,V_{\text{TIP}_1}}=\frac{R_\text{C}}{2\,r_e}$$

That's the same result, but done analytically on the schematic.

Case 2 -- An analytic method

If I understand your wording (perhaps I'm projecting), then in this case you want to assume that \$V_1=V_2\$. (Both bases are set to the same voltage value to start.) Then observe the output voltages, \$V_{\text{O}_1}=V_{\text{O}_2}\$. (Both collectors should possess the same voltage value as a result.) Then apply a change to both inputs such that \$\Delta\,V_1=\Delta\,V_2\$ and work out how both \$V_{\text{O}_1}\$ and \$V_{\text{O}_2}\$ change in response.

In this case, we have one \$r_e\$ in parallel with another \$r_e\$, so for the voltage divider purposes we can merge these into a single effective dynamic resistance of \$\frac12\,r_e\$.

I'm going to skip the KCL solution process to get \$V_\text{E}\$ and instead move immediately to the change in \$V_\text{E}\$ for a given change, \$\Delta\,V_1=\Delta\,V_2=\Delta\,V_{\text{TIP}_1}=\Delta\,V_{\text{TIP}_2}\$.

The voltage divider computation can be used to work this out:

$$\Delta \,V_\text{E}=\frac{R_\text{E}}{R_\text{E}+\frac12\,r_e}\cdot \Delta\,V_{\text{TIP}_1}$$

That will incur a small change in \$I_{\text{R}_\text{E}}\$ and therefore: \$\Delta\,I_{\text{R}_\text{E}}=\frac{\Delta \,V_\text{E}}{R_\text{E}}\$. Again neglecting base currents so that \$\Delta \,I_{\text{C}_1}=\Delta \,I_{\text{E}_1}\$, that amount is split equally between the emitter currents for \$Q_1\$ and \$Q_2\$. It follows that \$\Delta \,I_{\text{C}_1}=\frac12\,\Delta\,I_{\text{R}_\text{E}}\$.

That tiny change then once again also causes a tiny change in the voltage drop across \$Q_1\$'s \$R_\text{C}\$:

$$\begin{align*} \Delta\,V_{\text{O}_1}&=-\Delta \,I_{\text{C}_1}\cdot R_\text{C}=-\frac12\,\Delta\,I_{\text{R}_\text{E}}\cdot R_\text{C}=-\frac12\,\frac{\Delta \,V_\text{E}}{R_\text{E}}\cdot R_\text{C}=-\frac12\,\frac{R_\text{C}}{R_\text{E}}\cdot \Delta \,V_\text{E}\\\\&=-\frac12\,\frac{R_\text{C}}{R_\text{E}}\cdot\frac{R_\text{E}}{R_\text{E}+\frac12\,r_e}\cdot \Delta\,V_{\text{TIP}_1}\\\\&=-\frac12\frac{R_\text{C}}{R_\text{E}+\frac12\,r_e}\cdot \Delta\,V_{\text{TIP}_1} \end{align*}$$

Dividing both sides by \$\Delta\,V_{\text{TIP}_1}\$ and recognizing that \$\Delta\,V_{\text{TIP}_1}=\Delta\,V_{1}\$ find the common mode gain as:

$$A_{v_{cm}}=\frac{\Delta\,V_{\text{O}_1}}{\Delta\,V_{1}}=-\frac12\frac{R_\text{C}}{R_\text{E}+\frac12\,r_e}=-\frac{R_\text{C}}{2\,R_\text{E}+r_e}$$

If you examine your own answer and divide through by \$\beta\$ you'll find your answer remarkably similar to this one.

(The only real difference is the fact that I've neglected the base current. Feel free to re-add that back in to the above and work out the new result. But the difference is negligible and therefore beneath notice for these purposes.)

Summary

The above approaches are non-standard and not likely to be found in just the above way in some textbook. At the outset here I pointed you to a page that follows well-worn approaches for this problem and then I proceeded to avoid it.

I enjoy using these same standard approaches, myself. I don't mean to dissuade you from that path. I even provided a page to help, there. But I also want to show you that there are many other ways of seeing the same thing and that actively developing and continually refining this flexibility and creativity of mind is also important -- perhaps even more important. (I think it is much more important to be frequently insightful than it is to be frequently right about things. Insights can be tested and thrown away, when wrong. But they can take you to solutions no others may see, too.

So, as you learn the well-worn path, also force yourself to try and consider new ways to see the same problem. Develop seeing a problem from at least three different ways, if not more. Verify that all of them take you to a similar place. Learn which are good, throw the rest away. You will benefit from that effort. You may then be able to help others and point to solutions they can only see better once you have shown them how to see it.

It's worth some effort. I can say so from experience. Personal stories are important because they are true. And I'd tell you one or two, but it would be difficult to write them where the meaning I want -- to convey the advantages of seeing many paths -- comes across well and where I avoid unintended self-aggrandizing. I don't know how to write well enough to achieve that without a lot of effort, so I won't even start.

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With one input signal-grounded, the whole circuit works like a common-collector stage (output at the emitter of Q1) which is loaded by a common base stage (input at the emitter of Q2). Then, it is a simple task to find the differential gain (assuming identical transconductances gm1=gm2=gm): Ad=Ad1*Ad2

Ad1=gm(RE||rin)/[1+gm(RE||rin)] and Ad2=gm*Rc.

Simplification: with rin=1/gm (common base) and RE>>1/gm we arrive at:

Ad=gmRc/2=Rc/2re.

(Note that re=1/gm is a differential resistance, other than the external RE)

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