Help me understand: simple transistor action--SOLVED

[golden_labels]

The virtual resistance R_AGC represents a source of voltage, not any physical resistor. It is “created” by Q3 follower producing V_AGC.

That 100 Ω resistor is not used to determine amplifier’s gain. Its job is to provide DC path to ground for the Q3 follower.

[TimFox]

Quantitatively, for normal silicon small-signal BJTs, under normal conditions
g_m = dI_c/dV_be = I_{c, quiescent} /(26 mV)

[ledtester]

Reading through my textbook, Electronic Principles by Albert Paul Malvino (1979),

Quantitatively, for normal silicon small-signal BJTs, under normal conditions
g_m = dI_c/dV_be = I_{c, quiescent} /(26 mV)

Starting on page 194 sections 9-2 and 9-3 cover the relationship between the gain, r_e and I_E.

Note equation (9-3), equation (9-3a) and Example 9-1.

I'm looking at this edition:

https://archive.org/details/electronicprincimalv/

[tggzzz]

The virtual resistance R_AGC represents a source of voltage, not any physical resistor. It is “created” by Q3 follower producing V_AGC.

Voltage sources have zero resistance, by definition.

Q3 does indeed set the large signal bias point. However, when doing calculations and explanations keep the large signal bias point completely separate from the small signal amplification.

That 100 Ω resistor is not used to determine amplifier’s gain. Its job is to provide DC path to ground for the Q3 follower.

Correct.

The small signal amplification depends on the small signal parameter r_e, which is completely separate from any R_E.

[tggzzz]

Quantitatively, for normal silicon small-signal BJTs, under normal conditions
g_m = dI_c/dV_be = I_{c, quiescent} /(26 mV)

Yes indeed.

Plus for large \$ \beta \$, r_e ~= 1/g_m, and that is how the gain is varied.

For more detail, see any basic BJT textbook, or derivative such as https://engineering.purdue.edu/wcchew/ece255s18/ece%20255%20s18%20latex%20pdf%20files/ece255Lecture_12_Feb22_BJT_Small_Signals.pdf

[temperance]

Indeed, and I somehow knew it didn't sink in for Analog Kid in the other thread.

The thread in question references Victor Meldrew. But let's not mention the war again.

[tggzzz]

How on earth did you remember that thread?! 

I had long forgotten it.

[Analog Kid]

OK, as expected, this thread is veering straight towards the weeds. But I suppose that's the nature of the beast.

Folks are referring to "that other thread" of mine which relates to this one. I was hoping not to even find out what that thread was, at least until I got a reply to my answer to this question posed by @temperance, which is a very good one for me:

Less IE means less IC means less "A".

Yes, but why does "A" change.

You did indeed post a question which provides the details to answer the gain riddle.

I posted an answer not long after that, one that I am quite unsure of; the question of why the gain of Q1 changes because of the change in AGC voltage is one that would really help my understanding of these things.

So if someone would be so kind as to point towards that "other thread", I'd appreciate it. I honestly don't remember it. And yes, it could well be that I didn't get whatever I was supposed to get back there.

I now understand part of the original question I asked, the part that explains why the voltage at that node decreases with increasing AGC voltage. Thanks to those who helped me get that.

Now I would like to know why that changes the gain of that stage. And if that could be explained without getting into things like intrinsic emitter resistance, Coulomb's Law, Ebers-Moll, and quantum mechanics, that would be even better.

A simple, first-order explanation would suit me just fine. Of course, I don't get to steer this discussion in any particular direction: I get that. I'm just hoping for something I can wrap my poor brain around.

Thanks in advance.

[TimFox]

The simple explanation is what I posted about the small-signal gain (transconductance) decreasing with decreasing emitter current.
The quantitative formula I posted is a direct result of the I vs. V curve of a PN junction (Shockley’s Equation, q.v.).

[Analog Kid]

The simple explanation is what I posted about the small-signal gain (transconductance) decreasing with decreasing emitter current.
The quantitative formula I posted is a direct result of the I vs. V curve of a PN junction (Shockley’s Equation, q.v.).

Is this the formula?

g_m = dI_c/dV_be = I_{c, quiescent} /(26 mV)

That doesn't do much for me. Can that principle (the relationship between gain and I_E) be expressed without calculus?

Besides, I don't see I_E in that formula, at least not directly.

[TimFox]

I_e is slightly larger than I_c, the difference is I_b.
(Three-terminal device: no where else for it to go.)
The ratio of collector current to base current is often called “beta”, but that’s confusing since the ratio is not well-determined, nor is it constant.  It is far more reliable to consider the BJT as voltage controlled.
You don’t need calculus to consider the small-signal transconductance:  g_m is the ratio of the small change in collector current divided by the small change in base-emitter voltage that caused it, and it is proportional to the quiescent current about which the variation wiggles.
Unfortunately, to derive the transconductance from the diode equation requires computing a derivative.
That’s a simple derivative to calculate, since the derivative of an exponential is a similar exponential.

[Analog Kid]

So what it boils down to is that since the transistor gain (β) equals collector current divided by base current, and since collector current is equivalent--not equal to, but close enough for gov't. work--to emitter current, therefore gain goes down when emitter current goes down.

Is that good enough for a first approximation?

[TimFox]

Beta isn’t directly relevant here.
The important feature is the transconductance, which is the voltage-in, current-out gain of the BJT, for small variations about the (DC) quiescent current.
Since beta is reasonably high, typically 100, the collector current is only maybe 1% lower than the emitter current;  since beta is not well known for the individual BJT, you might as well ignore the difference between emitter and collector current.
My approximate equation comes from the exponential dependence of I on V for a PN diode and the usual physical parameters (including temperature) for the silicon diode.
The input voltage to the entire circuit gives an output current from the first transistor, which, in turn, drives the load resistors and the base of the next transistor.

[magic]

I would like to know why that changes the gain of that stage. And if that could be explained without getting into things like intrinsic emitter resistance

If not intrinsic emitter resistance than transconductance, pick your poison.

You will never know what's the voltage gain (not to be confused with β) of a common emitter amplifier (AGC or not) without knowing how Ic changes with minor variations in Vbe.

That's all there is to it.

[PGPG]

therefore gain goes down when emitter current goes down.
Is that good enough for a first approximation?

Your sentence means only that there is monotonic dependence of gm (gain) and Ie and this tells a little about that dependence.
You can assume gm dependence on Ie is not only monotonic but is proportional.
The simplified equation is gm[mS]=40*Ie[mA].
For Ie=1mA gm is 40mS. For Ie=2mA gm is 80mS.

In real life I have never designed any circuit having amplification be dependent on gm. I always prefer to have gain be specified by resistors.
If you have Rc=10k and Re=1k you have gain about 10.
Assume that you have Rc=10k and Ie=1mA. If Re=0 you will get gain about = gm*Rc = 40m*10k = 400. And it is maximum you can reach.
If with Re=1k you set gain to 10 it is enough smaller from 400 that you can assume only resistors set it.
What if you set in this circuit Re=100Ω and expect to get gain of 10k/100 = 100. 100 is much closer to 400 than 10 and this absolute limit to 400 will partially reduces 100 to a smaller value. Math allows to calculate it but what for.

[Analog Kid]

I would like to know why that changes the gain of that stage. And if that could be explained without getting into things like intrinsic emitter resistance

If not intrinsic emitter resistance than transconductance, pick your poison.

You will never know what's the voltage gain (not to be confused with β) of a common emitter amplifier (AGC or not) without knowing how Ic changes with minor variations in Vbe.

Isn't that putting the cart before the horse in this case?
Because the question--the question I've been asked to answer here--is how (or why) voltage gain decreases when IE (and therefore IC) decreases. I don't see what Vbe would have to do with that.

Again, let me be clear: I'm looking for a reasonable first approximation explanation, not an exhaustive treatise involving Ebers-Moll, Child–Langmuir or advanced particle physics.

Jesus Fucking Christ: Why is it so god-damned difficult to get a straight answer around here? Seems like no matter what the question is, the discussion goes straight to the most convoluted, Talmudic, rabbinical/pontifical treatise of how many electrons can dance on the head of a crystal lattice, where no one who doesn't have a fucking PhD in advanced physics dare not enter?

C'mon. Humor me. Pretend I'm a high-school student who just wants to know on the simplest level how this works.

Rant over.

[TimFox]

Last time:  for small signals.
The transconductance of a BJT is proportional to the emitter current.
The overall gain of the entire circuit is proportional to the transconductance of the first stage.
Many designs use emitter degeneration from an unbypassed series resistor to reduce the transconductance to a more predictable value, as PGPB posted.  His factor of 40 is the same as my factor of 1/(26 mV).
This AGC circuit stabilizes the gain by feedback from an output voltmeter to the emitter current.
To use your wordage, why is this simple description so god-damned difficult?

[tggzzz]

Jesus Fucking Christ: Why is it so god-damned difficult to get a straight answer around here? Seems like no matter what the question is, the discussion goes straight to the most convoluted, Talmudic, rabbinical/pontifical treatise of how many electrons can dance on the head of a crystal lattice, where no one who doesn't have a fucking PhD in advanced physics dare not enter?

C'mon. Humor me. Pretend I'm a high-school student who just wants to know on the simplest level how this works.

Rant over.

That's unnecessary. Many people have tried to help you.

I gave you a reference containing the key equations and insights. Have you read it? Which bit don't you understand?

Look at equation 4.4. Add that to your understanding of how I_E and I_C are controlled in your circuit, and you have the answer.

[temperance]

And if that could be explained without getting into things like intrinsic emitter resistance

Unfortunately, I don't see a way around understanding the intrinsic emitter resistance to explain why the gain of the circuit you posted can be adjusted by a change in bias current.

There are numerous topics started by you where people did their utmost best to find ways to explain the concept in almost every possible and impossible way in treads about current mirrors, emitter degeneration,...

OK, as expected, this thread is veering straight towards the weeds. But I suppose that's the nature of the beast.

The intent was to bring you back to a part you have trouble with. But the problem seems more fundamental.

The original question:

Which makes sense except for one thing: why does the AGC voltage get subtracted from the 2.5 volts at that node?

This is a problem with applying Ohm's law, understanding voltage and current sources and applying KVL. You will need that if you want to understand transistor circuits.

To get your thinking into the right direction: why is the resistor in the emitter of Q3 100 Ohm? What would happen if the resistor would be 10 R, 1 K?

Maybe this question helps to make it more visible:

Take a 9 V battery with a 1 K Ohm resistor in series. Under what conditions could the given circuit be considered a voltage source, and under what conditions might it be considered a current source?

Edit: the above is far away from the ideal voltage and current sources presented in circuit analysis books. But this non ideal sources are fundamental in understanding and developing circuits of any kind.

[magic]

Because the question--the question I've been asked to answer here--is how (or why) voltage gain decreases when IE (and therefore IC) decreases. I don't see what Vbe would have to do with that.

And not seeing it is the whole problem, because you want to know voltage gain without considering how input voltage affects the circuit. It's simply not going to happen.

Gain is output AC voltage divided by input AC voltage. Hopefully you can figure out from Ohm's law that AC output voltage is AC collector current times external collector resistor (22kΩ here).

But how does AC collector current depend on AC voltage in? That's precisely the definition of trans- fucking conductance. It has nothing to do with β. And transfuctance happens to be proportional to collector fucking current, something you would have learned in kindergarten if you paid attention. So if bias decreases, then fucking transconductance falls too and you get less AC fucking current for the same AC fucking voltage applied to the fucking base. And your gain goes down the shitter.

(edit: The above applies because base AC voltage is same number as base-emitter AC voltage, because emitter is held constant by that 470nF capacitor as far as AC is concerned. This, of course, requires the capacitor to be "large enough".)

Jesus Fucking Christ: Why is it so god-damned difficult to get a straight answer around here? Seems like no matter what the question is, the discussion goes straight to the most convoluted, Talmudic, rabbinical/pontifical treatise of how many electrons can dance on the head of a crystal lattice, where no one who doesn't have a fucking PhD in advanced physics dare not enter?

Because they think you are missing some details and try to elaborate on those details, instead of recognizing that you simply refuse to look at concepts which are fundamental to the problem you want to solve

[Analog Kid]

Referring to the same textbook I took the AGC example from:

Starting on page 194 sections 9-2 and 9-3 cover the relationship between the gain, r_e and I_E.

Note equation (9-3), equation (9-3a) and Example 9-1.

OK; equation 9-3a, which is simplified by ignoring r_E, is this:

\$\displaystyle A\cong\frac{r_C}{r'_e}\$

So gain is equivalent to the collector resistor divided by the emitter resistor, ignoring the transistor's intrinsic emitter resistance. (Correct me if I got that wrong.)

I don't understand this in the context of reducing the gain by reducing I_E:

Looking at the diagrams in the book, they show actual resistors for these values. Fixed resistances.
Does changing the voltage across the emitter resistor change its value? or is this just some notional mechanism?
In the original example we have a 2.2KΩ emitter resistor. I don't see how changing the voltage across this resistor (therefore changing the current through it) can change its actual resistance value. Or are we talking about some kind of virtual resistance value?
Aaaargh; the denominator here is the intrinsic emitter resistance (r'_e), not the actual emitter resistor.
So when I_E decreases, does r'_e likewise change? And if so, in order to reduce the gain (A), it would have to increase, correct?

[tggzzz]

Referring to the same textbook I took the AGC example from:

Starting on page 194 sections 9-2 and 9-3 cover the relationship between the gain, r_e and I_E.

Note equation (9-3), equation (9-3a) and Example 9-1.

OK; equation 9-3a, which is simplified by ignoring r_E, is this:

\$\displaystyle A\cong\frac{r_C}{r'_e}\$

So gain is equivalent to the collector resistor divided by the emitter resistor, ignoring the transistor's intrinsic emitter resistance. (Correct me if I got that wrong.)

I don't understand this in the context of reducing the gain by reducing I_E:

Looking at the diagrams in the book, they show actual resistors for these values. Fixed resistances.
Does changing the voltage across the emitter resistor change its value? or is this just some notional mechanism?
In the original example we have a 2.2KΩ emitter resistor. I don't see how changing the voltage across this resistor (therefore changing the current through it) can change its actual resistance value. Or are we talking about some kind of virtual resistance value?

It is explained in the reference I gave.

If you can't be bothered to spend a little time looking at that, I can't be bothered to spend a little time failing to help you.

[TimFox]

r’_e is an equivalent resistance equal to 1/g_m.
Just stick with my description for your circuit, using g_m.
Do you not understand transconductance?

The use of r_e is a useful concept when you add an unbypassed emitter degeneration physical resistor to decrease the transconductance, but your circuit has a bypass capacitor directly at the emitter pin.

[Analog Kid]

Do you not understand transconductance?

No, not really.

[TimFox]

As I posted, it is how much change you get in the output (collector, plate, etc.) current for a change in the input (base, grid, etc.) voltage for an active device.
Current/voltage is conductance, 1/ resistance.
It depends on the operating conditions.