What exactly is emitter degeneration anyhow?

[Analog Kid]

We see this term used all over the place in reference to common-emitter amplifier circuits.
But I really don't know what it means.
And don't tell me to look it up in Horowitz & Hill's The Art of Electronics: that book is not much help.
In an example of what can only be described as hand-waving, the first time this term is mentioned is this:

In this case the gain is -10,000/1000 or -10. The minus sign means that a positive wiggle at the input gets turned into a negative wiggle (10 times as large) at the output. This is called a common-emitter amplifier with emitter degeneration.

Nowhere else in the book do they define this term; you're supposed to get it by osmosis or something.

Now I understand that this has something to do with the fact that there's an emitter resistor, rather than having the emitter connected to ground. But I still don't really know what "emitter degeneration" means (nor why it's "degenerate" in any way). So an explanation would be appreciated here.

[Smokey]

As a current flows through that degeneration resistor it develops a voltage across it.  If the gate voltage relative to ground stays the same, then the voltage across the resistor is subtracted from the gate voltage which results in a lower Vgs, and the device being "less on".   
It's a common way to make parallel FETs share current better since if one FET takes more current than the others it's degeneration resistor will cause it be driven less hard and take less current.

[ejeffrey]

But I really don't know what it means.
And don't tell me to look it up in Horowitz & Hill's The Art of Electronics: that book is not much help.

Then look it up on Wikipedia: https://en.m.wikipedia.org/wiki/Common_emitter

Now I understand that this has something to do with the fact that there's an emitter resistor, rather than having the emitter connected to ground.

That's literally all it means.  You put a resistor there which reduces gain but increases linearity and reduces the sensitivity to gm variation.

The name doesn't matter.  It could just as well be called "emitter frobotization".  But apparently "degeneration" is an old fashioned word for what is now called negative feedback, decreasing or "degenerating" gain to improve other performance.  Doing this goes back to the earliest days of tube amplifiers before the development of external feedback networks or modern feedback theory.

[TimFox]

The old-school terms for positive and negative feedback are "regeneration" and "degeneration", respectively.

[Analog Kid]

OK, feedback: I understand one benefit of "degenerating" your emitter is negative feedback, which improves circuit performance (linearity, freedom from variations due to beta, etc.).

I'm still trying to wrap my head around two things: one, how this works, and two, why it works. That second one is a deep subject that I'd rather leave for later.

So taking this circuit fragment:



Help me to understand how that resistor causes feedback. My understanding (and this is just me sitting here without looking up anything online or in a book, because I want to really grok this) is:

o As current through the transistor increases (or temperature increases?), more current flows through R_E, which in turn
o Causes a (higher? lower?) voltage drop across the resistor, which in turn
o Decreases? the current through the transistor, thus stabilizing operation.

Or something like that.

Please don't mock me or tell me to go read some book: I have some kind of mental block that makes it difficult for me to apprehend such dead-simple things. I'd like to be able to look at a circuit and intuitively understand these sorts of mechanisms.

So can someone please correct what I wrote above, and explain, in simple terms (please, no Ebers-Moll!) how exactly this works? This seems to me to be a quite important and fundamental concept to learn. I know how crucial a thing negative feedback is, so I'm trying to understand it more fully.

[Bud]

As current flows through Re it creates voltage drop across Re which reduces the Base-Emitter voltage differential, forcing the transistor to reduce Collector-Emitter current You need to draw voltage polarity across Re to see this. The more current flows through Re the more positive Emitter becomes relative to Base, or equivalently the Base voltage decreases relative to Emitter. Reduction in B-E voltage reduces Ic  , reducing voltage drop across Re, until an equilibrium is reached.

[Analog Kid]

As current flows through Re it creates voltage drop across Re which reduces the Base-Emitter voltage differential, [...]

OK, back up a second: this is what I have trouble with, and your explanation doesn't really help, sorry.

So what you're saying is that as increased current flows through R_E the voltage drop across it increases, is that right? Meaning that the emitter is now more positive than it was before? And this reduces V_BE? For some reason it seems to me that this would reduce V_BE, not increase it, but apparently that's wrong.

See, I need a really, really simple and clear explanation for this to sink into my poor brain. This happens, which causes that to happen, which causes ...

[KE5FX]

Because the resistor is in the signal's return path, it drops some voltage.  Voltage that appears across the resistor is voltage that can't appear across the base-emitter junction.  Ignoring for the moment the whole religious argument about whether a bipolar is voltage- or current-controlled, the overall gain of the amplifier stage is lower as a result. 

Often, when you can make a transistor circuit's operating characteristics depend more on the characteristics of linear passive components than on the the transistor's own characteristics, it's considered a Good Thing.  That's what the resistor does.

[Analog Kid]

OK, more pictures time:



Prizes to the most simple, clear, direct  explanation of how R_E acts to create negative feedback here.
Sorry, statements like "Voltage that appears across the resistor is voltage that can't appear across the base-emitter junction" , while no doubt true, are about as clear as mud to me.

[Xena E]

OK, more pictures time:

(Attachment Link)

Prizes to the most simple, clear, direct  explanation of how R_E acts to create negative feedback here.
Sorry, statements like "Voltage that appears across the resistor is voltage that can't appear across the base-emitter junction" , while no doubt true, are about as clear as mud to me.

Because it subtracts from the base bias which you do not seem to be considering? Hint: forget V_be for the moment, consider it fixed.

(Or this thread is just a troll ?)

[magic]

I think I'm becoming tggzzz

o As current through the transistor increases (or temperature increases?), more current flows through R_E, which in turn
o Causes a (higher? lower?) voltage drop across the resistor, which in turn

I might suggest looking up Ohm's law in Horowitz&Hill or even Wikipedia and paying particular attention to signs of the numbers involved.

o Decreases? the current through the transistor, thus stabilizing operation.

This cannot be explained for as long as one wishes to completely ignore the effect of Vbe on Ic.

[RJSV]

Yeah,  like Bud says,  but restated a little different:
   Any extra current, through that emitter resistor will RAISE UP your voltage right at the emitter.
That then causes a reduction in the base to emitter input voltage,  in an active fashion, meaning that the base input (voltage) is actively reduced.  Not just the input current,  but also active by way of the amplified current (collector to emitter current as output of transistor.

   So you could say it causes the gain to work against itself,  with a self-limiting result.
Intuitively,  maybe a 'less wild' gain could be seen, as potentially 'more linear'.

   Otherwise,  left unchecked,  the temperature variations, during a day,  could be causing noticably fluctuating output levels,  aside from the linearity considerations.

[Xena E]

I think I'm becoming tggzzz

 

[Analog Kid]

OK, more pictures time:

(Attachment Link)

Prizes to the most simple, clear, direct  explanation of how R_E acts to create negative feedback here.
Sorry, statements like "Voltage that appears across the resistor is voltage that can't appear across the base-emitter junction" , while no doubt true, are about as clear as mud to me.

Because it subtracts from the base bias which you do not seem to be considering? Hint: forget V_be for the moment, consider it fixed.

(Or this thread is just a troll ?)

No, it is most definitely not a troll. I'm sincerely trying to understand this concept.

So when you say that an increase in the voltage drop across R_E "subtracts from the base bias", is that because V_BE really is fixed, so an increase in V_E (because of the voltage drop puts it closer to the base voltage? (Because the base is, what? about 0.7 volts higher than the emitter?)

I hope you can see my problem here: what I really need is a really, really simple explanation, perhaps with a diagram, showing which values move in which direction because of other values changing. No hand-waving: I'm not a dummy, but for some weird reason this is a difficult thing for me to grasp, although it may be second nature to you.

Help a guy out here.

[janoc]

You will want to watch these two videos:

https://youtu.be/Lr7MZQ6GIng

https://youtu.be/vtHCTJz0jr4

The entire deal with the biasing of the transistor and the negative feedback/degeneration resistor is well explained there.

[tggzzz]

I think I'm becoming tggzzz

I admire your choice of role model 

Caution: I'm considering changing my moniker to "Cassandra", in homage to the Greek legends

...
(Or this thread is just a troll ?)

You're not the first to have such thoughts

However, it could also be explained by someone who - for unknown reasons - can't understand Ebers-Moll, won't use Spice, and won't read the directly relevant parts of references (e.g. TAoE) people give him.

His learning curve will be long and slow, and will take a lot of time and effort by other people.

[Analog Kid]

You will want to watch these two videos:

https://youtu.be/Lr7MZQ6GIng

https://youtu.be/vtHCTJz0jr4

The entire deal with the biasing of the transistor and the negative feedback/degeneration resistor is well explained there.

Why did you delete your explanation? It was the best that was offered so far.
YouTube videos? Thanks but no thanks; I prefer text mode, like we're using here. I want the folks here who know a lot more than I do to explain things.

So my guess was correct: increased I_E causes increased voltage drop across R_E which causes decreased V_BE (contrary to what someone else posted here) which causes decreased I_E (and I_C). Good. I think my poor brain has finally absorbed that.

(And of course the B-E junction has to be forward-biased in order for the transistor to operate in the active region, correct?)

The upshot of all this being that this effect doesn't totally cancel the effects of amplification (as Horowitz and Hill put it, "a wiggle at the input gets turned into a larger wiggle at the output"), but it moderates it by reducing gain, which has all kinds of good effects.

Which is something else I still don't understand: how exactly negative feedback acts to improve things like linearity and dependence on beta. But as I said, I don't even want to get into that whole mess until I understand the mechanism thoroughly.

[SteveThackery]

Follow this step by step, and make sure you fully understand each step before proceeding.

1/ Current flowing out of the emitter causes a voltage drop across the resistor, such that the top of the resistor is positive with respect to ground.

2/ Current flowing into the base terminal travels through the transistor innards to the emitter terminal, and then through the resistor to ground.

3/ When current flows into the base, it must overcome the base-emitter voltage drop (0.6V) plus the voltage drop across the resistor on its way to ground.

4/ When the preceding circuitry drives the base more positive, more current flows into the base.

5/ The increase in the base current causes a bigger increase in the collector-emitter current.

6/ This increase in the collector-emitter current increases the voltage drop across the resistor, making the top of the resistor more positive than it was before.

7/ In 2/ above we said that the current flowing into the base must overcome the base-emitter voltage drop (0.6V) plus the voltage drop across the resistor on its way to ground.  But now the voltage drop across the resistor is bigger, so the total voltage drop the base current must overcome is bigger.

8/ The bigger voltage drop from the base to ground reduces the current flowing into the base.  This in turn reduces the collector-emitter current.

Therefore:

9/ For a given voltage on the base, the collector-emitter current is less than it would be without the resistor. This effect is known as negative feedback, or degeneration.

(I've skimmed over and simplified various details.)

OP: that's my best effort, which I'm adding to the other explanations. If you still haven't got it, I suggest you do two things. Firstly, go back to Ohm's law and study it until you have it totally nailed. Not just the equation, I mean you have a complete, gut-level understanding. Secondly, come back to us with specific questions about which detail of the circuit operation you still don't understand.

[Analog Kid]

^^^^^ Now we're getting somewhere. Yes, I think I've finally got it. Thanks for the detailed, step-by-step explanation.

(Unlike some other folks here who can't be bothered to take the time to do that, but instead diss me for asking questions and not following their protocols ...)

[Bud]

Meaning that the emitter is now more positive than it was before? And this reduces V_BE? For some reason it seems to me that this would reduce V_BE, not increase it, but apparently that's wrong.

Seems you were lost in your Increased/Decreased word salad 

More current through Re Increases voltage drop across Re, that voltage's Negative side is applied to the transistor Base through the bias circuit, Reducing the bias voltage  at the Base relative to Emitter.

[Xena E]

^^^^^ Now we're getting somewhere. Yes, I think I've finally got it. Thanks for the detailed, step-by-step explanation.

(Unlike some other folks here who can't be bothered to take the time to do that, but instead diss me for asking questions and not following their protocols ...)

Dont think anyone is trying to 'dis' you as you put it. I personally was intending to do a step by step walk through but as you'd gone off line its unfair to make these comments.

If contributors with questions are blatantly not prepared to read and digest standard texts and say so I'm sure you will agree it will appear to others they're either arrogant or fucking lazy.

My first post seemed to have you making a reply that was on the way to an understanding,  however these are all mental models of how things actually work, you do need to understand the fundamentals to have even a remote chance of how apply them, and trying to take each part in isolation won't work.

The easiest way to use BJTs as linear amplifiers is by employing emitter degeneration and fixed base bias voltage and proportioned colector load resistance, to provide predictable gain.

Steve Thackery's explanation of the specific question you ask is excellent, (apart from the expression "current flowing"), but thats only part of the picture.

In a previous employment I was mandated to help students study electronics but the ones who had the attitude that they wouldn't read the texts but just wanted someone to do their homework for them got shut off short order.

As for the troll comment, it is a good troll question, not in itself a bad thing, as even if you understand what your asking already it can help other readers who don't.

Regards
X.

[ejeffrey]

So when you say that an increase in the voltage drop across R_E "subtracts from the base bias", is that because V_BE really is fixed,

No.  For the purposes of analyzing circuit gain, consider Vb fixed.  Thats the input voltage to the amplifier referenced to ground.

Vbe is just Vb - Ve. So increasing Ve decreased Vbe.

[tggzzz]

If contributors with questions are blatantly not prepared to read and digest standard texts and say so I'm sure you will agree it will appear to others they're either arrogant or fucking lazy.
...
In a previous employment I was mandated to help students study electronics but the ones who had the attitude that they wouldn't read the texts but just wanted someone to do their homework for them got shut off short order.

There was a tradition on usenet, where it was easy to spot such questions in October and November. The responders had gentle fun by gave answers that were superficially reasonable, but anyone with the slightest knowledge would realise was wrong. The lazy and/or clueless submitted the answers, and their tutors instantly spotted what was going on.

Nowadays that technique is redundant, having been superseded by ChatGPT "answers".

I'm more than happy to help people who are prepared to put in the work to understand the subject more deeply.
I'm more than happy to point people in the right direction so they can avoid known blind alleys.
I'm not prepared to spoon feed lazy people who aren't prepared to do their homework.

The OP might care to see CharlotteSwiss' attitude; many people were delighted to help her. This thread https://www.eevblog.com/forum/beginners/what-an-oscilloscope-recommended-for-a-woman-passionate-about-electronics/msg3100087/#msg3100087 had almost 1000 responses over 4 years.

[coppercone2]

Lol someone bothering to ask a question about their home work... I Have a really bad feeling this is now considered work ethic. That would assume they pre-plan for at least 1 week before the assignment is due and bothering typing. You don't see that kinda thing anymore. That is the kind of initiative that gets you on the work management team. Refuses to work but DOES plan. manager 

[janoc]

You will want to watch these two videos:

https://youtu.be/Lr7MZQ6GIng

https://youtu.be/vtHCTJz0jr4

The entire deal with the biasing of the transistor and the negative feedback/degeneration resistor is well explained there.

Why did you delete your explanation? It was the best that was offered so far.
YouTube videos? Thanks but no thanks; I prefer text mode, like we're using here. I want the folks here who know a lot more than I do to explain things.

I have deleted it because it was redundant to what has been posted above.

I am not fan of Youtube for something like this either but these videos are very good, step-by-step walkthrough of a common emitter amplifier calculation. The first one focuses on biasing the transistor and the second one addresses the small signal behavior, including using the simplified transistor model to calculate the gain and the various component values and the various "rules of thumb" and simplifications that are commonly used but rarely explained.

So well worth the time.

So my guess was correct: increased I_E causes increased voltage drop across R_E which causes decreased V_BE (contrary to what someone else posted here) which causes decreased I_E (and I_C). Good. I think my poor brain has finally absorbed that.

That's correct.

(And of course the B-E junction has to be forward-biased in order for the transistor to operate in the active region, correct?)

Of course. Without that you have no collector current and no amplification either.

The upshot of all this being that this effect doesn't totally cancel the effects of amplification (as Horowitz and Hill put it, "a wiggle at the input gets turned into a larger wiggle at the output"), but it moderates it by reducing gain, which has all kinds of good effects.

Yes, it can't  totally cancel it - for that you would need to increase the emitter resistor value so that the voltage drop would drive the transistor to unity gain.

Which is something else I still don't understand: how exactly negative feedback acts to improve things like linearity and dependence on beta. But as I said, I don't even want to get into that whole mess until I understand the mechanism thoroughly.

The beta independence isn't really the work of the emitter resistor but the way you bias the base. The resistor's role is to set the gain of the amplifier stage, then the beta only needs to be larger than the final gain of the stage you want. The exact value won't matter because the base bias network + the emitter resistor will ensure the gain is fixed to the value you want.

The main role of the emitter resistor is to compensate for DC bias/collector current changes due to temperature. That's why it is commonly bypassed by a capacitor too - we don't want it to affect the AC signal amplification, only (undesired) changes  in the DC bias point.

The linearity improvement comes from this as well - when you bias the transistor in an amplifier, you typically put the DC bias point on the linear part of the characteristic because you want to minimize distortion. If the bias point moves due to the transistor getting warm or ambient temperature changes (or part tolerances/variations), it could land in the non-linear part of the characteristic. Same thing could happen even if the bias point remains on the linear part of the characteristic but the added input signal could end up on the non-linear bit already -> distortion. So you stabilize the bias point by adding this negative feedback to prevent this.

But do watch those two videos - Nick explains it really well and in an accessible way. All you need is Ohm's law, really.