How does Emitter bypass capacitor not affect the amplifier's working?

[gauthamtechie]

I have spent weeks analyzing biasing of transistors and I just about understood them to a basic extent.

From my understanding, the Small signal AC is varying the value of Ib and consequently Ic in the load line. If that is the case, then If i provide a Capacitor to short out The emitter resistor, then the AC will not cause any drop across Emitter Resistor and consequently the Operating point just remains there. Then how will there be a Voltage swing at the output? I am definitely missing something

 

[Nigel Goodwin]

It simply means the AC and DC operating conditions are different - essentially the gain of the stage is set by the collector and emitter resistors (the emitter resistor causes negative feedback). Bypassing it increases the AC gain as you're removing most of the negative feedback (there's still some, due to the internal emitter resistance).

 

[kubeek]

Re degenerates the current gain of the transistor. Thus the lower Re, the higher gain you get, up the point dictated by β and Rc.
Now take two situations, DC gain and AC gain. For DC, the gain is lowered by the Re, and thus lets you set the operating point without it being too much affected by variance in β.
For AC, the Re is effectively zero, so the amplifier has higher gain. Also see https://en.wikipedia.org/wiki/Common-emitter_amplifier

 

[Winterstone]

Both explanations above are, of course, correct.
Nevertheless, here comes another formulation of the mentioned effect:
An emitter resistor Re with a parallel capacitor Ce provides frequency dependent feedback.
That means: For very low frequencies (in particular for dc) you have sufficient feedback to stabilize the dc operating point and for frequencies far above the
corner frequency (set by Re||1/g and Ce) you have maximum gain without any feedback.

 

[MrAl]

Hi,

Some very good explanations already, i'd like to add just a little here...

There is a rough approximation to the gain of the common emitter amplifier. If the collector resistor is Rc and the emitter resistor is Re, the voltage gain Av is very roughly equal to:
Av=Rc/Re


That assumes that the amplifier has been designed properly. With a capacitor across Re we end up with a reactance from the emitter to ground Xe. As we increase the capacitor value, we lower Xe. And now the AC gain is very roughtly:

Av(AC)=Rc/Xe

So we can start to see that when we increase C we lower Xe, and since Xe is in the denominator of the gain equation we see the gain go up.

The difference here is that the gain for DC is roughly Rc/Re, while for AC it is roughly Rc/Xe. So we can maintain the bias point with Rc/Re yet get higher AC gain because of Rc/Xe. Note that we want to keep the DC bias point as constant as possible but we still want high gain for AC. We could not get that with just a resistor from the emitter to ground.

 

[Ratchit]

gauthamtechie,

I would advise you to read the full thread in the link below for maximum edification, but especially post #10. I am sure you will find it very ...interesting. This voltage-centric method of biasing was exposited by a well respected author of an electronics text, and member of a highly respected educational institution. It will also show you what parameter really controls a BJT. He confirms what I have been averring in this forum for a long time now. Ratch

https://cr4.globalspec.com/thread/68055/voltage-vs-current

 

[Winterstone]

........ With a capacitor across Re we end up with a reactance from the emitter to ground Xe. As we increase the capacitor value, we lower Xe. And now the AC gain is very roughtly:

Av(AC)=Rc/Xe

The difference here is that the gain for DC is roughly Rc/Re, while for AC it is roughly Rc/Xe. So we can maintain the bias point with Rc/Re yet get higher AC gain because of Rc/Xe.

Ohh - I think, this is an approximation that is "too rough" because with rising frequencies Xe is approaching zero and the gain goes to infinity.

Here comes the gain formula, which is to be used for varying Xe:

G=-Rc/(1/g + Xe)

Here you can see that the mentioned approximation is valid only as long as Xe>>1/g .
(Remember: the inverse of the transconductance g equals a resistor, which noemally is in the range 1/g=10...100 ohms, strictly depending on the selected bias point) .
Thus, for large frequencies we have

G=-g*Rc.

 

[ericgibbs]

Ohh - I think, this is an approximation that is "too rough" because with rising frequencies Xe is approaching zero and the gain goes to infinity.

Hi W,

A physical capacitors Impedance value, never goes to zero.

The Elliott people have an excellent record for accurate posted information.

**broken link removed**

Eric

 

[Winterstone]

Hi W,

A physical capacitors Impedance value, never goes to zero.

Hi Eric,

* Please quote correctly. Let me repeat that Xe is APPROACHING zero. Don`t you agree ? (I never have argued that it "goes to zero").
* Is the meaning behind your above comment that the claim from MrAl concerning the gain (Av(AC)=Rc/Xe) is correct? I cannot believe.
* Do you really take into account the parasitic ohmic resistance of a capacitor during design of a common emitter stage? This would be a very poor engineering approach.
Don`t you think that - even if you take the ohmic part of the capacitor into account - the value of 1/g will dominate above a certain frequency ?
* I am sure, you know that the formula as mentioned by me can be found in each relevant textbook. So - what is engineering contents and the intention of your reply?
Does it help the questioner to understand what`s going on?

With regards
Winterstone

 

[MrAl]

Ohh - I think, this is an approximation that is "too rough" because with rising frequencies Xe is approaching zero and the gain goes to infinity.

Here comes the gain formula, which is to be used for varying Xe:

G=-Rc/(1/g + Xe)

Here you can see that the mentioned approximation is valid only as long as Xe>>1/g .
(Remember: the inverse of the transconductance g equals a resistor, which noemally is in the range 1/g=10...100 ohms, strictly depending on the selected bias point) .
Thus, for large frequencies we have

G=-g*Rc.

Hi there Winterstone,


What part of "very rough" dont you understand

For reasonable values it's a good enough approximation to show what is happening in the circuit. If i wanted to get more particular i would have just analyzed the entire circuit which includes other various things. For a beginner i think it explains it pretty well. For a seasoned engineer like yourself you wont be happy until it is nearly exact.

If you like we can look at when the approximation fails but then we could look at when the beta assumption fails too. I was just trying to keep it simple for the OP. But i wont mind digging down into this more if you like, i might do that on my own anyway. Also, many of these "first projects" and first ideas deal with audio circuits where the frequency is limited to around 20kHz.

I guess all i was trying to show in as simple a manner as possible, was that the capacitive reactance makes the total resistance from emitter to ground look smaller and thus it raises the gain and the gain does go up when the Rc/Re gets bigger as Re gets smaller. So perhaps we can modify the gain equation a little to make it more appealing to your sense of snow.

 

[The Electrician]

If that is the case, then If i provide a Capacitor to short out The emitter resistor, then the AC will not cause any drop across Emitter Resistor and consequently the Operating point just remains there. Then how will there be a Voltage swing at the output? I am definitely missing something

There is another resistance in a transistor that is not shorted by the capacitor Ce. See for example the 4th figure on this page:



There is a resistor labeled REE in the diagram; this is the internal emitter resistance.

Also, see figure 5-6b here:

https://www.electro-tech-online.com/custompdfs/2013/05/05.pdf

The resistance labeled r'e is the same internal emitter resistance.

At frequencies high enough that Ce shorts out Re, the internal emitter resistance is NOT shorted and the ratio Rc/REE (or Rc/r'e) gives a good approximation to the AC gain of your circuit.

 

[ericgibbs]

* Please quote correctly. Let me repeat that Xe is APPROACHING zero. Don`t you agree ? (I never have argued that it "goes to zero").

Morning W

Its the gain goes to infinity statement which infers the Xc value [impedance] must be zero, which I am disputing.

is approaching zero and the gain goes to infinity.

I know you like precision in your answers.

Eric.

 

[gauthamtechie]

At frequencies high enough that Ce shorts out Re, the internal emitter resistance is NOT shorted and the ratio Rc/REE (or Rc/r'e) gives a good approximation to the AC gain of your circuit.

Thanks I get this now! Although here is something else I'm not getting: How does having a high Collector resistor compared to emitter resistor give me better gain? I am measuring my output from what is left out of the drop across the collector resistor. So having a higher voltage drop at the collector resistor should only give lesser voltage at the output but it is not so according the ratio I take. I don't seem to get this?

 

[Winterstone]

Hi there Winterstone,

What part of "very rough" dont you understand
.........
For reasonable values it's a good enough approximation to show what is happening in the circuit. If i wanted to get more particular i would have just analyzed the entire circuit which includes other various things. For a beginner i think it explains it pretty well.
.........
I guess all i was trying to show in as simple a manner as possible, was that the capacitive reactance makes the total resistance from emitter to ground look smaller and thus it raises the gain and the gain does go up when the Rc/Re gets bigger as Re gets smaller. So perhaps we can modify the gain equation a little to make it more appealing to your sense of snow.

Hi again MrAl,

Certainly I know that it is your intention to help the questioner as much as possible. And you can be sure that the same applies to me.
Nevertheless, I cannot agree with your approach - let me explain:

1.) At first, many books on electronics basics contain the formula for a simple gain stage without feedback G=-Rc*h21/h11=-g*Rc (g=transconductance)
Each beginner would be lost knowing only the rough approximation as given by you: G=-Rc/Xe.

2.) It is a good and important engineering practice to know under which conditions an approximation is valid (with acceptable errors).
This is connected with the basic mathematical rule that if you neglect one quantity you are allowed to neglect it only in comparison to another quantity.
That means: 1/g in the denominator of the gain expression may be neglected only if the condition 1/g<<Xe holds.
Of course this condition is not fulfilled with rising frequencies (because normally 1/g=10...100 ohms).
This is even true if Xe approaches a final value of less than 1ohm (see Eric`s comment)
In summary: From my experience as a lecturer for more than 25 years I consider it as very important - in particular for beginners - that he knows if (a) a formula is an approximation only and (b) under which conditions this approximation applies. (BTW: In electronics each formula is an approximation only - nothing is completely correct).

3.) Therefore: The gain expression
G=-Rc/(1/g + Xe)
contains the most important parameters for designing and understanding a basic amplifying stage in common emitter configuration.
(a) Without feedback: Xe=Re=0 and G=-g*Rc
(b) With frequency dependent feedback Xe=Re||1/wCe and
G=-Rc/(1/g+Re) (very low frequencies incl. DC) and G=-g*Rc (for frequencies far above the corner frequency wc=1/Ce*[(1/g)||Re] )
(c) This formula can even be used for the most common circuit configuration with signal feedback (only a part of Re=Re1+Re2 shunted by Ce):
with Xe=Re1+ Re2||1/wCe

I hope this clarifies things. I think, we shouldn`t simplify electronic principles/formulas not too much. This will not help beginners to understand basics - just the opposite is true.

Regards
W.

 

[Winterstone]

Its the gain goes to infinity statement which infers the Xc value [impedance] must be zero, which I am disputing.

Morning Eric,

of course, you are right (in principle). No doubt about it - there is no ideal capacitor. You can believe me that I know about this.
In the past, very often I have emphasized the fact that (a) no part and (b) no formula in the field of electronics is ideal or correct, respectively.

However, my point was if this really is an important information with regard to the subject discussed in this thread. More than that, I was afraid it could confuse the questioner.
You certainly will agree that during circuit design each engineer should try to find a solution that does not depend too much on parts parasitics.
To be honest: Up to now I never have seen a formula that contains the parasitic ohmic resistances of a capacitor. Did you?
More than that, this would be a very bad approach because there are some other parameters that normally are neglected, although their influence is much larger than the "forgotten" ohmic part of the capacitor.
(Examples: Uncertainties of beta, Miller effect, base width modulation/Early effect).

Regards
W.

 

[Winterstone]

The resistance labeled r'e is the same internal emitter resistance.

At frequencies high enough that Ce shorts out Re, the internal emitter resistance is NOT shorted and the ratio Rc/REE (or Rc/r'e) gives a good approximation to the AC gain of your circuit.

Hi Electrician,

in order to avoid confusion for beginners I like to point to the fact that the quantity you call "internal emitter resistance" is identical to the inverse of the transconductance g.
Thus

r´e=1/g..

Therefore, your approximation Rc/r´e is identical to the gain expression I have used g*Rc.
And - of course, this internal resistance is not bypassed by an external capacitor. Otherwise you would have shortened the controlling action of the base region.

W.

 

[Winterstone]

Thanks I get this now! Although here is something else I'm not getting: How does having a high Collector resistor compared to emitter resistor give me better gain? I am measuring my output from what is left out of the drop across the collector resistor. So having a higher voltage drop at the collector resistor should only give lesser voltage at the output but it is not so according the ratio I take. I don't seem to get this?

Gauthamtechie, you should consider the fact that the transistor acts as a voltage-controlled CURRENT source.
That means: The signal voltage drop across an external load resistor Rc is proportional to the value of this resistor.
Thus: Vout=-ic*Rc.

If you insert ic=g*Vin (eqation of the controlled source) you arrive at the given formula (without feedback).

 

[MrAl]

Hi again MrAl,

Certainly I know that it is your intention to help the questioner as much as possible. And you can be sure that the same applies to me.
Nevertheless, I cannot agree with your approach - let me explain:

1.) At first, many books on electronics basics contain the formula for a simple gain stage without feedback G=-Rc*h21/h11=-g*Rc (g=transconductance)
Each beginner would be lost knowing only the rough approximation as given by you: G=-Rc/Xe.

<snip>
Regards
W.

Hello again Winterstone,


You're just getting absurd here. When the internal emitter resistance is small compared to the external reactance we still end up with the formula i gave. That's why i stated "very rough" and it certainly emphasizes the point that the decreasing reactance increases the gain. Certainly the more exact formula is:
Gac=Rc/(re+Xe)

http://hyperphysics.phy-astr.gsu.edu/hbase/electronic/npncegain.html

but when re is taken to be much smaller than Xe we still see:
Gac=Rc/(Xe)

and that's good enough for now.

You say that you expect a beginner to get "lost" because of that formula, but then you expect a beginner to know what h11, h21, and g are without question. Maybe you are the one that is "lost" here because you dont seem to understand how approximations work

So lets hope that the OP knows what h parameters are.

 

[Winterstone]

Hello again Winterstone,


You're just getting absurd here.
..................
....................
Maybe you are the one that is "lost" here because you dont seem to understand how approximations work

MrAl, thank you for your nice and polite comment.

May I kindly ask you to say, which of my statements are "absurd"?
It is the first time in your last post to mention the role of re.
How should a beginner use your formula Rc/Xe for increasing frequencies?
How does he arrive at Rc/re ?
Why don`t you argue on a pure technical basis? What do you think is wrong in my arguments?

Regarding approximations: Up to now, I was of the opinion that approximations work as follows:
One quantity can be neglected in comparison to another one only if
(a) both are summed up or subtracted, and
(b) both are known in its magnitude.

If you are right, I was wrong for 45 years.

W.

 

[MrAl]

Hello again,


You are summing up this whole discussion as if this was the sole question that makes or breaks your entire career from start to finish. If you are wrong you have been wrong for almost 50 years. That's not the right way to look at this. It's a simple question, it does not make or break your entire career or make you look bad in any way even if you were completely and entirely wrong. You are much better than that and i always like to hear your take on these subjects.

One thing comes out here however, and that is that you just cant accept an approximation that is "good enough" to show a certain point of fact. It's just a simple concept, it's not the question of the century. The world isnt going to come to end. But i will try to explain my point of view a little better anyway so you might see what i am talking about.

Point of technical fact: if we include the 10 ohm re value when we have already an external emitter resistor value of 1000 ohms, the re makes a small contribution to the total resistance from the 'real' emitter to ground. Either we have 1000 ohms or we have 1010 ohms with and without the re being included. So if the collector resistor is 10k, we see a gain of either:
1. 10000/5000 (my approximation)
2. 10000/5010 (your approximation)

So my gain is 2 and your gain is 1.996, but that's not the end of it. We want to look at the AC gain next.

When the cap comes into play, we might design this stage such that the cap reactance looks somewhat like a resistor at a particular max design target frequency. If we wanted a gain of 10 at the max frequency, then we would have made the cap value appropriate to make the total resistance look like 1000 ohms. So now we have two more gains to look at:
1. 10000/1000 (my approximation)
2. 10000/1010 (your approximation)

So my approximation yields a gain of 10 and yours a gain of 9.9, not a heck of a lot of difference there.

BTW, in a real amplifier an additional resistor is added in series with the resistor that is already in parallel with the cap. This limits the upper frequency gain without depending on the internal re value as much.

Im sure we could find examples that have a more wide difference than that shown here. But this is the circuit i wanted to design and it worked just like that. So you can see how this approximation, being simpler than yours, gets the point across without having to dig too deep in to the theory of the transistor.

Later, more questions usually come up, and those are answered at the time they come up. So the student gets a basic grasp on what is happening and then can ask questions about more of the details as they will surely do anyway.

But dont make this look like an "I am right and you are wrong, or i am wrong and you are right" situation because it's not like that. You're approximation is better than my "very rough" approximation. Your's just requires more basic knowledge of the workings of the transistor to be able to apply than mine does. I was keeping it very simple and showing the same basic idea, while you got more into it and had shown a little more detail. So there is no "If you are right i was wrong for 4000 years" here