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

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Hello Ratch and Winterstone,


Ratch:
I agree with your last statement, that we dont want to reject the model when some operating condition is marginal to the method. However i was under the impression that it worked well and accurately for almost any operating point except maybe saturation. The posts on that other site didnt seem to suggest that there was a limit to the operating point as he even mentioned scaling re with collector current starting at 1ma. I would then be led to believe that if we had anything over 1ma it should work.
Also, the spice simulation seems to work just fine with 0.65 volts on the base, and also i have done a few designs where the base emitter voltage was assumed to be 0.65.
So then what would you say is the appropriate operating conditions?

Winterstone:
Well see we were calling it the "voltage control method" just to have something to call this method. It's not that we really wish to control the current, just to calculate a single operating point. The post linked to in Ratch's post clearly indicates we can do this because it even delves into multiple stage DC coupled amplifiers. Without being able to calculate the operating point it would be worthless for that.
Also, when a transistor part number is given in a schematic it does not have the "Is" value written next to it. It also does not have the Beta value written next to it. All it has is the part number written next to it or in a parts list. So when we see the circuit we see "2N2222A" but that's it. No Is, no Beta, but we still have to analyze it. So my point was if we have a hard time finding Is we have a hard time calculating the collector current using the exponential formula. The post linked to stated that the Beta is all over the place. So we dont know Beta exactly, but we dont seem to know Is exactly either.
 
Hello MrAl,

I must confess, that I feel a little helpless now because I really don`t know how to answer you.
This is because in your last reply I couldn`t find neither any answer to my statements regarding "control" vs. "determination" (post#36 and post#39) nor any new information from your side.
What is really your position regarding the meaning and the application limits of Shockley`s equation?

Thus, I kindly ask you again to answer the following question:

Can you agree with me that Shockley`s equation
(a) clearly proves that Ic can be externally controlled by the voltage Vbe,
(b) cannot be used to determine the actual value of Ic (for a BJT stage without feedback) in case of an unknown value for Is.
(you may remember: This was your request in post#25).

For my opinion, the whole discussion has been reduced to this key question and I am really interested to hear about your position.
Thank you and regards
W.
 
Hello,

You are given a 2N2222A transistor with a collector resistor of 1k, +10v power supply, and a base emitter voltage of 0.65v, emitter grounded, that's it.
Are you saying you have no way to calculate the collector current unless i tell you how to do it and/or supply you with more 'values' ???

I gave you the part number of the transistor, i gave you the only other elements in the circuit. You should not need anything else except the 2N2222A data sheet if you care to look at that. Either you know how to do this or you dont. Look on a web site if you dont.
 
I have spent weeks analyzing biasing of transistors and I just about understood them to a basic extent.

Well, as usual, the debaters have taken over this thread and have gone off into the weeds without answering your question. The question wasn't "how to design stabilized bias network for an NPN amplifier".

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
View attachment 73808

I took the liberty of coming up the following amplifier which uses stabilized bias, utilizing a 2n3904, a 4700Ω load resistor Rc, and a 9V supply. Note that the emitter resistor Re is 1000Ω, so you would expect the amplifier to have a gain of 4700/1000=4.7 if the emitter is not bypassed. I choose the bias network to set the collector voltage just slightly higher than 1/2Vcc, allowing for some DC drop across Re.

First, look at the schematic, and the LTSpice DC bias solution. Note that with the values I chose for the bias network, V(E) is ~1V and V(C) is ~5V, and Vbe = 0.652V.

To make the amplifier useful, the input signal needs to be AC-coupled to the base, which is why C1 is there. C2 is the emitter bypass. The reason the values where chosen as shown will be discussed later.

Next plot is a freq domain sweep of of the amplifier gain V(C)/V(in). Note that I'm plotting it on a logarithmic scale (similar to a Bode plot, except not in db). Note that I chose the input coupling cap C1 such that the frequency where the mid-band gain drops to ~2 is 1Hz.

I chose the emitter bypass capacitor C2 such that the frequency where the mid-band gain increases to ~1/2 of the ultimate bypassed gain to be ~1Khz, thereby exposing the mid-band region where the emitter bypass is not yet effective, i.e above the corner caused by C1 but below the corner caused by C2.

You would expect the mid-band gain to be Rc/Re= 4700/1000 = 4.7 [because I(Rc) ≈ I(Re)] and you see that at 10.158Hz (Cursor 1), the gain is ~4.5, just slightly lower than the expected 4.7. That is due to some intrinsic emitter resistance that is internal to the transistor.

Now look at cursor 2, which shows that the gain at ~81kHz (where the bypass cap is fully effective) is ~156, so the gain at any audio frequency would increase from 4.5 to 156 as you add an emitter bypass capacitor whose impedance at that frequency is small compared to Re.

Obviously, if you intended to make an audio amplifier which has reasonably flat response for voice or music frequencies, you would choose a much larger C2. Increasing C2 to 100uF moves the lower corner (where the gain drops -6db) to ~35Hz.

Finally, look at the last plot, which simulates the time-domain response of the amplifier with an input signal of 10mV at 100kHz. V(C) is > 100χ V(in), as expected. It is centered on a DC value of ~4.8V, as predicted by the DC bias solution. The (bypassed) emitter has only a small amount of ripple and is sitting at a DC level ~0.9V, also as predicted by the DC bias solution.

Hope this clears up the effect of bypassing the emitter resistor...
 

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You are given a 2N2222A transistor with a collector resistor of 1k, +10v power supply, and a base emitter voltage of 0.65v, emitter grounded, that's it.
Are you saying you have no way to calculate the collector current unless i tell you how to do it and/or supply you with more 'values' ???

I realize that your reply to my last question (which could be answered with a simple yes or no) contains no answer, but instead again a question,
which I have answered already three times in this thread (post#36 and #39 and #42). Not a very effective way to discuss.
OK- so I will give up since I cannot expect an answer from you.

I gave you the part number of the transistor, i gave you the only other elements in the circuit. You should not need anything else except the 2N2222A data sheet if you care to look at that. Either you know how to do this or you dont. Look on a web site if you dont.

Yes - no surprise. That`s the way you like to "discuss" (see also post#19).
We should stop it now.

Winterstone
 
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Well, as usual, the debaters have taken over this thread and have gone off into the weeds without answering your question. The question wasn't "how to design stabilized bias network for an NPN amplifier".

Hi Mike, I cannot fully agree to this conclusion. For my opinion, the initial question has been answered at the beginning of this thread (see post#1 to #4 and post #17.) - I must confess, however, not as detailed as in your contribution.
And - as you will have noticed - the rest of the thread was about another problem (raised by the member Ratchit), which was/is connected with the question voltage vs. current control of the BJT
Of course, it would have been better to create a new thread for this "discussion", which unfortunately was not really a fair "discussion".
It is not the first time for me to experience that a fair and pure technical discussion with some forum members is "difficult". I have some experience with many other technical and non-technical forums - surprisingly, this form of "debate" I was faced in this forum only.

You would expect the mid-band gain to be Rc/Re= 4700/1000 = 4.7 [because I(Rc) ≈ I(Re)] and you see that at 10.158Hz (Cursor 1), the gain is ~4.5, just slightly lower than the expected 4.7. That is due to some intrinsic emitter resistance that is internal to the transistor.

Mike, may I give a short comment to this?
To explain the difference between 4.7 and 4.5 you refer to "some intrinsic emitter resistance that is internal to the transistor". I know that there are some publications which are using such a terminology.
And - indeed, the input resistance in common-base configuration (at the emitter node) is identical to this resistor re.
However, according to my experience with many students and beginners I consider it as very important to explain that this resistance re is NOT any parasitic (unwanted) quantity (as it seems by using the wording "...some intrinsic...).

Instead, this value re=1/g clearly demonstrates the main action of the transistor - that is: To transfer a small voltage change at the input into a corresponding current change at the output.
And this is characterized by the Ic=f(Vbe) curve which has a slope at the selected bias point which is called transconductance g.
And - as you know - to develop the gain formula you have no other choice than to use the transconductance g.
Thus, it turns out that the gain with resistive feedback contains the term g [ GAIN=Rc/(1/g+Re) ].
Because 1/g is given in V/A we can allocate a new symbol to it: re=1/g.

I think, this is very important because
* One can clearly see the connection between re=1/g and the characteristic curve Ic=f(Vbe), and
* it becomes obvious that the designer can SELECT this value g=1/re (and thus determine the gain) because the slope is g=Ic/Vt (Vt=thermal voltage).


Now look at cursor 2, which shows that the gain at ~81kHz (where the bypass cap is fully effective) is ~156, so the gain at any audio frequency would increase from 4.5 to 156 as you add an emitter bypass capacitor whose impedance at that frequency is small compared to Re.
.

Using the transconductance g=Ic/Vt you are able now to calculate the gain for large frequencies (instead of stating simply "...whose impedance...is small compared to Re").

Gain=-g*Rc=-Rc*Ic/Vt=-4700*0.9*1E-3/26*1E-3=-162.

This result is in good agreement with the simulation (~156). I think, the main reason for the difference is the BJT internal output resistance that was not included in the calculation.

Thank you and regards
W.
 
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Hello again,


Yes Winterstone i will not answer your question. I asked a simple question of you and you can not answer it. So out of respect for the OP i will start another thread and see if anyone else can answer this simple question. This so we can leave this thread without discussing this side topic more.
 
Thanks everyone I appreciate it ! Although I realize I have quite a long way to go to understand all your posts .,
And Ratch, i did go through the thread and also have the student manual for "The Art of Electronics" . I appreciate your concern in helping me get a start on transistors.

I believe my understanding of all answers needs a little more learning from my side, although in post #11 The Electrician mentioned the internal resistance which doesn't get shorted out. So I took that as the answer to how the Small signal AC does cause a Wiggle to get amplified in Collector.
 
Hi there,

Sorry we got into a little side discussion there but we were trying to look at different ways to do this so you could benefit. I've started another thread to handle those queries however so we dont have to muddy up this thread with the side issues.

As you are probably finding out now, there is often more than one approach to a problem like this.
 
I believe my understanding of all answers needs a little more learning from my side, although in post #11 The Electrician mentioned the internal resistance which doesn't get shorted out.

gauthamtechie, you - as a beginner - who is interested to learn the "secrets" behind transistor operation, should also know the physical background of this "internal resistance".
To give you some further help, I like to ask you:
Do you know
* the approximate value of this resistance re, and
* if this value is constant or not?

W.
 
No I don't know the value, but a link to this post: https://www.electro-tech-online.com/custompdfs/2013/05/05.pdf in the electrician's answers used this way to find out the internal resistance: 25mV/Ie. So that is a very small resistance in every case. I don't know how they zeroed in on the 25mV across the internal Base-emitter region and the details.
Regarding constant, it looks like they're keeping the 25mV constant instead of taking a constant resistance. Now it strikes me why 25mV because even if they considered the forward voltage drop in Base-emitter it'd have to be 0.7 volts or so. Or is it changing with forward current after biasing ? I've read this someplace..
 
No I don't know the value, but a link to this post: https://www.electro-tech-online.com/custompdfs/2013/05/05.pdf in the electrician's answers used this way to find out the internal resistance: 25mV/Ie. So that is a very small resistance in every case.

Hi gauthamtechie, it is correct that re=25mV/Ie - however this is an approximate value only because the temperature voltage Vt=25mV can, of course, change with rising/falling temperatures.
However, the main point is that this value depends on the selected current Ic (which is nearly identical to Ie). And - as an example - for Ic=0.1mA we have a resitance of 25*1E-3/0.1*1E-3=250 Ohms.
I am not too happy with the invention of the term re. For my opinion and for the sake of a good understanding of the transistor principle it is better to use the inverse value: g=1/re.
I have explained this in my reply#46 in this thread and, therefore

I repeat the corresponding part of my reply#46:

Instead, this value re=1/g clearly demonstrates the main action of the transistor - that is: To transfer a small voltage change at the input into a corresponding current change at the output.
And this is characterized by the Ic=f(Vbe) curve which has a slope at the selected bias point which is called transconductance g.
And - as you know - to develop the gain formula you have no other choice than to use the transconductance g.
Thus, it turns out that the gain with resistive feedback contains the term g [ GAIN=Rc/(1/g+Re) ].
Because 1/g is given in V/A we can allocate a new symbol to it: re=1/g.


I think, this is very important because
* One can clearly see the connection between re=1/g and the characteristic curve Ic=f(Vbe), and
* it becomes obvious that the designer can SELECT this value g=1/re (and thus determine the gain) because the slope is g=Ic/Vt (Vt=thermal voltage).


With other words: The element re is NOT an internal resistance of the BJT. But it can be treated as such because of re=1/g. However, using only the term re the connection with the
characteristic curve Ic=f(Vbe) cannot be seen. And this makes it more difficult to understand the BJT`s operation.

Regards
W.
 
gauthamtechie,

Sorry I could not get back to you earlier, but I was on vacation.

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?

Think of it this way. Applying a proper vb signal voltage across Re through the base terminal will cause a definite emitter current swing. The emitter and collector current of a BJT operating in the active region are almost identical. Therefore, if the Rc resistance is 10 times higher than Re, the voltage swing across Rc will be 10 times what it was across Re, due to the collector and emitter currents being almost equal. That is, a signal voltage gain of Rc/Re will equal 10. So by making Re as small as possible, either by bypassing it with a capacitor or using the method expounded by Winfield Hill in the link I posted, we get the highest gain. Ask if you have any question about this.

Ratch
 
Gauthamtechie, in addition to Ratchit`s explanation I like to remind you (see also post#17) that the BJT works as a voltage-controlled CURRENT source (therefore: Vout~Iout).

W.
 
Hi I think I finally get it!!!!:) It is about the swing!
What I mistook was that If a Large Voltage dropped at Rc, then only what is remaining of the Vcc will be available across the output. But now It seems clear, only If I create a Big Enough Swing with a large Rc, will I be able to use a Vcc and get a bigger difference voltage which is Vout. Isn't that right?
 
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