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.