I would (and did) say, it depends on how you use it.
Vbe controls Ib , Ib controls Ic.
So voltage causes current but also current can cause voltage.
And the question was NOT if a voltage or a current is injected in the controlling node.
Are you really happy with such an answer? A device like a BJT allows for a current Ic that can be controlled externally.
And the question was NOT if a voltage or a current is injected in the controlling node.
The question was what happens physically in the base region! Which quantity influences the current flow - Vbe or Ib ?
But I don`t intend to start the discussion again. This was just to clarify the point of discussion.
Neither Vbe nor Ib, rather Ie is what controls Ic.
Hello Claude,
this assertion from you I could read already several times in this forum. I cannot remember one single book or any other serious text , which can support this (in my eyes: somewhat "surprising") opinion.
Up to now - you are the only person I know who is stating that Ic=alpha*Ie describes the control properties of a BJT. Of course, I can be in error and my opinion is of less importance.
Therefore, I promise you - I will change my mind if you answer the following question (and it is not the first time, I did ask you already at least twice, but without any response):
How do you explain the stabilizing effect without using the voltage Vbe, which an emitter resistor Re has on unwanted and unexpected changes resp. variations of the dc quiescent collector current Ic?
With regards
W.
Winterstone what do you want to stabilize?
If the DC gain is A and you change the Vbe by x volts, then the output voltage changes by A*x volts. So with a DC gain of 10 if the output voltage is 5.8v and Vbe goes down by 0.1v then the output goes down to 4.8v approximately. That's when biasing with a low impedance voltage divider and using an emitter resistor too.
The whole thing is a complete waste of time - it's dead simple:
A BJT takes current in it's base - so is current operated.
An FET doesn't take current in it's gate - so is voltage operated.
Easy as that, anything else is complicating things to try hide the truth.
But what does it matter anyway?.
Primarily, we are discussing the role of Re in stabilizing the design value for Ic. Perhaps you didn`t realize.
I think, each newcomer/beginner cannot understand how a BJT - if "current operated" - can be stabilized with a feedback voltage.
Could you explain?
W.
Hello Claude A.
Before answering - at least, I will try it - your posts#50 and #54. I like to present a feedback model in form of a closed control loop.
(I hope Eric will not close the thread before posting; creating "further arguments" is not the worst thing, is it? Is an "agreement" necessary to keep the thread open? )
I think, we all agree that the resistor Re provides feedback. Thus, why not create a corresponding control loop that can visualize what´s going on?
I have prepared a very rough (somewhat simplified) drawing with ic~ie, which shows two cases:
A) Control loop based on a voltage control model of the BJT with the corresponding loop gain LG(A)
B) Control loop based on current control model of the BJT - also with the corresponding loop gain LG(B).
Please see the attached pdf document.
Comment to case B): For a current control model I have found no other method than to build the difference of two currents through the base region. Other methods (if they exist) are welcome.
In this context I have used a base current ib2 driven by the emitter voltage Ve. The value of ib2 is determined by the input resistance seen at the emitter node, which is re=1/g. Thus: ib2=Ve/re=Ve*g
Result: When we compare the loop gain LG(A)=-g*Re with
*the known expressions for the amplifier input resistance at the base node rin=rbe*(1+g*Re), and
*the known expression for gain (with feedback) G=-g*Rc/(1+g*Re)
it can be seen that only the voltage control model (A) confirms the theoretical expressions (which are confirmed by measurements).
(Correction: Please read (1+g*Re) in the last line instead of (1-g*Re) )
In contrast, the loop gain for the current control case is larger by a factor beta, which is NOT in accordance with measurements.
Did I make a severe error?
Winterstone
In the foregoing, I have assumed that you know how the loop gain modifies the input resistance and the gain
*rin is increased by a factor (1-loop gain)
*gain is reduced by a factor (1-loop gain).
hi,
This Thread is going nowhere and is just being used as a platform to create further arguments as each new post appears.
As neither side is not prepared to change their position, common sense says the Thread should be Closed, as most likely agreement will never be reached.
E
Hello Claude A.
Before answering - at least, I will try it - your posts#50 and #54. I like to present a feedback model in form of a closed control loop.
(I hope Eric will not close the thread before posting; creating "further arguments" is not the worst thing, is it? Is an "agreement" necessary to keep the thread open? )
I think, we all agree that the resistor Re provides feedback. Thus, why not create a corresponding control loop that can visualize what´s going on?
I have prepared a very rough (somewhat simplified) drawing with ic~ie, which shows two cases:
A) Control loop based on a voltage control model of the BJT with the corresponding loop gain LG(A)
B) Control loop based on current control model of the BJT - also with the corresponding loop gain LG(B).
Please see the attached pdf document.
Comment to case B): For a current control model I have found no other method than to build the difference of two currents through the base region. Other methods (if they exist) are welcome.
In this context I have used a base current ib2 driven by the emitter voltage Ve. The value of ib2 is determined by the input resistance seen at the emitter node, which is re=1/g. Thus: ib2=Ve/re=Ve*g
Result: When we compare the loop gain LG(A)=-g*Re with
*the known expressions for the amplifier input resistance at the base node rin=rbe*(1+g*Re), and
*the known expression for gain (with feedback) G=-g*Rc/(1+g*Re)
it can be seen that only the voltage control model (A) confirms the theoretical expressions (which are confirmed by measurements).
(Correction: Please read (1+g*Re) in the last line instead of (1-g*Re) )
In contrast, the loop gain for the current control case is larger by a factor beta, which is NOT in accordance with measurements.
Did I make a severe error?
Winterstone
In the foregoing, I have assumed that you know how the loop gain modifies the input resistance and the gain
*rin is increased by a factor (1-loop gain)
*gain is reduced by a factor (1-loop gain).
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