How we can physically imagine transistor saturation current?

[LvW]

LvW, I think your two transistor example you described earlier is wrong (different beta, same gain). If there are different betas, there will be different gain unless there is exactly balancing different rbe. (since Ib = Vbe/rbe, Vbe is same for both devices, so Ib is the same for both devices, so, since beta is different, Ic is different). the gm you pointed to is a linearization.

No - I don`t think so.
If the DC current remains the same also the transconductance does not change (gm=Ic/VT) - and the voltage gain still is gmRc/(1+gm*RE) .
This is true for RE finite oand RE=0.
The only difference is the input resistance (smaller for larger Ic).
In contrast, a larger DC current Ic (same transistor, same beta) causes an increase in gm and voltage gain.

 

[meBigGuy]

I'll have to go back to the analog designer that told me that and understand why he said that gmRl was a useful linear approximation and that actually Ib = Vbe/Rbe. Do you agree that if Ib is the same, then the gains would be different due to different beta?

 

[Jony130]

Jony130, I am still in a learning process (who is not?) - and this is not ironic - but I really don`t understand the meaning behind ...will depend on beat value, not by much but it will ( β/(β+1 ) . Please help me improving my english knowledge.

Forget about it, I was using the wrong small-signal model and this is why I jump with this ( β/(β+1) factor. And I also forget that β/rpi = gm.
Av = Rc/re * ( β/(β + 1) = (β/(β+1)*re)*Rc = (β/rpi)*Rc and β/rpi = gm.
So for a constant gm and difference in β value the voltage gain will be the same. And the only thing that has changed is rpi.

As for the "control" what about phototransistor? Is phototransistor a voltage or current controlled device? Or may be a "light/photon" controlled device.

 

[LvW]

I'll have to go back to the analog designer that told me that and understand why he said that gmRl was a useful linear approximation and that actually Ib = Vbe/Rbe. Do you agree that if Ib is the same, then the gains would be different due to different beta?

For comparing the voltage gain values of two different transistors (different beta values) I think it is only "fair" to use the same DC quiescent current in both cases. For this purpose, the DC bias must be adopted correspondingly.
Now - in both cases we have the same transconductance gm and, hence, the same voltage gain (formula given in post#51).
(That`s what also Jony130 has stated above).
Of course, in both cases the base currents (DC as well as ac) and the dynamic input resistance h11 (y11) will be different. But this has no influence at all on the voltage gain between base and collector.

To answer yout last question: Yes, if we keep Ib constant and - at the same time - have different beta values we also have diffent Ic values and, hence, different transconductances gm (different gain). But why is gain different? Again, because the transconductance gm has changed. And gm is the slope of the Ic=f(Vbe) curve!
The current gain beta (as we have seen: not a good and appropriate term) does not appear at all in the gain formulas.

By the way: Speaking about gain, input and output resistances, we always are speaking about "linear approximations". This is always the case when we are analyzing non-linear circuits. This is not any kind of "drawback", but an unavoidable consequence of small-signal analyzes.

 

[LvW]

In fact, from the first pdf Claude just posted, on slide 11, it's titled " Bipolar Junction Transistors: basic operation and modeling… … how the base-emitter voltage, vBE, controls the collector current, iC"
I'd be happy if he could just say that Vbe controls Ie.

Yes - it seems that Claude is trying to convince us and others that the BJT is voltage controlled. Funny situation.

 

[meBigGuy]

For comparing the voltage gain values of two different transistors (different beta values) I think it is only "fair" to use the same DC quiescent current in both cases. For this purpose, the DC bias must be adopted correspondingly.

You lose me here. If I take two NPN transistors of different beta, each with 1K collector resistor to 5V, Emitter to ground, and bases connected together, their gain will be different. I think this because Vbe directly controls Ie, not Ic. That's what Ebers-Moll says.

 

[LvW]

You lose me here. If I take two NPN transistors of different beta, each with 1K collector resistor to 5V, Emitter to ground, and bases connected together, their gain will be different. I think this because Vbe directly controls Ie, not Ic. That's what Ebers-Moll says.

Big misunderstanding! May I repeat my first sentence from post#54?

For comparing the voltage gain values of two different transistors (different beta values) I think it is only "fair" to use the same DC quiescent current in both cases. For this purpose, the DC bias must be adopted correspondingly.

Hence, in my example both base nodes are, of course, NOT connected because both transistors need a different bias scheme (for different IB but equal IC and equal DC operating point). That is the reason, the input resistance will be different in both cases (however, without any influence on the voltage gain).

I think, for comparing two different -but similar - circuits we must define, of course, the operating conditions for ensuring a "fair" comparison. As another example, comparing the gain capabilities of a FET and a BJT, I think both amplifiers should have the same load resistor (RC resp. RD) and the same operating point. Otherwise a comparison is useless and without any value.
And the same applies in our case: For comparing the influence of beta on the voltage gain, both BJT units must have the same operating point - otherwise a fair comparison is not possible (remember: Even one single BJT unit has different gain values for different operating points).

 

[meBigGuy]

I misunderstood you way way back in the beginning. no problem. I get it.

Moving on:

In active mode, Ie is totally controlled by Vbe in a beautiful way that includes Ib. That just wraps this thread up completely for me. (I had it wrong in an earlier post)

Claude can post all he wants to regarding his misunderstandings of transient effects caused by internal capacitances.

 

[LvW]

Claude can post all he wants to regarding his misunderstandings of transient effects caused by internal capacitances.

Yes - he continuously complicates the discussion by introducing effects which may overshadow the basic question. Therefore, the whole "ELI the ICE man" story is useless - as far as our main subject is concerned. Or he is teaching us some commonplace truths (truisms), for example "Emitter current and base current, LvW, are NOT the same, trust me." And, of course, one could ask - why such silly questions?

 

[LvW]

Readers may remember my claims that it is not necessary to go down to molecular physics (charged carrier level) for explaining how the collector current Ic is controlled (voltage Vbe or current Ib). Yesterday, by accident, another outstanding and very simple example has come into my mind: Darlington transistor:

It is known that the combined output-to-input current ratio for the classical Darlington combination (intentionally, I do not use the misleading term "current gain") is the product of the corresponding β values of each transistor: β,d=(β,1*β2,).

Now, let`s compare the voltage gain of such a Darlinton array with a single transistor - identical to the one with ic/ib=β,2.
For a fair comparison, in both cases we again should allow the same DC operational point.

As a result, we will find that the voltage gain of the Darlington combination is only 50% of the value obtained with a single transistor - although the Darlington stage has a beta value β,d that is β,1 times larger than that of the single transitor (β,2).

The reason is obvious: One can show that for the transconductances of both devices we have gm,d=0.5*gm,2.
This clearly prooves that the BJT is voltage controlled because it is the the transconductances gm=dVbe/dIc that determines the Ic variations and, hence, the voltage gain.

 

[cabraham]

For comparing the voltage gain values of two different transistors (different beta values) I think it is only "fair" to use the same DC quiescent current in both cases. For this purpose, the DC bias must be adopted correspondingly.
Now - in both cases we have the same transconductance gm and, hence, the same voltage gain (formula given in post#51).
(That`s what also Jony130 has stated above).
Of course, in both cases the base currents (DC as well as ac) and the dynamic input resistance h11 (y11) will be different. But this has no influence at all on the voltage gain between base and collector.

To answer yout last question: Yes, if we keep Ib constant and - at the same time - have different beta values we also have diffent Ic values and, hence, different transconductances gm (different gain). But why is gain different? Again, because the transconductance gm has changed. And gm is the slope of the Ic=f(Vbe) curve!
The current gain beta (as we have seen: not a good and appropriate term) does not appear at all in the gain formulas.

By the way: Speaking about gain, input and output resistances, we always are speaking about "linear approximations". This is always the case when we are analyzing non-linear circuits. This is not any kind of "drawback", but an unavoidable consequence of small-signal analyzes.

"Voltage gain" is just that, Vout/Vin. The "gm" value is defined for the express purpose of computing transconductance and stage vo.tage gain. Beta OTOH is defined for computing current gain. Saying that beta does not appear in voltage gain computations is rather trivial. The input signal source outputs a current and a voltage. A part of the voltage is dropped across internal signal source impedance, coupling cap if used, and rbb' the base spreading resistance, and re the internal emitter resistance, and Re the external resistance. A part of it is dropped across Rpi the small signal equivalent resistance. That voltage across Rpi times gm gives the signal value of ic, i.e. ic=gm*vbe. The gm value represents the transconductance of the raw device. The transconductance of the stage is always less. If "Gm" is that of the stage, then Gm < gm.

Likewise, if stage current gain is "Ai", then Ai < beta. A part of the signal generator's current is diverted across bias resistors. The part which enters the base is ib. Since ib is less than the current out of the generator, the stage current gain must be less than the raw part beta.

But beta does influence voltage gain as well. A device with a higher minimum beta value allows the use of a lower emitter resistor, or higher value bias resistors. This results in less attenuation of the signal due to signal source impedance. In other words, gm only affects voltage gain, but beta affects current gain, and voltage gain as well. Maybe this weekend I will attach a comp sheet.

"Transient capacitive effects"" have no relevance?! To say that Vbe "controls" Ie/Ib is dead wrong. I attached the MIT, Ga Tech slides just to show how inconsistent sources are regarding which is the control variable. It has been stated as being Ie, Vbe, and Ib, all from reputable sources. Those who insist that Vbe controls Ie cannot accept Eli the ice man because it demolishes their case, so they just disregard Eli. To say that Vbe controls Ie flies in the face of logic. Barrie Gilbert has done more without an EE degree than most high school grads can do, but theory wise, he is no match for a graduate EE, esp one with an advanced degree. He has no understanding of semiconductor physics. I've read numerous papers by him, and he simply re-iterates the myth that "Ib is an undesirable consequence to be minimized". But he doesn't understand, as well as others, that w/o Ib there is no Vbe.

When charges move through a pair of wires, then reach the b-e junction, both Ib and Ie arrive as well as Vbe at the same time. In order to change the value of Vbe the depletion zone must undergo a change in charge density. How does that happen? By transporting charges through the b-e junction. But charges in motion are current, by definition. The theory that Ib and Ie are "effects" of Vbe is untenable.

Regarding my simulation plots, I am aware that simulators are not 100% reliable 100% of the time. But I've used scopes and probes (I & V) to view diode, bjt, FET, IGBT, and other waveforms. I can eventually post a video in the lab with equipment and display Ib/Vbe?Ie waveforms but that will be in the near future because it takes time. Anyway just 1 more point.

This dispute is not limited to bjt parts. The fundamental conflict is in the way different people view the relation between I & V. Please answer my voltage divider question from before. Schematic attached. The input is a 12 volt battery. A pair of 1.0 kilohm resistors form a divider. The output is across R1. Obviously each resistor drops 6.0 volts, neglecting interconnection resistances, small in comparison.

Does the 6 mA current in R1 determine the 6.0 volts across R1, vice-versa, or is it a circular relation? I controls V, V controls I, or circular? What is your answer? No point discussing the bjt, because this question is at the root of the issue. BR.

Claude

 

[meBigGuy]

I say the BJT is voltage controlled and am then told by Claude that is a Myth, that Ic is determined by Ie (whatever that means). He then can produce no math or references that support his conjecture, and in fact posts papers that say quite the opposite.

Is the following true for a BJT in the active region (actually, I know it is):
Ie = Ies (exp(Vbe/Vt) -1)

Claude's voltage divider question is only another obfuscation designed to lend credence to his unscientific, unsupportable position that base current is automagically sucked out of the input circuit because Emitter current started flowing and, because of that, the base voltage then increases.

Claude's misinterpretation of his transient simulation data is confounding his understanding of cause and effect. Capacitive effects within the transistor simply cause the appearance of current change without voltage change, but voltage can be 0 while dV/dT is a positive value. Look at the math for ANY RC circuit.

This does not boil down to "Does I cause V or vice versa?". If you say the BJT is current controlled, the the base current controls it. If you say it is voltage controlled, the base voltage controls it. Nowhere is there room for "the emitter current controls it".

I think it is time for "claude's pet BJT theory" to be put to an end. It is not suitable for this forum, and cannot be substantiated.

 

[LvW]

But beta does influence voltage gain as well. A device with a higher minimum beta value allows the use of a lower emitter resistor, or higher value bias resistors. This results in less attenuation of the signal due to signal source impedance.
Claude

Claude, as mentioned already - it seems not possible to discuss with you on a fair and scientific basis. You are not able to concentrate on the main subject - neglecting secondary or other unimportant points. For example, you are not able to discriminate between (a) physical priciples of a device and (b) some aspects of circuit design.
In this context, it is pure nonsense to state that "beta does influence voltage gain as well". Are we speaking about the problem "how to design an amplifier stage" ?
By the way - even this design-oriented comment is false: The choice of the emitter resistor value does not depend on beta but only on the desired feedback factor gm*RE.

(Fortunately, you have removed those parts of the original version of your reply dealing with Barrie Gilbert and his qualification. These disrespectful lines speak for themselves - and against the writer).

Finally, let me say that, in general, I like discussions about engineering problems. Understanding comes through questions and discussion - and one can learn a lot from it.
However, from this thread (starting with post#18) I can only learn how a discussion should NOT look like.