# How we can physically imagine transistor saturation current?

1. Jul 20, 2015

### goodphy

Hello.

I've known that saturation current occurs when VCE is less than VBE, which means, the base-emitter is forward-biased and so it is for collector-base.

For NPN transistor, electrons from emitter (majority carrier of emitter) is able to go up to only base, no further travels into emitter. The electrons from collector (also majority carrier of collector) can go only up to base.

So, all currents from both ends of the transistor finally arrives at the base and it should go through base terminal. It means base current is the most strong than emitter and collector current?

I've always though base current is weakest one. Where am I wrong?

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2. Jul 21, 2015

### Svein

I always think of saturation as the point where the collector current no longer depends on the base current.

3. Jul 21, 2015

### meBigGuy

What the heck does that even mean?

4. Jul 21, 2015

### goodphy

In this case of NPN transistor, base-emitter and collector-base are both forward-biased, thus we can expect that there are two currents, one is from emitter to base, other is from collector to base. (We have two forward-biased junction.) Then at the base, two currents are combined and only place they can go through is base-terminal. Thus base current is sum of two currents.

Do you think my picture of saturation current right?

5. Jul 21, 2015

### LvW

Sorry - I have deleted my post because it was not correct (sign error).
In the saturation region, the base current IB is larger than the expression IB=IC/beta..

EDIT/UPDATE:
At first, let me say that it is not easy to find a "visual" description of the effects which can be observed for saturation operation. Let me try the following explanation:

1.) For VCE=0 we have only one energy source in the circuit (VBE) and it is true that the base current IB is divided into two parts: IE and IC. In this case, IB is larger than IC. That means: The equation I have given in my first (deleted) post holds: IB=IE+IC. Note: IC is in a direction opposite to the classical one.

2.) However, as soon as VBE>VCE>0 the situation changes a bit. The reason is (a) that the base region is very thin and (b) the doping profile of the transistors. Because both pn junctions are now forward biased they are effective in a series connection and provide a rather small resistance. Therefore, even a rather small voltage VCE is able to allow a current between C and E. This is possible also because the emitted electrons from the emitter have enough moving energy to cross the very small base region and go to the collector.

3.) But - at the same time - the effect as described above under 1.) makes the collector current smaller as dictated by the formula IC=IB*beta (because there is a small current from B to C in the opposite direction).
Thus, the resulting current IC is smaller than IC=IB*beta which means: IB>IC/beta.

4.) Therefore, the saturation region is decribed by this condition: IB>IC/beta.

UPDATE 2:
Some additional considerations regarding 1:) and 2.) :
The described behaviour can also be observed with a closer look to the output characteristics IC=f(VCE) of a BJT: The various curves for different fixed values for IB resp..VBE do NOT cross the origin. For VCE=0 the current IC is slightly negative (collector part of the base current) and - as a consequence, we have IC=0 (both parts in opposite direction cancel each other) for a very small value of VCE (some tenth of mV).

Last edited: Jul 21, 2015
6. Jul 21, 2015

### meBigGuy

Nothing about it is right. In NPN, there is never current from emitter to base or from collector to base.

Please look at a formula that relates base-to-emitter current to collector-to-emitter current in the linear region. What is that ratio called? When that ratio is reduced to 10 by overdriving the base, most manufacturers quote that as saturation and measure Vbe, and Vce.

7. Jul 21, 2015

### LvW

meBigGuy - I think, you are a bit too harsh. For my opinion, the questioner only has mixed-up the direction of the electron movements and the direction of current (which is nothing else than a definition!), see his first posting.

8. Jul 21, 2015

### goodphy

Thanks for giving me very detailed comments!

I feel like I'm still going around the first place I starts. According to your answer, my original picture of saturation current, where both forward-biased junctions will give "individual" current towards to base from them and base current is simply sum of them, seems right. I would like to ask you confirm this once more.

I feel that I'm still staying to think about saturation mode of the transistor very simply as two diodes of opposite directions are series connected with base terminal at the center and both are forward-biased.

9. Jul 21, 2015

### LvW

* At first - as mentioned already by meBigGuy - your current directions are inverse. For npn transistors the base current goes from the base node to the emitter node (and also to the collector node in saturation mode). In the latter case, this base current (consisting of two parts) is the only current for the case VCE=0 only!.
* However, as I have tried to explain, for a finite value of VCE>0 (but VCE<VBE) there is another current from the collector to the emitter (like in "normal" linear mode). Hence, we have two currents through the collector node but in opposite directions. Therefore, the resulting current Ic is smaller as in linear mode - which means IC<IB*beta (identical to IB>IC/beta).
* Note that there is a collector current in the "normal" direction (from C to E) - even in case of rather small VCE voltages (VCE=0.1 or so). This is due to the very thin base region, the typical doping profile and excessive carrier density in the base region (due to VBE).
* Note also that this explanation is based on the physical fact that the BJT is a voltage-controlled device IC=f(VBE) ; (maybe you have learned something else, but a small current never can directly control a larger current).

10. Jul 21, 2015

### Jony130

11. Jul 21, 2015

### goodphy

12. Jul 21, 2015

### goodphy

13. Jul 22, 2015

### LvW

I hesitate a bit to "destroy" the picture you may have about the working principle of a BJT - but to tell you the truth: The explanation in the given reference is wrong!
Let me try to comment some parts of the text:

During normal operation, a potential is applied across the base/emitter junction so that the base is approximately 0.6v more positive than the emitter, this makes the base/emitter junction forward biased.

Correct. This clearly shows the dominatiing role of the voltage VBE. Without this voltage biasing nothing happens!

When the base emitter junction is forward biased, a small current will flow into the base. Therefore holes are injected into the P type material. These holes attract electrons across the forward biased base/emitter junction to combine with the holes.

Also correct. This constitutes the base current which cannot be avoided.

However, because the emitter region is very heavily doped, many more electrons cross into the base region than are able to combine with holes. This means there is a large concentration of electrons in the base region and most of these electrons are swept straight through the very thin base, and into the base/collector depletion layer. Once here, they come under the influence of the strong electric field across the base/collector junction. This field is so strong due to the potential gradient in the collector material mentioned earlier, that the electrons are swept across the depletion layer and into the collector material, and so towards the collector terminal.

Yes. Correct description.
(Question: Has anybody the feeling that the current IB is more than a "byproduct" only? If yes - for what reason?)

Varying the current flowing into the base, affects the number of electrons attracted from the emitter.
OK - lets assume that there is an increase in base current corresponding to 100 additional wholes. Hence, 100 additional electrons can recombine (and do not enter the collector). Does anybody see any amplification?

In this way very small changes in base current cause very large changes in the current flowing from emitter to collector, so current amplification is taking place.

Does anybody see any justification or explanation for the claim that a "very small changes in base current cause very large changes in the current flowing from emitter to collector" ? It is simply an assertion without any verification. And it is wrong! I know that there are many contributions (even textbooks) claiming that the BJT would be current-controlled. Perhaps - because the authors think that the (correct) relation IB=IC/beta is identical with a cause-and-effect relation. But that is not the case.
Finally: There are many proofs (not on charged-carrier niveau but on circuit level - to be measured!) that the BJT clearly is voltage controlled - based on Shockleys famous exponential equation IE=f(VBE).

14. Jul 22, 2015

### goodphy

I've also known that BJT is "typically" considered as current-controls. But I've not taken this seriously as current and voltage for PN junction is one-to-one relation. Let's say we ave very good current regulator which keeps current at certain level, When it is connected to silicon diode, The only way this current flows through the diode is that whole circuit behaves to build voltage across the diode about 0.6 V. Can somebody imagine current injection to the diode without voltage developed across it? Without proper voltage, the diode is nothing but insulator!

I'm just wondering voltage-control or current-control is really important manner in not only practical view but also theory.

15. Jul 22, 2015

### LvW

Hi goodphi - the mentioned arguments are not new for me. In this context, I very often hear the term "chicken-and-egg" problem.
For my opinion it is not - because: No current without a driving voltage. That means: Applying what we call "current source" to a BJT or a diode is nothing else than to have a (large) voltage source with a very large source resistance Rs which forms a simple voltage divider between Rs and the non-linear device (BJT or diode). Remember the classical scheme how the operating point of a diode is fixed using a voltage source and a resistor (exponential curve and resistor load line in a common diagram).
In any case, the non-linear device allows a current only according to the voltage across it (equilibrium between V and I).

To me, it is really funny (surprising) that everybody starts designing a BJT stage with VBE=(0.6..0.7V) - in order to "open" the transistor (with a certain DC quiescent current). However, suddenly any change in the output current should be caused by the base current and NOT by a change in VBE?
I cannot understand how somebody who knows
(a) how a pn diode works and
(b) who is able to use his knowledge properly
can think that the pn junction within the transistor behaves completely different.

Example: Compare the gain values for two common-emitter stages (of course, the same dc bias point): One with beta=100 and the other with beta=200.
Wouldnt you expext a larger voltage gain for the second case (if the BJT would be current-controlled)?
However, the gain is equal in both cases because it is the transconductance gm=dIc/dVbe that determines the gain value (that means: the cause is dVbe).
On the other hand, it is well known that the gain can be increased for a given design by increasing the DC quiescent current. Why?
How can this be explained using current-control? The explanation is that we just have an increased slope gm on the Ic=f(Vbe) curve.

I think, one must "stumble" over such contradictions.

Finally, yes - it is important to know the real physics behind the working principle of parts. Otherwise, no circuit inventions are possible and we can rely on 30 years old books.
(Current mirrors and diff. amplifiers do function only because of voltage control; the same applies to B. Gilberts famous translinear circuit concept).

16. Jul 22, 2015

### meBigGuy

17. Jul 23, 2015

### LvW

Yes - of course. The gain is Vc/Vbe=-gm*Rc. And gm is independent on beta. As you know, without Re feedback such a circuit is not practical.
However, the same applies in case of feedback: Gain Vc/Vin=-gm*Rc/(1+gm*Re).

I know Claude Abraham already from some other discussions. He very strongly always is claiming that the BJT would be current-controlled.
However, Ive got the impression that he is not open for counter arguments - e.g. from leading instititions in the US (Berkeley, Stanford, MIT,..).
Did you ever hear or read a proof or justification for current-control?

18. Jul 23, 2015

### cabraham

The original 1954 Ebers-Moll paper from Bell Labs modeled the bjt as emitter current controlling collector current. The emitter current is functionally related to Vbe, the base-emitter voltage, but the circuit model describes Ic as alpha*Ie. Berkely, Stanford, and MIT to my knowledge do not claim the bjt is VC at all. In the small signal model, aka "hybrid pi" model, the small signal collector current can be computed as "gm*vbe" or as "hfe*ib". If the signal ac swing is small enough, current or voltage, the equations hold very well. But for large signal analysis, or using a bjt as a switch, this "gm*vbe" model is grossly non-linear and not used.
I posted plots on another forum where the signal generator driving the 1-stage bjt amp operates at hundreds of kilohertz up to tens of megahertz. At these speeds it is apparent that the change in Ie precedes the change in Vbe. Ic responds immediately to Ie, and Vbe catches up after Ic already settled. Clearly it is Ie in control, not Vbe. One of those plots is attached here.
I've probed many circuits in the lab, be it switching power converters, LED drivers, motor drivers, using diodes and bjt. When the frequency is in the rf range, it is all too easy to see that emitter current precedes base-emitter voltage, and collector current trackes emitter current very precisely. For a diode in a SPC, the forward voltage drop lags behind the forward current. The fallacy in your argument is based on Shockley's diode equation.
1a) Id = Is*exp((Vd/Vt)-1) is just one way to express this relation. Another form is:
1b) Vd = Vt*ln((Id/Is)+1)
Form 1a is what most are familiar with, but it cannot be overstated that the voltage value across the diode, Vd, does not "control" forward diode current Id. The diffusion capacitance formed at the p-n junction is non-linear, but it displays the same "Eli the ice man" properties of caps. A change in current will **precede** a change in voltage, always. Your position is based on the theory that although Ie determines Ic, it is Vbe that ultimately controls Ie, which is a mere assumption based on intuition. Ie does not depend directly on Vbe. Form 1b of SE should be regarded as well. Every critic of current control insists that currents are controlled by a corresponding voltage. It is assumed but cannot be proven because it isn't so.
You once asked me about the control mode of a motor. I stated that motor speed is controlled by voltage, motor torque is controlled by current. Not jumping topics, but we must be precise as to what that means. You stated, correct me if I misquote you, that a motor is "controlled by voltage", with no conditions. I will not put words into anybody's mouth, but am I correct in presuming you mean that the current in the windings controlling torque is a function of terminal voltage so that torque is ultimately controlled by voltage, as well as speed?
Here is my rebuttal, a dc motor made of superconducting windings has zero R, thus zero V drop, but non-zero current. Current is seen to be related to torque. Now study a copper wound motor with a small R in the windings. Current is still related to torque, albeit a small forward V drop occurs due to winding R. But if we construct a motor with carbon wire, 100 times the R of copper, we see that the same current produces the same torque, with a higher voltage drop. THe voltage in the 3 cases is zero, small, and large, and 3 currents are the same. The torques are the same as well.
The extra voltage needed in the higher resistance windings does not contribute to torque, only to heat. The extra voltage results in heat dissipation, it is an undesirable loss.
Just because a voltage drop is inevitable with resistive windings does not mean that V "controls" I at all. Likewise in a p-n junction, the forward drop is incurred due to charge diffusion, drift, recombination, and so forth. Vd/Vbe is a DROP, not an EMF. Drops do not "drive current". Anyway I will elaborate for those interested. Best regards to all.

Claude

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19. Jul 23, 2015

### LvW

My first reaction was not to reply to the above contribution because there are no new arguments. I dont know why I cannot resist to reply.
Perhaps because I was engaged in teaching electronics for more than 25 years - and I never got tired to explain things again and again.

Claude, are you aware that your long and extensively contribution does not contain one single proof or justification for your claims?
Sometimes you repeat very well known facts about „functional relations“ (which never were put into question).
Or you speak about alternative calculations ("gm*vbe" or as "hfe*ib"), which do not proove anything.
Sometimes you speak about circuit models (which not necessarily reflect the real physical behaviour).
Or you manipulate Shockleys equation (solving for Vd) - do you really think this mathematical procedure has any physical meaning?
Ohh yes - and, finally, you cannot resist to mention „Eli the ice man“.
Not the best argument because during steady-state conditions a phase of -90 deg is equivalent to +270deg. Hence, it says nothing about cause and effect.

I have told you already (with examples) that - to justify the voltage-control picture - it is not necessary to go down to carrier physics.
We just need to analyze some properties of selected BJT circuits.

PS: If desired I can give you references from Stanford, Berkeley, MIT,...
For example: Berkeley University: ....IB is an undesirable but inevitable side effect of producing IC by forward biasing the BE junction....

20. Jul 23, 2015

### Jony130

I don't want to be picky or pedantic. But the voltage gain will depend on beat value, not by much but it will ( β/(β+1 ). But of course in general case I agree with your whole post.