# How a BJT Transistor works (base current version)

#### cabraham

Feb 12, 2015
63
It's simple the circuit you sent me has a reactive component in the CE path, a capacitor across R2. So the easiest thing to do was to insert a 100R in series with the capacitor. Got rid of the spike.

View attachment 18662
The circuit you sent me didn't have R7. Top plot without resistor and bottom plot with resistor.

View attachment 18660 View attachment 18661
So where is vbe? When I simmed the circuit in my original schematic, there was no spike at all. Please show Vbe, and we can discuss.
Claude

#### Arouse1973

Dec 18, 2013
5,178
That's because I thought the circuit you sent me was the one to simulate. You didn't mention about it being different. So forget the pulse for a minute, do you agree with the fact you can see a change in Ie which has a very similar slope to Vbe in my first plot.
Cheers

#### cabraham

Feb 12, 2015
63
Hi Claude,

"The ie current is the starting point where all begins. An increase in ie can only result in a likewise increase in ic as well as vbe. No way to skirt that. The ie is what absolutely determines ic."

How can a current like Ie be the „starting point“ ? I cannot follow your reasoning.
There is no current without potential difference (voltage).
Let the collector node unconnected and apply a positive base-emitter voltage (npn case). As a result, we have a current Ie=Ib - caused by Vbe - from B to E (exponential law), correct?
Now - apply some volts between C and E. What happens? The current Ie is now split into Ic and Ib. Thats all.
You simply cannot deny the role of Vbe because THIS voltage is the real starting point.
Question: Is there any reason to assume that - suddenly - Vbe should loose its control function?

EDIT: It is easy to demonstrate this behaviour with circuit simulation.
(a) Collector open, Vbe=0.67V: Ib=Ie=1mA ; (b) Vbe=0.67V, Vce=2V: Ie=Ic+Ib=1mA with Ib very small.

"The internal physics does not change by adding Re, Rc, Rb1, or Rb2."

Correct.
However, adding these elements certainly will distort the measurements/simulations which cannot be attributed to the „naked“ BJT alone.
Example: Emitter resistor Re influences Vbe as a result of the product Ie*Re. Hence, there is a control loop (output influences input) that influences the timing properties (which you were interested in).

"If vbe must change 1st before ie can change, what is the agent that changes vbe w/o incurring a prior change in ie? Can you answer that. "

I dont understand the question. Vbe is an external voltage (bias condition) which is altered due to a signal to be amplified.

"Early effect is something you should explain and detail why it supports your position."

Lets evaluate the characteristics Ic=f(Vce) for Ib=const. These curves are used to find the Early voltage as a measure for the slope of the various curves.
In words: We keep Ib=const and observe an Ic increase for a rising voltage Vce. And I agree to your explanation:

"When Vce increases, the depletion zone shifts and the base region gets narrower, the depletion zone has moved into the base region, reducing its width."

What is the consequence? Due to the reduction in width the electrical field strength increases and allows the current Ic to increase (more electrons can cross the depletion region).
Hence, it is the electrical field in that region that influences the current Ic. And which external quantity produces this field? Answer: The voltage Vbe.

"Tempco? How does that "prove" vbe is in control? "

The tempco of app -2mV/K tells us the following:
Rising temperatures cause an increase of the current Ic. In order to bring Ic back to its initial value (Ic=const) we must reduce the voltage Vbe by app -2mV/K.
For my opinion, this illustrates the controlling role of Vbe.
And note: This value of -2mV was verified by calculation based on charged carrier physics!

"You claim that the Ic rise causes a rise in voltage across Re, which is putting the cart before the horse. Then you claim that Vbe decreases as a result, then Ic decreases. But seriously, how can Vbe decrease before Ie/Ib decrease."

I agree with you that, in some cases, it is problematic to describe the correct timing sequence within a control loop.
However, thats not the point - and I am sure you will not deny that Re provides feedback (and Ic stabilization).
The main point is: Re causes a drastic increase for r,in at the base node - and from feedback theory we know that such an increase occurs for voltage feedback only!
Hence, the controlling quantity is a voltage between B and E. Is this conclusion false?

______________________

Finally, let me make a statement that sounds rather simple:
If somebody knows how a simple pn diode works (exponential relation between applied voltage and resulting current), how/why can he assume that - suddenly - the pn junction between base and emitter will behave completely different? Is there any good reason for such an approach?

LvW
My response to bold quote. I know the relation between I & V for a simple p-n junction diode. I have 37 years of hardware design experience, and am in the dissertation stage of Ph.D. Everybody with a BS has seen Shockley's famous diode equation. The fact that Ie and Vbe are related via Shockley is not contested. Here is your fatal error in thinking.

The diode forward voltage Vd, is not the cause of diode current Id. Search using key words as follows:
diode forward reverse recovery
Also you stated "applied voltage and resulting current". That in itself is a prejudice that many self taught people have. You apply voltage and then current is the result. In order to charge up the diode junction to Vd, current must pre-exist said voltage. I get weary saying this. I have more sims I will post involving diodes in switching power converters. I comes before V and determines V. I'll post them later. Good day.

Claude

#### Arouse1973

Dec 18, 2013
5,178
I'll be interested to see Professor Lutz (Lvw) reply on this one.

#### cabraham

Feb 12, 2015
63
That's because I thought the circuit you sent me was the one to simulate. You didn't mention about it being different. So forget the pulse for a minute, do you agree with the fact you can see a change in Ie which has a very similar slope to Vbe in my first plot.
Cheers
But the timing is the key. Please show Ie and Vbe in the same plot so we can see the sequence of events. A rising Ie results in a rising Vbe, we all know that. My point is that Ie settles on a plateau while Vbe keeps rising then settles. With your first sim, Vbe monotonically rose while Ie rose, fell, then rose again. Both cases prove that Vbe does not "control" Ie.

Just think about how charges transit through the junction. How can it be possible that Vbe changes before Ie? Vbe is the barrier potential formed by the two edges of the b-e depletion zone. The line integral of the E field that spans the depletion zone is the potential of the barrier. In order to change it you must transport more charge through the region.

If we have a Vbe, then suddenly cut the base and emitter current, Vbe will be held a short time. Diffusion capacitance needs time to discharge. Ib/Ie go to zero, Cd, the capacitance across b-e junction, hold the charge and voltage Vbe momentarily. Then these minority charge carriers in the depletion zone eventually recombine. The time needed is the carrier lifetime. Then Vbe decays to zero.

To increase Vbe, charges transported through the junction must first increase. This higher population density of charges crosses the junction, and some recombine, others continue to collector, increasing Ic immediately. The recombination ionizes more base atoms, likewise increased hole density ionizes more emitter atoms, and Vbe begins its ascent.

Vbe responds to an increase in Ie/Ib. Ic responds as well. Vbe and Ic are incidentally related, like Ib and Ic. This causality argument flies in the face of decades of observation of junction behavior. I will post diode sims in power converters when I dig up the files later. Thanks for your interest.

Claude

#### LvW

Apr 12, 2014
604
Claude - save your time. I do not need new simulations from your side.
However, to convince yourself you should repeat the simulation I have mentioned (see first part of my recent answer under "EDIT").
It clearly shows what happens when Vbe is changed and when Vce comes into play.

I do not know how to respond to sentences like "The diode forward voltage Vd is not the cause of diode current Id".
Unfortunately you didnt say what else could be the cause.
More than that, for my understanding a real discussion seems to be not possible; you didnt respond to any of my examples/claims/questions.
Although you state that "Shockley is not contested" you are denying the control property of Vbe and, at the same time, declare the current Ie as a "starting point".

You can believe me - since several years I have taken an interest in this issue because I was engaged in teaching analog electronics at a university
for more than 20 years - and, of course, I didnt want to give wrong explanations to the students.
I am aware that there are textbooks stating that the BJT would not be voltage-controlled, however I didnt see not a single proof for this assertion.
(In some cases, it is just the equation Ic=B*Ib that is used for verification - stupid!).

In all of my contributions I have mentioned many properties, examples and observations which clearly demonstrate and proof that an external control of Ic is caused by Vbe only - up to now without any counter-evidence.
Question to you: Can you give one single example or circuit or observation that supports your approach? Just a small circuit example that helps to explain what you mean ?

I am afraid - we will never come together and, finally, we should agree not to agree. OK?
Regards
LvW

Last edited:

#### Arouse1973

Dec 18, 2013
5,178
Thanks Claude. I'll look at this a bit later. It's late over here.
Cheers

#### cabraham

Feb 12, 2015
63
Here is a switching power converter simulation where diode D2 I & V are plotted. I will elaborate later at home tonight. For now, those interested can examine the waveforms and present questions/comments. BR.

Claude

#### Attachments

• smps diode fwd recov zoom in.jpg
195.1 KB · Views: 99
• smps diode fwd recov.jpg
129.4 KB · Views: 110
• smps diode rev recov zoom in.jpg
163.8 KB · Views: 107
• smps diode rev recov.jpg
165.7 KB · Views: 113
• smps diode sch.jpg
130.2 KB · Views: 102
• smps diode.jpg
174.5 KB · Views: 111

#### Arouse1973

Dec 18, 2013
5,178

I think I see what you are trying to explain here and I think we have to be very precise in the terminology we use which is where I think it's gone a bit wrong. What I think you are saying is that the voltage difference between the base and emitter terminal is not always an indication of the quantity of base emitter current and hence collector current. This is because of the time lag for the internal capacitances to, as you say charge or discharge. This is apparent at high frequency switching.

We all know transistors delay their turn on and turn off times to some degree due to parasitic components. So yeah I can see your point that you see that the Vbe needs to catch up while the collector / emitter current is doing something else than it should be. So yeah in that instance I would say that Vbe is indicating something different to what is happening inside the transistor. After all voltage difference across a component is just telling you the amount of energy that's being used for each coulomb of charge that's moving within an electric field of certain strength.

This will only become apparent at higher frequencies as you have demonstrated. And all the standard formula are for an ideal p-n junction and or BJT in a steady state, and yes the formulas work in steady state very well. But when you start switching voltages very fast the internal capacitance start to play a part in the base emitter current. These capacitance effectively bypass the normal operation of the p-n junction for a short amount of time which wouldn't be seen in steady state analysis.

Maybe we should change the way we answer the question of what controls the current in a p-n junction. Maybe we should just say an electric field or electrostatic potential or something else and not Vbe as it appears in your experiments to show something different.

I don't know if I got that quite right it that what you mean?

Cheers

#### cabraham

Feb 12, 2015
63
What controls current is the external network, which is usually designed to fix the emitter current at a set dc operating point, if we are discussing amplifier stages. If we use the bjt as a switch to turn on a fan, heater, relay, etc., then charge control modeling works best. The current control model, Ic = alpha*Ie, is only valid at low to moderate speeds, and only in the active region, not in saturation. Likewise for the other 2 equations: Ic = alpha*Ies*exp((Vbe/Vt)-1), as well as Ic = beta*Ib.

At high speeds, internal charge storage renders these equations inaccurate, and charge control models are most often employed. When I say that Ie controls Ic, the condition is that the device is not operating near its limits regarding speed.

Even at low speed, Vbe does not control Ie/Ic/Ib. It just happens that if you measure a network by using a scope, the delay is not visible unless you zoom in on an edge. Using a storage scope and pulsing a bjt at 1.0 Hz, if you zoom in on the edge transitions, it is visible that all 3 currents settle to their plateau value before Vbe does likewise.

I will elaborate if necessary. I'm at work. I was going to do a write up last night but worked late and conked out when IO got home. I'll do it tonight. Thank you and all others for your interest and feedback, best regards.

Claude

#### cabraham

Feb 12, 2015
63
Claude - save your time. I do not need new simulations from your side.
However, to convince yourself you should repeat the simulation I have mentioned (see first part of my recent answer under "EDIT").
It clearly shows what happens when Vbe is changed and when Vce comes into play.

I do not know how to respond to sentences like "The diode forward voltage Vd is not the cause of diode current Id".
Unfortunately you didnt say what else could be the cause.

The cause is the external source and network which the diode is connected to. How can a diodes forward voltage drop be the cause of its own current. Electrons/holes transported through a junction LOSE energy, i.e. a 0.65 volt forward drop indicates that for every coulomb of charge passing the junction, 0.65 joules of energy is LOST. The depletion zone barrier potential of 0.65 V is taking energy away from the charges and photons are emitted in the infrared band. The current times the 0.65 V is the power dissipated as heat. Vd, the forward voltage drop is not the cause of Id the forward current.

More than that, for my understanding a real discussion seems to be not possible; you didnt respond to any of my examples/claims/questions.
Although you state that "Shockley is not contested" you are denying the control property of Vbe and, at the same time, declare the current Ie as a "starting point".

Refer to my plots below, a p-n junction diode used in a switching power converter. Indictor current is continuous, when the FET shuts off the current continues through the diode. Current is the starting point, and the voltage drop builds up due to current. In the forward direction that is the case. In the reverse direction, diode is connected right across the input voltage supply. In this case the reverse current is dictated by the reverse voltage. In forward direction, forward current dictates forward voltage.

Shockley's equation is merely a relation between I & V, and does not imply which one controls the other. The network dictates that. My power converter example clearly shows that in a diode I can control V and vice-versa.

You can believe me - since several years I have taken an interest in this issue because I was engaged in teaching analog electronics at a university
for more than 20 years - and, of course, I didnt want to give wrong explanations to the students.
I am aware that there are textbooks stating that the BJT would not be voltage-controlled, however I didnt see not a single proof for this assertion.
(In some cases, it is just the equation Ic=B*Ib that is used for verification - stupid!).

In all of my contributions I have mentioned many properties, examples and observations which clearly demonstrate and proof that an external control of Ic is caused by Vbe only - up to now without any counter-evidence.
Question to you: Can you give one single example or circuit or observation that supports your approach? Just a small circuit example that helps to explain what you mean ?

I've been doing that since day 1.

I am afraid - we will never come together and, finally, we should agree not to agree. OK?
Regards
LvW

#### LvW

Apr 12, 2014
604
Claude - without going again into details I like to answer as follows:

Before presenting new circuits or alternative explanations it is established and legitimate practice to show why present theories are not sufficient (or even not able) for explaining some observed phenomena.
I think, therefore,

(1) you should be aware (I suppose you are) that your new findings are NOT in accordance with the established theory for the BJTs working principles (References: Stanford Univ., Berkeley Univ., MIT, Georgia Inst. of Technology, Sedra-Smith (book), Horowitz/Hill: Art of Electronics,...). More than that, there is a variaty of observable electronic phenomena (I have mentioned only some of them) which can be explained only under the assumption of voltage-control [Ic=f(Vbe)].
(My comment: In general, such disagreements are not a problem. In contrary, without forgetting established things we never would be able to discover new techniques.)

(2) However, in such a case you should demonstrate why/that these existing explanations are faulty/incorrect. In this context, it is necessary to show why existing observations and interpretations are wrong. Otherwise, your contributions are assertions only without any justification.

LvW

#### Arouse1973

Dec 18, 2013
5,178
What controls current is the external network, which is usually designed to fix the emitter current at a set dc operating point, if we are discussing amplifier stages. If we use the bjt as a switch to turn on a fan, heater, relay, etc., then charge control modeling works best. The current control model, Ic = alpha*Ie, is only valid at low to moderate speeds, and only in the active region, not in saturation. Likewise for the other 2 equations: Ic = alpha*Ies*exp((Vbe/Vt)-1), as well as Ic = beta*Ib.

At high speeds, internal charge storage renders these equations inaccurate, and charge control models are most often employed. When I say that Ie controls Ic, the condition is that the device is not operating near its limits regarding speed.

Even at low speed, Vbe does not control Ie/Ic/Ib. It just happens that if you measure a network by using a scope, the delay is not visible unless you zoom in on an edge. Using a storage scope and pulsing a bjt at 1.0 Hz, if you zoom in on the edge transitions, it is visible that all 3 currents settle to their plateau value before Vbe does likewise.

I will elaborate if necessary. I'm at work. I was going to do a write up last night but worked late and conked out when IO got home. I'll do it tonight. Thank you and all others for your interest and feedback, best regards.

Claude

This is basically what I am saying yes. But you say at low speeds of 1Hz, but you have to consider the rise time and not necessary the period of the square wave. The rising edge will contain lots of harmonics as you know. But I think from all the initial discussions this one and previous we had been talking about steady state and not transient analysis.

#### cabraham

Feb 12, 2015
63
Ω
Claude - without going again into details I like to answer as follows:

Before presenting new circuits or alternative explanations it is established and legitimate practice to show why present theories are not sufficient (or even not able) for explaining some observed phenomena.
I think, therefore,

(1) you should be aware (I suppose you are) that your new findings are NOT in accordance with the established theory for the BJTs working principles (References: Stanford Univ., Berkeley Univ., MIT, Georgia Inst. of Technology, Sedra-Smith (book), Horowitz/Hill: Art of Electronics,...). More than that, there is a variaty of observable electronic phenomena (I have mentioned only some of them) which can be explained only under the assumption of voltage-control [Ic=f(Vbe)].
(My comment: In general, such disagreements are not a problem. In contrary, without forgetting established things we never would be able to discover new techniques.)

TI, LInear Tech, On Semi, Fairchild, etc., all refer to bjt as current controlled. My findings are NOT NEW. All my course work from BE, to MS, to Ph.D. was taught by professors who modeled bjt as current controlled, or charge controlled at high speed and/or saturated switch operation. Your observable electronic phenomena do not disprove that Ie is in control of Ic. Vbe is incidental, and does not control anything. My plots are proof, and you cannot refute one iota of what my plots demonstrate.

(2) However, in such a case you should demonstrate why/that these existing explanations are faulty/incorrect. In this context, it is necessary to show why existing observations and interpretations are wrong. Otherwise, your contributions are assertions only without any justification.

Observe the power supply catch diode waveforms. Ask me any question. Show me how voltage drop controls current. Show me.

LvW

#### LvW

Apr 12, 2014
604
"Show me how voltage drop controls current. Show me."

Just to avoid misunderstandings (voltage "drop") - while speaking about "voltage control" I am always referring to an externally applied voltage (in this case Vbe).
I do not intend to list my examples again (see post 17, 22, 24, 33 and, in particular, the simualtion experiment in my post 38).
But what about YOU? Where are YOUR examples, observations and proofs showing how the BJT is current-controlled?

#### LvW

Apr 12, 2014
604
This will only become apparent at higher frequencies as you have demonstrated. And all the standard formula are for an ideal p-n junction and or BJT in a steady state, and yes the formulas work in steady state very well. But when you start switching voltages very fast the internal capacitance start to play a part in the base emitter current. These capacitance effectively bypass the normal operation of the p-n junction for a short amount of time which wouldn't be seen in steady state analysis.

Maybe we should change the way we answer the question of what controls the current in a p-n junction. Maybe we should just say an electric field or electrostatic potential or something else and not Vbe as it appears in your experiments to show something different.

Adam, do you think that the general working principle of the BJT depends on the applied frequency?

I rather think, we have the same situation as for ALL electronic parts:
They follow a certain working principle and they do what they are supposed to do (up to nearly 100%) - as long as no parasitics (always present!) come into play.
This applies to resistors, capacitors and - of course - also to BJT`s.
That means: For very large frequencies their "normal" and desired operation is degraded and disturbed by parasitic influences - however, this must not be interpreted as a change of the general operation principle.

#### cabraham

Feb 12, 2015
63
"Show me how voltage drop controls current. Show me."

Just to avoid misunderstandings (voltage "drop") - while speaking about "voltage control" I am always referring to an externally applied voltage (in this case Vbe).
I do not intend to list my examples again (see post 17, 22, 24, 33 and, in particular, the simualtion experiment in my post 38).
But what about YOU? Where are YOUR examples, observations and proofs showing how the BJT is current-controlled?

I gave numerous examples, and I intend to add one, a synchronous rectifying FET where the body diode picks up current after channel is switched off, later today. But I assure you that Vbe is NOT an externally applied voltage. Even w/o resistors, base region rbb' spreading resistance, and re intrinsic emitter resistance exists. The Vbe' is the junction potential barrier, and that is the Vbe in Shockley's equation. You must know that one never applies a voltage source directly across b-e junction? Do you believe that one can do that?

Claude

#### LvW

Apr 12, 2014
604
I gave numerous examples...
You must know that one never applies a voltage source directly across b-e junction? Do you believe that one can do that?
Are you joking? Do we speak about complete circuits - stabilized against tolerances and temperature sensitivity - or about a physical working principle?
Did you realize that in my simulation example (post 38) I have directly applied the voltage Vbe?

PLease, would you be so kind and tell me where I can find your "numerous examples"? I cannot find them in my post box (certainly my fault)..

PS1: Did you never (as a student?) record the transfer characteristics Ic=f(Vbe) of a BJT?
Together with my students, we have measured hybrid parameters and, of course, the most important property of the transistor: The transfer curve.

PS2: "TI, LInear Tech, On Semi, Fairchild, etc., all refer to bjt as current controlled. "
Yes - I know. Simply assertions without justification (because a false statement cannot be proofed).

Last edited:

#### cabraham

Feb 12, 2015
63
Are you joking? Do we speak about complete circuits - stabilized against tolerances and temperature sensitivity - or about a physical working principle?
Did you realize that in my simulation example (post 38) I have directly applied the voltage Vbe?

I advise against that in real world applications. A bjt is thermally unstable, subject to runaway in that case. Resistor needed or current source.

PLease, would you be so kind and tell me where I can find your "numerous examples"? I cannot find them in my post box (certainly my fault)..

PS1: Did you never (as a student?) record the transfer characteristics Ic=f(Vbe) of a BJT?

Ic=beta*Ib is a transfer curve
Ic=alpha*Ies*exp((Vbe/Vt)-1) is a transfer curve
Ic=alpha*Ie is a transfer curve.

I use all 3 of them.

Together with my students, we have measured hybrid parameters and, of course, the most important property of the transistor: The transfer curve.

PS2: "TI, LInear Tech, On Semi, Fairchild, etc., all refer to bjt as current controlled. "
Yes - I know. Simply assertions without justification (because a false statement cannot be proofed).
The Gilbert slide you cited is just that, assertion w/o proof. Just what do you ask in terms of proof? The sims I gave demonstrate that Ie is in control of Ic, not Vbe. What needs explaining. What would you consider as "proof"? Just asking.

#### LvW

Apr 12, 2014
604
The Gilbert slide you cited is just that, assertion w/o proof. Just what do you ask in terms of proof? The sims I gave demonstrate that Ie is in control of Ic, not Vbe. What needs explaining. What would you consider as "proof"? Just asking.

Yes - the slide I have cited long time ago does not contain a proof. But note: It is one single slide with (estimated) 10 words and 3 formulas - do you expect a derivation on a single slide,? Not a good counter example.!
If you are interested in the university papers with extensive derivations - I can give you the links.
My answer to your question: A proof has a theoretical part and a practical part to verify theoretical predictions. I suppose you will agree to that.
And all the observations and technical data I have mentioned earlier are resp. can be verified theoretically and in experiment!

As an example I enclose a very simple arrangement : A class-C push-pull amplifier.
What does the output show? A real mirror of the input-output exponential transfer curve Ic=f(Vbe).
I suppose, the cross-over distortions are known to you.
How do YOU explain the form of the output ?

#### Attachments

• Class_C.pdf
45.1 KB · Views: 112

Replies
11
Views
1K
Replies
13
Views
2K
Replies
7
Views
2K
Replies
2
Views
948
Replies
2
Views
1K