# Basic BJT Questions

Discussion in 'General Electronics Chat' started by ActivePower, Dec 25, 2013.

1. ### ActivePower Thread Starter Active Member

Mar 15, 2012
155
23
I was recently revisiting old lecture notes on BJT amplifiers and am a little stuck. I have been a out of touch with analog electronics for some time so I'd really appreciate it if someone stepped up and rectified anything here that I may have understood incorrectly.

Assuming a NPN BJT under CE bias:

1. A given amount of base current, say Ib, would be able to drive a corresponding amount of collector current, Ic and no more (assuming the B-E junction is forward biased and C-B is reverse biased).

2. In effect, for a given base current the BJT would limit the collector current to an amount specified by its gain, thereby acting as a current regulator.

3. As we begin to drive more and more base current into the transistor and correspondingly increasing the collector current, the collector-base voltage keeps dropping (rate of drop depends on the load(?)).

4. After this collector-base voltage hits a certain minimum value, the transistor can source no more collector current and is said to be saturated.

5. To design a circuit to reliably saturate a BJT, we need to forward bias the collector-base diode. For this, we need to drive enough current into the transistor to make sure that the output current is always able to provide enough drop across a pull-up resistor to ensure collector voltage less than base volts.

6. Why is the collector-base voltage drop less than the emitter-base volt drop. I understand the BE junction behaves almost like a diode and has ~0.7V of drop (silicon) whereas the CB junction has something like 0.4V-0.5V.

(Clearly both cannot have equal voltage drops as that would mean the net drop across the BJT is zero, ideally?)

7. It says here

Aren't the voltage and current interdependent? As soon as the BE voltage rises above the 0.7 V threshold, isn't there base current to drive the transistor into ON state?

8. Saturated transistors are said to act like closed switches, in that the voltage across them is almost zero and the current is maximum. I know I am missing something here because the characteristic curves show the collector current to be more in the linear region for a given base current value. Reference.

This was a long and quite possibly a trivial question but I've been struggling with these basics for some time and some help would go a long way.

Thanks.

2. ### antonv Member

Nov 27, 2012
149
27
Instead of "can source no more collector current" I would say the base current looses control of the collector current and collector current is now controlled by resistance between the collector and power supply, i.e. if the collector is connected directly to the supply then the transistor could destroy itself from over current.

Yes, but not linearly as with a resistor. A diode's forward current is related exponentially to its forward voltage and it is temperature dependent. However, due to the physics of the transistor, its collector current is pretty directly controlled by its base current and the base voltage settles where it needs to exponentially and temperature-wise.

Look closely at the curves; in the saturation region collector current becomes more and more dependent on collector voltage rather than base current, especially as you move to the left of the graph.

Nov 30, 2010
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4. ### LvW Well-Known Member

Jun 13, 2013
779
105
ActivePower, since you have indicated your goal to understand the transistor`s operation I think it is necessary first to correct your basic assumption regarding the relationship between base and collector current.

Do you really think that it would be possible that small current (Ib) can "drive" or control a larger current (Ic)? This is physically impossible.!
Like for a simple pn diode it is of course the base-emiiter voltage that controls the current - in this case the collector current Ic.
The principle task of the BJT is the same as for the tube (triode) or even the FET: A control voltage determines the amount of current that flows through the controlling region of the device.
The main difference between the BJT and the tube/FET - as seen from outside the transistor - is the fact that there is a small current through the controlling node (base current Ib).
And this current is a small (more or less fixed) part of the collector current, leading to the somewhat misleading term "current gain". But, in fact, it is not a "gain".
This was extensively discussed elsewhere (also in this forum). It is true that the current gain B (or beta) can be used during design of a transistor amplifying stage and - in this context - we can ASSUME that the collector current is related to the base current using the "current gain". However, this is not the physical truth.
Finally, if somebody is trying to really understand the operation of the transistor as an amplifying element within a more complex circuit, I think it is necessary to know the real physical background.
For example, the function of the well known current mirror can be explained only based on the voltage-control feature of the BJT.

5. ### cabraham Active Member

Oct 29, 2011
82
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Vbe is not what controls Ic, this is a heresy that just won't go away. In Dec. 1954, Drs. Ebers and Moll published their paper detailing the bjt as having Ic controlled by Ie the emitter current. The number of emitter electrons (npn) injected into the base controls the number collected by the collector.

Vbe changes after Ib and Ie change. The injection of charges into the b-e junction constitute Ib and Ie. When charges cross the depletion zone they form a barrier and Vbe is determined by doping density of materials, temperature, and current densities. Vbe is not what controls Ie or Ib.

This has been well discussed and every bjt OEM from On Semi to Fairchild to TI to Intl Rectif, etc., describes bjt device as current controlled.

Current mirrors operate on the basis that 2 differing devices with equal Ie will have near equal Ic. Ic = alpha*Ie. Vbe is not what forces the 2 collector currents to be equal. I've designed hundreds of current mirror circuits using discrete parts. Two identical devices in close proximity will not have matched Vbe specs. The one device that has a slightly lower Vbe will hog the current.

Placing emitter degeneration resistors in eac h emitter will force the 2 devices to share emitter currents and thus collector currents are equal. Two devices that are different part nos. at different temps w/o emitter resistors will not track. Their Vbe's are unmatched and one hogs current. Placing emitter resistors of sufficient value forces the 2 emitter currents to balance and the collector currents balance.

I can discuss the current mirror in detail if you wish, but I assure you that trying to place 2 bjt parts so that their matched Vbe's result in shared current is not easy. Even on a monolithic IC, there is some intrinsic emitter resistance "re" in the silicon which helps forcing the current to share. Ie is what keeps the mirror working, not Vbe.

Claude

6. ### LvW Well-Known Member

Jun 13, 2013
779
105
Hi Claude,
I suppose we know each other from a similar discussion in another forum. Thus, I do not intend to start this discussion again and to repeat all the arguments and references supporting the voltage control principle for BJT`s.
I think the problem in your justification becomes visible with your sentence "Vbe changes after Ib and Ie change". Thus, Vbe is the RESULT of changing Ib and Ie ???

I think, to understand the basic principle of the BJT`s operation it is sufficient to verify the common properties of pn diodes and bipolar transistors. In both cases it is the width of the depletion layer that controls the amount of current which can flow through it (Remember: Even the model created by Ebers and Moll - as mentioned by you - is based on two coupled diodes!).
And which electrical quantity controls the width of the depletion area? It is , of course the VOLTAGE across this region.
A current is ALWAYS the RESULT of a a driving voltage (possible exception: optical devices with energy supply from outside).
Some other parts of your contribution (Vbe matching and tolerances, emitter resistor) do not help, in my opinion, to answer the general question: Voltage or current control?

EDIT
Quote:
"The number of emitter electrons (npn) injected into the base controls the number collected by the collector."

May I finally ask - which electrical quantity determines the "number of emitter electrons injected into the base" ?

Last edited: Dec 27, 2013
7. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
790
Hopefully that proves to be the case.

This has been discussed on AAC & just about every where else in the electronics universe ad nauseam. The opposing views are never reconciled & probably never will be. Each "side" claims to have won the high ground. If the combatants wish to continue this again on AAC it would seem fair to the OP not to hijack this thread - rather start a new thread if the discussion is to continue.

8. ### LvW Well-Known Member

Jun 13, 2013
779
105
Hi t_n_k,

yes - I agree that we shouldn`t start a discussion with pro&cons again.
However - is it really a "hijack" of the OP`s thread to answer ("rectify") his statements like this:

I'd really appreciate it if someone stepped up and rectified anything here that I may have understood incorrectly.
Assuming a NPN BJT under CE bias:
1. A given amount of base current, say Ib, would be able to drive a corresponding amount of collector current
.

For my opinion, a formulation like this deserves some "rectificaton".

LvW

9. ### ActivePower Thread Starter Active Member

Mar 15, 2012
155
23
Thanks for all the answers! I looked over the characteristic curves once more and it did help answering some of the questions.

As to the Vbe vs Ib debate (although I really don't think I am qualified to comment), I did look up a few websites trying to make sense of it all. The transconductance and the transfer characteristics here both show the output current being dependent on the input volts as well as current.

I also found a StackExchange question which said something along the lines of ".. different explanations fitting different situations" (can't seem to find the link right now)

Remark: Also found an explanation for Q6 but I still don't fully understand it. A little help would be great here.

Thanks!

10. ### LvW Well-Known Member

Jun 13, 2013
779
105
The output current Ic is controlled by the input voltage Vbe. However, because the (input) base current Ib also is related to Vbe (of course!), there is a relation also between Ic and Ib. For example, Ib can be used as a fixed parameter to draw the output characteristics Ic=f(Vce). But note there is another set of curves with Vbe as a parameter.

Where can I find Q6?

11. ### ActivePower Thread Starter Active Member

Mar 15, 2012
155
23
Sorry, it's this one in the original post.

12. ### cabraham Active Member

Oct 29, 2011
82
32
There is no "Vbe vs. Ib" debate at all. All electrical devices, except a pure short or open circuit, cannot have I w/o V, nor V w/o I. Period. In order to increase Ic, Ie must first increase. But to increase Ie, an external source, like a mic, antenna picking up waves, photodiode, guitar pickup, etc., must input an increase in the signal. The increased Ib and Ie propagate through the conductors, then reach the b-e junction. The increased Ie results in a direct increase in Ic, i.e. more electron emission (npn device) results in more electron collection.

Of course as the increased electron emission takes place, more charges interact with acceptor ions in the base and increased ionization happens, resulting in a slightly increased depletion layer width. This depletion width is not what makes Ie or Ic change, the external source does that. Vbe is the integral of the depletion E field over distance (width). This Vbe changes as a result of Ie changing, it is not what makes Ie change.

The theory that in order to change Ie/Ic in a bjt, you must first change the depletion width is utter nonsense. No semicon phy texts support this at all. In every text, even in circuit design texts w/o delving into physics, the b-e junction is modeled as a hybrid-pi equivalent network. The b-e junction for small signal operation is a parallel resistance r_pi, and a diffusion capacitance. As such, vbe is the voltage dropped across said junction, R in parallel w/ C. The ie will always change before vbe because current in a resistor is in phase w/ voltage and in a cap current leads voltage. At very low frequency the current is nearly all in the resistor. A scope plot of Ie vs. Vbe will reveal very small phase difference at say 1.0 or 10 kHz freq. But raising the frequency to 10 MHz or higher results in more cap current. The phase between Ie and Vbe is clearly visible w/ Ie leading.

All of this can be settled in a lab w/ instruments. Those who claim that Vbe changes, then depletion width changes, resulting in Ie/Ic changing are flying in the face of physics. How does one change vbe w/o injecting current through the b-e junction? If vbe is what controls ie/ic, how does vbe change before a change in ie/ib. It is a capacitive junction, only after current changes can voltage change.

Re current mirrors, I've designed many. I assure I am right when I say that 2 devices w/ equal Vbe values may or may not have equal Ic values. Vbe matching is hard to do, since Ies for 2 devices vary with specimen and temp. Should one device hog the current it gets hotter and hogs even more, even if Vbe values are equal.

But if we place degenerating resistors in the emitters (also called ballast resistors), the mis-matches in Ies or Vbe are mitigated, and as one device draws more current the voltage drop across the resistor increases. The 2 devices share emitter currents despite mis-matches in temp, Ies, etc.

Since Ic = alpha*Ie, if we match Ie for 2 parts, alpha is typically 0.98 to 0.998, maybe 0.95-0.98 for power bjt, then Ic will match within a percent or two. It is always Ie that controls Ic. As far as Ib being a by-product goes, we can make the same argument for Vbe.

In order to change Ic, we must change Ie so that the no. of electrons collected corresponds to no. emitted. If we emit 1000 e- per picosecond and collect 990 per psec, and we wish to collect 1040 e- per psec, we MUST increase the no. of e- emitted, namely 1050 e-/psec. The signal source must output more current.

In the process of emitting more e- from emitter, inevitable more holes are injected from base to emitter, increasing Ib. But also, more ionization occurs when these extra holes recombine in emitter edge near depletion zone increasing depletion width ans charge. A few extra carriers recombine in base increasing local E field as well. As a result Vbe increases. The same "by-producr" argument can be made for Vbe as well.

I've repeated ad nauseum for over a decade on many forums that the current which controls Ic is emitter current Ie, NOT base current Ib. Yet the Ib straw man cards keep getting played.

The reason some don't except current control is because they believe every current is ultimately driven by a voltage except for optical stimulus. It's hard to convince them that a battery, or generator produces I & V in unison, neither driving the other.

Finally, the transconductance of a bjt is one parameter of interest, but so is current gain. An active device, bjt or FET, offers gain in both I and V greater than unity. The current gain of the raw device, beta, is the upper limit on the amp stage overall gain. For a single bjt stage, if beta is 100, the stage current gain cannot exceed 100 broad band (an L-C resonant can, but only narrow band near fres). Likewise, the raw device transconductance gm = Ic/Vt, is the upper limit for stage transconductance. At Ic = 1.0 mA, Vt = 25.7 mV, gm = 39 mS (mA/V). The stage transconductance gain must be less than this value.

To LvW, no personal attack here, just curious. What subject did you teach when you were teaching engineers? What type of R&D did you do, and in what industries? No offense, I am just curious as to the source of your claims. I've been an EE for 35 yrs, near done on my PhD, worked w/ hundreds of MS/PhD EE personal on defense, aerospace, automotive, commercial, optical, power, analog, digital, etc. I never met an EE who put forth the stuff you do.

Is the stuff you post taught in uni? What is your source other than non-peer-reviewed web sites? No confrontation, but it's fair to ask for sources. BR.

Claude

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13. ### #12 Expert

Nov 30, 2010
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Well, that clears it up a lot of misunderstandings.

14. ### LvW Well-Known Member

Jun 13, 2013
779
105
@ Claude:
For my opinion, your very long contribution contains a lot of statements which cause more questions than to clarify things. But I do not intend to jump into the discussion again. (You will remember the exchange of opinions we had some month ago. Thus, your arguments are not new for me).
However, as you have mentioned my name directly, I will ask only some questions arising from your text (perhaps interesting for all who - at first sight - feel satisfied with your explanations):

Quote: In order to increase Ic, Ie must first increase. But to increase Ie, an external source, like a mic, antenna picking up waves, photodiode, guitar pickup, etc., must input an increase in the signal. The increased Ib and Ie propagate through the conductors, then reach the b-e junction. The increased Ie results in a direct increase in Ic, i.e. more electron emission (npn device) results in more electron collection.

For my opinion, a rather vague formulation (...an external source ...must input an increase in the signal).
Which signal (voltage or current?) at which node?
Which electrical quantity (not „signal“) at which of the three nodes causes Ie to change? Why so unspecific?

Quote: This Vbe changes as a result of Ie changing, it is not what makes Ie change.

Again, what makes Ie changing? I suppose, during this discussion, everybody has a simple grounded emitter configuration in mind.
Therefore: Vbe is the result of changing Ie? And which effect causes a change in Ie ? No information.

Quote: The theory that in order to change Ie/Ic in a bjt, you must first change the depletion width is utter nonsense.

Nonsense? What happens with the depletion layer in a simple pn diode?
Which effect causes the known exponential relationship between V and I?
And why do you think that this does not apply to the B-E junction?
Don`t we observe the same exponential relation for a BJT?
Is it only a fortunate coincidence?

Quote: How does one change vbe w/o injecting current through the b-e junction? If vbe is what controls ie/ic, how does vbe change before a change in ie/ib.

You are asking how Vbe is changing its value?
In each transistor amplifier (common emitter configuration) the input voltage does influence Vbe. Sorry for this simple answer.
Each and every BJT amplifier design is based on a B-E dc voltage of 0.65-0.7V (surprisingly, a value known from the pn diode).
As a result, we have a certain quiescent current Ie resp. Ic.
And everybody knows that a larger voltage results in a larger current.
How can you say that it is NOT the voltage Vbe that influences the currents?

More than that, what about the temperature influence on Ic?
There is the well-known value of app. -2mV/K. This number gives the change of Vbe that is necessary to keep Ic constant.

And it is a calculated value - based on voltage control!
How does this BJT property - and the physics behind this key figure - fit into your explanation?

Quote: Re current mirrors, I've designed many. I assure I am right when I say that 2 devices w/ equal Vbe values may or may not have equal Ic values. Vbe matching is hard to do, since Ies for 2 devices vary with specimen and temp. Should one device hog the current it gets hotter and hogs even more, even if Vbe values are equal.

Does this matching problem help to answer the main question?

Quote: I've repeated ad nauseum for over a decade on many forums that the current which controls Ic is emitter current Ie, NOT base current Ib.

And what controls Ie??? (If we speak of „control“ we speak about an external control signal).

Quote: To LvW, no personal attack here, just curious. What subject did you teach when you were teaching engineers? What type of R&D did you do, and in what industries? No offense, I am just curious as to the source of your claims. I've been an EE for 35 yrs, near done on my PhD, worked w/ hundreds of MS/PhD EE personal on defense, aerospace, automotive, commercial, optical, power, analog, digital, etc. I never met an EE who put forth the stuff you do.
Is the stuff you post taught in uni? What is your source other than non-peer-reviewed web sites? No confrontation, but it's fair to ask for sources.

I do not intend to answer these personal-oriented questions. But I am sure, you know that I am not alone.
I am afraid, just the opposite is true: Can you give a peer-reviewed reference that can support your view?

Here are some „other than non-peer-reviewed“ sources which support my position:
W. Shockley (patent description), Lessons from Berkeley University,
Lessons from Stanford University, Book Sedra-Smith (chapt. 5),
University of California (San Diego, Lecture notes),
Book Horowitz-Hill, Winfield Hill (Co-author, forum contribution),
Barry Gilbert (known world-wide as an inventor of numerous new circuits).
_______________________________________________________

Sorry - my answer is longer than intended.
LvW

Last edited: Jan 1, 2014

Oct 29, 2011
82
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So is mine.

Claude

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16. ### LvW Well-Known Member

Jun 13, 2013
779
105
Claude, thank you for the long answer.
So - everybody who is interested can create his own view on the discussed subject.
However, one question remains:

Quote: I can post some app notes from On Semi, Fairchild, Intl Rectif, etc. later today.

Are application nodes from manufacturers (more or less design-oriented) the only references you can list? Are application nodes - according to your experience - reliable documents as far as theory is concerned? (Remember: You were asking for "peer-reviewed" documents). And - even more important - do those notes justify their claims?
LvW

EDIT: Sorry for being late with the following supplement - it took some time to find the file.
It is a 30-page document from Berkeley explaining in detail how the BJT works

http://www.eecs.berkeley.edu/~hu/Chenming-Hu_ch8.pdf

Excerpt from page 318: An undesirable but unavoidable sideeffect of the application of VBE
is a hole current flowing from the base, mostly into the emitter. This base (input) current, IB,
is related to IC by the common-emitter current gain, βF

Last edited: Jan 1, 2014
17. ### ian field AAC Fanatic!

Oct 27, 2012
6,499
1,187
There's a pretty good Ferranti transistor applications manual from the mid 70s - I think it might be on archive.org.

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