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Transistor Base/Collector Current mystery

  1. May 10, 2007 #1
    Hi,

    I've been learning about transistors lately through the Horowitz & Hill book. I'm surprised to see that when used as a current source the collector voltage only changes with load voltage with an ideal transistor.

    When I say I'm surprised I mean that in non-ideal models the base current seems to vary with changes in collector voltage/current. I believe this is related to something called the Early Effect and changes in hfe caused by collector voltage change.

    So my question is how can collector voltage affect base current? I thought base current dominated everything and set the emitter current, therefore I thought it would be totally independant of Vc and Ic.

    Would I be right in thinking of a transistor used as a current source as a 'current-stifler' from the collector's perspective? In the respect that regardless (assuming non-saturation) of Vc's magnitude the current stays set, and in a real model this doesn't quite hold true? Is it possible to bugger a transistor by setting the current too low and collector voltage too high?

    This is probably the most interesting thing I've discovered since I was a young child. I had no idea electronics was so absorbing!
     
    Last edited: May 10, 2007
  2. jcsd
  3. May 10, 2007 #2

    berkeman

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    I don't think you have the cause and effect right. The collector current is set by the base current, and yes, the Early Effect causes Ic to vary slightly with Vc.

    Think of how a curve tracer works when scanning a BJT. The base current is set at some value, and Vc is swept (for example from 0V to some higher voltage less than BVceo), and then Ib is stepped up some, and Vc is swept again. The result of this repeated stepping and sweeping is the characteristic "fan" shape plot that we're used to picturing when thinking of BJTs and FETs.
     
  4. May 12, 2007 #3
    So one could think of the collector current as being set by the base voltage. Is that about right?
     
  5. May 12, 2007 #4
    Yes Adder Noir, I think you've got it. Base voltage decides emitter to collector current. You do not want very much base current, only a voltage. The base potential acts like a variable barrier to the emitter to collector current. I like to think of it in plumbing terms in that the base is like a valve that controls the flow of the emitter to collector current. Too much base current equals a burned out transistor. The reason I used the plumbing anology is that I was taught tube theory and they were sometimes described as valves and even had the schematic reprsentation of a V for valve. Transistors are not that different at the macro level.
     
    Last edited: May 12, 2007
  6. May 12, 2007 #5
    Quality, thank you :redface:
     
  7. May 12, 2007 #6
    The key to understanding Early effect is as follows. The regions in a bjt known as "base, collector, & emitter" are not exact in terms of where one stops and another starts. Take the base and collector regions. This junction is reverse biased. A "space charge zone" exists in between the base and collector. This region is also called the "depletion region", or "transition zone" . The width of this region is not an exact value. It is strongly influenced by Vbc, the base-collector voltage, and by temperature.

    At room temp, with the base-emitter junction forward biased, the b-c space charge region has a specific width. If you examine the characteristic curves, Ic vs. Vce, with Ib as a parameter, you will see a family of curves. When Vce is more than 1 volt or so, the collector current flattens out to an almost horizontal line, approximating constant current source behavior. It is not, however, perfectly horizontal, but sloped upward. That is, for a given base current, if Vce is increased while Ib is held constant, Ic increases. Thus Vce, or more exactly, Vcb, the reverse bias potential on the b-c junction, influences Ic slightly. This is Early effect.

    Early effect is due to "space charge region modulation". As Vce (Vcb) increases, the b-c depletion region, or space charge region gets wider. This is the case with any reverse-biased p-n junction. When the b-c depletion region increases in width, it impinges upon both the collector region and the base region. The base region is very very thin to begin with. When the depletion zone extends into the already thin base region, the base region becomes even thinner. Semiconductor physics text will explain this in detail along with detailed math.

    When the base gets thinner, the forward current gain, hFE, increases. A thinner region contains the same density of dopant atoms, but now with reduced volume. For a given forward bias, and electric field strength, the number of carriers injected from the base into the emitter decreases due to the reduced volume of the base region. But the number of carriers emitted from the emitter hardly changes. Thus the same value of emitter current can be attained with a smaller value of base current, so that hFE has increased. But, the data sheet characteristic curves are generated by holding base current, Ib constant. The emitter current, Ie, will increase with the same value of Ib due to increased hFE inevitable with increased Vce.

    In a nutshell, Ic is strongly determined by the forward bias on the b-e junction, and weakly influenced by Vce.
     
  8. May 13, 2007 #7
    That's an amazing reply thanks a bundle. That's being printed off and is going in my notes!

    I see, just what one would need for a good current source.
     
  9. Feb 12, 2010 #8
    Why does the same slanting of output characteristics curve not occur for CB configuration...(essentially,the same things are happening in bothe CE and CB)
     
    Last edited by a moderator: Feb 12, 2010
  10. Feb 12, 2010 #9

    sophiecentaur

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    Just reading that question 'from cold' it strikes me that you are sort of assuming the transistor 'knows' where it is. The characteristics if devices will have to apply, always, under the same driving conditions. 'Flaws' in the transistor characteristics (departures from the ideal) will have different levels of effect in different configurations but the transistor still has to be doing the same thing in response to what you do to it.
     
  11. Feb 12, 2010 #10
    Base current Ib, base-emitter voltage Vbe, & emitter current Ie, are all needed to obtain collector current Ic. Ib/Vbe/Ie are mutual & inter-related. Which one "sets Ic" is a chicken-egg vicious circle. All 3 are involved.

    The key word is "same". The slant on the curves is much less for CB configuration. This is becuse CB has the base at a fixed potential wrt ground, and the input signal is at the emitter. Thus the transfer ratio is given by:

    Ic = alpha*Ie.

    The CB current gain, alpha, changes very little w/ Early effect. If Early effect increases beta from 100 to 120, the corresponding increase in alpha is much smaller. Since alpha = beta/(beta+1), then a beta increase from 100 to 120 results in an alpha increase from 0.9901 to 0.9917. Beta increased by 20%, but alpha increased by a mere 0.165%.

    Thus the slope of the CB curve is almost horizontal.

    Did I help?

    Claude
     
  12. Feb 13, 2010 #11
    Looking at this intuitively....in CB, the input is the Ie,the output is the Ic...now,increasing the Vbe,the emitter-base (E-B) current increases....due to the early effect,the base region has become very small....so the Ib decreases,resulting in a corresponding increse in Ic....but Ic is basically a part of Ie (Ie=Ic+Ib),and since Ib has reduced in magnitude,Ic is now a bigger fraction of Ie than usual...so now,alpha is bigger.
    (But I guess my explanation still doesn't explain why the change in alpha is much smaller than the change in beta).

    On the other hand,in CE,the input is the Ib,and the output is the Ic...on increasing the value of Vce,the reverse bias on the C-B junction increses,causing the early effect....due to this,there is a large reduction in Ib,and a corresponding increase in Ic...thus,as I said before,the Ic is now a larger fraction of Ie than what it would be wothout the early effect....thus there is a certain change in alspha (it incresases slightly)...but since Ic is now much bigger than Ib,the value of beta is much bigger....is this right?

    (looking at all this in the intuitive way really helps me to understand,please refer to this issue in 'intuitive' terms).
     
  13. Feb 15, 2010 #12
    But I did explain it, in post #6 above. Please reread post #6, and then if you need clarification, let me know. As I stated in post #10, the reason alpha changes to a much lesser degree than beta, is simply because the math relation is

    alpha = beta/(beta+1). Set beta to some starting point, 100, 80, 125, whatever. Compute alpha. Now increase beta by 10%. Recompute alpha, and it will have changed very little. It's just the nature of the beast.

    "Intuitively" is hard to do with semiconductor physics. The thermal energies, bandgaps, fields, covalent bonding, doping, carrier concentrations, saturation velocities, etc. are too much to intuitively grasp. One needs to study the principles, do the math, and rely on the semiconductor producers' info since they know it best. That is my advice.

    Claude
     
    Last edited: Feb 16, 2010
  14. Nov 1, 2011 #13
    I have a small question. How can you Hold the base current constant. I mean because of base-width modulation the base current is bound to change right?
     
  15. Nov 1, 2011 #14

    Averagesupernova

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    I guess the obvious answer would be to feed the base with a current source. Do you have a specific configuration in mind? Most transistor circuits are about configuration to get the desired result.
     
  16. Nov 1, 2011 #15

    sophiecentaur

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    For example, a 100V source, connected to the base via a 10MΩ series resistor would put 10μA into the base (as near as you could measure) whatever Vbe turns out to be (with the emitter grounded).
    There are more convenient ways of doing it but brute force and ignorance can give you a 'current source'.
     
  17. Nov 1, 2011 #16
    Ok...! Actually i just thought that considering a common emmiter configuration...if at all Vce is increased...keeping Vbe a constant. Then the base width decreases...so there should be some decrease in the Base current. Thats what i thought.

    Also i had a question....that if u consider Early Effect in a common Base configuration then with increase in Vcb and keepin Vbe constant, the emmiter current increases..How?
    How can Vcb affect the emmiter current at all?
     
  18. Nov 1, 2011 #17
    How can you keep VBE constant? A constant voltage source directly across the b-e junction will likely destroy the bjt. A bjt cannot withstand "voltage drive", that is why a bjt is always "current driven".

    To view Early effect, you can hold IB constant & vary VCE. Or you can hold IE constant & vary VCE, for the common base configuration. The current gain here is alpha, not beta. Alpha will change very little due to Early effect as I stated above. BR.

    Claude.
     
  19. Nov 2, 2011 #18
    Ok.
    Please have a look at this
    "http://hyperphysics.phy-astr.gsu.edu/hbase/electronic/ietron/npnce.gif" [Broken]
    SO just tell me , theoretically....according to the figure
    1)In a CE configuration, VBB is constant. If i increase Vcc which inturn increases Vce (Early Effect) will the base current change? Ic will increase im sure...but what will happen to the base current?
     
    Last edited by a moderator: May 5, 2017
  20. Nov 2, 2011 #19
    The base current, IB, will remain constant to a good approx. The base current is determined by (VBB - VBE) / RS. Increasing VCC, which consequentiually increases VCE, has negligible influence on IB. Did this help?

    Claude
     
    Last edited by a moderator: May 5, 2017
  21. Nov 3, 2011 #20
    I was thinking the same thing! But cudnt assure myself.
    Thanku :)
     
  22. Nov 3, 2011 #21
    Excellent explanation.
    You should write a book on BJT.
     
  23. Sep 14, 2013 #22
    Could you explain this more? Why drive a bjt by a voltage source will destroy it?
     
  24. Dec 15, 2013 #23
    How Ib affects Ic?

    After a lot of searching and thinking, i have come up to the result that:

    "As we know size of Base is very small as compared to Emitter of transister. So when we increase Vbe, for example 1 volt rise in Vbe. Then suppose it causes 1A rise in Ib. We also know I=q/t. 1 Ampere current mean, 6.2x10^18 electrons will be injected from emitter to base region. But if base is 20 times smaller than the emitter,it will have twenty times less number of holes to accept those electrons. All other electrons will swept into collector region. So Ic will increase by 19 times the Ib. So gain HFE is 19. Ic=19xIb. And the relation of Ie=Ic+Ib is also clear before us, when we have seen that if 20 electrons are injected from emitter to base, then 19/20 of electrons have moved to collector and 1/20 have moved to base".

    Hopping the illustration will be useful.
     
  25. Dec 16, 2013 #24
    Ie = Ies*exp((Vbe/Vt) - 1); but remember that Ies is a strong non-linear increasing function of temperature, Ies(T). This is the Ebers-Moll equation, E-M.
    As temp increases, Ies increases greatly, while Vt increases to a lesser extent.
    If a voltage source is connected directly across b-e junction, Ie is given above by the Ebers-Moll equation, E-M. Ie multiplied by Vbe is the b-e junction power, Pbe = Ie*Vbe. Since power is now being dissipated, the local junction temperature must increase, so that Ies increases as well.

    Ies*exp((Vbe/Vt) - 1) is now larger than at first due to temp increase resulting in Ies increase. Power now increases since Vbe is constant with Ie increasing. The temperature will rise higher since power has increased, resulting in larger Ies, resulting in larger Ie, resulting in larger Pbe, resulting in larger power, higher temp, higher Ies, higher Ie, etc. Thermal runaway will take place and the junction will overheat. Although the increase in Vt tends to reduce power, the increase in Ies is much greater so that temp increases until junction is destroyed.

    But drive the b-e junction with a current source we get the following relation per E-M equation:

    Vbe = Vt*ln((Ie/Ies) + 1).

    Now Pbe = Ie*Vbe just as before, so that temperature rises when power is dissipated. Vt does increase slightly with temp increase, but Ies increases much more. But Ies is in the enominator so that Vt increases a little while Ie/Ies decreases a lot. Overall Vbe decreases as temp increases until equilibrium is reached.

    With current drive increasing temp results in a decrease in power so that the temp increase converges to a steady state value. As long as this temp is less than the junction limit, all is well. But with voltage drive, the increase in temp can run away w/o limit so that the device cannot survive.

    I hope this helps.

    Claude
     
  26. Dec 16, 2013 #25
    The rise in Vbe is not what causes the rise in Ib. The junction is a non-linear capacitance, which means that Ib must change first before Vbe can change. A good text on bjt will explain transistor action at a detailed level. Hfe is not related to the simple volume ratios of the regions. The base is vary thin, typically 1 micron or less and it is doped much lighter than the emitter, in order to get high injection efficiency.

    Again, the OP asked about Early effect, and I believe we should not stray from this topic. Basic bjt operation can be discussed in a dedicated thread. A search will turn up a few good discussions on this forum from recent years.

    Claude
     
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