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I've a question concerning the function of a transistor in the saturation state. In a Common Emitter transistor (n-p-n), at the saturation state (when collector current maximum), both junctions are forward biased. My question is how can the electrons coming from the emitter region and enter the collector region (through the collector-base junction) constitute the collector current (you have to note that it is maximum here)?

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  • \$\begingroup\$ One thing to perhaps note is that the \$V_{BC}\$ forward bias under saturation isn't ordinary diode forward bias. Its voltage is \$V_{BE}\$ minus \$V_{CE}(sat)\$. I suspect that "forward bias" is just a way of saying that the B voltage is higher than C, not that it's a physically identical phenomenon to diode forward bias. Anyway, you can't understand the BJT as two diodes, because simply soldering two diodes to make a three-terminal device does not give you a transistor. \$\endgroup\$ – Kaz Feb 22 '13 at 0:24
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First lets look at what happens just before saturation.

Condition Just before saturation.

During forward bias of n-p-n transistor base-emitter junction is forward biased and the collector-base junction remains reverse biased.

You can have a look at a circuit in when a transistor is in forward biase and see that collector ( n region ) is always at a higher potential than base ( p ) region. period.

Next, please don't consider two diode back to back model of npn transistor that is given as an example in many books ( i have read those in my college days ).

The most important part to consider is that the base is a single silicon. If you consider the two diode model then we have two metal ( consider it as very very highly doped n type silicon wraffer ) - silicon junction in base. So its far from being an actual n-p-n transistor. Its more like n-p-N-p-n something.

So coming back to n-p-n transistor. We have LARGE collector which is averagely doped and a very thin base which is lightly doped and an emitter which is highly doped.

Now when a transistor is forward biased.

Since Base-Collector junction is reverse biased, we have no current flowing through it.

Now when Base Emitter junction in forward biased electrons in emitter junction is pushed towards base. Reason, the increased negative voltage at emitter terminal and a increased positive voltage greater than the potential barrier at Base.

If you remember emitter was highly doped and base was lightly doped as well as thin. Thus many electrons which entered base region finds itself unable to find any holes to combine with. Some gets attracted to the positive Base terminal ( making up for base current Ib, remember the direction of flow of electrons and current is opposite ).

But a large number of electrons are still in base region. So what do they do? :3

Since the base layer is thin, and neighbouring collector region is large and moderately doped they cross the c-b junction into collector region. And then they push the original electron in colector region out of collector terminal. Thus it results in collector current Ic.

If you apply Kirchof's Current Law, we can easily calculate Ie = Ib + Ic.

The jumping of electrons from base to collector can be hard to visualize if base is large. But if you imagine base getting smaller and smaller and smaller, it seems more likely to happen ( and it does happen ).

Here is a excellent article to read about n-p-n transistor's working. There are few figures in that article which may keep you interested :)

Now what happens when transistor is working in saturation.

Condition after saturation.

As you have stated the transistor's base-emitter junction is forward biased and the collector-base junction is also forward biased.

Now here is the important part. I had mentioned that Electrons where able to jump the potential barrier when it was in normal mode ( amplifying mode ). Now after saturation that potential barrier vanished. So it got much easier for all extra the electrons pulled up by base's potential to cross the barrier. If its getting much easier for the electron means resistance has decreased, so the potential drop across collector and emitter has decreased.

Also please note that The Emitter is still at a lower potential than Collector. So the electron continues to flow from emitter to collector.

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  • \$\begingroup\$ I think you are missing the point of the question. He is asking about saturation. In that case, the collector voltage is lower than the base voltage, unlike what you stated. B-E may be 700 mV and C-E only 200 mV, for example, which leaves the collector 500 mV lower than the base. \$\endgroup\$ – Olin Lathrop Feb 21 '13 at 13:21
  • \$\begingroup\$ thanks for bringing it up, i agree i gave a wrong answer, gotta edit it right away. Thanks @Olin Lathrop \$\endgroup\$ – D34dman Feb 21 '13 at 14:16
  • \$\begingroup\$ i understand how the electrons enter the base region(from the emitter), but how can these electrons enter the collector region(from the base) when the collector-base junction is forward biased, preventing the motion of the electrons into the collector region? \$\endgroup\$ – vpp Feb 21 '13 at 14:54
  • \$\begingroup\$ ( this is the reason why i asked you to stop thinking of two diode topology). Let me put it in another way. During saturation, Now all the holes in base is filled up with electrons agreed? and then there are extra electrons in base agreed? then what does it make base, a N type semiconductor ? so we have essentially a N-P-N becoming a N-N-N semiconductor where electron flows from lower potential to higher potential agreed? \$\endgroup\$ – D34dman Feb 21 '13 at 15:13
  • \$\begingroup\$ So you're saying that, the electrons, having entered the base have to flow to the collector inevitably (forward or no forward bias)? \$\endgroup\$ – vpp Feb 21 '13 at 15:50

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