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i have come accross this circuit. it is simple switch circuit as per i know.

generally, in all switching circuit, emitter is connected to ground and collector is connected to high level. i am not getting why is it so? is there any special purpose?

ashish
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  • because it's much easier to have a higher base voltage than the emitter? Which determines the ability to turn it on? Usually that's for low-side switching anyway. – KyranF Dec 17 '14 at 13:55
  • In that diagram, you have Q1 backwards it seems.. – KyranF Dec 17 '14 at 13:56
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    Shouldn't Q1 be a PNP transistor? –  Dec 17 '14 at 13:59
  • @BrianDrummond yeah, that's the alternative fix to what I said lol. Good point – KyranF Dec 17 '14 at 14:07
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    @BrianDrummond It (MMBTA56) is a PNP, but the schematic symbol is wrong. – Spehro Pefhany Dec 17 '14 at 14:07
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    ... apparently it is PNP(MMBTA56). So the schematic has been lazily drawn with the wrong symbol. Q2 (MMBTA06) is correct (NPN). @Spehro ... we overlapped. –  Dec 17 '14 at 14:07
  • The question is very vague. But, if you are asking whether transistors can be reversed as far as emitter and collector goes, please see here: http://electronics.stackexchange.com/questions/29756/bjt-in-reverse-active-mode-of-operation – user57037 Dec 17 '14 at 17:06

2 Answers2

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Transistors are not really symmetric. If you turn NPN turned around is still NPN, but that is an over simplification. The emitter typically has a much higher doping level than the collector.

You want high doping in the emitter because the higher doping level will give you a higher current gain. It injects more electrons into the base if there are lots of free electrons. This is why the emitter is tied to power or ground if you are amplifying a small signal; you want the BE current to be amplified into a larger CE current.

You don't want high doping on the collector at the base junction because that reduces the breakdown voltage, but light doping is not very conductive so you want the bulk of the collector to be strongly doped in regions away from the base to improve maximum current.

To sum up:

  • NPN is really (N+)(N-) P (N+) for C B E
  • If you plugged it in backwards, it might function but at lower gain and capacity.
Jeff Bell
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  • The diagrams I've seen of transistors also have a geometry that favors the "normal" usage, with the emitter surrounded by the base which is in turn surrounded by the collector. I don't know to what extent inertia is applicable to the "holes" used in PNP transistors, but at least in NPN I think some of the gain comes from the fact that if electrons that go from the emitter to the base keep going in the same direction, they'll hit the collector. Electrons going from the collector into the base, however, might "miss" the emitter. – supercat May 18 '15 at 16:12
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Usually to operate BJT in active region Base-Emitter Junction must Forward Bias and Base-Collector Junction must be Reverse Bias

In above circuit if you assume it as PNP BJT. then Base is at lesser voltage level compared to Emitter and higher compared to collector

JigarGandhi
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