i want to ask a basic question..
Que: For the transistor to be operated as amplifier (CE case) emitter junction should be forward biased and collector junction should be reverse biased.
but how the collector of the transistor is able to sense the input signal since the junction is reverse biased i.e. no current is able to flow from emitter to collector?
the two junctions are ideally independent.....in short how the reverse biased collector junction is able to sense the input signal and amplify it?
please explain analytically considering the basic definition of the transistor(transfer resistor).
waiting for the reply...
The c-b junction is indeed reverse biased, meaning only reverse leakage current exists, if nothing else is going on. If the b-e junction is unbiased, Ic is quite small, where Vbc is large. But if the b-e junction is forward biased, holes flow from base to emitter, & electrons flow (drift) from emitter to base (npn polarity). A forward biased p-n junction has a small forward voltage & a large forward current. But electrons drifting from emitter to base, enter the base region which is very thin. There is an electric field from base to collector, and since the b-c junction is reverse biased, the polarity of the b-c E field is such so as to attract the electrons entering the base from the emitter.
The electrons emitted by the emitter, do not recombine with holes in the base region except for very few, but rather, nearly all are yanked into the collector by the reverse biased c-b junction's E field. The b-c junction is originally reverse biased, meaning only minority charges drift, resulting in low current. Holes in the collector & electrons in the base drift due to E field in b-c region. But these are minority carriers since collector is n type, & base is p type silicon.
The forward biased b-e junction supplies an abundance of electrons from emitter towards base. This is natural as the electrons are majority carriers in emitter region, in plentiful supply. Thus the current in the collector is provided by the emitter which emits majority carrier electrons from its n type material, through the base region, where these electrons are minority carriers due to base being p type material, then they continue into the collector where they are again majority carriers. So the forward bias on the b-e junction results in collector current in the b-c junction due to emitter ejected electrons crossing the ultra-thin base region & becoming collector electrons.
The equation for this transistor action is I[sub]C[/sub] = alpha*I[sub]E[/sub]. This works best if alpha is close to unity in value. If the base region was wide, say 1.0 mm, then alpha would be near zero, & the number of electrons collected is very small, most electrons emitted from emitter would stop in the base due to recombination, never reaching the collector. But if the base is very thin, say 1.0 micron, then nearly all emitter electrons would transit through the base & reach the collector.
A thick base results in little to no transistor action, & such a structure is nothing more than a pair of back to back diodes, with common anode. The thin base makes transistor action possible, providing current amplification. Have I helped? I will elaborate if requested. Best regards.
thanks for the reply..
it means that during the C-B reverse bias the base is almost covered with the depletion region and the electrons from the emitter are able to jump from emitter to collector...
am i right?
Yes, but I would not say the base is "almost covered by depletion region". The c-b depletion region extends into the base partially. If Vcb is increased, the depletion region extends further into the base, reducing the base volume, increasing beta the current gain. Do a search on "Early effect" to get details. Anyway, the base region being so thin results in most electrons emitted from emitter traveling through base into collector. Holes from base move into emitter & recombine with electrons. That is the long & short of it. Best regards.
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