CBSE Class 12 · 2007 · SemiconductorsLong Answer

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Draw the symbolic representation of a
(i) p-n-p and
(ii) n-p-n transistor. Why is the base region of a transistor thin and lightly doped? With a proper circuit diagram, show the biasing of a p-n-p transistor in common base configuration. Explain the movement of charge carriers through the different parts of the transistor and show that .

Solution

Symbols: in both p-n-p and n-p-n symbols the emitter arrow points in the direction of conventional current — outward for n-p-n and inward (towards the base) for p-n-p.
The base is thin and lightly doped so that few carriers recombine there, letting most reach the collector and giving a high current gain.
Common-base p-n-p biasing: the emitter-base junction is forward biased by and the collector-base junction is reverse biased by .
Carrier movement: the forward bias injects holes from the emitter into the base; about 5% recombine with electrons in the thin base (forming ), while about 95% diffuse across and are swept into the collector by the reverse-biased junction (forming ).
By charge conservation at the emitter, the total emitter current equals the sum of the collector and base currents:

CBSE Class 12 Board · 2007 · Subjective

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