For a normal transistor, the emitter current can be given by:
Correct answer: B. IE = lC + IB
- A. IE = IC
- B. IE = lC + IB
- C. IE = IB
- D. None of these
Explanation
After the application of the biasing voltage to the transistor, the emitter-base junction is forward-biased, so the emitter injects a large number of electrons into the base region. These free electrons in the base can flow in either of two directions. They can either flow out of the base to the positive terminal of VBB (the base-biased voltage which forward-biased the emitter-base junction) or they can be attracted towards the collector because of battery VCC( current collecting voltage which reverse-biased the collector-base junction). In a normal biased transistor, an electronic current IE flows from the emitter into the base. A very small part of it, the current IB, flows out of the base, and the rest of it IC flows out of the collector. Hence, we can say that IE = IC + IB.
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A p-n junction forms when p-type and n-type semiconductor regions meet, creating a depletion layer, barrier potential and rectifying action. The work covers forward and reverse bias, junction current, diode characteristics, and how a diode converts alternating current into pulsating direct current through half-wave and full-wave rectification.
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