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Question
With the help of a neat diagram, explain the working of the npn transistor?
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Solution
Working of n-p-n transistor:
- The majority of charge carriers in the emitter of the n-p-n transistor are electrons.
- A typical biasing of a transistor is shown in figure (a). In this, the emitter-base junction is forward biased while the collector-base junction is reverse biased.
Figure (a) - At the instant when the EB junction is forward biased, electrons in the emitter region have not entered the base region as shown in figure (b).
Figure (b) - When the biasing voltage VBE is greater than the barrier potential (0.6 – 0.7 V for Si transistors), many electrons enter the base region and form the emitter current IE as shown in figure (c).
Figure (c) - These electrons can either flow through the base circuit and constitute the base current (IB), or they can also flow through the collector circuit and contribute towards the collector current (IC).
- The base is thin and lightly doped, the base current is only 5% of IE.
- Electrons injected from the emitter into the base diffuse into the collector-base depletion region due to the thin base region. When the electrons enter the collector-base depletion region, they are pushed into the collector region by the electric field at the collector-base depletion region. The collector current (IC) flows through the external circuit as shown in figure (d). The collector's current IC is about 95% of IE.
Figure (d)
From the figure, we can conclude that IE = IB + IC Since the base current IB is very small we can write IC ≈ IE.
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