Question
Explain the working of the PNP transistor?
✓
Answer
Working of p-n-p transistor:
vi. The base is thin and lightly doped, the base current is only $5 \%$ of $I _{ E }$.
vii. Holes injected from the emitter into the base diffuse into the collector-base depletion region due to the thin base region. When the holes 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 ( $l _{ C }$ ) flows through the external circuit as shown in figure (d). The collector's current $I _{ C }$ is about $95 \%$ of $I _{ E }$.
Figure (d)
From the figure, we can conclude that $I_E = I_B + I_C$ Since the base current IB is very small we can write $I_C ≈ I_E.$
- The majority of charge carriers in the emitter of the p-n-p transistor are holes.
- 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, holes in the emitter region have not entered the base region as shown in figure (b).
Figure (b) - When the biasing voltage $V_{BE}$ is greater than the barrier potential $(0.6 – 0.7 V$ for Si transistors), many holes enter the base region and form the emitter current $I_E$ as shown in figure (c).
Figure (c)
vi. The base is thin and lightly doped, the base current is only $5 \%$ of $I _{ E }$.
vii. Holes injected from the emitter into the base diffuse into the collector-base depletion region due to the thin base region. When the holes 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 ( $l _{ C }$ ) flows through the external circuit as shown in figure (d). The collector's current $I _{ C }$ is about $95 \%$ of $I _{ E }$.
Figure (d)
From the figure, we can conclude that $I_E = I_B + I_C$ Since the base current IB is very small we can write $I_C ≈ I_E.$
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