Question 3.1: In the npn transistor of Fig. 3-1(a), 10^8 holes/μs move fro......

Schaum’s Outline of Electronic Devices and Circuits [1089587]

In the npn transistor of Fig. 3-1(a), $10^8 \text{holes}/μ\text{s}$ move from the base to the emitter region while $10^{10} \text{electrons}/μ\text{s}$ move from the emitter to the base region. An ammeter reads the base current as $i_B = 16 μ\text{A}$ . Determine the emitter current $i_E$ and the collector current $i_C$ .

3.1

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The emitter current is found as the net rate of flow of positive charge into the emitter region:
$i_E = (1.602 × 10^{-19} \text{C/hole})(10^{14} \text{holes/s}) – (-1.602 × 10^{-19} \text{C/electron})(10^{16} \text{electrons/s})$
$= 1.602 × 10^{-5} + 1.602 × 10^{-3} = 1.618 \text{mA}$

Further, by KCL,
$i_C = i_E – i_B = 1.618 × 10^{-3} – 16 × 10^{-6} = 1.602 \text{mA}$

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