Chapter 3

Q. 3.1

Q. 3.1

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

Step-by-Step

Verified Solution

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}