Question

Consider an NPN transistor amplifier in common-emitter configuration. The current gain of the transistor is 100. If the collector current changes by 1 mA, what will be the change in emitter current

• A. 1.1 mA
• B. 1.01 mA
• C. 0.01 mA
• D. 10 mA

Answer 1

Answer: (B)

1.01 mA

Answer 2

Current gain $\beta = \frac{\Delta i_{c}}{\Delta i_{b}} \Rightarrow \Delta i_{b} = \frac{1 \times 10^{- 3}}{100} = 10^{- 5}A$

= 0.01mA.

By using $\Delta i_{e} = \Delta i_{b} + \Delta i_{c} \Rightarrow \Delta i_{e}$= 1.01 + 1 = 1.01mA.