Topic: Electronic Devices
An ideal p-n junction diode in series with a 100 $\Omega$ resistance is forward biased such that the forward current flowing through the diode is 100 mA. If voltage across this circuit is instantaneously reversed to 20 V at time instant $t=t_0$, then the reverse current flowing through the diode at time instant $t=t_0$ is approximately given by
- A. 0 mA
- B. 200 mA
- C. 100 mA
- D. 2 mA
Answer: B. 200 mA
Explanation: Due to minority carrier charge storage in the p-n junction, the diode cannot turn off instantaneously. At the exact instant the voltage is reversed ($t=t_0$), the diode acts momentarily as a short circuit. The initial reverse current is limited entirely by the series resistor, giving $I_R = V / R = 20\text{V} / 100\Omega = 0.2\text{A} = 200\text{ mA}$.