Calculate the current through the circuit and the potential difference across the diode shown in figure. The drift current for the diode is 20 µA.

#### Solution

(a) From the circuit diagram, it can be said that the diode is reverse biassed, with applied voltage of 5.0 V.

Under reverse bias condition,

Current in the circuit = Drift current

So, the current in the circuit is 20 µA.

(b) Voltage across the diode will be equal to the voltage of the battery minus the voltage drop across the 20 ohm resistor.

\[\Rightarrow V = 5 - iR\]

\[ \Rightarrow V = 5 - (20 \times 20 \times {10}^{- 6} )\]

\[ \Rightarrow V = 5 - (4 \times {10}^{- 4} )\]

\[ \Rightarrow V = {10}^{- 4} (50000 - 4)\]

\[ \Rightarrow V = 49996 \times {10}^{- 4} \]

\[ \Rightarrow V = 4 . 9996 V \cong 5\] V