A p–n junction diode is connected in series with 4 V battery and 35 Ω resistor. A drift current 10 μA flows through the diode. What is the nature of biasing of the diode and the potential drop across the diode?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → Reverse bias, 3.99 V
A drift current of only a few microamperes is the tiny current carried by minority carriers across a reverse-biased junction; a forward-biased diode would carry milliamperes. So the diode is reverse biased, and almost the whole battery voltage appears across it.
\( V_R = IR = (10\times10^{-6}\,\mathrm{A}) \times 35\,\Omega = 3.5\times10^{-4}\,\mathrm{V} \)
\( V_{\mathrm{diode}} = 4\,\mathrm{V} - 0.00035\,\mathrm{V} = 3.99965\,\mathrm{V} \)
The drop across the diode is practically the full 4 V; Option (1) gives it as 3.99 V, the only reverse-bias value close to 4 V.