The barrier potential developed in a diode is 0.5 V, then the value of the current in the circuit shown below is:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 40 mA
The barrier potential is the potential difference across the depletion layer of a p-n junction; a forward-biased diode conducts only after this much voltage is used up across it, and the rest of the emf drives the current through the resistor.
In the figure the longer plate of the cell (positive terminal) is on the left, so the current goes up the left side and enters the diode at its triangle (p) side. The diode is forward biased.
\( V_{R} = 4.5\,\mathrm{V} - 0.5\,\mathrm{V} = 4.0\,\mathrm{V} \)
\( \displaystyle I = \frac{V_{R}}{R} = \frac{4.0\,\mathrm{V}}{100\,\Omega} = 0.04\,\mathrm{A} = 40\,\mathrm{mA} \)