Electricity: Question 7

Worked solution

Step 1: Recall the defining feature of a diode

A semiconductor diode is designed to conduct current in essentially one direction only. This makes it strongly non-ohmic: its resistance is very different in forward bias compared with reverse bias, so the I–V graph is not a straight line through the origin.

Step 2: Forward bias behaviour

When forward-biased, very little current flows until the potential difference reaches a threshold (turn-on) voltage (of the order of a fraction of a volt, depending on the material). Below this threshold the diode has a very high resistance. Once $V$ exceeds the threshold, the current rises rapidly for further small increases in $V$. The diode’s resistance becomes low.

Step 3: Reverse bias behaviour

When reverse-biased, the diode presents a very high resistance across the whole range of voltages tested, so the current remains close to zero regardless of how large the reverse voltage is (within the range considered here).

Step 4: Why the other options are wrong

• B: a diode is not ohmic. Its resistance is not fixed, and it does not conduct symmetrically in both directions.
• C: this reverses the correct behaviour. It is in forward bias (not reverse) that the current rises rapidly once the threshold is passed.
• D: the whole purpose of a diode is that it does not conduct equally in both directions; in reverse bias its resistance is very high, not low.

Final answer

• Option A: forward bias gives near-zero current below a threshold voltage then a rapid rise; reverse bias gives near-zero current throughout.