Reverse Bias And Junction Breakdown
A junction diode is connected so that its n-side is held at a higher potential than its p-side by an external battery of moderate voltage. Which description best matches the diode's behaviour in this connection?
Select the correct option:
Solution
It is reverse biased and allows only a tiny reverse saturation current
The bias condition of a diode is set by the polarity of the applied voltage relative to the junction. Connecting the n-side to the higher potential and the p-side to the lower potential is reverse bias, because this polarity pulls majority carriers away from the junction and widens the depletion region. The internal barrier is thereby reinforced, so almost no majority-carrier current flows. Only a very small reverse saturation current persists, arising from thermally generated minority carriers swept across the junction, and it is nearly independent of the applied reverse voltage until breakdown is reached. The current saturates because it is limited not by the applied voltage but by the rate at which minority carriers are thermally generated, so once every available minority carrier is being swept across, raising the voltage cannot increase the flow. This is why the reverse characteristic appears almost flat over a wide voltage range. The forward-bias option is wrong because that requires the p-side to be at the higher potential. The short-circuit option is wrong; a reverse-biased junction presents a very high resistance, not zero. The oscillation option is wrong because a simple biased diode has no mechanism to switch states on its own. As a consistency check, this high-resistance, low-current reverse behaviour is exactly what allows a diode to block current in one direction and act as a one-way valve.
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About This Question
- Subject
- physics
- Chapter
- semiconductor electronics
- Topic
- reverse bias and junction breakdown
- Difficulty
- Medium
- Year
- 2025
Solution
Correct Answer:
It is reverse biased and allows only a tiny reverse saturation current
The bias condition of a diode is set by the polarity of the applied voltage relative to the junction. Connecting the n-side to the higher potential and the p-side to the lower potential is reverse bias, because this polarity pulls majority carriers away from the junction and widens the depletion region. The internal barrier is thereby reinforced, so almost no majority-carrier current flows. Only a very small reverse saturation current persists, arising from thermally generated minority carriers swept across the junction, and it is nearly independent of the applied reverse voltage until breakdown is reached. The current saturates because it is limited not by the applied voltage but by the rate at which minority carriers are thermally generated, so once every available minority carrier is being swept across, raising the voltage cannot increase the flow. This is why the reverse characteristic appears almost flat over a wide voltage range. The forward-bias option is wrong because that requires the p-side to be at the higher potential. The short-circuit option is wrong; a reverse-biased junction presents a very high resistance, not zero. The oscillation option is wrong because a simple biased diode has no mechanism to switch states on its own. As a consistency check, this high-resistance, low-current reverse behaviour is exactly what allows a diode to block current in one direction and act as a one-way valve.
This medium difficulty physics question is from the chapter semiconductor electronics, covering the topic of reverse bias and junction breakdown. It appeared in the 2025 exam.
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