Class 12 Physics Chapter 14: Semiconductor Electronics: Materials, Devices and Simple Circuits case study question with answers
One case-based question of the kind the board asks: a passage and a table, then 3 parts worth 4 marks in all, with the answer and the marking scheme for each. Written from the NCERT chapter.
| V (V) | 0.05 | 0.10 | 0.15 | 0.20 | 0.25 | 0.30 | 0.35 |
|---|---|---|---|---|---|---|---|
| I (mA) | 0.0 | 0.1 | 0.3 | 1.0 | 4.0 | 10.0 | 18.0 |
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Answer: (a) germanium
The current becomes appreciable near 0.2 V, the threshold voltage of germanium; a silicon diode would carry almost no current below about 0.7 V.
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Answer: (a) 2.7
rd = ΔV/ΔI. Between 0.20 and 0.25 V: 0.05/3.0 mA ≈ 16.7 Ω. Between 0.30 and 0.35 V: 0.05/8.0 mA ≈ 6.25 Ω. Ratio ≈ 2.7. (b) inverts it; (c) assumes equal steps of V give equal resistance.
Answer and marking scheme
The reverse current is a drift of minority carriers near the junction swept across by the field. Even a small reverse voltage sweeps all of them, so the current is limited by the minority-carrier concentration, not by the voltage: it saturates at a few μA.
- reverse current is a drift of minority carriers (1)
- limited by their concentration, so nearly independent of voltage (saturation) (1)
Answer and marking scheme
Forward bias lowers the barrier to V₀ − V. At small V only the few carriers in the uppermost energy levels can cross, so the current is negligible. Beyond the threshold the barrier is low enough that, for each small increase in V, many more carriers have the energy to cross, so the current rises steeply (exponentially).
- small V: barrier V₀ − V still high, only a few carriers cross (1)
- beyond threshold many more carriers cross, so current rises exponentially (1)