Electronic Devices – JIPMER Physics Study Notes
Scope – Semiconductor devices that form the building blocks of modern electronics:
- Diodes (PN‑junction, Zener, Schottky)
- Transistors (BJT, MOSFET, JFET)
- Basic rectifiers & power‑conversion circuits
- Small‑signal models & common amplifier configurations
These notes are written for quick self‑study and revision before the JIPMER exam.
1. Simple Explanation
| Device | What it is | How it works (basic principle) |
|---|---|---|
| PN‑junction diode | A p‑type and n‑type semiconductor joined together. | When forward‑biased, majority carriers cross the junction, producing current; when reverse‑biased, depletion zone widens and blocks current. |
| Zener diode | A heavily doped PN junction that allows controlled reverse conduction. | Operates in the breakdown region (reverse‑biased) to maintain a nearly constant voltage. |
| Schottky diode | Metal‑semiconductor junction with low forward drop. | Provides very fast switching due to lack of minority‑carrier storage. |
| BJT (Bipolar Junction Transistor) | Two PN junctions in series (emitter–base–collector). | Current at one terminal (base) controls larger currents at the other two (collector & emitter). |
| MOSFET (Metal‑Oxide‑Semiconductor FET) | Gate insulated from channel by oxide. | Gate voltage controls channel conductivity; useful for high‑impedance input. |
| Rectifier | Circuit that converts AC to DC using diodes. | A half‑wave or full‑wave rectifier lets only one polarity of the input through. |
2. Key Points
| Concept | Symbol | Important Relations / Numbers |
|---|---|---|
| PN‑junction depletion width | (W) | (W \propto \frac{1}{\sqrt{D}}) (where (D) is doping). |
| Diode I‑V characteristic | (I = I_S \left(e^{\frac{qV}{n k_B T}} - 1\right)) | (I_S) – saturation current; (n) – ideality factor (≈1–2). |
| Forward voltage drop | (V_F) | ~0.7 V for Si, ~0.3 V for Ge, ~0.2 V for Schottky. |
| Reverse breakdown (Zener) | (V_Z) | (I_Z = I_S (e^{qV_Z/kT} - 1)), but effectively constant (V_Z). |
| BJT current relations | (I_C = \beta I_B), (I_E = I_B + I_C) | (\beta) (current gain) ≈ 50–300. |
| MOSFET threshold | (V_{th}) | (I_D = \frac{1}{2}\mu C_{ox}\frac{W}{L}(V_{GS}-V_{th})^2) (saturation). |
| Rectifier output | (V_{DC} \approx V_{peak} - V_D) | For half‑wave: (V_{DC} = \frac{V_{peak}}{\pi} - V_D). |
| Average rectified voltage | (V_{avg}) | Half‑wave: (V_{avg} = \frac{V_{peak}}{\pi}), Full‑wave: (V_{avg} = \frac{2V_{peak}}{\pi}). |
| Power dissipation in diode | (P = V_D I_D) | Keep (P) below rated power (e.g., 1 W). |
3. Worked Example
Problem:
A 12‑V AC mains supply (RMS) feeds a half‑wave rectifier using a silicon diode (forward drop 0.7 V). A series resistor (R = 1,\text{k}\Omega) limits current. Find:
- Peak forward current through the diode.
- Average DC output voltage across the resistor.
- Power dissipated in the resistor and the diode.
Solution
-
Peak voltage of AC source
[ V_{\text{peak}} = \sqrt{2},V_{\text{RMS}} = \sqrt{2}\times12 \approx 16.97 ,\text{V} ] -
Peak forward voltage across diode + resistor
[ V_{\text{peak,drop}} = V_{\text{peak}} - V_D = 16.97 - 0.7 \approx 16.27 ,\text{V} ] -
Peak diode current (Ohm’s law across (R))
[ I_{\text{peak}} = \frac{V_{\text{peak,drop}}}{R} = \frac{16.27}{1000} \approx 0.0163 ,\text{A} = 16.3 ,\text{mA} ] -
Average (DC) output voltage across (R)
For a half‑wave rectifier, the average of the sinusoid over the conducting half‑cycle is: [ V_{DC}