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Semiconductor diodes and applications

Electronics and Communication Engineering (B.E./B.Tech) · Circuits and devices · free study notes for exams, placements and interviews

In short

A semiconductor diode is a single pn junction that conducts easily in one direction and blocks in the other, so it is the basic one-way valve of electronics. Everything from a mobile charger to an inverter starts by rectifying 230 V, 50 Hz ac with diodes, and diodes also clip, clamp, multiply, protect and steer voltages. University exams and GATE (EC/EE/IN) ask the diode equation, small-signal resistance, load-line and model-based analysis, rectifier performance figures and wave-shaping circuits; interviews at power-electronics and hardware firms (Delta, Secure Meters, Luminous, Havells, Bosch, Tata Power, Centum, BEL, DRDO labs, semiconductor design services) ask the same things against datasheet reality. This note covers the ordinary pn-junction diode and its circuit applications; Zener, Schottky, varactor, LED and photodiode behaviour is treated with the special devices, and junction formation with semiconductor physics.

Explained

The pn-junction diode and its V-I characteristic

The diode equation, resistance and capacitance

Diode models and circuit analysis

  1. Ideal-diode model — on: short circuit; off: open circuit. Use for quick analysis and when supply voltage is much larger than 0.7 V.
  2. Constant-voltage-drop (CVD) model — 0.7 V (Si) battery when on. The standard exam model.
  3. Piecewise-linear model — V_γ in series with r_f; needed when the drop and slope both matter.
  4. Exponential (large-signal) model — solve the Shockley equation with the loop equation iteratively or graphically; used when accuracy at low current is needed.
  5. Small-signal model — r_d (plus C_D or C_T) about a dc bias point, for detectors, limiters and bias compensation networks.
  6. Load-line method: write the loop equation V_S = I·R + V_D, draw the straight line (intercepts V_S on the voltage axis and V_S/R on the current axis) on the measured characteristic; the intersection is the Q-point. Slope = −1/R.
  7. Assumed-state method for multi-diode circuits: assume each diode on or off, solve, then check that every "on" diode has positive current and every "off" diode has negative voltage; if a check fails, change the assumption.
  8. Series and parallel diodes: in series the drops add; in parallel, the diode with the lower drop hogs current, so power diodes are paralleled only with current-sharing resistors or matched devices.

Rectifiers

Filters, ratings and practical selection

Wave-shaping and other applications

Key points

Formulas and examples

I = I_S [ eV_D /(n V_T) − 1 ] — Shockley equation; I_S = reverse saturation current (A), n = ideality factor (1-2), V_D = diode voltage (V)
V_T = kT/q ≈ 25.85 mV at T = 300 K — k = 1.38×10⁻²³ J/K, q = 1.6×10⁻¹⁹ C, T in kelvin
V_D = n V_T ln(I/I_S + 1); ΔV per decade of current = 2.303 n V_T ≈ 60 mV (n = 1, 300 K)
R_static = V_D / I_D ; r_d = n V_T / I_D (Ω; r_d ≈ 26/I in mA)
C_T = C_T(0) / (1 + V_R/V_bi)m (m = 1/2 abrupt, 1/3 graded); C_D = τ I /(n V_T)
Half wave: V_dc = V_m/π, I_dc = I_m/π, V_rms = V_m/2, γ = 1.21, η = 40.6%, FF = 1.57, PIV = V_m
Full wave: V_dc = 2V_m/π, I_dc = 2I_m/π, V_rms = V_m/√2, γ = 0.482, η = 81.2%, FF = 1.11
γ = V_r(rms)/V_dc = √[(V_rms/V_dc)² − 1]; η = P_dc/P_ac = I_dc²R_L /(I_rms²(R_L + r_f))
Capacitor filter (full wave): V_r(pp) = I_dc /(f_r C), f_r = 2f; V_dc ≈ V_m − V_r(pp)/2; γ = 1/(4√3 f C R_L), f = supply frequency
Load line: V_S = I_D R + V_D ; with CVD model I_D = (V_S − 0.7)/R ; with PWL model I_D = (V_S − V_γ)/(R + r_f)
% Regulation = (V_no-load − V_full-load)/V_full-load × 100

1. Diode equation, both ways. A silicon diode has I_S = 1 pA, n = 1, T = 300 K (V_T = 25.85 mV).

2. Load line and model choice. 10 V source, R = 1 kΩ, Si diode.

3. Bridge rectifier with capacitor filter (typical 230 V/15 V bench supply). Secondary 15 V rms, 50 Hz; C = 1000 µF; R_L = 500 Ω.

4. Half wave versus full wave, same transformer. V_m = 20 V, R_L = 100 Ω, ideal diodes.

QuantityHalf waveFull wave / bridge
V_dc20/π = 6.37 V40/π = 12.73 V
I_dc63.7 mA127.3 mA
V_rms10.0 V14.14 V
Ripple factor γ√((10/6.37)²−1) = 1.21√((14.14/12.73)²−1) = 0.482
P_dc = V_dc²/R_L0.405 W1.62 W
Efficiency η40.6%81.2%
Ripple frequency (50 Hz mains)50 Hz100 Hz
PIVV_m = 20 V2V_m = 40 V (centre-tap), V_m = 20 V (bridge)
TUF0.2870.693 / 0.812

5. Clipper and clamper on the same input. v_in = 10 sin ωt, Si diodes (0.7 V).

Where it is used

Interview questions

Common mistakes

Related topics

Semiconductor physics: energy bands, carriers and junctions · Special diodes and optoelectronic devices · Oscillators and power supplies · BJT and MOSFET · Operational amplifiers · Network theorems and circuit analysis

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