Diode characteristics, rectifiers and clippers and clampers

Why it matters

Every instrument that runs from the mains starts with a rectifier, and every sensor input that must survive a wiring fault has a diode clamp or clipper in front of it. Diode circuits are also the simplest place to practise the piecewise-linear reasoning (which device is ON, which is OFF) that you will reuse for transistors and op-amps.

Key ideas

The p-n junction diode. A diode conducts readily when the anode is more positive than the cathode (forward bias) and passes only a tiny leakage current when reversed. Its current follows the Shockley equation: current grows exponentially with forward voltage, so for a silicon diode at room temperature the voltage across a conducting diode stays close to 0.6–0.7 V over a wide current range. In reverse, the current saturates at I_S (nA for small silicon diodes) until breakdown.

Diode models. Choose the simplest model that answers the question:

• Ideal diode: short circuit when forward biased, open circuit when reverse biased.
• Constant-drop model: a battery V_γ (≈ 0.7 V for Si, ≈ 0.3 V for Ge) in series with an ideal diode.
• Piecewise-linear model: V_γ plus a small forward resistance r_f.
• Small-signal model: around a DC operating point the diode looks like a resistance r_d = η·V_T / I_D (about 26 Ω at 1 mA for η = 1).

Temperature. At fixed current the forward voltage of a silicon diode falls by about 2 mV/°C, and the reverse saturation current roughly doubles for every 10 °C rise.

Rectifiers. A half-wave rectifier passes one half-cycle; its output repeats at the supply frequency. A full-wave rectifier (centre-tapped with two diodes, or a bridge with four) passes both half-cycles, so the output repeats at twice the supply frequency, the DC value doubles and the ripple falls. The centre-tapped circuit has only one diode drop in the path but each diode must withstand a peak inverse voltage (PIV) of 2V_m; the bridge has two diode drops but each diode sees a PIV of only V_m.

Capacitor filter. A capacitor across the load charges to the peak and discharges through the load between peaks. When the ripple is small, the peak-to-peak ripple is set by the charge drawn per period, so it is inversely proportional to f, R_L and C.

Clippers remove the part of a waveform above (or below) a chosen level. A shunt clipper puts a diode in series with a reference battery V_R across the output; when the diode conducts, the output is pinned at V_R + V_γ. Clampers (DC restorers) use a capacitor and a diode to shift the whole waveform up or down so that one extreme sits at a reference level; the shape is unchanged, only the DC level moves. For a clamper to work, the R_L·C time constant must be much longer than the signal period.

Formulas

I_D = I_S·(e^(V_D/(η·V_T)) − 1) — I_S reverse saturation current (A), V_D diode voltage (V), η ideality factor (1–2), V_T = kT/q ≈ 25.9 mV at 300 K.

r_d = η·V_T / I_D — small-signal (dynamic) resistance (Ω) at DC current I_D (A).

V_dc = V_m / π (half-wave), V_dc = 2·V_m / π (full-wave) — ideal diodes, resistive load; V_m is the peak of the output voltage (V). With real diodes use V_m − V_γ (half-wave, centre-tap) or V_m − 2V_γ (bridge).

V_rms = V_m / 2 (half-wave), V_rms = V_m / √2 (full-wave).

Ripple factor γ = √((V_rms/V_dc)² − 1) — 1.21 for half-wave, 0.482 for full-wave.

η_max = 40.6 % (half-wave), 81.2 % (full-wave) — maximum rectification efficiency P_dc/P_ac.

V_r(p-p) ≈ V_p / (f·R_L·C) (half-wave), V_r(p-p) ≈ V_p / (2·f·R_L·C) (full-wave) — capacitor filter, small ripple; f supply frequency (Hz), C (F), V_p peak output (V).

V_dc ≈ V_p − V_r(p-p)/2 — DC output of a capacitor-filtered rectifier.

Worked examples

Example 1 (standard). A half-wave rectifier is fed with a 10 V peak sine. The silicon diode has V_γ = 0.7 V and the load is R_L = 1 kΩ. Find the peak and average load voltage, the average load current and the PIV.

1. Peak output: V_p = V_m − V_γ = 10 V − 0.7 V = 9.3 V.
2. Average output: V_dc = V_p / π = 9.3 V / π = 2.96 V.
3. Average current: I_dc = V_dc / R_L = 2.96 V / 1 kΩ = 2.96 mA.
4. During the negative half-cycle the diode is OFF and the load carries no current, so the diode sees the full source: PIV = V_m = 10 V.

Answer: V_p = 9.3 V, V_dc = 2.96 V, I_dc = 2.96 mA, PIV = 10 V.

Example 2 (GATE level). A bridge rectifier is fed from a 12 V (rms), 50 Hz transformer secondary. Each diode drops 0.7 V. A 2200 µF capacitor filters the output into a 500 Ω load. Find the peak-to-peak ripple and the DC output voltage.

1. Secondary peak: V_m = √2·V_rms = 1.414 × 12 V = 16.97 V.
2. Two diodes conduct at a time: V_p = V_m − 2V_γ = 16.97 − 1.4 = 15.57 V.
3. Full-wave ripple: V_r = V_p / (2·f·R_L·C) = 15.57 / (2 × 50 × 500 × 2.2×10⁻³) = 15.57 / 110 = 0.142 V.
4. DC output: V_dc = V_p − V_r/2 = 15.57 − 0.071 = 15.50 V (load current ≈ 31 mA).

Answer: V_r ≈ 0.14 V peak-to-peak, V_dc ≈ 15.5 V.

Example 3 (clamper). A 5 V peak square wave drives a series capacitor with a shunt ideal diode whose cathode is at the output and anode at ground. The diode conducts whenever the output tries to go below 0 V, so the capacitor charges until the negative extreme sits at 0 V. The output swings from 0 V to +10 V: same 10 V peak-to-peak shape, shifted up by 5 V.

Common mistakes

• Using V_m/π for a bridge but forgetting that two diode drops (not one) are in the conduction path.
• Quoting PIV = V_m for the centre-tapped full-wave rectifier; it is 2V_m because the non-conducting diode sees the whole secondary.
• Using the supply frequency instead of twice the supply frequency for full-wave ripple.
• Confusing clippers (remove part of the waveform) with clampers (shift the whole waveform).
• Applying the clamper result when R_L·C is comparable to the period; the shift then decays and the output is distorted.
• Mixing peak, rms and average values in the same equation.

For GATE IN

Expect short numericals on average and rms output, ripple with a capacitor filter, PIV, and the output waveform of a clipper or clamper given a reference battery and V_γ. Questions often hide the trick in which diodes conduct, so always assume a state, solve, and check the assumption. Practise sketching transfer characteristics (v_o against v_i) of clipper circuits and computing r_d from the bias current.

Quick check

1. What is the ripple frequency of a bridge rectifier on a 50 Hz supply?
2. A half-wave rectifier with ideal diode has V_m = 31.4 V. What is V_dc?
3. What PIV must each diode of a centre-tapped full-wave rectifier withstand?
4. What is the small-signal resistance of a diode (η = 1) carrying 2 mA at 300 K?
5. A shunt clipper has V_R = 3 V and V_γ = 0.7 V. At what output level does it clip?

Answers: 1. 100 Hz. 2. 10 V. 3. 2V_m. 4. About 13 Ω. 5. 3.7 V.