Voltage Multiplier

A single rectifier and capacitor can only produce a DC voltage lower than the AC peak. A voltage multiplier uses diodes and capacitors in a cascade to produce DC voltages that are multiples of the peak AC input, with no transformer step-up needed. CRT televisions used Cockcroft-Walton multipliers to generate 25 kV from a 1 kV flyback stage.

Core Concept

A voltage doubler uses two diodes and two capacitors. During the negative half-cycle, D1 conducts and charges C1 to Vm. During the positive half-cycle, the AC source voltage Vm adds to the voltage on C1, so D2 sees a voltage of 2Vm and charges C2 to 2Vm. The output across C2 is double the peak AC voltage.

A Cockcroft-Walton multiplier chains multiple such stages. Each pair of diodes and capacitors adds another 2Vm to the output. Four stages give 4Vm, eight stages give 8Vm, and so on. The circuit is simple, requires no transformer, and can generate very high voltages with small, inexpensive diodes and capacitors.

The penalty for voltage multiplication is current. A voltage doubler can supply only a fraction of the current a simple rectifier provides from the same source. Output regulation is poor: the output voltage drops quickly under load because each capacitor must transfer charge through the diode chain. For the same reason, voltage multipliers are only used in low-current applications such as photomultiplier tube bias supplies and ion generators.

Key Equations

Ideal output voltage (no load): V_out = 2n * Vm where n is the number of doubler stages and Vm is the peak AC input.

Output voltage under load (approximate): V_out_load = 2n*Vm - I_L * (2n^3 + 3n^2 + n) / (6*f*C) — shows strong load dependence.

PIV per diode: PIV = 2Vm in a standard half-wave multiplier.

Ripple voltage in an n-stage multiplier: Vr = I_L * n*(n+1) / (2*f*C)

Given:
AC input: Vm = 100 V (peak)
Voltage doubler (n = 1, 2 stages total: D1,C1,D2,C2)
C1 = C2 = 10 µF = 10e-6 F
f = 50 Hz
Load current I_L = 5 mA = 0.005 A

Why this formula:
Doubler gives 2Vm at no load; load causes a voltage drop proportional to I_L/(f*C).

No-load output:
V_out = 2 * Vm = 2 * 100 = 200 V

Voltage drop under load (simplified for n=1):
Dropout = I_L * 3 / (2 * f * C)
         = 0.005 * 3 / (2 * 50 * 10e-6)
         = 0.015 / 0.001
         = 15 V

Output under load:
V_out_load = 200 - 15 = 185 V

Ripple voltage:
Vr = I_L * 1 * 2 / (2 * f * C)
   = 0.005 * 2 / (2 * 50 * 10e-6)
   = 0.01 / 0.001 = 10 V (peak-to-peak)

PIV per diode:
PIV = 2 * Vm = 200 V

Final Answer:
No-load output = 200 V, Loaded output = 185 V, Vr = 10 V, PIV = 200 V

Exam Tip: The output of a voltage doubler is 2Vm, not 2*V_rms. Vm is the peak AC voltage, not the RMS value. If the AC source is 100 V RMS, then Vm = 141 V and the doubler gives 282 V, not 200 V. Also, PIV per diode = 2Vm in a half-wave multiplier. For a full-wave doubler (different topology), PIV = Vm. GATE distinguishes between these two topologies.

Key Properties

• Ideal output voltage = 2n * Vm for an n-stage Cockcroft-Walton multiplier. Each stage contributes 2Vm.
• Output voltage drops significantly under load. Multipliers are only practical for low-current loads (microamps to low milliamps).
• PIV per diode = 2Vm in a half-wave (one-diode-per-half-cycle) multiplier. Use high-voltage diodes like 1N4007 (1000 V) or specialized HV diodes for kV-level designs.
• Ripple increases with number of stages n and load current. Both degrade with more stages.
• Applications include photomultiplier tube supplies (100 V to 2000 V at microamp currents), Geiger counter HV bias, and electrostatic precipitators.
• Full-wave voltage doubler uses two diodes and two capacitors with a center-tapped arrangement. It gives lower ripple than the half-wave version.

Quick Revision

• Voltage doubler: 2 diodes + 2 capacitors. Output = 2Vm at no load.
• Cockcroft-Walton multiplier: n stages. Output = 2n*Vm at no load.
• PIV per diode = 2Vm in half-wave multiplier.
• Output drops under load. Only suitable for low-current applications.
• Ripple Vr = I_L * n*(n+1) / (2*f*C). Increases with stages.
• Vm is peak voltage, not RMS. Output = 2 * sqrt(2) * V_rms for a doubler.
• Used in CRT HV supplies, photomultipliers, ion generators.
• Exam trap: Calculating output as 2 * V_rms instead of 2 * Vm (= 2 * sqrt(2) * V_rms). This underestimates the output by a factor of sqrt(2) = 1.414.