Voltage multiplier

From HvWiki

A voltage multiplier is a circuit that using a series of diodes and capacitors raises AC voltage without a transformer.

CW Multiplier Tower, courtesy [www.BNL.gov]

Charge-Pump Voltage Doubler

The Charge-Pump voltage doubler will output DC current at two times the voltage of the AC input. Voltage seen across each capacitor is equal to the AC input recified to DC. With a 120VAC input, each capacitor will see at least 170V (see peak and RMS) Each diode should be rated the same. However, the voltage across the capacitors cumulatively will be 2 times the AC input, DC. Hence, Vout.

Charge Pump Doubler

Waveforms

Note: These are simulated values. For this demonstration:

• Load = 1000ohms
• Capacitors = 1000uF
• Input = 120VAC (RMS)

Voltage Input (120VAC)

Current Waveform of D1 and D2

Voltage across C1 and C2

Voltage across D1 and D2

Output Voltage (+ of C1 to - of C2)

The above simulations are done well within the safe output of this doubler. However, what happens if we were to overload it? It will actually limit it's own current, but with one problem. Under heavy load, the voltage across C1 and C2 approaches 120VAC (170 peak). If we're using a polar capacitor such as an electrolytic, then it may cause it to fail. Electrolytics may very well take this stress for a few cycles, but any more than that and you've got a problem. A diode can be put across the 'lytics from negative to positive so that they never experienced reverse voltage. Another solution to the problem, and suggested if the doubler is to limit it's own current, is to use bipolar capacitors.

Voltage across C1 under heavy load (Load = 1ohms)

Cockroft-Walton Voltage Multiplier

Sometimes abbreviated "CW"

Cockroft-Walton Schematic: The illustrator forgot something. The very last wire to the right should be a more diode pointing up toward the "HV out" following the convention of the previous diodes. This configuration will work, but the last capacitor in the string is useless.

This mulitpier can output any voltage that you wish, provided you have the number of stages.

Maximum voltage can be found by

However, this is optimum, and often draw down by the load current. "Nstages" or the number of stages can be found by counting the capacitors and dividing by 2.

To find the actual voltage under load, you use this equation:

• I is the current draw of the load in amps

• f is the frequency of the AC wave in hertz

• C is the stage capacitance in Farads

• n is the number of stages

Then subtract.

Expected voltage stress on the diodes and capacitors will be:

And will be greatest when cascade is unloaded, as load increases, stress should decrease.

Other References

Blaze Labs - Good information on design and function of CW multipliers.

Maxim-IC.com - "DC/DC Conversion without Inductors." - Voltage multipliers in digital circuit applications.