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Voltage multiplier

Electrical circuit power converter

  • 10 min read
  • 6 chapters

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Through Electrical network, Alternating current and Voltage

Voltage multipliers can be used to generate a few volts for electronic appliances, to millions of volts for purposes such as high-energy physics experiments and lightning safety testing. The most common type of voltage multiplier is the half-wave series multiplier, also called the Villard cascade (but actually invented by Heinrich Greinacher).

Operation

Assuming that the peak voltage of the AC source is +Us, and that the C values are sufficiently high to allow, when charged, a current to flow with no significant change in voltage, then the (simplified) working of the cascade is as follows:

  1. going from positive peak (+Us) to negative peak (−Us): The C1 capacitor is charged through diode D1 to Us V (potential difference between left and right plate of the capacitor is Us).
  2. going from negative peak to positive peak: The voltage of C1 adds with that of the source, thus charging C2 to 2Us through D2 and discharging C1 in the process.
  3. positive to negative peak: Voltage of C1 has dropped to 0 V by the end of the previous step, thus allowing C3 to be charged through D3 to 2Us.
  4. negative to positive peak: Voltage of C2 rises to 2Us (analogously to step 2), also charging C4 to 2Us. The output voltage (the sum of voltages of C2 and C4) rises until 4Us is reached.

Adding an additional stage will increase the output voltage by twice the peak AC source voltage (minus losses due to the diodes ‒ see the next paragraph).

In reality, more cycles are required for C4 to reach the full voltage, and the voltage of each capacitor is lowered by the forward voltage drop (Uf) of each diode on the path to that capacitor. For example, the voltage of C4 in the example would be at most 2Us - 4Uf since there are 4 diodes between its positive terminal and the source. The total output voltage would be U(C2) + U(C4) = (2Us - 2Uf) + (2Us - 4Uf) = 4Us - 6Uf. In a cascade with n stages of two diodes and two capacitors, the output voltage is equal to 2n Us - n(n+1) Uf. The term n(n+1) Uf represents the sum of voltage losses caused by diodes, over all capacitors on the output side (i.e. on the right side in the example ‒ C2 and C4). For example, if we have 2 stages like in the example, the total loss is 2+4 = 2*(2+1) = 6 times Uf. An additional stage will increase the output voltage by twice the source voltage, minus the forward voltage drop over 2n+2 diodes: 2Us - (2n+2)Uf.

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