# 90-diode-voltage-multiplier-circuit

Diode voltage multiplier circuit given in figure bellow.
Simulate diode voltage multiplier circuit.

Diode voltage multiplier operates as follows: during first negative half period of the input sine voltage, diode D11 is forward biased, while diode D12 is blocked. Capacitor C11 is charging from right hand side, i.e. electrical potential of the point A goes to +VCC-VD during first negative half period of the input signal.

During next positive half period of the input signal, diode D11 is blocked, while diode D12 is forward biased. Charged capacitor C11 can be treated as voltage source charged on +VCC-VD, and together with input voltage, we have two voltage sources in series. At the peak of the input voltage, capacitor C12 is charged to 2VCC-2VD.

When second negative half period appears on the input side, voltage on C11 drops, but at negative peak of second half period C11 is charged again on VCC-VD, C12 is charged to 2VCC – 2VD and C21 is also charged to 2VCC – 2VD.

Theoretically, if all components would be perfect, one full section containing two capacitors and two diodes (Ci1, Ci2, Di1, Di2) will double input peak voltage VCC on every half period. That is why it is also called half-wave voltage doubler. So at the and of diode voltage multiplier circuit, output voltage is approximately Nâˆ™2VCC (if we neglect VD to VCC). This circuit is also called diode charge pump. It is also important to see what is the voltage level at the peek of every half cycle of the input voltage.

In practice, it takes more cycles for the output to reach Nâˆ™2VCC. Diodes must have break down voltage of at least 1000V (1N4007 for example), for safety reason, if mains is on 220V effectively. Output resistor value should be chosen depending on output voltage level, but for outputs of several thousands kV, output resistor should be not less then 30MÎ©, but in this application output resistor is as higher as better. Capacitors should be in range from 10nF to 100nF.