In the tutorial about Rectifiers, we saw that the DC output voltage being controlled by the rectifier is at a value below that of the mains input voltage.

The **Voltage Multiplier**, however, is a special type of diode rectifier circuit which can potentially produce an output voltage many times greater than of the applied input voltage.

Although it is usual in electronic circuits to use a voltage transformer to increase a voltage, sometimes a suitable step-up transformer or a specially insulated transformer required for high voltage applications may not always be available. One alternative approach is to use a diode voltage multiplier circuit which increases or “steps-up” the voltage without the use of a transformer.

**Voltage multipliers** are similar in many ways to rectifiers in that they convert AC-to-DC voltages for use in many electrical and electronic circuit applications such as in microwave ovens, strong electric field coils for cathode-ray tubes, electrostatic and high voltage test equipment, etc, where it is necessary to have a very high DC voltage generated from a relatively low AC supply.

Generally, the DC output voltage (Vdc) of a rectifier circuit is limited by the peak value of its sinusoidal input voltage. But by using combinations of rectifier diodes and capacitors together we can effectively multiply this input peak voltage to give a DC output equal to some odd or even multiple of the peak voltage value of the AC input voltage. Consider the basic voltage multiplier circuit below.

The above circuit shows a basic symmetrical voltage multiplier circuit made up from two half-wave rectifier circuits. By adding a second diode and capacitor to the output of a standard half-wave rectifier, we can increase its output voltage by a set amount. This type of voltage multiplier configuration is known as a **Full Wave Series Multiplier** because one of the diodes is conducting in each half cycle, the same as for a full wave rectifier circuit.

When the sinusoidal input voltage is positive, capacitor C_{1} charges up through diode D_{1} and when the sinusoidal voltage is negative, capacitor C_{2} charges up through diode, D_{2}. The output voltage 2V_{P} is taken across the two series connected capacitors.

The voltage produced by a *voltage multiplier* circuit is in theory unlimited, but due to their relatively poor voltage regulation and low current capability there are generally designed to increase the voltage by a factor less than ten. However, if designed correctly around a suitable transformer, voltage multiplier circuits are capable of producing output voltages in the range of a few 100’s to tens’s of 1000’s of volts, depending upon their original input voltage value but all with low currents in the milliamperes range.

As its name suggests, a **Voltage Doubler** is a voltage multiplier circuit which has a voltage multiplication factor of two. The circuit consists of only two diodes, two capacitors and an oscillating AC input voltage (a PWM waveform could also be used). This simple diode-capacitor pump circuit gives a DC output voltage equal to the peak-to-peak value of the sinusoidal input. In other words, double the peak voltage value because the diodes and the capacitors work together to effectively double the voltage.

So how does it work. The circuit shows a half wave voltage doubler. During the negative half cycle of the sinusoidal input waveform, diode D1 is forward biased and conducts charging up the pump capacitor, C1 to the peak value of the input voltage, (Vp). Because there is no path for capacitor C1 to discharge into, it remains fully charged and acts as a storage device in series with the voltage supply. At the same time, diode D2 conducts via D1 charging up capacitor, C2.

During the positive half cycle, diode D1 is reverse biased blocking the discharging of C1 while diode D2 is forward biased charging up capacitor C2. But because there is a voltage across capacitor C1 already equal to the peak input voltage, capacitor C2 charges to twice the peak voltage value of the input signal.

In other words, V(positive peak) + V(negative peak) as on the negative half-cycle, D1 charges C1 to Vp and on the positive half-cycle D2 adds the AC peak voltage to Vp onC1 and transfers it all to C2. The voltage across capacitor, C2 discharges through the load ready for the next half cycle.

Then the voltage across capacitor, C2 can be calculated as: Vout = 2Vp, (minus of course the voltage drops across the diodes used) where Vp is the peak value of the input voltage. Note that this double output voltage is not instantaneous but increases slowly on each input cycle, eventually settling to 2Vp.

As capacitor C2 only charges up during one half cycle of the input waveform, the resulting output voltage discharged into the load has a ripple frequency equal to the supply frequency, hence the name half wave voltage doubler. The disadvantage of this is that it can be difficult to smooth out this large ripple frequency in much the same way as for a half wave rectifier circuit. Also, capacitor C2 must have a DC voltage rating at least twice the value of the peak input voltage.

The advantage of “Voltage Multiplier Circuits” is that it allows higher voltages to be created from a low voltage power source without a need for an expensive high voltage transformer as the voltage doubler circuit makes it possible to use a transformer with a lower step up ratio than would be need if an ordinary full wave supply were used. However, while voltage multipliers can boost the voltage, they can only supply low currents to a high-resistance (+100kΩ) load because the generated output voltage quickly drops-off as load current increases.

By reversing the direction of the diodes and capacitors in the circuit we can also reverse the direction of the output voltage creating a negative voltage output. Also, if we connected the output of one multiplying circuit onto the input of another (cascading), we can continue to increase the DC output voltage in integer steps to produce voltage triplers, or voltage quadruplers circuits, etc, as shown.

By adding an additional single diode-capacitor stage to the half-wave voltage doubler circuit above, we can create another voltage multiplier circuit that increases its input voltage by a factor of three and producing what is called a **Voltage Tripler Circuit**.

A “voltage tripler circuit” consists of one and a half voltage doubler stages. This voltage multiplier circuit gives a DC output equal to three times the peak voltage value (3Vp) of the sinusoidal input signal. As with the previous voltage doubler, the diodes within the voltage tripler circuit charge and block the discharge of the capacitors depending upon the direction of the input half-cycle. Then 1Vp is dropped across C3 and 2Vp across C2 and as the two capacitors are in series, this results in the load seeing a voltage equivalent to 3Vp.

Note that the real output voltage will be three times the peak input voltage minus the voltage drops across the diodes used, 3Vp – V(diode).

If a voltage tripler circuit can be made by cascading together one and a half voltage multipliers, then a **Voltage Quadrupler Circuit** can be constructed by cascading together two full voltage doubler circuits as shown.

The first voltage multiplier stage doubles the peak input voltage and the second stage doubles it again, giving a DC output equal to four times the peak voltage value (4Vp) of the sinusoidal input signal. Also, using large value capacitors will help to reduce the ripple voltage.

Then we have seen that **Voltage Multipliers** are simple circuits made from diodes and capacitors that can increase the input voltage by two, three, or four times and by cascading together individual half or full stage multipliers in series to apply the desired DC voltage to a given load without the need for a step-up transformer.

Voltage multiplier circuits are classified as voltage doubler’s, tripler’s, or quadrupler’s, etc, depending on the ratio of the output voltage to the input voltage. In theory any desired amount of voltage multiplication can be obtained and a cascade of “N” doublers, would produce an output voltage of 2N.Vp volts.

For example, a 10-stage voltage multiplier circuit with a peak input voltage of 100 volts would give a DC output voltage of about 1,000 volts or 1kV, assuming no losses, without the use of a transformer.

However, the diodes and capacitors used in all multiplication circuits need to have a minimum reverse breakdown voltage rating of at least twice the peak voltage across them as multi-stage voltage multiplication circuits can produce very high voltages, so take care. Also, voltage multipliers usually supply low currents to a high-resistance loads as the output voltage quickly drops away as the load current increases.

The **Voltage Multiplication Circuits** shown above, are all designed to give a positive DC output voltage. But they can also be designed to give negative voltage outputs by simply reversing the polarities of all the multiplier diodes and capacitors to produce a negative voltage doubler.

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Thanks for your innovative ideals.I want to design a 220-240/50Hz VAC to 10000VDC convert.Can you please suggest the best design topology for this.

please let me know how to select the component values based on load, load regulation required etc. Can you please give exactly diode name and numbers. Together with voltage capacity with capacitor uF.

Regards,

R.P.Singh

which type of capacitor and diode and their rating are used for the 8 stage cockroft walton voltage multiplier circuit?????????????????

What if you have DC input? How do you get something like a square wave or sine wave to input to the doubler?

Can you provide the mathematical proof of all the voltage multiplier.

I NEED YOUR HELP TO DRAW USING MATLAB FOR VOLTAGE DOUBLER CIRCUIT WITH INPUT AC.

THANK YOU

I want dc-dc voltage doubler

Please send to me the circuit to my email

Thank you

thank you so much for good explaining

Hello.

There’s one important thing you probably have to make clear about this cct, which will help ppl understand why it may not work in their setup:

Generally, ppl connect the output of this cct to some resistance (or other load) that DRAWS CURRENT. Hence, the AC voltage source that feeds the input, should be capable of “sourcing” the required amount of current. To make things more complicated, I’ll say that it also should be capable of “sinking” the same amount of current during the negative half-cycle, for this cct to work properly.

The examples of AC voltage sources that are NOT good for the purpose:

– Output of say 7805 voltage regulator, chopped with the P-MOSFET (can not sink current)

– Open-collector or open-drain output (can not source current)

– Oscillating general output pin of the microcontroller (current supplying capability is generally too low)

– Output of the signal generator (current supplying capability is generally too low)

The examples of AC voltage sources that ARE good for the purpose:

– Push-pull pair capable of supplying the required load current (actually, twice the required load current, as the output capacitor is only charged during the positive half-cycle)

– High load capacity output pin of the microcontroller (this is normally stated in the datasheet of the uC, and such pins are used mainly for direct LED drive)

Lovely share!!

Thanks for providing informative stuff especially the voltage tripler which is generally used in television receiver to supply high voltage!!

Keep sharing!!

I have built this circuit with the same diode and capacitor value but still not working. Even if I connect it directly to a transformer or through an astable multivariator circuit. Any help!

There are only two diodes and two capacitors, so if you have connected them correctly you should see double the supply voltage, (minus two diode volt drops) at the output, 2Vrms. Try using capacitors over 1uF in value, the higher the better.