An Article from Blaze Labs
One of the cheapest and popular ways of generating high voltages at relatively low currents is the classic multistage diode/capacitor voltage multiplier, known as Cockcroft Walton multiplier, named after the two men who used this circuit design to be the first to succeed in performing the first nuclear disintegration in 1932. James Douglas Cockcroft and Ernest Thomas Sinton Walton, in fact have used this voltage multiplier cascade for the research which later made them winners of the 1951 Nobel Prize in physics for "Transmutation of atomic nuclei by artificially accelerated atomic particles". Less known is the fact that the circuit was first discovered much earlier, in 1919, by Heinrich Greinacher, a Swiss physicist. For this reason, this doubler cascade is sometimes also referred to as the Greinacher multiplier.
Unlike transformers this method eliminates the requirement for the heavy core and the bulk of insulation/potting required. By using only capacitors and diodes, these voltage multipliers can step up relatively low voltages to extremely high values, while at the same time being far lighter and cheaper than transformers. The biggest advantage of such circuit is that the voltage across each stage of this cascade, is only equal to twice the peak input voltage, so it has the advantage of requiring relatively low cost components and being easy to insulate. One can also tap the output from any stage, like a multitapped transformer. They have various practical applications and find their way in laser systems, CRT tubes, hv power supplies, LCD backlighting, power supplies, x-ray systems, traveling wave tubes, ion pumps, electrostatic systems, air ionisers, particle accelerators, copy machines, scientific instrumentation, oscilloscopes, and many other applications that utilize high voltage DC.

The polarity of the high voltage output is simply determined by the configuration of the diodes, as shown in the diagrams above.
The Cockcroft Walton or Greinacher design is based on the Half-Wave Series Multiplier, or voltage doubler. In fact, all multiplier circuits can be derived from its operating principles.
It mainly consists of a high voltage transformer Ts, a column of smoothing capacitors (C2,C4), a column of coupling capacitors (C1,C3), and a series connection of rectifiers(D1,D2,D3,D4).
The following description for the 2 stage CW multiplier, assumes no losses and represents sequential reversals of polarity of the source transformer Ts in the figure shown below.
The number of stages is equal to the number of smoothing capacitors between ground and OUT, which in this case are capacitors C2 and C4.
- Ts=Negative Peak:C1 charges through D1 to Epk at current ID1
- Ts=Positive Peak:Epk of Ts adds arithmetically to existing potential C1, thus C2 charges to 2Epk through D2 at current ID2
- Ts=Negative Peak:C3 is charged to Epk through D3 at current ID3
- Ts=Positive Peak:C4 is charged to 2Epk through D4 at current ID4.
Output is then 2n*Epk where N = number of stages.

Several cycles are required to reach full voltage. The output voltage closely follows the curve of an RC network as shown above. R is the output impedance of the ac source, whilst C is the effective dynamic capacitance of the CW multiplier. This charging occurs only upon switching on the CW multiplier from a discharged state, and does not repeat itself unless the output is short circuited. Most common input AC waveforms are sine waves and square waves.
Three stage CW multipliers, commonly known as tripler, were used in most of the early B&W and colour TV's. The voltage drops rapidly as a function of the output current. In some applications, this is an advantage. The output V/I characteristic is roughly hyperbolic, so it serves well for charging capacitor banks to high voltages at roughly constant charging power. Furthermore, the ripple on the output, particularly at high loads, is quite high.
Increasing the frequency can dramatically reduce the ripple, and the voltage drop under load, which accounts for the popularity of driving a multiplier stack with a switching power supply. A clever way to reduce ripple is to implement a full wave voltage doubler as shown below. This effectively doubles the number of charging cycles per second, and thus cuts down the voltage drop and ripple factor. The input is usually fed from a centre tapped ac transformer or MOSFET H-bridge circuits.
You can find more detailed information regarding this subject by visiting Blaze Labs
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A voltage multiplier is a great way to create a high voltage DC power supply. It is very easy to generate high voltages from easily available components. The next section contains information on where to buy the components and how to connect them. It also gives details of sources of high voltage power supplys which will run from batteries.
This information is for educational purposes. Replicating this project is done so entirely at your own risk. More DIY Voltage Multiplier





The information provided here can not be guaranteed as accurate or correct. Always check with an alternate source before following any suggestions made here.
*urgent...reply soon*
There are many of these circuits available used for powering flourecent tubes or cold cathodes.
Each stage of the voltage multiplier will theoretically double the input voltage. A nice Small inverter that will fun on 9VDC is the cold cathode power supply shown in the High voltage section of our shop. this outputs about 800V from a 9V battery, so adding three multiplyer stages would give you about 2400V.
If you have further questions, you can use the forum, as it will send you an email when you recieve a reply.
You can also find more information in the DIY Voltage Multiplier section.
If you would like to post larger images, please use the forum.
I could guess that either the components are not connected correctly or that they have been damaged.
If you are connecting this multiplier directly to the mains then it is posible that your diodes are blown. Normally something such as an isolation transformer would used to limit the current flowing from the mains.
Without any current limiting its possible that the diodes were exposed to a surge current beyond what they can tollerate. This happens because the uncharged capacitors will briefly draw a large current when the device is first connected to the mains supply.
Thank you!
I am designing a voltage multiplier circuit. I started off by building a 2-stage half wave voltage multiplier circuit using NTE517 (5kV) diodes and 1000pF (15kV) capacitors. I connect the circuit to a variable transformer and tuned the variable transformer to 10Vac. My result for my 1st stage is 5.57Vdc and 2nd stage is 3.72Vdc. Why is it that the output voltage is decreased? Are the types of diodes and capacitors used unsuitable? Could you recommend me a suitable type and value of diodes and capacitors?
Thank you.
There is a response to your enquiry on the DIY Voltage Multiplyer Page.
I am woundering how would be posible to build a voltage divider with diods and capacitors or may be with vacuum tubes.?
Thanx
alhege@hotmail.com
You can learn more about a voltage divider in the electronics section.
Could I use something as simple as this, attached to the mains (240V) to get the voltage I require? (say, 5-7kV). I'm trying to do it as simply as possible.
Great site, by the way, just discovered it today. I'll be back!
-mu
You could either use a step up transformer so you would need less multiplier stages, or you could just use a small isolation transformer to limit the current available.
Whole idea of using multiplier is to avoid xformer!
I am designing multistage DC generator to produce 10kV,1mA,and ripple factor of 5%,my questions are how do u come with the values of capacitors,the output from the transformer is 1kv,since the output for every stage is 2nV,does it mean i need to have 5 stages?
For an ideal voltage multiplier...
V(out) = 2 x NumberOfStages x PeakInputVoltage
..but this does not account for the voltage drop across each diode, so you could end up with much less. The output voltage will also drop proprtionaly to the load on it.
You can find more detailed info and calculations on this page.
Can someone tell me if there is a good way to measure the current from a half wave CW multiplier? the output is about 30kV and I want to measure the current to regulate an SG3524 chip.
Best regards,
Erwin
The output Voltage of a multiplier drops off quickly with current drawn. If your load is constant then you coul possibly use a voltage divider and a few sums.
Mike:
It depends on the components you use to make it. Use big capacitors for bigger current.
i want to build negative ion generator
i only have little knowledge in electricity can you make an instruction how to make a 12v dc negative ion generator.. i want the many -ve ions produced... my porpuse is to make an pika shoes found in this link (http://afrotechmods.com/cheap/negativeiongenerator/pikashoe7.htm)
i hope you will reply...thanks
It consists of an inverter and a votage multiplier like shown on the DIY voltage multiplier page.
That project just looks like a bad idea as you are likley to hurt yourself or others.
im very sorry that i have so manyy question because i'm not expert in electrical...
thanks....
it is ok to make an -ve ion generator without an inverter just voltage multiplier, 12v battery and a neddle?
The current is limited by the impedance of the capacitors at the frequency of the input voltage. If you know the impedance, you can use ohms law to calculate current.
I'm looking at using a BXA 12553 inverter for a HV power supply. The datasheet says output voltage is min 600 V to max 800 V, and output current min 4 mA to max 7 mA. What happens if I try and draw > 7mA (the effective short cct to ground across the uncharged capacitors -> current surge)? And what about when the current draw drops < 4mA during charging? Will the output keep a steady voltage (about 700V) below the rated 4mA min, and can it cope with > 7mA?
You should direct your query to the manufacturer of the device for further information.
the 1st question : its inductance is 101.6 milliH is this value enough ?
the 2nd :i use i caps beside it as impeders ,& cuz of this am facing a paradox in calculating intensity in the whole circuit and inside the inductor.the calc. current in the circuit is 0.4A while it's calculated to be 0.002A inside the inductor , how could this be ?
The last thing you said is not a proper sentence. If your 1st language is English then you need to try harder. If not then I guess it's fair enough, but I can only answer if you make sense.
and is 0.1016 of a Henry enough
by the way am british
The calculations you need go beyond the scope of this article on voltage multipliers i'm afraid.
Simplistically you can calculate the reactance of your inductor by using the formula X = 2pifL.
This is 2 times pi (3.142) times the frequency times the inductance of the coil. This page may help.
how can I best feed a CW stack, 10kv would be nice. Not much current needed only to cope with leakage. Funds limited!
You should be getting more than 10kV from one of those coils with 18V input. Maybe your driver needs adjusting.
You might be able to feed a CW with two ignition coils in anti-parralel. This would give you a better AC waveform. See comments in the ignition coil driver page for info.
The load impedance 'seen' by the transformer varies with time. When the tank capacitor is empty, the load at that instant will be very high. As the capacitor charges the load will decrease and the voltage will rise. Calculate load impedance of a capacitor (reactance) using 1 / (2*pi*f*C).
Thank you
I have several large old Zenith flybacks with 4,200 turns of AWG # 42 on the secondaries. Should produce 10-15 kV.
Regards,
D.C. Cox
tks.
I've mad a negative ionizer, but it's power is too low, i mean the quantity of air in movement is too low, i want to increase that quantity, any advice ?
I replaced the rectifiers 3 times and each time one or more of them shorts out after between 1 and 3 discharges. The caps are all fine.
I am assuming that when the cap discharges, the current is very high (>1A for short time). But since the rectifiers are reverse biased I did not think they would be affected. If anything, I would think that the caps might over heat.
Is it possible that the discharge of the output cap is causing the voltage to be unequally distributed across the 3 stages and therefore causing one or more of the rectifiers to breakdown?
Any ideas or suggestions?
Thanks and regards, Murray
Also all the diodes will have slight differences in timing due to manufacturing tolerances. This could mean that one diode takes most of the current or voltage for a brief moment causing it to blow. Try placing one of your 22M resistors in parallel with each diode to even out any voltage differences.
Thank you!
I would like to build DC to DC
converter with input 12V and output 2KV. Any circuit or website that i can refer to?
Thank you and have a nice day.
Best Regards
James Lee
With the NST centre tapped to earth will the output be fully negative or half the voltage below zero.
What you need to do is to decrease the resistance of your cell by increasing the plate area, adding more plates, or bringing the plates closer together. Ideally your cell should draw too much power when connected directly to a battery. You would then control the gas production rate with a pulse width modulation circuit.
As i have one circuit of mosquito zapper input uses 2AA cells. output 1300V Dc . i am having the problem of
the transformer failure or transistor failure. Is there any protection for transformer & transistor . Kindly provide me the details with circuit. AS in our circuit uses only one transistor 965 & ferrite transformer.
I am thinking of making a battery powered SSTC (running off mosfets and things). Here is the link for the page:http://stevehv.4hv.org/SSTC6.htm
I want to run it off 12V 7A. I need 12V for the 555 Timer IC, and 50V to drive the Primary. Do you think i should use:
an inverter/DC pulse circuit to drive a transformer
or
a voltage multiplier (for which i would need an inverter)
Would using these methods make the current too low for use?
Thanks :)
I want to know if voltage multiplier make use of Transformer? if yes,what type of transformer does it use?and what should be the primary turns and secondary turns of the transformer?
Thanks and God bless you.
I need to design the high voltage diodes used in this multiplier. How can you design the diode which passes 50Kv voltages in it.
I have seen some ultra fast diodes website but i am not really understand it. Do i need to use seires of diodes to genrate one diode?
Thanks
Vivek
You would loose too much power in the conversion process. Anyway, if you step up the voltage, the current goes down proportionally, therefore it will not help you.
mike Okoh,
No, it does not need a transformer, just an AC input.
Vivek,
You can put diodes in series, but it can be unreliable as they wont always switch at exactly the same time. To avoid this place a high ohm resistor in parallel with each diode.
thanks for all your help re building btw !!
i am making a ion generator with cockroft walton voltage multiplier by using 22n/630V cap. and in4007 diodes...but my circuit is not working.how do i measure output? i am using 24 diodes and caps to generate -8100 volts? is it right? please help me...
my input voltage is 240 volts from the mains. when i measure the voltage from second stage i am getting 600volts but from third stage tge voltage is decreasing to 500 volts.
i want to ask that from the theoretical formula when we want to calculate the ripple voltage and also voltage drop on load the formula is
∆V_total=q/C n(n+1)/2=I_1/fC n(n+1)/2
δV_total=q/C (2/3 n^3+1/2 n^2-n/6)=I_1/fC (2/3 n^3+n^2/2-n/6)
i want to ask that the current, I use in this 2 formula is how to find out? got any formula use to find I? Or I is the specific given in design a circuit?
Thanks
Pulsing is not important for killing them, but it is usually a product of the way the voltage is stepped up.
Is the output positive or negative?
Is the input positive or negative?
Do i need to make 2 of them, 1 for the positive and 1 for the negative?
Help!
Thanks.
Input is AC.
Output is DC voltage between output and ground. Polarity depends on how you build it.
I intend using a 33W source (transformer) to power it. Do u think that will be sufficient to drive the circuit? and also if it is safe in the case of human contact with the grids?
Please give me any possible recommendations urgently. Thanx. Great site by the way! Really educative.
You will have to do the calculations yourself for how much current it will draw. I just don't have time to work it out for you.
SmeeAgain,
I don't know what you are asking me, if you are even asking anything.
Maybe you should post a photo of your own experiment and a schematic so I can understand what you are doing.
does the Dioda N4007 can be used?
i'm making a negative ionizer with 30 .01uF 1Kv caps and 30 1N4007 diodes, with input of 220VAC. I'm having an issue with low output. I have yet to measure the output directly, but I have measured the voltage over each diode. The first diode reads out as 220V but decreases after that, and the final diode reads out as 50V. I've rebuilt the circuit several times being extra careful with ensuring the connections are correct, but I still get the same results. There seems to be an output, but I feel that with 30 caps/diodes the output should be higher. Any ideas of whats going on and how I can fix this?
Thanks
I have build a ionizer from a ccf lamp and 4 stages of diodes and caps. As I didn´t have hv diodes I used 2 4007 in series on each stage. It seems ok, I can feel the ion wind at the end. Do you think that could be ok in the long term?
But, if I try to make a spark the ccf blows after some seconds. Could that problem be solved with a high resistor? what do you mean when you say that the resistor has to be in the input? is it before the ccf or is it between the ccf and the multiplier? should I use also the fuse? if the ccf lamp was 15w, will a .5 A fuse works, or I should use a minor one?
Yours is the only site I have found excelent answers. Thanks!
This will not work. The multiplier is charging capacitors and will smooth out your modulation signal significantly.
The CCF circuit either can't tollerate the voltage spke created when you make a spark or it can't tollerate the surge current when it tries to recharge the empty capacitors. Try using smaller capacitors and/or put a resistor between the CCF and the multiplier input.
I plan to build a 240VAC mains voltage supplied negative ion generator (To use for Oduor reduction). The circuit is as per your negative voltage multiplier with 30 stages(Output 14kV?) but with a couple of 10Mohm resitors to limit the touch potentail in line just before the DC outut pins. Could you possibly offer any advise on selecting the capacitor uF (I have seen circuits using .01uF and also circuits using .22uF Caps) (My understanding is that the higher value cap will allow higher current draw and offer more ripple smoothing but my question is will more this smoothing aid the Air Ionisers perormance?
Where R is the resistor value needed, V is your calculated peak output voltage, and I is the current you want to limit to.
When you use high ohm resistors in a hv circuit, you need to make sure that there is no possility of sparks just jumping over the surface of the resistors and thus bypassing any saftey precautions.
I have the SG-LCR2M resonate signal generator. Using the Blaze Resonate Multiplier calculator I find that 4.7 micro farad capacitors in 24 stages can accomplish this.
Question: Can I use a florescent Light balast with 120V input and 277V output as a power source to replace the SG-LRC2M and also reduce the number of stages of the multiplier.
thank you in advance.
I want to be able to tap from the multiplier at different stages though, using switches. For example, tapping out at stage 6 would isolate stages 7-10 and redirect the current to the output terminals, and I don't know quite what kind of (affordable) switches to use. If the switch is "inactive" (for example, simply connecting stage 6 to stage 7 and beyond), it's only exposed to the initial voltage/current, but if it's "active" (from stage 6 to output terminal), it has to be able to withstand the entire voltage produced by however many stages are being used. Any ideas?
If you are using a variac, there's no need to try to switch different voltages.
I can get small arcs (3mm - 4mm) when connecting the two AC connections of the Inverter but when i connect one of them to a ground connection the PSU cuts out.
Could this be the overload protection on the PSU or another issue?. The Inverter only puts out around 30ma so I am assuming that it should not be drawing too much current form the PSU.
I would recommend making and testing just one capacitor first so you can be sure it can tolerate the voltage you will apply to it.
Thanks again for your help and
Happy New Year!
Do you have any ideas how to insulate the plate connected to the voltage multiplier to stop the electron discharge but still keep the electric field intact and so usefull to accelerate electrons.
Thanks for any comments
Any gas in the tube will seriously impede the electron flow and will cause deflections that may cause them to miss the hole.
I run this from a 240 to 110v Ac 50ma converter. I intend to make a voltage multiplier using this as a power source purely for the hell of it. My question; will 10kv capacitors work with an 8.5kv input and if so which diodes should be used?
Could running the transformer at UK mains voltage damage it or other componebts that may be inside, the whole transformer is ecapsulated in a pitch like substance in a steel case so I cannot see any other components that may be hidden in there.
10kV cap will work with an 8.5kV input. Generally though it would be advisable to use components rated for twice the supply voltage.
Well, I want to make a high voltage supply to charge h/v capsacitors.I need to generate 40kV, at around 40mA. Is the Voltage multiplier good enough or capable for such an application?
What about pulsed power supplies?
My input will be the 230VAC 50Hz, from the mains.One phase.
Do you think a pulsed power supply will do better?
Also any idea on how I can have floating voltage with a multiplier(like +20 and -20, and not 40 to ground)?
for testing inverters i need a variable DC voltage from 100V up to 1000V, i will use a combined design which consists of first a up transformation followed by a CW multiplier, i was thinking about using the rectified 100 Hz ripple or producing PWM pulses, what are the (dis)advantages of both? thx
questions are:
1: if i build this style of volt amp what kind of and how many capacitors/diodes should i use that can take the power?
2:are there any problems with the components relative to the power?
3: any suggestions or advice?
Can I use the capacitor from an old Electric Fan that have a voltage rating of 400v and have a capacitance is 1.5uF to make this voltage multiplier?
I think there is enough information between this page and the DIY Multiplier page for you to work out what is needed.
It is important for you to understand what you are doing so that you can be fully aware of the dangers.
please link me to any person who will help me design voltage multiplier, using 360v/200uf and 6amps/600v diode.
This should produce 1,200volts of dc when 120vac is being put.
Thanks and God bless you.
Hello, I have constructed a voltage multiplier that does not work. I am using a microwave oven transformer to power the multiplier, and I have used 1N4007 diodes, 3 of them connected in series for a 3000v rating, and 3kv ceramic disc capacitors. I constructed 120 stages, but I have also tried just connecting 30 stages and still have not got anything to work.
Here is a picture of my multiplier. What am I doing wrong?
Thanks in advance.
I can't see were you are connecting your transformer or where your output terminal is.
Sorry, the previous photo shows the output end of the multiplier.
Here is another photo. In this photo the left side is the input side of the multiplier.
Is that a string of resistors from output to GND? What value are they? I would suggest first trying with just 1 or two stages to make sure that your components are not faulty. Placing diodes in series like this is not very reliable.