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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.

• T_s=Negative Peak:C₁ charges through D₁ to E_pk at current I_D1
• T_s=Positive Peak:E_pk of T_s adds arithmetically to existing potential C₁, thus C₂ charges to 2E_pk through D₂ at current I_D2
• T_s=Negative Peak:C₃ is charged to E_pk through D₃ at current I_D3
• T_s=Positive Peak:C₄ is charged to 2E_pk through D₄ at current I_D4.
  Output is then 2n*E_pk 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

A DIY

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

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Questions and Comments on the Voltage Multipliers

Neil	Wednesday, 12th July 2006 12:29pm - No.94
Hi! I wanted to know how many steps would i need to make a multiplier to convert 9V DC to about 2kV DC. what components would i require other than capacitors and diodes. *urgent...reply soon*
RMCybernetics	Wednesday, 12th July 2006 2:46pm - No.95
First of all you will need some sort of inverter. This will convert the 9V DC to a Higher voltage AC 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.
Oscar	Wednesday, 2nd August 2006 11:26pm - No.140
Hello! I've seen in the web voltage multipliers with for example 15 or 20 stages, so you can connect to 220 V and get about a few kV (ionizer for example). I tried with six stages to start with typical 1N4007 diodes and capacitors of 47 nF and 400 V, but in my case at the second stage the voltage starts to decrease and in the sixth stage I only have a hundred of volts. What's the problem? Can you help me? It's better to have a input with many volts or it doesn't matter for the voltage drop along the cascade.
Oscar	Wednesday, 2nd August 2006 11:27pm - No.141
The second photo.
RMCybernetics	Thursday, 3rd August 2006 1:45am - No.142
Hi, It is hard to say what is wrong with your device. I can't see from your photos how your circuit is configured. 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.
Oscar	Thursday, 3rd August 2006 4:21pm - No.145
Ohh! I think it works! I forgot to connect also the ground terminal. When I did that, I got a little spark (only 6 stages). Very good. Thank you!
Li Yun Lim	Tuesday, 15th August 2006 3:24pm - No.165
Dear Sir/Madam, 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.
RMCybernetics	Tuesday, 15th August 2006 7:14pm - No.168
Li Yun Lim, There is a response to your enquiry on the DIY Voltage Multiplyer Page.
Alex	Thursday, 21st September 2006 5:25am - No.306
Dear Sir, 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
RMCybernetics	Thursday, 21st September 2006 4:16pm - No.309
A voltage divider is usually made from just resistors. Sometimes capacitors are also used. Diodes and valves are not used unless you need to switch or rectify the current. You can learn more about a voltage divider in the electronics section.
Mr. Underperson	Friday, 6th October 2006 7:52am - No.380
I was thinking of making some Ozone, like this... http://www.emanator.demon.co.uk/bigclive/ozone.htm using two meshes at either side of a piece of glass. 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
RMCybernetics	Friday, 6th October 2006 2:11pm - No.381
You can make a multiplier run directly from the mains, but I would reccommend using a transformer inbetween the mains and the multiplier circuit. This would limit the current and thefore reduce the chance of seriously dangerous power levels on your circuit. 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.
John L. Winters	Sunday, 25th February 2007 5:09pm - No.812
I have built several ionizers from 4007 diodes and 4.7 uF electrolytics. It is not required to use isolation x former. Just limit current surge at mains. Also, fuse is good to include. Don't for get current limit at top too! Whole idea of using multiplier is to avoid xformer!
Abel	Monday, 5th March 2007 7:15pm - No.844
Hi there 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?
RMCybernetics	Monday, 5th March 2007 8:31pm - No.845
larger capacitors will provide more smoothing. 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.
Mante	Thursday, 22nd March 2007 4:07pm - No.954
Iam designing multistage DC generator to produce 10kV,1mA,and ripple factor of 5%,it has an input of 2kV.Ive already found out it has two stages. My question is why is it that the value of the first capacitance has to be twise the smoothing capacitance?
RMCybernetics	Thursday, 22nd March 2007 5:40pm - No.956
It doesn't have to be. Can you describe what you mean by smoothing capacitance and the 1st capacitance, I'm not sure I understand your question.
Bulle	Tuesday, 8th May 2007 8:38am - No.1233
Hi there, 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
Mike	Friday, 25th May 2007 3:03am - No.1336
How much current can a 10x full rectified voltage multiple circuit delivery on the final output stage?
RMCybernetics	Friday, 25th May 2007 1:15pm - No.1338
Bulle: 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.
chinwin	Wednesday, 25th July 2007 3:17pm - No.1503
hi 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
RMCybernetics	Wednesday, 25th July 2007 3:30pm - No.1504
A negative ion generator is just a high voltage DC supply with sharp points or a grating at the output. 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.
chinwin	Thursday, 26th July 2007 10:39am - No.1510
thanks....for me it is not bad idea because my porpuse is to perform magic... electrcity can give a nice effect....your site is great... if i will be famous in my place...you are one who help me...hehehehehe
chinwin	Thursday, 26th July 2007 11:12am - No.1511
what is an inverter? how can i do it? im very sorry that i have so manyy question because i'm not expert in electrical... thanks....
RMCybernetics	Thursday, 26th July 2007 1:43pm - No.1512
Google is your friend and Wikipedia has anwers to many basic questions.
chinwin	Friday, 27th July 2007 3:12pm - No.1514
hello again it is ok to make an -ve ion generator without an inverter just voltage multiplier, 12v battery and a neddle?
RMCybernetics	Friday, 27th July 2007 3:42pm - No.1515
No, this wouldn't work. You need an AC input for a voltage multiplier. You can buy tiny ones meant for powering cold cathode tube lights inside PC cases. They turn 12 VDC into about 1000V AC. You could then stick a couple of multiplier stages onto that.
koko	Friday, 17th August 2007 1:01am - No.1603
how can i grab current from the mains by homemade transformer safely for such a multiplier,and how could i limit the current by caps ,i mean wat impedence should give how much amps
RMCybernetics	Friday, 17th August 2007 10:06am - No.1604
You can't. Homemade transformer + mains = not safe. 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.
Kevin	Friday, 24th August 2007 7:16am - No.1650
Could I us my 15000 volt 30 millamp neon sign transformer for input power for a voltage multiplier circuit?
RMCybernetics	Friday, 24th August 2007 9:54am - No.1651
Yes. Any AC supply will work
joe	Monday, 27th August 2007 6:50am - No.1660
I want to try and build this generator so i went to RadioShack to get parts they don,t have any 15kv or 20kv capacitors and no high voltage diodes where do i get the parts to make this device?
RMCybernetics	Monday, 27th August 2007 11:12am - No.1664
We have some high voltage diodes. We often have the caps too, but not at this time. Maybe eBay will have them.
Tim Mappo	Monday, 3rd September 2007 9:02am - No.1689
Help with an inverter! 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?
RMCybernetics	Monday, 3rd September 2007 5:04pm - No.1690
There is usually a capacitor connected in series with the output of CCFL inverters which limits the current. It may also be worth adding a small resistor for protection. You should direct your query to the manufacturer of the device for further information.
moaza	Saturday, 20th October 2007 6:58pm - No.1996
i have made a variac on my own : 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 ?
moonrider	Wednesday, 24th October 2007 2:25pm - No.2034
would the value of the caps/diodes really matter for example i have 0.1uf caps and 4007 diodes. would this circuit drive a tesla coil? and im guessing that the more caps/ diodes you incorperate the more out put voltage and the more input voltage twice gain on the out.
RMCybernetics	Wednesday, 24th October 2007 2:39pm - No.2035
The value of the capacitors will determine the max current drawn from the PSU and the peak current when discharging the multiplier. 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.
moaza	Thursday, 25th October 2007 9:42am - No.2046
this is what am really facing but i've recently from a physicist that i should calculate the current in vector manner(he said when ac source is connected to a cap and inductor in series there is a whole different for calculating current i.e.phase must be put in consideration) ,how could i do that? and is 0.1016 of a Henry enough by the way am british
RMCybernetics	Thursday, 25th October 2007 2:56pm - No.2048
Sorry moaza, I was talking to moonrider about the poor English, yours is fine. 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.
moaza	Tuesday, 30th October 2007 11:42am - No.2078
great site very thankful...
Richard S	Wednesday, 12th December 2007 2:54pm - No.2329
I wish to build a 60kv supply for powder coating. I have 20 20kv diodes and 20 20kv caps. also one of your spark coils which produces about 10kv from a 555 circuit switching a 2n3055 with 18v but as this is pulsed dc it is no help? how can I best feed a CW stack, 10kv would be nice. Not much current needed only to cope with leakage. Funds limited!
RMCybernetics	Wednesday, 12th December 2007 6:49pm - No.2332
Ignition coils are not great for CW multipliers. The output is not DC but it is mostly biased in one direction and not sinusoudal unless is is driven at resonance with a capacitor in parallel with the input. 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.
Rudi	Thursday, 10th January 2008 1:32pm - No.2428
helo! first of all, great site! second of all, I'm not sure how u guys determine the rating of the capacitor and diodes used for voltage multiplier. In my case, Im using 2 fly swatter circuits, i put them in series i get a healthy approx 1400volts.. i want the voltage to be doubled, tripled and so on.. but how do i determine the rating of the diodes and capacitor to be used ? thx =)
RMCybernetics	Friday, 11th January 2008 7:50pm - No.2430
Your question has been answered already. See the comments on the DIY Voltage multiplier page.
Thomas	Wednesday, 13th February 2008 11:56pm - No.2521
Hello, I'd like to use this transformer as a base for a voltage multiplier. It say Sec: 9 mA 999V Sec Uo 3130V. I assume this means 3130V with no load? I made a voltage divider with 9 4M7 resistors and got 217 V between each resistor-thus 1953 V at that load. Is this correct? I am planning to use three of these to get about 6kV to drive a Tesla coil - will this work? Many thanks Thomas
RMcybernetics	Thursday, 14th February 2008 2:58pm - No.2522
Yep, all looks ok to me.
Thomas	Thursday, 14th February 2008 7:11pm - No.2524
What does the 999V refer to? The maximum voltage at full load? Is it possible to calculate the load of my tesla coil - I assume that if the load is too high, the voltage will drop too low and the coil will stop working - then start again as the load drops?
RMCybernetics	Thursday, 14th February 2008 11:09pm - No.2525
I would think that the 999V is typical RMS voltage at 9mA. 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).
Jo	Tuesday, 4th March 2008 3:55pm - No.2573
Hello, im attempting to make an amplitude modulator for an ultrasound signal. Im hoping to use a 60KHz carrier with a 600Hz signal. Ive got a 9V Vcc rail, but that can be changed. Is there any chance you could suggest some values for capacitors and diodes if i were to use this circuit? Thank you
RMCybernetics	Tuesday, 4th March 2008 9:52pm - No.2574
Sorry, no. Why do you want to use a voltage multiplier for your AM?
Barry	Sunday, 9th March 2008 6:47pm - No.2596
Could the voltage (and frequency) output from a generator be elevated using a voltage doubler array? Once brought to the appropriate level, then introduced to a capacitance storage of some type before it's introduced to a more traditional load? I think the capacitor would not "see" it as a significant load, allowing less loss of power. Possible?
RMCybernetics	Monday, 10th March 2008 11:18am - No.2601
The voltage would be increased, the current would be decreased. The frequency component would be lost as it is converted to DC in the process.
chek	Saturday, 5th April 2008 9:05am - No.2696
hi there, nice site...i had a question, i need to design a 15kV CW multiplier with 240 ac Voltage input..i used 10nF capacitors and 4007 diodes in 23 stages as calculated..as the current limitation, i use 10kohm resistor..but when i turn ON the circuit, the resistor blown...can u give any opinion why the resistor blown and any suggestion on how to overtake this problem?
RMCybernetics	Sunday, 6th April 2008 7:54pm - No.2698
Is the resistor on the input or output? If the resistor is overheating then I would guess you are using one not rated for enough power.
chek	Tuesday, 8th April 2008 3:31pm - No.2701
the resistor is on the input..i have change the resistor with 10kohm 10w resistor..thanks for your opinion...
chek	Monday, 14th April 2008 4:56am - No.2714
i have construct he CW multiplier with 23 stage...the input is 240V ac and my desired output is 15kV..there is some problem in selecting the load resistor...can you give any opinion?and what was the main consideration in selecting the resistor?is there any simulation program that i can use to simulate the CW multiplier?
RMCybernetics	Monday, 14th April 2008 11:52am - No.2716
You can find design info an calculationes here.
D.C. Cox	Monday, 12th May 2008 11:19pm - No.2758
did you ever use your pulse power modeulator to drive a flyback xmfr? 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
RMCybernetics	Tuesday, 20th May 2008 2:29pm - No.2771
Yes, they work well for this.
nahum	Wednesday, 21st May 2008 9:18pm - No.2776
hi first great site just found today. i try to do a voltage multiplier feeds from a H bridge generate VAC in the order of 12V@50A i need around of 60v@15A and i don't know how... if a multiplier work to get te current i need? and what would be the most suitable frequency and the value for the capacitors.(formulates) tks.
Nanda	Saturday, 31st May 2008 6:37pm - No.2793
Hi sir I am trying to use a half wave multiplier for a RF harvesting experiment with 6 stages and am using the 1N5822 diodes. In the experiment we are trying to recive a 6 MHz signal using the folded dipole antenna. Before connectng the antenna we could receive 0.6 volt peak to peak.but after we connect to our circuit the voltage dips. Is the multiplier shorting or antenna. Can u tell us how to calculatethe dynamic impedance of the multiplier circuit. We are using 10 uF capacitors at each stage. How can we match the impedance of our antenna to the multiplier cicuit??
Mina Fikri	Sunday, 1st June 2008 9:12pm - No.2796
Hi,, 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 ?
RMCybernetics	Monday, 2nd June 2008 11:30am - No.2797
More power / higher voltage / current. Optimized electorode design could help too
Murray	Saturday, 14th June 2008 3:32am - No.2826
Hi, I hope you can help me with this problem. I have one of those Tennis Racket electronic bug zappers that runs off 2 C Batteries. The circuit is fairly straight forward. There is a 2 transistor oscillator into a small transformer that produces approx 350 V on the output which then goes through 3 stages of conventional half wave rectification - two capacitors and 2 rectifiers per stage. The chain of 3 stages ends up producing 1500VDC across which is placed a 0.1 Uf cap and a 22M resistor as a load. It works fine for mosquitoes and very small flies but I needed a little more zapping power. So I wired a 0.5Uf 2KV capacitor across the 0.1 Uf cap. I get one or two nice zaps and then one or more of the 6 rectifiers goes short. (All of the caps are rated at 1KV except the output which are 2KV; all of the rectifiers are 1A 1000v 1N4007). 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
RMCybernetics	Monday, 16th June 2008 1:10am - No.2835
At the moment the output zaps, it is possible a voltage spike occurs which is larger than the rating of your diodes. 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.
Murray	Monday, 16th June 2008 4:13pm - No.2850
I tried the resistors but with no luck. I decided to double up the rectifiers to ensure 2KV breakdown. That solved the problem. Thanks & Best Regards, Murray
New guy	Friday, 18th July 2008 11:09am - No.2914
The input of the multiplers must be AC?Can i switch to DC input ? As one of my project input is DC12V and output required is 2KV.Or is there another circuit that can generate this kind of voltage? Thank you!
RMCybernetics	Friday, 18th July 2008 2:50pm - No.2915
Needs to be AC. You can buy a DC-DC converter fo those sorts of voltage quite cheaply.
New guy	Monday, 21st July 2008 2:09am - No.2923
Thank for the help. I would like to bulid my own DC to DC converter with that kind of output. Can you suggest any circuit or web site?
James Lee	Tuesday, 22nd July 2008 6:06am - No.2924
To whom may concern; 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
abe	Sunday, 27th July 2008 12:47am - No.2948
I want to use a neon sign transformer(15000 volt 30 ma) to feed a cockroft walton cascade to get a negative voltage. With the NST centre tapped to earth will the output be fully negative or half the voltage below zero.
TmK	Tuesday, 12th August 2008 3:07pm - No.2992
Hi, I wanted to know if i can just use a full wave rectifier with the AC transformer and hook the caps to it, instead of placing a diode with each cap? Thanks
RMCybernetics	Wednesday, 13th August 2008 2:56pm - No.2993
No, That would not increase the voltage.

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