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Need more voltage than your supply can deliver? This is the device to build.
Based upon the Villard doubler, cascaded in multiple stages, this circuit will multiply a small AC voltage into a huge potential.
With a 6000 VAC Oil Burner Transformer and just three stages, you can multiply the potential from 6000 to 51,000 VDC!
Also, the higher the voltage ratings on your components, the higher the input voltage can be and therefore, more bang for your buck.
One side benefit of a stack like this is you can tap off various voltages from stage to stage. In this example, three outputs of 17000, 34000, and 51000 volts are available. Nice!
Based upon the Villard doubler, cascaded in multiple stages, this circuit will multiply a small AC voltage into a huge potential.
With a 6000 VAC Oil Burner Transformer and just three stages, you can multiply the potential from 6000 to 51,000 VDC!
Also, the higher the voltage ratings on your components, the higher the input voltage can be and therefore, more bang for your buck.
One side benefit of a stack like this is you can tap off various voltages from stage to stage. In this example, three outputs of 17000, 34000, and 51000 volts are available. Nice!
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Signing UpStep 1The Theory
Each stage of this multiplier is a doubler circuit and is made up of (2) Capacitors and (2) Diodes.
On the positive half-cycle of the input, the capacitors charge in parallel to the peak value of the voltage presented through the forward biased diode. The other diode is reversed biased. Each cap charges to 8484 peak volts.
On the negative half-cycle of the input, the forward biased diode becomes reversed biased while the reversed biased diode becomes forward biased. This effectively connects the capacitors in series allowing them to discharge into the load at the output. The result is a doubled voltage presented to the load or 16968 peak volts.
Each connected stage adds its potential to the total output.
To calculate the expected voltage at the output with a given input and number of stages, plug the numbers into this formula:
Eout = (2 x Ein) x S x 1.414
Eout is the Output Voltage, Ein is the Input Voltage, and S is the number of stages in your design. I used a 6,000 VAC Oil Burner Transformer for my input and built 3 stages.
Eout = (2 x 6000) x 3 x 1.414
Eout = 12000 x 3 x 1.414
Eout = 36000 x 1.414
Eout = 50,904 volts
On the positive half-cycle of the input, the capacitors charge in parallel to the peak value of the voltage presented through the forward biased diode. The other diode is reversed biased. Each cap charges to 8484 peak volts.
On the negative half-cycle of the input, the forward biased diode becomes reversed biased while the reversed biased diode becomes forward biased. This effectively connects the capacitors in series allowing them to discharge into the load at the output. The result is a doubled voltage presented to the load or 16968 peak volts.
Each connected stage adds its potential to the total output.
To calculate the expected voltage at the output with a given input and number of stages, plug the numbers into this formula:
Eout = (2 x Ein) x S x 1.414
Eout is the Output Voltage, Ein is the Input Voltage, and S is the number of stages in your design. I used a 6,000 VAC Oil Burner Transformer for my input and built 3 stages.
Eout = (2 x 6000) x 3 x 1.414
Eout = 12000 x 3 x 1.414
Eout = 36000 x 1.414
Eout = 50,904 volts
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I think your answer is no because the 1n4007 does not have a fast enough switching speed. At high enough frequency like in the kHz range the diode would conduct backwards. Here a youtube link i saw this at. It starts at 3:02
http://www.youtube.com/watch?v=cyhzpFqXwdA&feature=endscreen&NR=1
I have a transformer, an OBT, with an RMS of 6000v, and youd think, hey! This is exactly what the design calls for! But Lo and behold, Im looking at it in my hands, and it says on the label: "PRI: 120V, 60Hz, 35VA SEC: 20Kvpk, 35mA, Midpt Grnd"
If the RMS is 6kV, how can the peak be 20kV? And can I still use this with 12kV rated diodes? Or should I get different ones rated for like, 25kV?
For the record, the OBT is a Beckett 51771U, A-type oil burner transformer, and everywhere Ive checked, its 6kV RMS
I emailed the company, and they told me that they use half the sine wave for the ratings, but I found out that isn't completely true.
20kV peak is actually a peak-to-peak value of the transformer using the full sine wave, negative peak to positive peak. The RMS value is found using only peak value, which uses only half the sine wave. So that leaves me with 10kV peak. If you go by burningsuntech's math (which I am inclined to believe), the peak value will be 30% higher than the RMS value, or the RMS is 60% of the peak value. 60% of 10kV happens to be 6kV. Problem solved!
I'm not an electrical engineer or anything, but I've had a bit of experience with high voltage as of late.
I think that you meant to say that you will be using an INPUT of about 200-600V, and if so, your 2kV-5kV capacitors should work. I assume that your INPUT power is in AC, because if its not, you're out of luck.
I believe that your capacitors will work because their voltage rating only needs to exceed the peak voltage of your INPUT (200-600V r.m.s. = 282-848V peak).
Unfortunately, for your very small input voltage, you will need A LOT of stages to get all the way to 20kV. If your input is 200V, you'll need about 35 stages! If your input is 600V, you will need 11 or 12 stages.
I finished my multiplier, and hooked it up to a 7,500 volt neon sign transformer. I used the output to power an ion "thruster," but the results were disappointing. Using the HV outout of an old CRT monitor, the thruster moves vigorously. Using the multiplier/NST combo, I got a little bit of thrust, hissing, and a stream of ions, but not enough to make the thrusters turn. Since the output of the CRT is around 25,000-27,000 volts, I suspect I am not getting as much from the multiplier. According to calculations, I should be getting 60,000 volts, but I don't think I am. Since I am using recycled diodes, maybe some of them are defective? I checked the doorknob caps and found them within spec. I did not check the diodes.
Paul
The (max) result would be output of 63,630 V DC?
thanks,
Paul
Give the man a kupie doll.
One change I would maKE in the build is to use parafin instead of oil in the multiplier case. Its easier to use and it is easier to seal the container. I left mine open with no problems and was able to get it to maximum voltage
You might consider using a higher amperage Variac as well. Mine was too close to tolerance with the transformer running at 2.25 amps and my variac was for 2.5 amps. A bit too tight.
Have fun.
RA
The input Ground goes to a real ground, like a pipe in the back yard?
The low voltage (relatively speaking) input is going to come from an NST--the NST has two output lines; which one do I use, and what do I do with the NST line I don't use? Just isolate it?
The output is + DC, correct?
Thanks,
Paul
The output of the high voltage section of a tv or monitor is too high. Typically, they output from 25,000 to 50,000 volts in the microamp range. The current won't be a problem. But the design of my HVM is for low inputs up to 10,000 volts. The project would fail on the first try.
To overcome the problem, you would need to change the values of the caps and diodes for it to withstand the much higher voltage of a tv or monitor output.
Raving Apache
what's the best way to discharge the capacitors after use?
Short them to ground.
RA
I have a 9030v, 30 ma neon sign transformer. I see the 20kv capacitors should be enough to handle the voltage, but the diodes may not be enough, particularly since the HV03-12 reverse voltage is 8400, possibly leading to a lot of smoke. I found some 15000v, .30 forward current, 100ns recovery time rectifiers (part number Z150UFG) at http://www.voltagemultipliers.com/html/selection_guide_hv_diodes.html. How do these look to build this for my transformer?
Thanks!
bs1500
Looks like they will work fine.
Good luck
RA
Here is a link that will explain it better:
http://en.wikipedia.org/wiki/Cockcroft%E2%80%93Walton_generator
You have to know that this is a PASSIVE device. There are no active components in it that cause a change in the voltage like a transistor or voltage amplifier would.
It depends on the input to do the switching for it; that is, supply a changing voltage to cause the device to work. DC does not change, therefore this device will not work on pure DC. However, pulsating DC does change and therefore this device will work on pulsating DC as well as AC. The output of a battery is pure DC. It will not make this device work.
Thats why I keep saying that the input can be AC or pulsating DC.
RA
1. Would a disposable camera flash be a good input (minus the capacitor)
2. what is the output like
Thanks.
The flash camera oscillator would work but there is a problem with the amount of current that it produces as well as with the frequency it operates at.
You would need to modify this multiplier by changing the caps to .01 mfd @ 10Kv and the diodes should be IN4007 which have a PIV of 1000 Volts.
The multiplier shown is for 60 Hz while the flash camera oscillator is around 15 Khz. That is a big difference and needs to be compensated for.
So leaving off the cap is a good idea. Btw - This device requires either AC or Pulsating DC to work. It will not work by pumping a voltage spike through it.
For output characteristics, go here: http://www.allaboutcircuits.com/vol_3/chpt_3/8.html
RA
I started with 125vac, and made a 6 stage multiplier.
(2 * 125) *6 * 1.414 = 2121
When I tested it I got 467vdc.
Or does that equation only work with the same components that you used?
When you get into your higher voltages, is there is there a way to check that voltage, instead of having to depend only on a math equation?
Thanks
Sorry for the late reply. A partial explanation of the CW multiplier can be found here: http://en.wikipedia.org/wiki/Cockcroft%E2%80%93Walton_generator .
A stage consists of two caps and two diodes. Therefore you would have 12 caps and diodes in your network.
Also, the lower the voltage, the higher the value of the caps in order to offset the impedance of the network at 60 Hz . Alternatively, you can raise the frequency of your input to compensate for the impedance.
As far as being able to measure the output of higher voltages, thats not possible without a high voltage probe or a bank of high voltage resistors calculated so you can use Ohm's Law to figure a more accurate reading. Otherwise, math will have to do. BTW - the output of this device is pulsating DC, therefore your meter will only read effective voltage, not peak voltage as the formula implies.
Thanks for the post.
RA