| Dr T | Saturday, 29th April 2006 8:24pm - No.28 |
| I made one just like it using the cold cathode PSU. If i stood on a chair (so I was insulated from ground) I could hold the high voltage output without getting shocked. I could then point at things in my room and my finger tip would glow with loads of purple sparks!! Certain objects I pointed at would also glow too.
It didn't make my har stand on end but I could certainly feel it trying. Cool! |
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| Li Yun Lim | Tuesday, 15th August 2006 3:29pm - No.166 |
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.
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| RMCybernetics | Tuesday, 15th August 2006 7:12pm - No.167 |
 From you description I would guess that the problem is that you are using kV rated diodes with only a very low voltage AC input (10V). Most high voltage diodes will not work correctly unless you are applying at least 100V. The NTE517 diode is a high power component and is unsuitable for a low voltage circuit.
Also make sure you are measuring the stage voltages between you ground connection ant the correct place on the circuit like shown in this diagram.
Are you trying to step up a low voltage source to HV DC? If you could explain your objective, I may be able to give you some more advice.
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| Li Yun Lim | Thursday, 17th August 2006 10:03am - No.173 |
Dear Sir/Madam,
I would like to step up from 220V to 15kV. What types/models of capacitors and diodes would you recommend me to use to build a voltage multiplier?
Thank You. |
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| RMCybernetics | Thursday, 17th August 2006 4:28pm - No.174 |
The components you have mentioned should work OK if you are applying a high enough voltage to the input. If you want to be able to adjust your variac (variable transformer) from as low as 10V then it may be better to use diodes rated for lower voltages.
Your componets are only exposed to the stage voltage so if you are only applying 220V AC the components only need to be rated as such. An example diode might be a 1N4007 or something similar in the series. These are only small so if you need to pass higher currents you should select a larger equivalent. |
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| Li Yun Lim | Friday, 18th August 2006 8:57am - No.177 |
Dear Sir/Madam
But the maximum rms voltage IN4007 diode is 700V only and my output voltage is 15kV. Can I still use IN4007 diode? Will it affect or burn the diode? Can you recommend me the rating/type of capacitor to use if I want to get the output voltage up to 15kV?
Thank you. |
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| RMCybernetics | Friday, 18th August 2006 1:30pm - No.178 |
The high voltage output from a multiplier is accross all the components in series so you don't need your components rated for the output voltage as it is shared between them.
Your componets are only exposed to the STAGE VOLTAGE so if you are only applying 220V AC the components only need to be rated as such.
700V is more than enough since you are only applying a maximum of 220V AC to each component.
The top diode may show 15kV relative to the ground, but the the voltage across the diode between its two pins will be only the stage voltage.
The same logic also applies to your capacitors. The capacitors you have will probably work ok, but ones rated for lower voltages will have a larger capacitance for the same sized component. |
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| Li Yun Lim | Monday, 28th August 2006 7:16am - No.201 |
Dear Sir/Madam,
May I know is there any other way that I could measure the readings of my high voltage multiplier, other than using a voltage divider?
Thank you. |
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| RMCybernetics | Monday, 28th August 2006 12:38pm - No.202 |
A voltage divider is the most reliable method, and this is the method used in commercial HV meters.
There are other methods you could use but they will not give you a voltage reading, just a relative scale.
An 'gold leaf electrometer' could be made using thin foil instead of gold leaf. This would give an indication of relative charge.
You could also make a 'FET electrometer' but the by far the best way is to use a voltage divider. |
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| Chin Weng | Saturday, 16th September 2006 5:37pm - No.286 |
Dear Sir/Madam,
May I know what are the ways to step up a 12Vdc battery to 16Vdc, other than using a transformer?
Thank you. |
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| RMCybernetics | Saturday, 16th September 2006 6:35pm - No.287 |
| You can use a "Step Up Voltage Regulator" like the LM2577T, but these still usually need a small inductor (coil) and a capacitor coneected to it. |
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| Weng | Sunday, 17th September 2006 4:01am - No.289 |
What is the circuit diagram? How do I connect it?
Thank you. |
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| RMCybernetics | Sunday, 17th September 2006 5:45am - No.290 |
The datasheet for the LM2577T has several example circuits.
You can also see an example here.
You can also buy ready made units designed for use as an in car laptop charger
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| MatriXBandit | Saturday, 4th November 2006 10:29am - No.452 |
In doing research on voltage multipliers I've run across information that seems to be conflicting and I was hoping you could clear this up for me.
Here's a few references that seem to be in conflict:
From your site:
http://www.rmcybernetics.com/science/high_voltage/voltage_mult.htm
"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..."
From the forum here:
"Your componets are only exposed to the STAGE VOLTAGE so if you are only applying 220V AC the components only need to be rated as such. 700V is more than enough since you are only applying a maximum of 220V AC to each component."
From another site:
http://www.powerlabs.org/cascade.htm
"The output voltage (Eout) is nominally the twice the peak input voltage (Eac) multiplied by the number of stages"
From your forum:
"...this outputs about 800V from a 9V battery, so adding three multiplyer stages would give you about 2400V."
If your site is correct then V_out = V_in * num_stages
If the other site is correct, then V_out = V_in * 2 * num_stages
The other thing in question is whether the components need to be rated for just the input voltage, or twice the input voltage.
Specifically, what I'd like to do is to use a 10kv AC input source to power a full-wave multiplier cascade so the theoretical output voltage is 1 million volts. Obviously this is a large undertaking so I need to really know how many stages and what rating the components need to be.
Another question I have is that since there is a dramatic price difference between ~4kv capacitors and ~10kv capacitors, is it perfectly acceptable to substitute multiple capacitors in series for a larger rated capacitor? If so, does this also work with diodes?
Thank you very much for your help, please drop me an email matrixbandit(at)gmail.com if you post a reply to this. |
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| RMCybernetics | Saturday, 4th November 2006 12:56pm - No.453 |
 I can see why this could be confusing. The problem is that The input is AC and the output or the voltage across the capacitors is DC. The AC voltage and DC voltages are not directly comparable.
You should take the values as rough guides. It is always a good idea to use components that are rated higher than the voltage / current you are expecting to expose them to.
AC voltage is usually given as an RMS (root mean square) value, which is a sort of average over a full cycle. When using a half wave multiplier your capaciors only see half of the peak to peak voltage, and therfore do not necessarily need to be rated to a DC equivalent of the RMS input voltage.
The diodes on the other hand could be exposed to the full peak to peak voltage when AC cycle is of opposite polarity.
With the full wave multiplier all components should be rated to tollerate the full peak to peak voltage of the AC input.
Yes you can connect capacitors in series to increase the voltage tollerance, but the capacitance will be reduced.
You can also connect diodes in series to increase the voltage tollerance, but there can be problems caused by slight differences in recorvery times of each diode used. It would definatley be better to use adaquatley rated diodes. |
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| jaxz | Wednesday, 15th November 2006 11:21am - No.474 |
| hi, what type of oil is recommended to use to provide HV insulation for the voltage multiplier circuit? thanks! |
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| RMCybernetics | Wednesday, 15th November 2006 1:20pm - No.475 |
Mineral oil is usually used, as it is less flamable than others.
You can use other oils like cooking oil, but these can have lower ignition temperatures and, may also go moldy if exposed to open air for a long time. |
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| John | Wednesday, 13th December 2006 7:06am - No.591 |
| will the output of a 555-based transformer driver circuit be suitable for the input to a multiplier cascade? or is a true AC-sinewave needed? |
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| RMCybernetics | Wednesday, 13th December 2006 9:45am - No.592 |
A true AC sinewave is not neccesary, but it does need to be and AC input of some sort.
Pulsed DC won't work, but you can use a 555 circuit to generate AC. |
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| Marc | Tuesday, 19th December 2006 5:41am - No.610 |
will used motor oil work as insulation, do the caps need tp be rated for peak input voltage or output voltage, and how much does your average 120 volt flourscent light ballast generate?
Marc |
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| RMCybernetics | Tuesday, 19th December 2006 9:35pm - No.612 |
Used motor oil is not reccomended. The oil will bw full of carbon and metals particles which could reduce the insulating properties of the oil.
As stated above - "....all components should be rated to tollerate the full peak to peak voltage of the AC input."
There's no average flourescent light ballast as far as I'm aware. I think you are misunderstanding the purpose of them anyway. They are not used like transformers for changing voltage, but they simply limit the current to the flourescent tube. Therfore you put 120V in, then you get 120V out, but with a shifted phase. |
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| Thomas | Sunday, 31st December 2006 9:12pm - No.652 |
I am trying to build an ion generator using your design (and also refering to http://www.emanator.demon.co.uk/bigclive/ioniser.htm)
I used 26 1N4007 and 26 10 nF, 1kV capacitors plugged on the main (220V - and am very careful!) but only get a very very small spark (3 mm) between DC out and Ground (using 2 needles). I could also charge dust specks etc. Any reason this is not generating an "ion wind"? many many thanks - Thomas |
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| RMCybernetics | Monday, 1st January 2007 4:40am - No.655 |
It sounds like your device is working ok. With 13 stages and 220V AC input, the output should be around 2.5kV. This voltage will only cause sparks to jump a few mm at most.
The ion wind generated from a single needle will only be very small @ 2.5kV. Multiple needles will help, but only up to the point at which they draw the maximum output current of your multiplier.
The amount of ion wind generated is proportional to the current flowing through the needles. To increase the current flowing through a single needle, the voltage applied to it needs to be increased.
To increase the current available from a voltage multiplier, the capacitance of the capacitors needs to be increased.
If you increase the size of your capacitors significantly, it may be neccessary to use a larger diode than a 1n4007. Maybe a P600J would be able to handle some increase in current. |
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| Amit | Monday, 1st January 2007 11:53am - No.656 |
Could you please put some light on the capacitors to be used for this?
MLC or Polyester as both have high voltage ratings. Also, a Radial, axil and potted box types are available. Which gives better results?
I would like to use it for 220 V AC, 50 Hz.
I also take this oppurtunity to wish you and your near & dear ones a very happy new year and a successful too. |
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| RMCybernetics | Monday, 1st January 2007 5:57pm - No.658 |
Thanks, Happy new year. The type of capacitor dielectric or casing is not of great importance here. Those factors are only generally considered when using the caps in signal processing circuits or an accurate PSU.
I've used resin dipped ceramic types because they are compact, low cost, and come with a range of high voltage tollerences. |
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| Thomas | Monday, 1st January 2007 6:04pm - No.659 |
Thanks a lot! I thought the 1N4007 would allow 1A, are we likely to have more current than that? And how could we measure the current safely on this circuit?
- great site by the way, will try to build your Tesla coil as well! |
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| RMCybernetics | Monday, 1st January 2007 6:55pm - No.661 |
See the 1n4007 datasheet for more info.
EDIT:-----
If the capacitors used are too large then when the inpult votage is applied there will be a low impedance to the flow of current. If the polarity forward biases the diode it will cause a large voltage drop across the diode (because the diode and cap are basically shorting out the supply at that instant of time) thefore causing it to draw lots of current and heat up.
The 1n4007 will withstand 1A contiuously and more if it is only for brief moments.
-----
The current flowing into the circuit is limited mostly by the impedance (capcative reactance) of the capacitors. The AC input passes directly through a capacitor at the input end and you can calculate its reactance with the following...
X = 1 / (2 * pi * f * C)
Where X is Reactance in Ohms
pi is 3.142
f is frequency
C is capacitance
So at 50Hz with a 10nF capacitor there will be 318310 Ohms of impeadence to the flow of current.
Roughly speaking (ignoring RMS conversions etc) you can calculate the maximum curent flowing through a single 10nF capacitor using Ohms Law.
I = V / R
= 220 / 318310
= 0.69mA
Although this does not give you the total current flowing through your diodes, it will hopefully give you an idea of how the scale of the different variables will effect some of the elements in the circuit.
The simplest way to measure the current is using a standard multimeter. with the mutimeter in current mode, its leads can be connected in series with a diode to see how much current flows through it. Obviously all power should be switched off and all capacitors dischraged before connecting or adjusting the multimeter. |
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| Susan | Wednesday, 17th January 2007 7:00am - No.702 |
| How do I calculate the total current? thanks |
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| RMCybernetics | Wednesday, 17th January 2007 10:31am - No.703 |
| The input and output current will vary with time and loading conditions. It can be difficult to calculat and it depends upon a lot of factors. The info above should help and you can also find more complex calculations on the Blaze Labs website. |
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| Debs | Thursday, 1st February 2007 3:50pm - No.730 |
Hello,
I was interested in building a Voltage multiplier for a power supply, that will have roughly 208 VAC going in, and need 800 VAC going out. So I'm guessing I will need a Voltage Quadrupler.
Is there a schmatic diagram available for this? And what kind/how many of Diodes and Capacitors will I need?
Thanks |
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| RMCybernetics | Thursday, 1st February 2007 4:59pm - No.731 |
A voltage multiplier / quadrupler wil give you a DC output, so I guess it will not be suitable for your application. You can learn more on the voltage multipliers page.
A simple transformer would be the common (and more efficient) approach to AC-AC voltage conversion. |
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| Santa | Wednesday, 7th February 2007 10:41pm - No.747 |
 Well that was a fun project. Thankyou. |
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| P.G. | Saturday, 31st March 2007 6:26pm - No.1027 |
| There is a very high voltage output and the circuit is simple but effective! |
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| Paul | Tuesday, 17th April 2007 8:40pm - No.1179 |
hello.
i've been searching the internet for days for proper instruction for making an ionizer...
and then i came across your website.
could you please tell me what kind of capacitors and diods i need to create a voltage multiplier with 12VDC input and 15kV negative output.
i'd really appreciate if you could draw a scheme for me on how it should be connected.
Thank You. |
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| RMCybernetics | Tuesday, 17th April 2007 10:18pm - No.1181 |
The schematics and info on components are allready on this page. Have a look through the previous comments and links.
Basically you need some sort of inverter to convert 12V DC to an AC voltage. This ac voltage is then fed to the voltage multiplier as shown in the diagrams. |
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| Bulle | Thursday, 26th April 2007 4:02pm - No.1204 |
Hi,
Is there a way to measure current at the input of the CW-multiplier? I have seen this method at high voltage power supplies. Does anyone know how this is done? |
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| RMCybernetics | Monday, 30th April 2007 12:02am - No.1209 |
You just place an ammeter in series with input power source.
If the voltage is too high vor your ammeter, the alternative is to place a low value resistor in series, measure the voltage drop accross it, and calculate the current using ohms law. |
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| drBuddy | Saturday, 2nd June 2007 5:29am - No.1361 |
Hi.
I enjoy this site and am enjoying learning about voltage multipliers. Your explanations and examples have helped me understand the multiplier itself, but what I'm having a problem with is understanding it's connection with the mains, on the source end, and the load on the output end.
For example, I have been studying all the pics and descriptions I can find, including the info at:
http://www.emanator.demon.co.uk/bigclive/ozone.htm
http://www.emanator.demon.co.uk/bigclive/ioniser.htm, but I don't understand if the hot and neutral and ground are connected to the multiplier, how does the tubes in the ozone generator example, get supplied from the mains neutral, as stated?
Also, if there is only one output from the DC end of the multiplier, and assuming that is neg. voltage, where does the positive voltage come from (i.e.: how is the circuit completed for anything other than an ionizer)?
I'm asking because I am interested in making my own ozone generator by using a mains powered voltage multiplier instead of a transformer, and it will be similar to the one here:
http://www.emanator.demon.co.uk/bigclive/ozone.htm
Thank you for this website and all the info.
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| RMCybernetics | Friday, 15th June 2007 2:56pm - No.1396 |
The output circuit is completed by the ionized air. Once ionized, the air mollecules become mobile charge carriers much like the electrons in a piece of wire.
Most of the ions are quickly neutralized by other air mollecules therefore you don't often see then reach the ground. |
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| Neomar | Thursday, 21st June 2007 1:34pm - No.1417 |
| Does anybody know how to calculate the rms current of the diodes and capacitors on a Voltage Multiplier circuit? (dee6neg@joinville.udesc.br) |
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| chinwin | Wednesday, 25th July 2007 3:32pm - No.1505 |
hello can someone give me a simplier instruction to build an strong 12v dc negative ion generator
i want to build pika shoe found at this link
(http://afrotechmods.com/cheap/negativeiongenerator/pikashoe7.htm) |
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| RMCybernetics | Wednesday, 25th July 2007 3:37pm - No.1506 |
This is as simple as it gets. Apart from buying a ready made one!
Don't make "strong" HV device and connect it to you body. You will probably die or be injured like this guy |
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| chinwin | Saturday, 28th July 2007 6:02am - No.1516 |
| my diode is only 100v...what shall i do? |
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| RMCybernetics | Saturday, 28th July 2007 2:54pm - No.1518 |
| You will need to buy some High Voltage Diodes |
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| highvoltage | Saturday, 4th August 2007 4:12am - No.1530 |
what is the uf on the compasitors?
(sorry for the misspelling) |
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| RMCybernetics | Saturday, 4th August 2007 7:21pm - No.1535 |
| The capacitance value used just depends upon how much current you want it to deliver. Smaller capacitance means smaller current. |
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| Mad scientist | Thursday, 16th August 2007 4:58am - No.1599 |
| Can you tell me some(part number) of the diodes that can handle about 15-20kV? I want to use this multiplier circuit with 15kV input. |
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| chinwin | Thursday, 16th August 2007 2:57pm - No.1600 |
| hi!!i want to make a pika shoe how many diode should i use??? |
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| RMCybernetics | Thursday, 16th August 2007 3:29pm - No.1601 |
2 diodes and 2 capacitors make a single stage of a half wave multiplier. The output voltage (very roughly) is proportional to the input voltage x the number of stages.
If you use a cold cathode inverter which converts 9-12V DC into about 1kV AC, 10 stages will give you about 10kV DC at the output.
It is important to note that it is the current that kills and not the voltage of an electric shock. The capacitors used in a voltage multiplier store energy and can therefore release large pulses of current. Smaller capacitors are cheaper and safer. It would also be wise to connect a current limiting resistor on the output (about 10M ohms). |
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| RMCybernetics | Friday, 17th August 2007 10:11am - No.1605 |
Some HV diode numbers:
BY8418 - 22kV
BY8414 - 17kV
We also have some 20kV Diodes available |
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| chinwin | Saturday, 18th August 2007 3:31pm - No.1618 |
| thanks...so you mean in my project i will use HV diode and low voltage capacitor |
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| RMCybernetics | Saturday, 18th August 2007 6:45pm - No.1619 |
| No. All the components need to be rated to at least the peak input voltage. If you put 1kV AC in, all the capacitors and the diodes need to be rated for at least 1kV. |
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| chinwin | Tuesday, 21st August 2007 3:46pm - No.1633 |
| thanks for your reply....how could i calculate the negative ion output?....i decided to make 10 stages...(20 diodes and 20 capacitor)... my inverter is not for cold cathode but it still convert 12v to 220 v ac... |
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| RMCybernetics | Tuesday, 21st August 2007 9:38pm - No.1634 |
| The actual number of -ve ions would be hard to calculate. It would be a factor of voltage, current, air temperature, humidity, air flow, and the shape and size of the electrode which is in contact with the air. |
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| Peter | Tuesday, 21st August 2007 11:01pm - No.1635 |
Hi can you help me? I am desperately looking for advice. I would like to build high voltage power supply. The input is 220V AC (from 150peak - 300W 12V to 220V inerter)
I have lots of electrolytic capacitors 400V , 470uF. I am going to build following voltage multiplier, But I found it difficult to establish diodes rating, what sort and what rating of diodes will be the most suitable for that solution? (10Amp 600V???)...
I know only I have to use rectifying diodes rated twice the current, but how to find out what current will be in the following caps.?
1.1200V DC (3x400V Capacitors)
2. 2000V DC (5x400V Capacitors)
3. 2400V DC (6x400V Capacitors)
4.3200V DC (6x400V Capacitors)
.I would appropriate any advice
Please specify the diode ratings and type
Thank you
Regards |
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| chinwin | Wednesday, 22nd August 2007 2:26pm - No.1639 |
hi!!!
what is the number of diode and capacitor you recommend to use in my project about the ionizer that attach myself....what kind of needle i must use and how many needles i gonna use? thanks |
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| RMCybernetics | Wednesday, 22nd August 2007 2:57pm - No.1640 |
I'm afraid I can only recommend that you don't connect any high voltage device to your body. Especially a home made one.
Trying to charge your body will be quite lossy (charge will leak off into the surroundings) so this means that there will be a continuous flow of current through your body. It will only be small but it's probably not that good for you.
A sparse (widely spaced) wire mesh or just the stripped and spread end of stranded copper wire in front of the fan would be adequate as an ionizer.
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| chinwin | Wednesday, 22nd August 2007 3:32pm - No.1641 |
hello...i want to make pika shoe that found in this link
(http://afrotechmods.com/cheap/negativeiongenerator/pikashoe7.htm
and im gonna use a homemade ionizer as substitute for 12v dc negative ion generator and put an inverter as what you told lately
in this link the instruction for making the ionizer
(http://www.emanator.demon.co.uk/bigclive/ioniser.htm)
pls i am an magician electricity is a big help
what is the no. of capacitor and diode should i use...pls help me...12v dc negative ion generator is not available in our country so i decided to make a homamade one.... |
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| Peter | Thursday, 30th August 2007 10:42am - No.1676 |
Hi I have got quantity x10 capacitors 400V, 470uF.I would like to build full wave voltage multiplier.
The input is from 150W(max peak300W) Inverter 220V AC(usually gives 208V AC)
I would appriepriate your advice ,what type and ratings of rectifying diodes should i use for that project?
Thank you |
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| SailorThor | Thursday, 13th September 2007 11:07pm - No.1709 |
Regarding the quote below... There are some problems!!! The 1.1V specification is for a forward biased diode. Meaning the current is flowing through the diode the "proper" direction. And, the rating is 1A, which leads to your 1.1W of power dissipated in the diode, within the 2.5W maximum spec. All hunky-dory so far. HOWEVER, once you launch into the 220V stuff you are off a bit. Applying a 220V forward biasing signal will result in no more than 1.1V of forward drop across the diode, with the other 218.9V being dropped by whatever else is in the circuit. Even under these circumstances, the diode can pass up to 1A. Now, make it a reverse biasing 220V signal, and the diode will pass essentially zero (0) current, and will dissipate essentially no power. (There is always a leakage current, but it is quite tiny.) Your ohm's law analysis is flawed. The 220V signal is irrelevant, so long as it is less than the RMS specification of the diode, because the diode is only dissipating power when forward biased (and at that no more than 1.1 Watts), and when leaking reverse current (certainly far less than 1 amp). In fact, the reverse current with associated power dissipation is to be considered negligible from the thermal standpoint, so long as the reverse voltage does not exceed the specification on the datasheet.
To suggest that the 1A diode can only pass 0.0114 amps just because the signal applied to the overall circuit is 220V is ridiculous.
QUOTE:
1.1V @ 1A is eqivalent to 1.1W of power. The absoloute maximum power rating for the 1n4007 is 2.5W.
If you were to apply 220V @ 1A to the diode this would equal 220 Watts! Your diode would not lke that at all.
To see the max current you can use with a 1n4007 you can use Ohms Law. If P=2.5 and V=220 then the current would be 0.0114 Amps.
Cheers,
SailorThor |
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| RMCybernetics | Friday, 14th September 2007 2:13pm - No.1712 |
Thanks for pionting this out. I think this part was quite badly explained. The diode will drop 1.1V if the current is limited to 1A, but at the first instant if the diode is froward biased by the input, the majaority of the voltage will be dropped across the diode because it is essentially acting as a short circuit until the capacitor has charged up a little.
QUOTE:"To suggest that the 1A diode can only pass 0.0114 amps just because the signal applied to the overall circuit is 220V is ridiculous."
Of course. It should have been made clear that this was refering to the brief moment where the voltage appears between the diodes terminals when forward biased.The original post has been updated. |
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| Peter | Monday, 17th September 2007 11:28pm - No.1733 |
Hi I have got quantity x10 capacitors 400V, 470uF.I would like to build full wave voltage multiplier.
The input is from 150W(max peak300W) Inverter 220V AC(usually gives 208V AC)
is that correct way to find out?
I am not sure... i get Reactance for capacitors 470uf 400V
1/(2x3.1415x50x0.000470)=6.7725
from Ohms Law
I=32.49Amps???
P = 7260W???
What do you think?
I would appriepriate your advice ,what type and ratings of rectifying diodes should i use for that project?
Thank you |
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| RMCybernetics | Tuesday, 18th September 2007 10:20pm - No.1736 |
470 microfarads is pretty large. Your calcs show the current in one cap if it were placed across the supply. You would be charging 10 capacitors in parallel which would probably trip out your inverter. You would need to limit the current drawn from it by using a smaller capacitor in series.
A common diode for a multiplier would be the 1n4007 but you would need to limit current to under 1A. |
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| Peter | Wednesday, 19th September 2007 12:44pm - No.1738 |
| Is it mean that would not be possible to find use of my 400V 470microF Capacitors?So the biggest capacitors I can use are with less 15microF? Am I right?Then I get less than 1.03686Amp with 220V input?Is it matter what voltage of capacitors I will use what is the relationship of that with input voltage?If i am assuming correctly Voltage mutiplier connects capacitors in series not in parallel so the capacitance will be reduced significantly along C=1/C1+1/C2+1/C3...the same as Amps |
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| RMCybernetics | Wednesday, 19th September 2007 3:27pm - No.1740 |
Capacitors like these are usually used for creating a large current pulse.
The current you calculated would only be drawn for a brief instant because the capacitors will charge up and draw less and less current. This current would only be continuous if the output of your multiplier were shorted. You should use only very small capacitors for a multiplier, especially if you are not experienced with such devices. Nano or pico farads would be much safer.
The voltage rating of your capacitors must be at least that of your supply voltage (220V)
A multiplier charges the capacitors in parallel and discharges them in series. From the output pov, the capacitance is calculated as 1/Ctotal = 1/C1 + 1/C2 ..... From an input pov, the capacitance total is the combination of all added together. |
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| Peter | Tuesday, 25th September 2007 10:43pm - No.1755 |
Again the last question is how to select proper voltage of the diodes for voltage multiplier?
For Example if you goin to use:
220 input AC and Capacitors 400v 100uF,reactance is 32Amps so Max current will be 8 amps.
what is the max diode Rev. Voltage we are going to choose?is 600V enought?or do we have to count capacitors as series I am bit confused about your last reply ('A multiplier charges the capacitors in parallel and discharges them in series') and sum the total voltage of the multiplier? of course to find out proper nad safe Reverse Voltage of the Diodes? Thank you again
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| RMCybernetics | Wednesday, 26th September 2007 10:35am - No.1759 |
| See this message. Before building yourself a multiplier I suggest you draw or print out a schematic and try to visualise how the current would flow through the circuit. |
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| motash | Thursday, 25th October 2007 9:29am - No.2045 |
| if you use a square wave vin to your transformer do you get a square wave out? i guess it would run a multiplier as a square wave is like a half sine wave. |
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| RMCybernetics | Thursday, 25th October 2007 2:59pm - No.2049 |
| No. A voltage only appears on the output of a transformer when the input current is changing. A square wave input would give you a high frequency pulse on the output during every hi-low transition of the input waveform. |
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| joe random | Sunday, 25th November 2007 9:12am - No.2222 |
how well will this work if powered by a microwave oven transformer (est 6 KV)
and using 10 stages for a HV lifter but i am guessing that the caps and diodes will be a bit hard to get. |
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| RMCybernetics | Monday, 26th November 2007 2:23pm - No.2232 |
MOT is approx 1.5kV. The are incredibly dangerous because of high current output! I recommend not using one.
We have HV capacitors and diodes if you need them. |
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| Engr. natalie fatima morales | Wednesday, 12th March 2008 8:46am - No.2617 |
i'm very curious about the behavior of the capacitor and other parameters regarding the circuit.
Is the voltage drop across each capacitor varies at every stage?
i made 2 units of voltage multiplier. this is an air ionizer, One is using this components: 104J/630 V Mylar Capacitor, with a IN4007 diode.
The other unit has a capacitor rated 104K/630V Mylar capacitor with the same rating of diode as the first unit.
I'm really begging to you, will you help me for some computation proving for suitable rating of the components to be used? thanks! good day |
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| RMCybernetics | Wednesday, 12th March 2008 12:39pm - No.2619 |
| The voltage drops are the same at each stage. I can't suggest values for you. They depend upon what you want from it and what you power it with. Read through all te comments on the pages about voltage multipliers for some examples people have used. |
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| raz | Wednesday, 26th March 2008 10:38am - No.2648 |
hi... i am suppose to build a power supply using voltage multiplier
my task is to achieve around 1kV
but I am only getting around 500+V
so what i wanna ask is if my capacitor voltage rating is 50V, will it limit capability of voltage multiplier?
thx alot |
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| RMCybernetics | Wednesday, 26th March 2008 1:52pm - No.2653 |
| The 50V rating of your caps only limits the input voltage you can use. Add more stages to get more voltage. |
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| illuzion | Monday, 31st March 2008 4:36am - No.2669 |
Hi I am building a 80kV multiplier. I am looking for approx 1A output, thats right you heard me, 80kW. I plan on using the popular UX-FOB diodes, and I am looking for fastest rise time possible. How can I calcuate capacitor size required for this output? Also how can I calculate the rise time to 80kV based on various different combinations of input voltages to the multiplier & stage combinations. I also plan on running this faster than 60hz. Sure I could do it by trial and error, but I was hoping to save some time.
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Cyber Circuits
Surplus
Items
Custom
Electronics & Software
Educational
Resources 
Science
Lab 
Using
a voltage multiplier is a great way to make a high voltage DC power supply.
It is very easy to generate high voltages from easily available components.
You
can see what high voltage static electricity from this device does to
a piece of one way window film in the violent
discharge experiments section. There are microscope images of the
aftermath and a video clip of the explosive action!
The
capacitors and diodes can be arranged in a variety of ways. The half wave
method is the easiest as it requires fewer components, but a full wave
circuit will perform better. If you just want to get one working as soon
as possible the the half wave method would be adequate. The circuit diagrams
below indicate how the components should be arranged.
For
safety and improved performance the voltage multiplier should be placed
in a protective casing, such as a PVC pipe filled with oil. The image
on the left shows two protruding screws used for the AC input connection,
and he other image shows polished coin used for the high voltage output.
By using Polymorph to
seal the ends of the pipe, it can be filled with oil to prevent corona
leakage from the internal connections. A more sturdy method would be to
fill the pipe with epoxy resin, but this may be difficult with compact
component arrangement.
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