A Homemade Voltage Multiplier (HV DC PSU)
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.
This page contains information on where to buy the components and how to connect them. It also gives details of sources of mini high voltage power supplys (inverters) which will run from batteries.
WARNING: Very High Voltage Device!
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!
For efficiency a voltage multiplier should be powered from a source that is already a relatively high voltage. There are a variety of small battery operated high voltage power supplys available. Many lighting devices contain inverters for powering vacuum tubes such as, florescent lights, cold cathode lights and plasma globes. These types of devices usually run from 12V DC and can output voltages up to around 20kV AC.
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Mini cold cathode tube PSU - ~1kV
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Plasma Globe PSU - ~15kV
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The capacitors and diodes required for the multiplier can be purchased from our shop.
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.
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Component
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Max Voltage
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Source
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1 - 30kV
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1 - 30kV
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The schematics above will output a positive DC voltage relative to the ground (GND). If a negative output is required then the polarity of the diodes should be reversed. you can learn more about how a voltage multiplier works, by visiting the voltage multiplier page.
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.
Example Experiments
A homemade voltage multiplier is perfect for powering an EHD thruster (aka Lifter). An EHDT can be made from just aluminium foil, sticks, and fine wire. To learn how , see the ElectroHydroDynamic Thruster page.
Using freezer spray (used by plumbers) you can grow ice crystals on the HV output with interesting results.
For more Simple Experiments with static electricity see the Experiments Section
The information provided here can not be guaranteed as accurate or correct. Always check with an alternate source before following any suggestions made here.
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.
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.
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.
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.
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.
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.
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.
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.
May I know what are the ways to step up a 12Vdc battery to 16Vdc, other than using a transformer?
Thank you.
Thank you.
You can also see an example here.
You can also buy ready made units designed for use as an in car laptop charger
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.
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.
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.
Pulsed DC won't work, but you can use a 555 circuit to generate AC.
Marc
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.
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
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.
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.
I've used resin dipped ceramic types because they are compact, low cost, and come with a range of high voltage tollerences.
- great site by the way, will try to build your Tesla coil as well!
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.
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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.
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
A simple transformer would be the common (and more efficient) approach to AC-AC voltage conversion.
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.
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.
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?
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.
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.
Most of the ions are quickly neutralized by other air mollecules therefore you don't often see then reach the ground.
i want to build pika shoe found at this link
(http://afrotechmods.com/cheap/negativeiongenerator/pikashoe7.htm)
Don't make "strong" HV device and connect it to you body. You will probably die or be injured like this guy
(sorry for the misspelling)
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).
BY8418 - 22kV
BY8414 - 17kV
We also have some 20kV Diodes available
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
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
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.
(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....
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
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
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.
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
A common diode for a multiplier would be the 1n4007 but you would need to limit current to under 1A.
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.
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
and using 10 stages for a HV lifter but i am guessing that the caps and diodes will be a bit hard to get.
We have HV capacitors and diodes if you need them.
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
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
than a magnetic transformer, i wil designe it for 3 phase, i have a big 1950s signal generator and a collection of 29 old valve radios i shal send pics later
My plan is to drive the Greinacher (CW) Multiplier with as little power as possible so a CMOS 555 (astable 2kHz) will trigger a fast-response-time capacitor driven photo-MOSFET NAIS(AQV258) cathode switching the output of an EMCO G30CT (Center-Tapped 1500Vmax) DC to HVDC converter. My intention is not to discharge the circuit but rather to use the potential field it generates. For this reason I chose 470pf 15kV Caps and 1N4007 Rectifiers. Using your ball park calculations puts it at 8.9mA which should be okay. I am not sure how the resonant coil in series with the CW input would affect this.
My questions are:
Will I get by with a second AQV258 for the negative output of the G30CT or is an H-bridge the answer? I was thinking of a both a positive and a negative half wave multiplier for each of the G30CT outputs. Does this seem feasible?
i have confusion in selecting current rating of diodes in voltage multiplier 100 kv. i want output current of multiplier 100 micro amp. please help me.
the voltage over a multiplier vil drop if ther is to many steps, it can be corected by butting a resonator in top or middel of it
Thanks for answering my quetion,i still need your help.Please,can you send me the schematc diagram of voltage multipler,with the capacitors values?
Thanks and God bless you.
im about to build a 10 stage half wave multiplyer as an add-on for the plasma globe psu with possable switches at every 2nd stage for a varing output. would the high voltage capacitors (30kV) and Diodes (30kV) be able to take the power at the higher stages? would i need to worry much about cooling? what switching gear would you suggest? oh and what do you guys reacon the output ampage of the PSU?
thanks and a bonny christmas to all.
I would like to generate a high voltage ( 10kv) low Milli amperage device for producing an "Electret self charging capacitor". These devices need a constant hv low amperage device attached to them during production. Could you please advice on the number and values of components needed
If your plasma globe power supply is being damaged, it is possibly overheating due to being overloaded. You could place a 10M resistor on the output of the multiplier to limit the current.
I was wondering if it were possible to use a bunch of stacked A23 12v batteries (connected for more current), to power some sort of pulsed dc or frequency generator to step up the 12v to hv through an auto ignition coil in order to drive a halfwave series multiplier?
There are a few experiments Id like to preform and these experiments require odd ways of powering a multiplier hehe
"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."
Other site:(http://www.kronjaeger.com/hv/hv/src/mul/)
An n-stage cascade produces 2n x Up output voltage.
Please explain.
My circuit is identical to that one: a voltage quadrupler on the transformer's output, resulting in around 1500V DC (not bad when powered by 2 AA batteries). The capacitors are type 223 and I can't tell the diode type (1N4007's?).
I was thinking about adding an additional doubler, multiplying the transformer's output 6 times instead of 4 (there's no room inside the handle to add a transformer to multiply cap discharge by 100-500 times via a neon bulb/spark gap and I want to keep it simple, but effective), then using a 9V battery to power the circuit.
Questions:
Would I need to insulate the circuit (with silicone or so) to prevent sparks jumping between the diodes/capacitors on the circuit board (I tried adding a 9V battery from the start & created a big spark on the circuit board while charging up the capacitors)?
And how can I control the current at the output (need about 3 mA)?
And if I want a very rapid cycle at the output, I can simply use smaller caps for quicker charge/discharge times, right?
I want the device to work like a stungun, incapacitate, not kill a person.
The capacitors in the circuit are 2 type 2J333J and 2 type 2J223J. They are not all 4 identical.
I have 2 102Z/2kV capacitors. Can these be used for the additional stage in the multiplier circuit?
Smaller capacitors will give faster spark rate but each spark will be of lower current.
Regards,
Des
I am trying to find a transformer (auto ignition coil or other) that will step up my 12v from the pulsed dc circut to 6k volts for the multiplier. BUT it seems i cant find one for that low of output voltage haha. Any suggestions?
The easiest way would be to use a lower input voltage.
also do you ship overseas because I am in America and I dont know how the currency transfers lol
Another possibility would be to use a high frequency. If you use one higher than the coil is designed for, the inductance of the coil will limit the current and therfore reduce the output voltage.
Yes we ship worldwide. If you go through the checkout process, the conversion to USD will be shown before you have to commit to payment.
I need to purchase about 20 or so 15kv capacitors from ya for a 200k multiplier. Any idea on when you will get more? Also how large are these capacitors as in size not capacitance from the photo they look pretty small but I cant be sure...
is the ionizer can help healt care?
Is the ionizer hazardous to other electronics equipments?
Is the ionizer like your design applied or similar with " Dr JISM healt care equipment "?
Thanks
Many thanks for your web site.
Thanks,
Jack
When making sparks you will cause voltage spikes which can damage the dimmer circuit. You need to protect it using a MOV.
Rolando,
You should use a diode with a higher voltage rating. More information including calculations can be found here
Jack,
The voltage is between the output and ground. AC input, DC output. Please read the page linked above for more information.
I want to build a negative ion generator. I got a bunch of components from a friend. 1N4007 diodes and 1KV 0.1uF film capacitors. I'm thinking about building the full wave rectifier since it will require less stages to get the voltage up to a proper level.
There are a few aspects I'm not sure about:
I was thinking about taking a board, sticking nails into it and solder the components to that board since building this on a VERO board is probably a bad idea since the small gaps between the tracks will cause arcing at some stage.
Second thing, what resistor should I place to limit the currect? I know this is related to output voltage but how do I calculate that anyway? (I have no electrical background at this time)?
3rd, the amount of caps in this circuit means a big in-rush currnet, how should I limit these with a resistor at the input?
4th (and last). I would like to test this thing before connecting to 220V. I have a 65-0-65 transformer. Can I hook the two 65 outs to one side and the 0 to the other side and connect the input to the wall ground (since the transformer has 2 inputs). will that work?
Thanks and keep the good work on the site, great info here!
The resistor is calculated using ohms law:
R = V/I
If you are just making an ion generator then using smaller capacitors would be the best way to limit current.
Yes, you could use the transformer to test it.
(American socket voltage/frequency)
What does the capacitance of the caps effect? More capacity, more power or what?
I want to be cheap. Why pay more?
Do I need to limit current at entry?
What current will it consume so I can fuse it with the right fuse.
How do I know the current consumption of this device? So I can pick the fuse. Do I need to limit it with a resistor.
What does the capacitance of the capacitor effect? More uF, more what?
I want to be cheap so I don't want much overkill.
Thanks.
I am working on a project for fencing grazing animals by use of invisible fencing. GPS positioning is used. All necessary equipment is inside a box connected to a collar. When the animal is moving outside the boundary, a warning sound is emitted, followed by a electric shock.
This shock is why I am posting this question. I am using a DC voltage from a LiIon battery (~4v) this voltage is turned into 600v AC by use of a photoflash transformer. A voltage multiplier is increasing this voltage to about 10kV. This circuit will be powered in less than a second at a time.
Everything OK so far.
But do you think this circuit could be integrated into the same PCB as the other components? (microcontroller, GPS module and so on)
Or should this HV circuit be separated, and placed into a separate housing of some kind?
How to find the amperage rating of the cap?Is not clear for me.
I have reading somewhere ,to find this,
must conect the cap with an ampermeter in series at the power source to find how much current will flow through cap.
I tried this with an electrolitic cap ,and hocking it in one way at the dc power source,no reading on ampermeter ,in another way my cap has exploded .Conecting it at the Ac source ,I think the cap will blow up anyway
Here is their words;
"
These plans use the old Diode and Capacitor Method, Turn 120 vac. into 25,000 to 50,000 volts DC
earth shaking power! Amperage depends on what size capacitors you use. The higher the micro fara
the cap, the higher the amperage rating. Use an AC amperage meter on the input wire, once you find
how much amperage the cap is rated at, use diodes 2 x’s higher that rating, Example: 200 uf x 360 v
photoflash capacitors will use about 3 to 4 amps max, so you will want to use diodes rated at 6 to 8 a
400 to 450 volts. If you exceed the rating you can cause a fire hazard or a cap or caps can blow up!"
Do you know more about this metod (by conecting the cap at the power source directly?)
Maybe I missed something from that paper.Please give me more details,(to help me to understand), about how to find the current rating of an capacitor using the metod described above or/and another useful method .
Thank you
The idea is to using capacitors that I have around in my backyard workshop and not to buy others,and increase the output amperage....
Also,however, will be interesting to know if is possible .
For one stage of Voltage Multiplier is using 2 capacitors and 2 diode right? I am having a problem after it successful step up on first stage and second stage it decreased. my capacitors is 104(630V)adn diode is 1N4007. The input is using center-tapped transformer, and which provide 15VAC. Is it i am having same problem with mr. Li Yun Lim?? The diode cant work nicely because of the low input voltage??
Thank you.
i would like to ask again, hope sir/madam can reply me as soon as possible because it is urgent. For the last two post, u answer this "Also for the input you should be using the centre tap and one of the other outputs from the transformer."
what do you mean in this sentence? can you explain me again? thanks
And one more thing, the load is add to output for calculate or measure the output current? measure the voltage drop across the resistor, use the value before minus the the voltage after across and then calculate with ohm's law??
thx for replying
9mA2 x 220k = 18W.
If your resistor is not rated for enough power, it will melt.
Please see this page, for more such calculations.
and tested the output on an oscilloscope and it worked fine (approx. 7.1kHz), but when I connected a 2 stage voltage multiplier circuit to the output, there was only about 7V between the multiplier circuit output and ground. Are my capacitances too high for that frequency? Or could it be that my 9V battery that's powering the 555 timer is too weak a power source? Or is the 555 timer alone not enough for my plan? I am planning on using a more powerful battery later on by the way, but figured a 9V would suffice for testing for now. Any help you can give would be greatly appreciated :)
Are you accounting for the voltage drop rating of your diodes? Each diode will drop some volts which may be significant when you have such a low starting voltage.
I'm using 30 1N4007 diodes and 10nF, 2k capacitors for half wave arrangement, and the source is 240v AC.
Actually I am just following the circuit schematic from this website.
http://www.brighthub.com/engineering/electrical/articles/78220.aspx
Now the problem is my circuit can't step up to kilovolts but it will keep decreasing to few volt at the output.. does anyone know what's going on?
http://en.wikipedia.org/wiki/File:Stacked_Villard_cascade.svg
It should gain 2-5 VAC into zener-limited 13,6VDC to charge battery. It comes into a very small wind turbine..
I'm using 0,15V forward voltage scottchy diodes, but I don't know how to calculate appropriate capacitance for capacitors.
I tested it with 1uF caps and on idle it gives 40VDC, but short circuit current is only 2mA, which is 1/100 of generators short circuit current.
I assume that bigger capacitors would pass through bigger currents, but how can I calculate right capacitance?In a case where:
Frequency is 100Hz
generators internal resistance is 16ohm
short circuit current 200mA
I want to maximize short circuit current after multiplier and also current that circuit draws from gen. So can capacitance be calculated like this or how?
X = 1/(2*pi*f*C)
C = 1/(2*pi*f*X), where X is the same as gens internal resistance.
C = 100uF
I made a voltage multiplier from an ionizer circuit for car(that circuit got DC input around 2 to 7 VDC, got DC - AC converter and transfomer to amplify the voltage )
Initialy it has 4 stages ( 8 capacitors) and I added 4 more stages.
I used Multimeter to measure Voltage across capacitors and here is the result:
C1:80V C2:140V C3:118V C4:103V C5:90V C6:82V C7:75V C8:69V
It is understandable due to lost energy.
But when I measure accross C1C2C3C4, it is 286V (not equal to 80+140+118+103). And Voltage C1234 > C12345 > C123456 ?? why does the voltage drop?
Could you suggest anyway to still increase the voltage? If it like that, the maxium voltage along the whole stages is just between C1 and C4 ( 286V ) only
Thanks
Another thing I notice is that you are using small 2kV capacitors and two 1kV diodes in series. The small capacitors will be inefficient, and the 2 diodes will just cause excess voltage drop.
You don't say what the output of your inverter is, but assuming it is about 80V (from your C1 measurement), then there is no need to have diodes and capacitors rated to withstand 2000V.
It says in the article...
"The schematics above will output a positive DC voltage relative to the ground (GND). If a negative output is required then the polarity of the diodes should be reversed. you can learn more about how a voltage multiplier works, by visiting the voltage multiplier page."