Hello chaps,
I have a project here. I need to build a small voltage doubler circuit from which I will then expand to suit my needs.
The problem I'm having is that the basic one outputs 50Vdc whereas I have 240Vrms in. In this case I expect a minimum of 500Vdc. (cant post my drawing for now but I'm using this replica: http://en.wikipedia.org/wiki/File:Greinacher_circuit.svg but without the tx; using mains at 244vrms)
I am using scavenged capacitors(disc ceramic capacitors -- the small blue ones) and diodes(FR107) for a single stage half wave multiplier.
I have tried changing the diodes and using brand new 1N4007 whereas I have been through all 30 capacitors I have at my disposal. Non of them appear broken and testing a few has shown nothing contrary to my prior findings. All diodes are in superb condition.
No component is getting hot and neither are there shorts or funny smells.
Any ideas where I'm going wrong?

I need to finish it ASAP like today!
Thanks in advance.

 

What is the value of your capacitors?

 

Feedback and Suggestions Forum for providing feedback and suggestions about All About Circuits, including corrections to the e-book. This forum is not for getting help with technical questions.

This was neither feedback nor a suggestion, so I have moved it to a more appropriate forum.

 

What is the value of your capacitors?

I'm using pairs: each pair is 1000pF, 10000pF and 2200pF.
Both sets of diodes have a forward voltage of 1000 with diode vdrops at 0.7 and 1.1
for N series and FR diode respectively.

 

Why the different values?
You will need capacitors of at least 0.1μF rated at 1kV.

 

The pairs here I used were to determine whether the caps were bad...the multiplier is a single stage. I am switching at ~50Hz.

You will need capacitors of at least 0.1μF rated at 1kV.

Ok, how does the 0.1uF 1kV cap influence the output, given that those I used are also 1kV with exception of the 222 caps which are 2kV?

 

The size of the capacitor will determine how much current you can draw from the output of the voltage doubler. What is your current load?

 

The current output is not of concern now. I have turned to the multiplier to explicitly scrub it off for the application I have in mind. If I'm to guess, 0.1A at most. Otherwise the voltage will do. I need bags of voltage irrespective of how much current will be available. Think of it this way: I need a electrostatic field for my app.
My intention is to switch this tension to my load via a Enhancement N-FET. This way, I further limit my load's current consumption.

 

Someone help please... where am I wrong in this?
I've been wondering: what effect does a charged non-polar capacitor have on a circuit if its polarity is suddenly reversed/charged with a voltage whose polarity is opposite to the capacitor's?

 

0.1A = 100mA is a lot of current to take from a voltage doubler with such low value capacitors. What are you using to measure the voltage?

 

Normal multimeter.
This current is an estimate. I haven't attached the load yet.
I can't get the multiplier to work well in the first place. This is where I'm stuck.
I can't move on.

 

1,000pF = 1nF = 0.001uF
10,000pF = 10nF = 0.01uF
2,200pF = 2.2nF = 0.0022uF

You really need much larger caps. They do need to be non-polarized. As already mentioned, voltage doublers are not good for much current; above a few mA, you'll really need something more like a transformer or a switching power supply.

 

As already mentioned, voltage doublers are not good for much current; above a few mA, you'll really need something more like a transformer or a switching power supply.

I am in need of high voltage and not current for this project.

need something more like a transformer or a switching power supply.

What's the difference between connecting the device directly to mains and having the above quoted? I mean is not the output the same, a sinewave?

 

Hi Sgt,

How have you been keeping? Good to see you still keeping in touch.
btw, your last line is off by a factor of 10.

 

your last line is off by a factor of 10.

Me or sgt?

 

I am in need of high voltage and not current for this project.

OK. However, your capacitors are so small that the charge is lost due to the turn-off time of the diodes. Get some larger non-polarized capacitors. If you have a hard time finding non-polarized caps, you can use a pair of same-value electrolytic caps back-to-back (both negative terminals together or both positive terminals together) that are rated for a higher voltage than required; if the capacitance of both are the same, then the resulting capacitance will be 1/2 of the rating.

What's the difference between connecting the device directly to mains and having the above quoted? I mean is not the output the same, a sinewave?

Do not connect a project directly to mains power; a transformer must be between the mains and your project. A transformer provides galvanic isolation from mains power, and effectively limits the amount of power that can be transferred to your project.

We do not support discussion of transformer-less power supplies; they are inherently unsafe, and there is nothing one can do to make them safe. Transformers are a remarkable bargain compared to the cost of a funeral.

MrChips,
Thanks, nice to "see" you, too.

btw, your last line is off by a factor of 10

Yep, thanks - I fixed it. Too many things taking my attention this morning...

Me or sgt?

Me.

 

Me.

@vk
I've removed the theory I had in mind.

 

In the mean time, I have a theory I'ld like to share:
I think that the mains supply from the grid isn't a true sinewave; I mean once after the step down distribution tx.
See, a sinewave will oscillate from + to - at the stated frequency. However, past the tx, because of the neutral line it ceases being one because that line isn't part of the input from the high voltage overhead lines.
Use the magnetic induction and EMF knowledge. Take the positive section of the cycle. A induction event takes place on both coils of the primary and distribution (secondary) coil. When the cycle heads towards 0v on its way to the negative cycle section, the magnetic field on both coils collapses and a EMF is generated in the direction opposite to the input. In this case of the distribution/secondary, the EMF follows the Live wire just as the predecessing voltage of the previous cycle followed. So regardless of the input being a sinewave, the output is rather a rippled unidirectional AC with no sine.
To prove this, I've observed animals stand beside one of these dist. tx close to the ground wire that stems from the neutral line.
If the sine wave polarity-changing A.C. theory holds absolute true, I'd have had roast mutton for dinner.
They stood there for some time (I found them there, I left them there) and not one smoked into oblivion. @ 50Hz, they should have been pulsed or electrocuted at least 25 times for every second.

Based on this theory, might be the reason all half-wave multipliers I've seen on the net have a tx on the input side, in addition to safety as sgt. stated above.
Do'no about this...please confirm...

Your "theory" is arrant nonsensense,& I suggest you go back to fundamentals to learn how Electricity really works!--or try some other hobby/occupation.
Maybe you can freak out Mechanics,or Boatbuilders instead!

In the meantime, I'll see about the tx. Hope to post tomorrow.
I don't have a proper stepup or coupling tx. I fear reversing the small one I have and risk shorting the mains because most secondary coils of tx measure a short circuit on the multimeter.
so I'll build one of them circuits from the 555 sinewave generator examples and reverse the little thing onto it. Hope it works.

Thanks sgt, MrChips.

I'll update soon if not start another thread...

Please dont!!

 

Ok.
Was just a thought in the distro chain relative to the grid input.

 

@sgt
I tried a simulation with exact values of my caps at 2200pF and the output is what I expect.
I'm trying with polar caps and 12vrms input. I'll see what's what.