High voltage DC to low voltage DC converter. How to?

Hello,
Assuming a high voltage capacitor charged (10-15KV). How can I convert this to say 12v or so for charging a lower voltage capacitor?

Not quite sure what you've gotten yourself into there, but could you use a stepdown transformer and rectifier?

barry said:

Not quite sure what you've gotten yourself into there, but could you use a stepdown transformer and rectifier?

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You cannot use a transformer at DC. I need to charge a low voltage capacitor from a high voltage charged one. Will a simple voltage divider do? I am afraid about the resistor losses

If you discharge the cap rapidly into a transformer you can do this (It used to be done automobiles all the time...). You COULD use a voltage divider but you are right-you are going to have a lot of losses, or else a long charging time. You haven't really explained your application, so it's hard to give you any really good advice.

Voltage conversion can be performed by charge pumps (only theoretically in this case, I presume) or converters involving inductors respectively transformers.

The first practical question arising from this problem: Which HV switches do you have at hand?

FvM said:

Voltage conversion can be performed by charge pumps (only theoretically in this case, I presume) or converters involving inductors respectively transformers.

The first practical question arising from this problem: Which HV switches do you have at hand?

Click to expand...

I do not have such a high voltage semiconductors (if they exist) currently in the lab.
I see you are proposing a transformer to convert the pulse from the rising edge (0v at the beginning to HV when the capacitor is connected to the primary) to a pulse of lower voltage. In this case the pulse does not always discharge the HV capacitor almost completely but I think that is fair enough.
How do you propose the connection from the HV capacitor to the transformer to be done?
For example a breakdown device like a HV neon tube or a spark gap, so that when the capacitor is fully charged, it automatically passes the pulse into the transformer?
I am worying about the falling edge of the pulse in the latter case, which will unduce a negative voltage on the secondary. On the other hand, if the HV capacitor is let to be discharged through the primary, the pulse presented on the primary should be more like a reverse sawtooth, which induces a high positive voltage on the secondary and a much lower negative voltage on it.

The puspose of the application is to charge a HV capacitor electrostatically and use this energy to gradually charge a low voltage capacitor/battery etc.

Hi is it like collecting lightning voltage??

First of all there is no HV cap or LV cap, They are just a rating of voltage which they can withstand.. Consider this If you have 10uF 10V charged capacitor what will be the final voltage when a 20uF capacitor is connected parallel??

Venkadesh_M said:

Hi is it like collecting lightning voltage??

First of all there is no HV cap or LV cap, They are just a rating of voltage which they can withstand.. Consider this If you have 10uF 10V charged capacitor what will be the final voltage when a 20uF capacitor is connected parallel??

Click to expand...

By saying HV cap, I mean for example a 20KV max capacitor, that has been gradually charged to 10KV or so (using an electrostatic charger to gradually charge it). I need to transfer some of this capacitor's energy to a lower voltage capacitor, to be able to power lower voltage devices later on.

I think electrostatic machines like van der graaf generator, Kelvin lord generator etc produce high voltages so an ordinary low voltage capacitor could not be used directly on these machines.

Before going further along this line of thinking, what capacitance is the HV being stored in? I ask because you mention Van Der Graff and the likes whch produce very high voltages but have extemely small charge holding. What you are trying to do is transfer the charge from a small capacitor to a large one so the formula V=Q/C applies but the initial 'C' may be tiny compared to the final one. Ignoring the practicality issues, if you make a guess at a Van Der Graff sphere having a capacitance of say 10pF and to make any usable energy you want to store it in 1,000uF, the voltage would be dropped by 1,000,000,000 times (10^9). Of course that assumes that the HV source can sustain it's output under load and in most 'static generator' models that wouldn't happen.

Brian.

betwixt said:

Before going further along this line of thinking, what capacitance is the HV being stored in? I ask because you mention Van Der Graff and the likes whch produce very high voltages but have extemely small charge holding. What you are trying to do is transfer the charge from a small capacitor to a large one so the formula V=Q/C applies but the initial 'C' may be tiny compared to the final one. Ignoring the practicality issues, if you make a guess at a Van Der Graff sphere having a capacitance of say 10pF and to make any usable energy you want to store it in 1,000uF, the voltage would be dropped by 1,000,000,000 times (10^9). Of course that assumes that the HV source can sustain it's output under load and in most 'static generator' models that wouldn't happen.

Brian.

Click to expand...

So there is no practical way of storing the energy out of such machines in a way that it will be useable in low voltage applications? (even if many sequencial "pulses" are needed to charge a lower voltage storage, like a capacitor)
I have seen a circuit used in a similar application https://peswiki.com/index.php/OS:Radiant_Energy_Antenna_System and I thought it might be able to adapt to electrostatic machines.

A different approach would be to use a motor-generator, with a high voltage electrostatic motor driving a more normal low-voltage generator - perhaps a repurposed slot-car motor, or similar.

No, that article is absolute nonsense from start to finish. Avoid articles that tell you:

Electricity found in nature is usually D.C. Although D.C. actually can and usually does have a very high (Millions of cycles per second) frequency.

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The way it works, if you are lucky enough to get any results at all, is static discharging from the antenna causes the neon lamp to flash over, fire the SCR and discharge the capacitor into the battery. In practice, the charge on the antenna would be incredibly tiny and almost certainly wouldn't be enough to overcome leakage currents in the capacitor le alone strike a neon, it could also be either polarity or change randomly. In the first instance, to get any significant charge on the antenna it would have to be in a very dry atmosphere, even air humidity would stop it working.

Here are other ideas I've heard of, all equally crazy:

1. Two metal plates are held apart by say 10mm to form a capacitor. A 1.5V battery is connected across them to charge the capacitor up then disconnected again. Next, move one plate so the gap between them becomes wider. The theory says that if 'C' is reduced and the charge 'Q' is still the same, the voltage must increase. Theoretically, if they are an infinite distance apart the voltage across them will also be infinite.

2. A 'a crystal set' radio receiver drives a voltage multiplier. when tuned to a local radio station you could generate many KV at the end of the generator.

3. Similar to the article you posted, use idea 2 but remove the tuned circuit, now it should multiply the voltage at all frequencies and therefore be more efficient.

As an experiment, use a normal DVM or analog multimeter, spread the probe wires apart so the meter is in the center of a dipole. Now measure the voltage and current to see what is actually present in the 'air'. Whatever you measure is the same energy that would be picked up by the schematic in the article.

Brian.

If you transfer energy between capacitors of different voltage without storage inductors, charge is preserved but energy is lost corresponding to the voltage ratio.

A "loss-free" charge pump can be made by switching capacitors between series and parallel circuit. Apart from the problem of finding a suitable switch, this would require many capacitors and switches for a large voltage ratio.

Yeah, Switching the circuit is good Idea, Consider about making the original cap into switchable circuit. If a separate capacitor network used still there will be some energy loss...
Choosing a low value of cap in parallel will reduce energy loss but will need lot of switching arrangement....