Continue to Site

# charging HV capacitor using low voltage

Status
Not open for further replies.

#### neazoi

hi all,
Is there any way to fully charge a high voltage capacitor at its full potential using a low voltage sourge?

I do NOT mean using an inverter circuit that converts DC into AC and then step it up using a transformer.

For example to have a 1uF/25v capacitor charged using a constant source of 5v.
At the end, the voltage of the fully charged capacitor will be 25v or 5v ????

If you charge a capacitor up with 5 volts, then 5 volts will be the maximum voltage the capacitor will charge to.
You are not getting confused with the maximum working voltage are you?
If you have a capacitor rated 1uF/25V, the 25 volts refers to the maximum working voltage the capacitor can operate with without breakdown.
It is common practice to select capacitors with a higher breakdown voltage for a safety margin in a circuit.

btbass said:
If you charge a capacitor up with 5 volts, then 5 volts will be the maximum voltage the capacitor will charge to.
You are not getting confused with the maximum working voltage are you?
If you have a capacitor rated 1uF/25V, the 25 volts refers to the maximum working voltage the capacitor can operate with without breakdown.
It is common practice to select capacitors with a higher breakdown voltage for a safety margin in a circuit.

Of course I do not ged confused about the max working voltage of the capacitor.

Is there any transformerless way of "collecting" a higher voltage from a lower voltage source? (slowly charging capacitors or so?). I do not care about the time of the capacitor charging, it may be days or weeks, as long as the final cap voltage is higher than the source.

I have seen some "energy collectors" on the internet that collect energy from long wire antennas and charge high voltage capacitors. these wires do not have high voltages at a specific time, but they can charge a hv capacitor in a relatively long time period.

I do not know of a way of doing this without using some kind of active circuit.
You can make a very simple switcher using just a small inductor, a diode and something like a 555 timer to drive it.

You can do it without a transformer if you use a voltage multiplier. There is an example at http://www.atv-projects.com/LNB_Voltage_Booster.html but the basic principle will work for any input voltage. Note that the most 'boost' you can get with this kind of multiplier is (supply voltage - (2 * Vf)) where Vf is the forward voltage drop in the diodes. You can add extra diodes and capacitors to get more voltage but the drop in the diodes will also accumulate so there will be diminishing efficiency.

Brian.

As the topic reads "HV" if you are interested in really high voltages in KV range, the solution is to use things like Marx generator. Google will fetch many results and you can pick one to suit your need.
Raoof

The idea is to achieve something like 3V from an input of 100-200mV but I wanted to see if voltage "amplification" can be done using no extra power. I guess I try to break the laws of physics

assume the objective is to create a large voltage difference accross a load when you have only access to a low voltage , but potentially large current supply.
If your supply voltage is an AC or time varying voltage , you can use a variation of the voltage multiplier circuit to create a much larger output voltage - This involves using a sequence of diodes and capacitors in a cockcroft - walton multiplier. - put in either cockcroft walton or voltage multiplier into google and you should find info.

If your supply voltage is only a DC. e.g the output of solar cell, then the options are;

1. Use an invertor circuit to convert the DC into an AC voltage and use step one. above.
2. apply the DC current into a large inductor and periodically switch off the current flow - THe Back EMF will produce a larger voltage which can then be stored on the capacitor. ... THis is basically a Switch Mode power supply.
3. apply the DC current into a set of capacitors each of equal value and connected in parallel. Then disconnect them and connect them in series. .... This sounds silly but can be done efficiently using switches - either mechanical or electronic . The output voltage is then the input voltage multiplied by the number of capacitors used.
4. It may be possible to have multipleDC sources , each of just 200mV , If 20 of these can be wired in series, then you will get yout voltage amplification? - Again this is what is done with thermocouples in order to improve the power output. ( I read once that the Soviet troops during WW 2 used thousands of thermocouples in series connected in series with the hot junctions in an oil lamp and the cold junctions in snow in order to produce voltages for their radios.

I have seen one of these lamps in a museum and I did not know how these were called! big THANKS!

the charge-in-parallel/discharge-in-series seems to be the most effective way, since the input voltage is DC a few hundrents milivolts. The device consums as much power as an orginary watch, so power handling is not a problem.
The only tricky point is to have a circuit that could operate in such low voltage that could make this switching..
this is the difficult part in this approach.. any ideas?

You decided to put the question in a general way and finally revealed the 100-200 mV requirement. If you intend a technical solution for a real world problem, this changes a lot.

As a first remark, it's strictly spoken not understandable, why you focus on switched capacitor and exclude inductive converters. At least, you didn't tell a technical reason. Either, it's a unfounded preliminary decision, or you have more unsaid requirements.

Secondly, step-up converters with input voltages down to 300 mV are commercially available. They possibly don't fit your requirements but at least tell about feasibility. Lower input voltages most likely require different circuit concepts. But, the problem should be discussed in terms of actual required voltage and current.

Yes of course you have got a point there.

The output power should be capable switching on a device like a LED, so 3v DC 1-2mA should be more than enough. 0.5mA at 2v could be acceptable too.

The input voltage is DC which varies between 100mV - 700mV (depended on the environment conditions) but the current is very low, a few uA I guess (I am not able to measure it).
So the problem should be quite complex because of the low input current I guess.

You are right, at these low voltages/currents the solution cannot be generalized.

All I am trying to do is to have a means of deriving a higher voltage using this low power low voltage source.
Imagine the condition where a tiny solar panel stays out on a cloudy day and cannot produce much voltage. But you need to somehow get a higher voltage.
The timing is not important factor, so a capacitor that slowly charges up to 2v seems to be a good solution...?

The approach seems to attpemt violating conservation of energy which can not be. At every conversion level high to low or low to high in any element or paramter, effeciency will always be less than one. What comes out (effectively) will be less than what you put it. Of course 1 mV can be converted to hundreds of volts trying comples designs but the net energy will be less than what has been put in.
Raoof

Status
Not open for further replies.