Welcome to EDAboard.com

Welcome to our site! EDAboard.com is an international Electronics Discussion Forum focused on EDA software, circuits, schematics, books, theory, papers, asic, pld, 8051, DSP, Network, RF, Analog Design, PCB, Service Manuals... and a whole lot more! To participate you need to register. Registration is free. Click here to register now.

charging HV capacitor using low voltage

Status
Not open for further replies.

neazoi

Advanced Member level 5
Joined
Jan 5, 2008
Messages
3,721
Helped
13
Reputation
26
Reaction score
9
Trophy points
1,318
Location
Greece
Activity points
34,113
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 ????
 

btbass

Advanced Member level 5
Joined
Jul 20, 2001
Messages
1,897
Helped
438
Reputation
880
Reaction score
287
Trophy points
1,363
Location
Oberon
Activity points
12,861
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.
 

neazoi

Advanced Member level 5
Joined
Jan 5, 2008
Messages
3,721
Helped
13
Reputation
26
Reaction score
9
Trophy points
1,318
Location
Greece
Activity points
34,113
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.
 

btbass

Advanced Member level 5
Joined
Jul 20, 2001
Messages
1,897
Helped
438
Reputation
880
Reaction score
287
Trophy points
1,363
Location
Oberon
Activity points
12,861
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.
 

betwixt

Super Moderator
Staff member
Joined
Jul 4, 2009
Messages
14,871
Helped
4,860
Reputation
9,738
Reaction score
4,643
Trophy points
1,393
Location
Aberdyfi, West Wales, UK
Activity points
126,579
You can do it without a transformer if you use a voltage multiplier. There is an example at https://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.
 

ark5230

Advanced Member level 3
Joined
Jun 29, 2009
Messages
858
Helped
163
Reputation
324
Reaction score
140
Trophy points
1,323
Location
India
Activity points
6,143
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
 

neazoi

Advanced Member level 5
Joined
Jan 5, 2008
Messages
3,721
Helped
13
Reputation
26
Reaction score
9
Trophy points
1,318
Location
Greece
Activity points
34,113
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 :p
 

dobelmpk

Newbie level 1
Joined
Nov 13, 2009
Messages
1
Helped
0
Reputation
0
Reaction score
0
Trophy points
1,281
Location
ireland
Activity points
1,297
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.
 

neazoi

Advanced Member level 5
Joined
Jan 5, 2008
Messages
3,721
Helped
13
Reputation
26
Reaction score
9
Trophy points
1,318
Location
Greece
Activity points
34,113
Thanks for your answer!

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?
 

FvM

Super Moderator
Staff member
Joined
Jan 22, 2008
Messages
48,299
Helped
14,233
Reputation
28,727
Reaction score
12,925
Trophy points
1,393
Location
Bochum, Germany
Activity points
279,669
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.
 

neazoi

Advanced Member level 5
Joined
Jan 5, 2008
Messages
3,721
Helped
13
Reputation
26
Reaction score
9
Trophy points
1,318
Location
Greece
Activity points
34,113
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...?
 

ark5230

Advanced Member level 3
Joined
Jun 29, 2009
Messages
858
Helped
163
Reputation
324
Reaction score
140
Trophy points
1,323
Location
India
Activity points
6,143
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.

Part and Inventory Search

Welcome to EDABoard.com

Sponsor

Top