Capacitors to increast volts(does it increase current?

Thread Starter

JoshAldursHull

Joined Feb 22, 2022
3
iv had an idea to line up + - capacitors polarized capacitors(the ones that hold charge after disconnecting them.

so im trying to charge them individually so if i had 4 capacitors and one battery if i connect the plus+ of the battery to the plus of all the capacitors and the - the all of the capacitors how can i disharge with a circuit. not having the plug the battery to it then recconect all the capacitors in serires by hand ?
 

Papabravo

Joined Feb 24, 2006
19,607
If your scheme could increase the voltage and the current, that would mean increasing the power available and there is no circuit that can do that. I'm not sure what your purpose is but it would require something with considerable complexity.
 

crutschow

Joined Mar 14, 2008
31,128
Capacitors can be used to momentarily increase current and/or voltage with proper connection of switches, but they can't increase the power over what the battery provides as Papabravo noted, (and will actually be less due to the inherent inefficiency of such a configuration).
Is that what you were thinking?
 

Ya’akov

Joined Jan 27, 2019
6,856
The battery produces power which can be integrated over time to create a higher voltage and current but not more total power.

In spite of the pitfalls, let's take the water analogy.

If a river is filling a reservoir over a few days, opening the flood gates on the reservoir could release the total amount of water that took the river days to provide in a much shorter time. And the current of the water, or the pressure of it, or both could be increased.

But, unless you empty the reservoir at the same rate the river filled it you will not be able to provide it for as long. You will only be able to provide the flowing water at higher energy for as long as the reservoir is full, and it will drain much faster than it was filled.

The battery charging a capacitor is very much like this. A practical example is a speedlite (in camera photo flash). Typically they use 4AA cells to charge a capacitor in a few seconds which, if fired at full intensity flashes much brighter than any lamp the battery could power directly, but for a tiny fraction of the flash time.

Relatedly, you can choose to trade current for voltage, or vice versa, but the accounting in all these cases is very strict. Not only can't you get more out than you put in, you can't even get out as much. This is called "efficiency" and it is unfailingly less that 100% and usually not even close to it.
 

Ya’akov

Joined Jan 27, 2019
6,856
How is a voltage doubler related to the question except as an aside?

Relatedly, you can choose to trade current for voltage, or vice versa, but the accounting in all these cases is very strict. Not only can't you get more out than you put in, you can't even get out as much. This is called "efficiency" and it is unfailingly less that 100% and usually not even close to it.
 

PaulNewf

Joined Mar 24, 2020
16
Sounds like you want to make a switched capacitor charge pump or voltage doubler. The MAX232 IC uses caps to get 9V from 5V, have a look at those. There are simpler ICs to just boost voltage without the RS232 drivers.

Depends on your target output voltage: are you aiming for 12V or 120V or 1200V? That totally changes the components.

Also is this continuous output, or just occasional bursts/pulses/flashes. A bit of info on your projects helps get better solutions.

If pulses your caps can gather power slowly and let out in a burst. Strobe and camera flash circuits worth looking at.

If continuous then Pi=Po*Efficiency%, so if voltage up then current down. DCDC converters do this quite well (many modules and ICs simplify this).

Paul
 

Ya’akov

Joined Jan 27, 2019
6,856
Sounds like you want to make a switched capacitor charge pump or voltage doubler. The MAX232 IC uses caps to get 9V from 5V, have a look at those. There are simpler ICs to just boost voltage without the RS232 drivers.

Depends on your target output voltage: are you aiming for 12V or 120V or 1200V? That totally changes the components.

Also is this continuous output, or just occasional bursts/pulses/flashes. A bit of info on your projects helps get better solutions.

If pulses your caps can gather power slowly and let out in a burst. Strobe and camera flash circuits worth looking at.

If continuous then Pi=Po*Efficiency%, so if voltage up then current down. DCDC converters do this quite well (many modules and ICs simplify this).

Paul
I think you missed the part where he wanted to increase voltage and current.
 

Thread Starter

JoshAldursHull

Joined Feb 22, 2022
3
Capacitors can be used to momentarily increase current and/or voltage with proper connection of switches, but they can't increase the power over what the battery provides as Papabravo noted, (and will actually be less due to the inherent inefficiency of such a configuration).
Is that what you were thinking?
if you charge them individually then place them in plus to minus + _ + _ it dooubles the volts , think it increases current ? im not sure havnt tried it yet.
 

Ya’akov

Joined Jan 27, 2019
6,856
if you charge them individually then place them in plus to minus + _ + _ it dooubles the volts , think it increases current ? im not sure havnt tried it yet.
You can never increase power which, measured in Watts, and is Volts multiplied by Amps. In fact, you can only decrease it because no system is 100% efficient, or even very close in practice.

You can trade off voltage for current and vice versa, but again suffering a loss in power. Because current has a time element, it is one Coulomb of electrical charge per second moving past a particular point, you can integrate charge by trading time. Like charging up a battery, where the charger cannot do in the same time what the charged battery can do, or in your case a capacitor.

No matter how you store the energy, paying for it with time, you will never be able to increase the power you put in. So, you can increase the voltage or current in realtime, trading one for the other and suffering an overhead loss in the process, or you can integrate the power by spending time, then have increased power for a shorter time but the total power can’t change and time is the cost to do that.

If you have a power source that can provide 1V at 1A you have 1W to work with. No matter what you do you will always end up with <1W once you use it to power something due to unavoidable inefficiencies in real systems.
 
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