I was trying to see if using a higher current in a flash circuit would make the cap charge faster and yield shorter wait times between flashes (for photography purposes) but I haven't seen any data on a capacitors maximum input current.
if I increase the current to the capacitor will it charge faster?
- Thread starter cvangordon
- Start date
Is there a way to find the max current that can be input to a given capacitor? I haven't seen that info anywhere.
??? You don't force more current into the capacitor. A lower resistor value allows more current, charging faster. T = R x C. "I" isn't in the equation. Yes, a capacitor with a lower ESR will help, but not a lot.
T = (R + ESR) x C
Okay ... T = (V/I) x C
But can we deliberately increase current without changing resistance values (R or ESR).
How do you calculate T with a constant current source instead of a resistor? The constant current source is effectively a constantly varying resistance, isn't it?
Will that fit the needs of the original poster? A constant current source instead of a resistor?
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Is it being recharged from a battery or a main power source?
If its battery power there is a small power inverter circuit that steps up the battery voltage to the 300 or so volts the flash bulb needs to operate and that is the part of the system that limits your flash cycle time. Not the capacitor itself.
Typically the capacitors used for Xenon or other similar flash bulbs have fairly low ESR values given the bulbs require a fairly high but brief multi tens of amps current through them to give them their brightness.
So from my estimation it's the charging circuit not the capacitors internal resistance that is the charging time problem.
Yeah that was my thought actually. Just to use a slightly more powerful transformer that would output a higher current. I'm not sure if a higher voltage would do anything more amps is what I was wondering about. That's why I was asking how much a capacitor can typically handle since I don't want to do 2 amps instead of 1 and end up with little bits of shrapnel in my head.
You want to charge at twice the current cut the resistance in half.
If it can put out several tens of amps in a short pulse it can obviously handle that going back in.
The only limit on that is the repetition rate of which the materials around the flash tube will likely melt before the capacitor gets dangerously hot.
Over voltage would be the bigger concern is a unregulated power source was being used. That's what blows up capacitors.
No doubt you don't want to start messing around with your flash unit. Well, go to a drug store where they process film. Ask if they have any disposable cameras with flash units inside them. I'm talking about cameras that people have brought in for processing, not NEW disposable cameras. They'll probably be willing to GIVE you a couple if they have them. And they come with a single AA battery inside - often still containing significant charge. Since the price is dirt cheap - messing with and destroying one is certainly no big deal. And you'll learn something about the timing circuit that charges up the capacitor. Just remember that the capacitor can hold a LOT of charge. Certainly can be unpleasant to an errantly placed finger. Or if you happen to touch the trigger with one finger and with the other hand make contact with a high voltage. COULD EVEN BE LIFE THREATENING. So be VERY careful when messing with flash units.
The capacitor is charged through an inverter/multiplier. (probably not the correct terminology, but the voltage from a double A battery can be multiplied and create a charge in the hundreds of volts). If you listen carefully you can actually hear the whine as the capacitor charges up. Now, here's a part I DON'T KNOW - if you change the resistance of the circuitry that charges the capacitor you MIGHT get a faster charge. But I suspect that the circuit is already tuned to give you the best performance. It's been my experience that the limiting factor is the state of charge of the battery itself. If the battery is weak it will take longer for the circuit to reach flash potential.
Maybe the answer isn't in changing the resistance to up the current but maybe it lies in the power supply itself. I have no idea what will happen if you power that cheap disposable flash unit (meant to operate on a single AA battery) if you use TWO AA batteries. Maybe the circuit can't handle that much voltage. Would you hook the batteries in parallel? Or should they be hooked in series? All questions I don't know the answer to. But if you're going to experiment with a flash unit in that capacity (over powering it) then I'd suggest wearing a protective face shield - should something suddenly and spontaneously give up the ghost. Wearing long sleeves and gloves might be a good idea too in order to limit any possible shrapnel.
Whatever you do - don't fail to consider safety requirements. Since there are probably no defined safety precautions - err on the safe side. Certainly, two AA batteries in parallel won't OVER voltage the system. You'll just have greater capacity. Putting the batteries in series, however, CAN present a problem as some of those components may be rated for a specific voltage and if you double that voltage you might come close to - or exceed the device's rated capacity.
Final words: Be safe. Experiment with a disposable flash unit. And be safe. Seriously! That cap can hold a lot of juice. And flash caps are GOOD at giving back what it has stored in a very fast way.
The capacitor is charged through an inverter/multiplier. (probably not the correct terminology, but the voltage from a double A battery can be multiplied and create a charge in the hundreds of volts). If you listen carefully you can actually hear the whine as the capacitor charges up. Now, here's a part I DON'T KNOW - if you change the resistance of the circuitry that charges the capacitor you MIGHT get a faster charge. But I suspect that the circuit is already tuned to give you the best performance. It's been my experience that the limiting factor is the state of charge of the battery itself. If the battery is weak it will take longer for the circuit to reach flash potential.
Maybe the answer isn't in changing the resistance to up the current but maybe it lies in the power supply itself. I have no idea what will happen if you power that cheap disposable flash unit (meant to operate on a single AA battery) if you use TWO AA batteries. Maybe the circuit can't handle that much voltage. Would you hook the batteries in parallel? Or should they be hooked in series? All questions I don't know the answer to. But if you're going to experiment with a flash unit in that capacity (over powering it) then I'd suggest wearing a protective face shield - should something suddenly and spontaneously give up the ghost. Wearing long sleeves and gloves might be a good idea too in order to limit any possible shrapnel.
Whatever you do - don't fail to consider safety requirements. Since there are probably no defined safety precautions - err on the safe side. Certainly, two AA batteries in parallel won't OVER voltage the system. You'll just have greater capacity. Putting the batteries in series, however, CAN present a problem as some of those components may be rated for a specific voltage and if you double that voltage you might come close to - or exceed the device's rated capacity.
Final words: Be safe. Experiment with a disposable flash unit. And be safe. Seriously! That cap can hold a lot of juice. And flash caps are GOOD at giving back what it has stored in a very fast way.
