How do you know the dischage amps of a capacitor?

ErnieM

Joined Apr 24, 2011
8,377
The current you can get out of a cap is actually quite simple to predict mathematically, it is just:

I = V / R

Yep, ohms law. V is the capacitor voltage, R is the sum of the load resistance plus the internal resistance of the cap. Oft you can ignore the internal resistance (though I doubt this at 30A).

It may be hard to find ratings to handle this pulse: it is essentially a short circuit.

Also, since the voltage on a capacitor decays exponentially, so does your current.

If this was my project I would be looking for a different power source, even a battery.
 

ericgibbs

Joined Jan 29, 2010
18,738
hi Ernie,
He has not told us what the forward voltage drop is of the LED.

If its a White type, it could be ~3.2V, so he will not discharge the cap below this voltage, there will be only about 1.8V available from a 5V charged cap.
Eric
 

AfdhalAtiffTan

Joined Nov 20, 2010
120
The current discharged by capacitor can be calculated with both V=IR and I=ie^r.

But it is still pretty hard to estimate as current drawn by LED is not linear.

The best way would be to control it by using more precisely timed digital circuit, like 555 or any mcu.

Just curious, are you trying to build a strobe light?
 

Thread Starter

PulseLED

Joined Jan 24, 2014
27
Thankyou greatly for the replies guys.

So there's no real easy way to determine the discharge current as you can't readily measure the resistance of a LED? heres the LED specs:

CREE
XLamp XT-E LED

Size (mm x mm)3.45 x 3.45Maximum drive current (A)1.5Maximum power (W)5Light outputUp to 456 lm @ 5 W, 85°CTypical forward voltage @ 350 mA, 85°C (V)2.85Viewing angle (degrees)115

So the 5v 30A 1us was only round abouts figures, if it were 30A tapering off over 3us it would be fine. So I'm trying to do it as simple as possible, avoiding 555's and the like. From my research these high power LED's should be able to handle 30x the max current is pulsed for such a short time.

Yes I'm trying to make a very fast strobe, not to the intensity of a normal strobe (I think that's out of reach) but enough to generate some image.
 

ErnieM

Joined Apr 24, 2011
8,377
I looked at Cree's data sheet but cannot find the 30A 1uS pulse you mention. Can you link to that data sheet?

From what I see is the max current is 1.5A continuous, no mention about pulse.
 

Austin Clark

Joined Dec 28, 2011
412
Check-and-guess is probably your best option. Trying to calculate this out, even with the datasheet (which probably lacks the information you need), will be very difficult if not impossible.
 

ronv

Joined Nov 12, 2008
3,770
When I model it I can see 2 problems:

Vf goes to about 8 volts at 30 amps. So 5 volts won't be enough.
The other is the ESR(equivalent series resistance) of the cap. The one you show has an ohm or 2 so can't supply 30 amps. You can use ceramic caps.

It may turn into a bomb- A flash of light and a wisp of smoke.:cool:;)
 

Thread Starter

PulseLED

Joined Jan 24, 2014
27
I looked at Cree's data sheet but cannot find the 30A 1uS pulse you mention. Can you link to that data sheet?
What I saw wasn't specific to the Cree LED that I have, rather another 1.5A LED driven at 30A pulse or more, here is the article http://iopscience.iop.org/0957-0233/21/7/075402/

I'm ok with destructive testing until I get the results I'm after, but really don't even know where to start, what capacitor type, voltage and capacitance?

I don't claim to know much, but I just had a random thought. Could you discharge a photoflash capacitor (330v 800uf) into a transformer to drop it down to low voltage, increase the amperage then into a LED. It would be great if I could just convert a camera flash unit.

The problem I'm trying to overcome is the afterglow of a xenon flash creating the longer then wanted pulse of light. I know this is normally done with an air flash although tat presents a lot of problems in itself.
 

Thread Starter

PulseLED

Joined Jan 24, 2014
27
What is the fastest speed of your camera?
What are you actually trying to do?
Xenon flash bulbs will only go as low as 40us or there about's. The faster the flash the less blur on fast moving objects and 40us just is too slow to capture a bullet.

Anyway unless someone stops me I'll pull the little 1.5v->330v transformer from a disposable camera, flip it around, hook it up to an LED and discharge the camera capacitor into it.. and see what happens :confused:
 

ronv

Joined Nov 12, 2008
3,770
So what your looking for is something on the order of 10 usec. so you can see it every 1/3 of an inch or so?
It's getting easier. I think I would just use say a 12 volt power supply and several LEDs in a series parallel arrangement to get the light I wanted. You can over drive them a bit if you like to keep the numbers down, but I'm betting you want like 10 Usec. on and 10 off, so I wouldn't go for more than 50%. You can just use a standard power supply and a big FET switch to turn them on and off.
I have to go now and most of the day tomorrow, but expand on the requirements and I'm sure someone will help.
 

MrChips

Joined Oct 2, 2009
30,621
Xenon flash circuits don't work like that.
The xenon flash bulb is non-conducting until a high voltage pulse is sent to the trigger.
This causes ionization and avalanche breakdown of the xenon gas. The charge from the storage capacitor is dumped through the flash bulb once this happens.

You cannot simply replace the flash bulb with an LED in such a circuit.
 

KL7AJ

Joined Nov 4, 2008
2,229
The current you can get out of a cap is actually quite simple to predict mathematically, it is just:

I = V / R

Yep, ohms law. V is the capacitor voltage, R is the sum of the load resistance plus the internal resistance of the cap. Oft you can ignore the internal resistance (though I doubt this at 30A).

It may be hard to find ratings to handle this pulse: it is essentially a short circuit.

Also, since the voltage on a capacitor decays exponentially, so does your current.

If this was my project I would be looking for a different power source, even a battery.
The problem is, most capacitors don't give maximum current specs UNLESS they're designed for high current, such as R.F. vacuum caps.
Eric
 

Thread Starter

PulseLED

Joined Jan 24, 2014
27
Xenon flash circuits don't work like that.
The xenon flash bulb is non-conducting until a high voltage pulse is sent to the trigger.
This causes ionization and avalanche breakdown of the xenon gas. The charge from the storage capacitor is dumped through the flash bulb once this happens.

You cannot simply replace the flash bulb with an LED in such a circuit.

Oh ok, I understand. Would a way around this be to use the high voltage pulse instead of directly from the cap? or alternatively use the trigger voltage to trip a IGBT/mosfet allowing the capacitor to dump?
 

ronv

Joined Nov 12, 2008
3,770
Forget about the cap. While you may put one on the output of the supply to help provide the pulse current you can simply turn the LED on and off with a FET.
While LEDs do put out more light with higher current they are less efficient past their optimum so you are better off with more LEDs after a point. How often you want to pulse it and how much light you need would take it from guessing.
 

Thread Starter

PulseLED

Joined Jan 24, 2014
27
Forget about the cap. While you may put one on the output of the supply to help provide the pulse current you can simply turn the LED on and off with a FET.
While LEDs do put out more light with higher current they are less efficient past their optimum so you are better off with more LEDs after a point. How often you want to pulse it and how much light you need would take it from guessing.
I only want a single brief pulse, to take a single photo. I bought a couple MOSFET and IGBT on ebay to experiment with, but have never used them before.
 
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