Building an LED microflash

I am having a little fun thinking around corners.

I have a huge array of LEDs and I want to pulse them on and off quickly (on ce).

A trip switch will trigger the LED flash briefly.

I need help with the circuit design. I'll try to show what I need done...

The LEDs are all in parallel and are 3v. The input voltage will be from a 3 v DC supply (a coin cell).

A capacitor will charge and a charge indicator light will light up (could b e a red LED). The LEDs remain off.

A switch closes the circuit over the capacitor causing the capacitor to dis charge 3v into the LED array which glows briefly. I will have to work out t he right capacitor value myself. When the capacitor has discharged, no more current flows across the LED array regardless of whether the switch is now open or closed.

Can it be done simply and can anybody help with a diagram?

Reply to
Zephod Beeblebrox
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Just watch the diodes down the left-hand side. ;)

Cheers

Phil Hobbs

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Reply to
Phil Hobbs

Please, I'm serious. This is for a personal project. I do need help - real help, not jokey, snooty time-wasting stuff.

Reply to
Zephod Beeblebrox

Are you claiming that Zephod Beeblebrox is, seriously, your real name?

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John Larkin                  Highland Technology Inc 
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Reply to
John Larkin

(Context restored.) Fair enough, but you're the one who picked the jokey nym, not me. Serious folks on SED generally use their real names.

Putting all the LEDs in parallel will probably work OK for pulsed use, though it's fairly far from optimal since there's nothing much to ensure equal current sharing.

Knowing how many "a huge array of LEDs" means would help.

Knowing how long "briefly" is would also help.

Knowing what you're actually trying to do, e.g. make something cool looking, doing time-lapse digital photography, or whatever, would help as well.

And since you're asking for somebody to design something for you, you might be better off in sci.electronics.basics. (I've cross-posted and set followups there.)

It's okay if you don't know the answers in great detail, but a bit of context would help us to help you.

Cheers

Phil Hobbs

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Dr Philip C D Hobbs 
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Reply to
Phil Hobbs

Ok. Continuing in sci.electronics.basics

Reply to
Zephod Beeblebrox

once).

3v DC supply (a coin cell).

be a red LED). The LEDs remain off.

ischarge 3v into the LED array which glows briefly. I will have to work out the right capacitor value myself. When the capacitor has discharged, no mo re current flows across the LED array regardless of whether the switch is n ow open or closed.

There should be no problem at all. We manufacture high-intensity white ligh t LED lamps (with simple built-in charger) with LEDs all in parallel. The parallel connection provides the high intensity. Basically, there should be a dropping resistor for each LED, and the driving voltage should be bit high, say 6 Volts, to take into account the voltage drop in the dropping resistor. For added protection, add a high Watt resistor that will limit the total current into the parallel array. That is, suppose 20 LEDs are in parallel, each running at the nominal current of 20mA. So the high Watt resistor would limit the total current that flows into the array, and each individual low Watt resistor will limit the current into the LED connected to it. The calculations are simple. Hope that helps.

Reply to
dakupoto

How large is huge? In here it tends to mean advert sized display boards.

How short is briefly? 1ps, 1ns, 1us, 1ms 1s

Coin cell isn't going to hack it as white LEDs need a bit more voltage than that to reach their full brightness. Having the LEDs in parallel isn't great either as the weakest one will get most of the current.

You need about 4v across one to get maximum light output output 20mA and for a single pulse of current you can overdrive them somewhat provided you don't exceed manufacturers maximum ratings. I'd suggest trying two coin cells in series or better a pack of 4 AA cells at 6v.

See this basic tutorial which may help you

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With pulse drive the lead inductance will save you from disaster. On DC the weakest LED will suffer from too much current and the stronger ones will be dim. Might not be too bad if they are all the same batch and you weed out any obvious stragglers but not in general a good idea to do this without a small series resistor in series with every one.

I'd use a green one - looks brightest for least current. But how long does it have to recover between flashes?

Use a changeover switch so that the capacitor is switched from the battery to the LED bank.

Battery --R-- -----Leds | \ | | _|_ | | ___ C | |__________|________|

Try R=1k and C=1000uF as a first guess.

The charge indicator LED wants to go across the capacitor with a largish series resistor and maybe a diode in series so that it only lights up when the capacitor is fully charged.

If you describe what it is for then you might get a much better answer. Ring flash for macro photography designs in amateur electronics magazine EPE have much more sophisticated drive, timing and trigger.

Basically using LEDs for flash requires running them aggressively close to their maximum permitted drive voltage and power dissipation.

These days you can buy single batwing LEDs good for 7W continuous dissipation on a decent heatsink.

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Regards, 
Martin Brown
Reply to
Martin Brown

Count the heads ;-)

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"Design is the reverse of analysis" 
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Reply to
Fred Abse

The calculations are indeed simple. 1 joule for 1 microsecond is 1 megawatt.

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John Larkin                  Highland Technology Inc 
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Reply to
John Larkin

I think that's a bit misleading, John.

1 joule (a watt-second) _per_ microsecond is 1 MWatt. 1 joule _per_ second is one watt.

But, that's just my opinion.

Cheers, John S

Reply to
John S

Don't be like JF and start mincing words. You knew what I meant by "for."

The OP wants to make a microsecond photoflash from his LEDs, and the concept has serious numerical problems.

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John Larkin                  Highland Technology Inc 
www.highlandtechnology.com   jlarkin at highlandtechnology dot com    
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Reply to
John Larkin

IIRC an actively-quenched flashtube can get down to the few-microsecond range. There's an afterglow, of course, but the peak is big enough that that doesn't matter much. Doc Edgerton did fine with .303 bullets, which are a teensy bit faster than air guns.

Cheers

Phil Hobbs

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Dr Philip C D Hobbs 
Principal Consultant 
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Reply to
Phil Hobbs

I think a simple triggered spark gap in air can get you down to the microsecond range, if you keep the electrical loop tight.

Early LIDAR used a spark gap in an old searchlight reflector, and a PMT, to bounce light off clouds.

Wiki, as usual, knows...

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10 joules, 20 megawatts.
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John Larkin                  Highland Technology Inc 
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Reply to
John Larkin

Yes, but maybe to OP did not. Perhaps I did not phrase my reply in a tactful manner.

I can't disagree with that.

Reply to
John S

Your LEDs will take maybe 50 mA for a few microseconds? If there's a dozen of 'em, the capacitor will have to supply 50 mA x 5 us of charge, and droop by less than or equal to Vmax - Vswitch-Vload after those five microseconds. So, it'll take a fast switch (probably NMOS power FET), and a capacitor that can handle microsecond pulses.

A fast switch with higher voltage and lower current (LEDs in series) and an inductor to store energy, is the more usual strobe-light design.

Reply to
whit3rd

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