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Ultra Short, Ultra Fast LED Flash

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Marco:
Lets do some numbers, lets say you want 10 kW for 1 us. That's 0.01 Joule, lets round that up to 0.1 Joule to keep the current approximately constant for the pulse ... also a relatively safe amount of energy.

Lets say you use 100V string of LEDs (LED filaments would work for that) so you need 100A of current, so you need 20 uH of inductance, of course with a SRF>>1 MHz. That's a spicy inductor. Compare that to storing 0.1 Joule at 100V in a capacitor, that's a 2 dollar electrolytic capacitor in singles.

ogden:

--- Quote from: ajb on March 08, 2019, 03:42:50 pm ---You've missed the boat.

--- End quote ---

Yes, kind of. :) Did read only first datasheet.


--- Quote ---This will establish the desired current in the inductor, then when you want a pulse of light, you turn the PWM transistor off, which will allow the inductor current to flow through the LEDs instead.

--- End quote ---

I am afraid that this is where you are missing the boat. When load is shorted, regulator provides let's say 0.1V @ 10A  = 1W power. When shunt is open then regulator shall suddenly jump to 20V @ 10A = 200W power (or whatever is operating voltage of the LED's). Voltage of inductor cannot jump instantly, it is against it's formula V = L(di/dt). Sure/agreed - stopping/starting converter is even slower. Inductor can't compete with capacitor no matter how you twist it. I did not find any real, measured LED current PWM shunt waveforms for this case. If you know any or can measure - please provide, would be nice to see.

ajb:

--- Quote from: ogden on March 08, 2019, 07:43:38 pm ---
--- Quote from: ajb on March 08, 2019, 03:42:50 pm ---You've missed the boat.

--- End quote ---

Yes, kind of. :) Did read only first datasheet.


--- Quote ---This will establish the desired current in the inductor, then when you want a pulse of light, you turn the PWM transistor off, which will allow the inductor current to flow through the LEDs instead.

--- End quote ---

I am afraid that this is where you are missing the boat. When load is shorted, regulator provides let's say 0.1V @ 10A  = 1W power. When shunt is open then regulator shall suddenly jump to 20V @ 10A = 200W power (or whatever is operating voltage of the LED's). Voltage of inductor cannot jump instantly, it is against it's formula V = L(di/dt). Sure/agreed - stopping/starting converter is even slower. Inductor can't compete with capacitor no matter how you twist it. I did not find any real, measured LED current PWM shunt waveforms for this case. If you know any or can measure - please provide, would be nice to see.

--- End quote ---

You're neglecting the energy stored in the inductor, and that the defining characteristic of an inductor is that it resists changes in current, which is the whole point.  We're literally talking about a boost converter minus the output capacitance and with a larger inductor.  Boost converters clearly work, including constant (ish) current LED boost drivers, so there's no reason that this won't too.  Of course there's plenty of room left to argue about the practicalities, costs, and other relative design tradeoffs of the inductor vs capacitor based solutions, and a lot of that comes down to how closely you need/want to regulate the LED current, and how much energy is involved.  The OP said 75W * 500ns, while for some reason Marco has decided to make the problem much harder by assumed 10kW for 1us, which is 266 times as much energy.  (I have no idea how much energy is actually required to get the results the OP is after, but chances are it's somewhere between 75W and 10kW...)

ogden:

--- Quote from: ajb on March 08, 2019, 09:21:41 pm ---You're neglecting the energy stored in the inductor

--- End quote ---

Oh, suddenly it's me who neglects energy stored in the inductor.  :-DD
Look, unless you show real-world shunt-PWM waveform that disproves what I say, we do not have any further progress in this discussion.

Marco:
In retrospect I decided that 75W isn't really a replacement for an open air spark gap ... it's not going to be enough to expose a normal camera.

He's going to need some serious power, keeping the energy to supply that power circulating in an inductor is awkward. For a couple ms it's not so bad, but when you want to do it continuously to be ready for a trigger you need so much copper to keep the losses down. Meanwhile a capacitor can just sit there minding its own business doing nothing.

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