Crutschow has me beat six ways to Sunday on just about anything with a volt in it, so I'm just dropping in for a moment.

I wanted to say you probably can't buy .12 Henrys that doesn't have 1/3 of an ohm in it. However this turns out, you are going to want escape routes for the current in case of an LED failure. I would try to arrange that so any failure would result in an emergency shut down of the whole operation because a sudden change of 10,000 watts makes things explode. Meanwhile, 10,000 ohms in series with 1000 uf requires several seconds to charge up. (You aren't quite finished polishing this.)

 

Hey thanks ronv and crutschow, great circuits. It looks like you both are using the same software, LTSpice? I'm thinking I should probably invest in that as well. Can you specify real-world parameters of capacitors and inductors, such as ESR?

Yes, I use LTspice. It's a free download from Linear Technology. You can include real world parasitic parameters. Just right click on the component to add them.

Edit: I add the LTspice .asc file to my previous post if you want to simulate it.

 

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However this turns out, you are going to want escape routes for the current in case of an LED failure. I would try to arrange that so any failure would result in an emergency shut down of the whole operation because a sudden change of 10,000 watts makes things explode. Meanwhile, 10,000 ohms in series with 1000 uf requires several seconds to charge up. (You aren't quite finished polishing this.)

An open failure is not generally a problem in my circuit, since the current to energize the inductor goes through the LEDs.

Yes, the charging time is long with that resistor but that was just for the simulation. As I previously noted his charging circuit (whatever it is) will likely be much faster.

 

Since efficiency is part of the plan, I'm wondering about the optimality of the choice of 360V. Would some other value, higher or lower, improve overall efficiency? I'm assuming the OP has to step up from his battery to achieve that.

All I'm suggesting is that it's worth taking a look at the whole system from battery to LEDs

 

crutschow, if M1 is shut off before the circuit has dissipated all of the energy, I'm guessing it's likely there would be energy stored in both the capacitor and inductor. Does D6 serve to provide a path for the inductor to dump it's energy back into the capacitor, and keep it there?

edit: That can't work the way I thought... I am at a loss as to what happens to any leftoever energy in the inductor if M1 is switched off, causing an open circuit. Might the inductor ramp up the voltage until it overcame M1's breakdown voltage?

 

wayneh, I agree, I am going to take a hard look at options for charging circuits, capacitors and inductors (and their ESRs), etc. to determine an efficient combination that will provide the light output I need.

 

crutschow, if M1 is shut off before the circuit has dissipated all of the energy, I'm guessing it's likely there would be energy stored in both the capacitor and inductor. Does D6 serve to provide a path for the inductor to dump it's energy back into the capacitor, and keep it there?

edit: That can't work the way I thought... I am at a loss as to what happens to any leftoever energy in the inductor if M1 is switched off, causing an open circuit. Might the inductor ramp up the voltage until it overcame M1's breakdown voltage?

D6 allows the inductor to dump all its energy into the LEDs. Otherwise the circuit would oscillate with a damped sinewave from the LC tank.

Your last statement is correct. If M1 is cutoff before all the inductor energy is dissipated then the inductor will indeed increase the voltage until M1 (or something else) breaks down for the time it takes the inductor energy to dissipate. So you want to make sure the pulse is longer than the inductor discharge time (or add a zener across M1 from drain to source to prevent M1 from being zapped).

 

ah I see, the D6 keeps the inductor from re-charging the capacitor.

hmm, I think that I'll almost always have to switch off M1 before the inductor dissipates all of its energy, because of the long tail of the decay. I wouldn't be able to afford wasting it -- would it be feasible to somehow dump that remaining energy back into the capacitor?

 

If you connect a reverse-biased diode across the inductor (cathode to capacitor) then the LED pulse will terminate when the capacitor voltage reaches zero. The rest of the inductor energy is then returned to the capacitor (with a long decay tail) through the diode.

Also a smaller inductance will reduce the pulse width.

Edit: I just realized that you don't need an added diode to terminate the pulse, just remove the cathode of D6 from ground and connect it to the inductor output. That gives about a 20ms pulse with 0.12H.

 

not sure what you mean, D6's cathode isn't connected to ground, its anode is. If you left the D6's anode at ground, and connected its cathode to the inductor output, the inductor would fight with both M1 and D6 when M1 switched off. If you moved D6's anode from ground to the inductor output, then it would create a short for inductor?

Okay, because the pulse width is going to vary between 1 and 10ms depending upon user input, and because the long decay tail through the LEDs is undesired, M1 will nearly always be switched off before the capacitor has emptied completely. What if you removed D6 altogether, and switched the inductor output to ground at the same time M1 is switched off. That would create an LC tank, with the capacitor continuing to empty completely into the inductor, with the inductor then dumping the energy back into the capacitor, but reverse biased. What if we waited another cycle for the capacitor to discharged/charged again, but this time in the correct bias for the charging circuit to be able to top it up. The losses would hopefully be minimal, depending on their ESR.

Edit: damn, just realized that the capacitor is going to be an aluminum electrolytic, which can't be reverse-biased. Upon switching off M1, would it be feasible to switch the inductor input to ground, and inductor output to capacitor input? I've no idea, but that seems tricky.

 

I think what you have to do is to magically switch the inductor so that its "in" side is going to Gnd and its "out" side (not marked that way on the diagram--the side that drives the LED string) goes to the + terminal of the capacitor. That way, the surplus energy pushes current from Gnd up into the capacitor, ready to be used in the next cycle. It might be simpler to make L1 into a transformer, and use the other winding for the energy-saving function.

 

Yeah I thought that seemed kind of magical too... I don't know much about transformers -- if L1 was a transformer with, say, the primary winding connected to the LEDs, what would happen when M1 is switched off? Rather than busting through M1, would the stored energy be able to escape through the secondary winding somehow?

 

Yes, that would be the idea. The two windings together will attempt to keep the magnetic flux in the core constant, so a change in current in one, will induce a voltage in the other. But you'd have to think about what the secondary winding would do under all conditions--when the current is building in the primary, when it's declining normally, and when the LEDs are shut off with inductive energy still left.

 

I think Carl missed that there are 10 strings of 100 LEDs in series. But having said that I'm still looking for ideas.

 

ronv I've been thinking about your comparator circuit... I like the idea, but it seems that if you had a bunch of them floating over their own set of LEDs, every comparator would see a similar voltage drop and end up shorting out 1/3 of the total LEDs simultaneously (over 300 of them), instead of a more desirable successive shorting (i.e. the 100th LED on each string, followed by the 99th, 98th, etc.).

What if there were several comparators looking at the total capacitor output voltage, with different trigger thresholds? e.g. say the forward voltage is 350V, and we want to successively short out 3 LEDs per string (~10V forward voltage). Comparator1 waits for the capacitor voltage to drop to 340V, then shorts out LEDs #100,99,98 on each string using a big IGBT; Comparator2 waits for the voltage to drop to 330V, shorting out LEDs #97,96,95; and so on. I know those numbers don't quite add up (more for illustrative purposes), but that would short out 30 or so LEDs per step. Might even be able to reduce that number, depending upon the accuracy of the comparator.

Maybe those are excessively high voltages for a suitable comparator to deal with but, depending on the accuracy of the comparator, perhaps a voltage divider with high-tolerance resistors would make it work. Any thoughts?

Edit: I'm kind of thinking now that that sort of successive shorting would cause the capacitor voltage to jump around too much (due to the changes in resistance), making it impossible to tune all of those comparators properly. Maybe If comparator1 looked at the total voltage across LEDs #1-100, Comparator2 looked at LEDs #1-97, etc.?

 

Maybe, but lets pin down a couple of things first. Do you know what LED you want to use? If not, what color? I'm guessing you are using like 3 watt LEDs but pulsing them for the 10 watts?

 

Experimenting with a few different ones, but yeah, generally 3 watt LEDs pulsed for 10 watts

 

not sure what you mean, D6's cathode isn't connected to ground, its anode is. If you left the D6's anode at ground, and connected its cathode to the inductor output, the inductor would fight with both M1 and D6 when M1 switched off. If you moved D6's anode from ground to the inductor output, then it would create a short for inductor?

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I misspoke. I meant move the anode to the output of the inductor. It does create a short circuit (transferring the energy back to the capacitor) but only after the capacitor voltage becomes less than the output voltage.

I'd show you the simulation but I'm on a different computer that doesn't have the simulator.

 

Hey crutschow, I tried simulating it (see attached), the capacitor pushed current through the inductor and LEDs until it was close to empty (didn't go to 0V, instead it must have stopped when voltage became lower than the total LED forward voltage), upon which the inductor continued pushing current, but not through the LEDs, instead shorting to D6 with a very gradual current drop off. It doesn't seem that the inductor ever manages to transfer the energy back to the capacitor, instead slowly dissipating it through the small resistance of D6. I tried a shorter pulse, to cut off the cap before it completely discharges, and a similar thing happens, both capacitor voltage and inductor output voltage are nearly equal, around 230V with cap voltage a little lower -- inductor current again has a very gradual drop-off, instead of charging the capacitor.

Edit: hmm, I guess that even if the inductor did charge the capacitor with its leftover energy, having D6 connected that way prevents the inductor from sucking the cap dry, which was the reason for the inductor in the first place.

Edit: just had a thought. Say the pulse width is to be varied between 1 and 10ms, and we choose the inductor to have ~0.01H, making the RL half-cycle ~10ms. Then, to be able to choose different pulse widths, would we be able to just switch the capacitor out of the circuit temporarily, before it completely discharges? Could that change the resonance frequency of the circuit, and result in the inductor driving the LEDs for a shorter pulse width?

 

What???? That scheme won't do anything. Not anything useful, at least. It's just a freewheeling circuit that dumps the energy by warming up the inductor and diode.