lm317 as current regulator

No, it's the differential across it that's limited to 40V.

You probably need a resistor for each string, to equalize currents. And there's another reason you need extra resistance in the circuit; your regulator will get too hot. Even assuming all the led strings have equal current and voltage, look at the circuit -- you drop 40 volts across the led's (12 X 3.3) and 1.2 volts across the resistor, leaving 18.8 nominal across the terminals of the lm317t. At one amp that's 18.8 watts, way to much power unless you have one hell of a heatsink. So use a resistor in each led string. It will equalize currents among the strings and get rid of extra heat. You want to drop about 15 volts. Exactly how much depends on how stiff your power supply is. But for now we'll say 15 volts. Now since you want 25 mA in each string, you will get a 15 volt drop with about 600 ohms (or nearby standard value) resistor in each string. Use a half watt or one watt resistor. If your power supply sags with an amp load, you could start with a 560 ohm resistor and see how that works. If your supply doesn't droop you can go for more than 600 ohms. The next standard value is 630. The higher you go with the equalizing resistors, the cooler your regulator chip will run, so try different values until you find the right one.

Thanks kell, I have decided to abandon the current mode regulator function and go with the lm317hvt regulator in voltage normal regulation. The regulators can handle 1.5 amp and my led board I designed is split in half, as to one side of 12 X 40 string then the other mirror image to that. So a total of 960 leds. The reason I abandoned the current mode is after reading what I typed in I realized that by doing current reg mode, If 1 or 2 strings of 12 opened, It would over power the others and the whole board would burn out. So Voltage mode it is. I am going to run the lm317hvt (2 x parallel) to supply the 2 amp desired current. This project is for an led light source to make an lcd projector. Started out making a board to run at 102 vdc from mains direct regulation, and decided for safety reasons to shy away from that foolishness. This seems as far as voltage and current demand to be more logical. Lower voltage - higher current , that is why firstly looking at the 102 volt, as I could run 1024 leds with only 800 ma. So this is middle of the road on voltage versus current. If anyone wants my pcb designs, I will send to then once I get them tweaked just right. The led board is 8" X 8". JTT.

How much does your 60V power supply vary its voltage?

Your main need would be to avoid wasting power, if you're considering dropping

As at least one earlier poster said, you'll want a single current limiter resistor for each series string of LED's. If you try one LED at the maximum current you want it to take, measure the voltage. Better to do this empirically if you want to cut it close to save power.

LED voltage drops will vary, but not much, so each string's resistor should not need to drop more than maybe a fifth of what any single LED will drop. Your single LM317 current regulator (and it's control resistor) need only bear the difference between minimum and maximum expected supply variation, plus up to one LED's worth of voltage drop, plus its own total voltage drop at 1 amp. (A low dropout type might save another watt at 1 amp).

Setting the optimum number of LED's is easy, start with one string of a few more than you need, and cut back one at a time till the string lights up with the lowest expected supply voltage applied. LED voltage drops are so well behaved that if your PSU is stable to within less than two LED drops, you can probably do it right with simple caclulations and not worry about overcurrent or waste.

More: I just saw the post about overcurrent if any series string went opencircuit. I agree, voltage control is safer. You can still work that inst the context of my main post though. Why set to exactly 12x40 array? The connection array need not be the same as the physical one. Any inconvenience might well be more than offset by the saving in power. You can regulate closer to 60V even though the LM317 has a differential limit of around 37V, you put a zener of 30V or 40V in the adjust network's connection to ground to get to a safe operating range on a

But if you're drawing 1A with a drop of 40V you'd better check if the chip can handle 40W!

If the specs say it can handle 1.5 amp with a differential of 37 volts, would that not mean it can handle 55.5 watts. ?? My circuit will have a differential of 20 volts at 1 amp, which means it will disapate 20 watt to heat. I have a fairly good sized heatsink for it so all should be well. My transformer can source around 4 amp, so it also should stay fairly cool. I am designing it and pcb in a rectangle shaped caseing with a cooling fan on one end to dirct airflow over it all. Other than the led's, I have a second winding with 17 volts to run a lm7812 for running the lcd electronics.

No. :) You might get 1.5A, OR 37V safely handled, butthere are things like heat dissipation to consider. Even then, I doubt the very best preparations will keep the junction temperature low enough if you try for both at once.

Please read my other posts, I've tried to address this 20 volt drop. It's a great waste that could be easy to avoid. I've probably made a gaff or two but nothing uncorrectable. The main weakness of my case might be in assuming excellent consistency in the voltage drops of your LED's, but if you're wanting

If I were making this panel, I'd use a different transformer, or a cheap effcient power converter if I could get one, to make 140 volts or more. If I used an old valve transformer which could put out 280 - 300 VAC there need only be 6 chains of 80 LED's, easy to wire in your array. Each would have a single LM317T configged as current regulator. This way you need 6 of them, but you don't need to waste power in series resistors, and most of the current will be passing usefully through LED's, and each regulator will be running at around

Small extra point about safety:

My point is, that if you're running 60 VDC you're going to need protection such that running 300 VDC isn't going to be any worse if you do it safely.

Oh boy. You really need to settle on a design.

Don't ever use two LM317's in parallel, without equalizing resistors. In this case, do not parallel the LM317's at all. Use a 3 amp LM350 - one heatsink, simpler design.

Assuming your 60 volt DC supply is tightly regulated and will maintain 60 volts under a 2 amp load, use an LM350K mounted on a good heatsink - Mouser part # 567-641-A will work - and two 25 watt resistors in series between the +60 and the input pin on the LM350. One is 3 ohms, the other is 3.3. Each one is 25 watts. Don't change this - it is to spread the heat.

That will drop the voltage at the input to the LM350 to 60 - 12.6 or 47.4 volts with a 2 amp load. Keeps some of the heat out of the LM350.

Set the LM350 for 44 volts out, with 240 ohms for R1 and

You *still* have a bucket of heat in a small area with the LEDs: 960*.025*3.3 = 79.2 watts

Finally, what are you using for your 60 volts DC source? The design above assumes a well regulated 60 volt DC source capable of at leats 2 amps. You know what "assumes" means.

Ed

The transformer is from an old hifi receiver which gave 60vdc after the bridge on the caps. It is center tapped, but not using the center. My led board was designed as 2 banks sharing a center conductor, so I could feed each side separately. This let me have 1 amp per side, and that is why the lm317 being 2 of them sharing a single adj connection, but feeding each side independently. I worded it wrong last post. If I were to add a 10 ohm 10 watt resistor from the regulator output to feed the led string, that would help on the dissipation of the regulator- right. It would drop 10 volts at 1 amp on each the resistor and the regulator. There is going to be forced air cooling on the led board as well as this power supply.

Your right, and it can sneak up on you. Got zapped good from a meter calibrator supply once. It was set at 1000 volt too. Worse yet is radio, had the top of of an old military tube radio and leaned on the top edge and zap, made me buzz all day.

I will be very carefull with the danger there..

There is also the LM317HVK which has a 63V or thereabouts input voltage, made in TO-3 and TO-220 formats.

The 1.5A output quoted for the LM317 is only for the TO-3 version IIRC, the TO-220 build has a lower output current capacity unless you are well below its safe operating area.

Peter

-- Peter & Rita Forbes Email: snipped-for-privacy@easynet.co.uk Web:

Yes.

Don't white LEDs typically have a 4-5V Vled? You really will need 60V to drive this panel.

Nope. Why do you need an lm317t anyway? Just stick a current limiting resistor in front of the panel to limit the amount of current through it.

My question is how bright will the LEDs be if you only run 1A through the panel.

Will you have enough voltage to drive 12 white LEDs at 42V? What's the forward voltage drop of these LEDs?

Also do you have individual access to each string? The truth of the matter is that each string really needs its own current limiting resistor.

BAJ

Update: Advice taken from replies. Pcb revised to 17 led per string, with a 20 ohm .5 watt resistor per string. At 3.5 volt per led fv drop that is 59.5 volt drop per led string, with the remaining .5 volt on the 20 ohm resistor which sets the Current per string to 25ma and therefore has the resistor dissipating .0125 watt. The resistor being .5 watt will not even get warm. All power from the supply of 60 volt is used for producing light instead of just heat. If this is correct, I will not even need a regulator. Total Current draw for the led's will be

How do you know the transformer is capable of 3+ amps? (You need more than 2 amps ac to get 2 amps DC in this set up.) How much capacitance? Under what load did it give 60 volts?

You can do that if you want, at higher cost. The LM350K on a single heatsink is lower cost and takes less space than 2 LM317K's on two heatsinks.

Yes, and it would screw up the regulation. In this case, with a fixed load, that's not a problem, provided the +60 is tightly regulated. But you'd get the same benefit with the resistors on the input of each LM317, and would not screw up the regulation, again assuming tight regulation of the +60 volts. If the +60 volts sags appreciably (more than 3 volts) under load the resistors have to be changed or eliminated. And if it sags more than 13 volts, you can't set the regulators for 44 volts output.

No. The resistor would drop 10 volts at 1 amp. The regulator would drop Vin-Vout at 1 amp.

If the +60 volts is tightly regulated and you set Vout to 54 volts, the regulator will drop

By tightly regulated, I mean that there is very little variation in the +60 volts from no load to full load, with full load being 2 amps.

Forced air cooling on the power supply is good. With the design I posted it will work fine. I cannot predict with accuracy what will happen to the LEDs - just applying forced air cooling may not cool them enough.

Regarding your 60 volt supply: you'll be using at least 120 watts in your LEDs and regulators. (2 amps at 60 volts) Your 60 volts supply has to be capable of at least that.

Ed

It's pretty good :) You got the idea of getting the LED's to use all available voltage overhead, reducing the need to get rid of waste heat. If your PSU is very well behaved, you don't need the regulator. Just bear in mind that your resistor is only dropping half a volt! While that's efficient, it's a lot closer than my posts advised, so your supply is going to have to be VERY well behaved. I think you'll find that the line voltage will vary more. 0.5V in so is 1 V in 120V, or 2V in 240. You can be damn sure that the average mains supply is going to drift more than this.

I still advise using s valve (vacuum tube) transformer for around 300V to get strings as long as 40 LED's, and replacing each resistor with an LM317 current regulator. You only need 6 of those then, and the performance will be far better and far easier to control.

I'd been thinking about the heat from the panel itself, btw.. I don't know what the efficiency is for those LED's. maybe 15% or so? So not all power in will be heat out, so at least one previous poster gave wrong figures for heat. I guess it could be 70 watts lost, from a likely panel size of 9x30 cm. The odds are that you won't need a fan, if you can get decent convection around the LED's.

Not really. In fact, it's a disaster. The only regulation in evidence is the transformer; the only supply in evidence is the transformer, diodes and capacitor bank. Look at worst case: Vf 3.3, line voltage 130. Vary the line voltage up to 130, and his 48v transformer will put out 52v. Allowing a volt drop for each diode at ~2 amp, that's 50 * 1.414 or 70.7 volts. At Vf 3.3, 17 LEDs drop (17*3.3) 56.1 volts. That means 13.9 volts across his 20 ohm resistor, for 695 mA.

Disaster - almost 10 watts for the 1/2 watt resistor and more than 27 times the target current through the LEDs. And with 56 strings (960/17)= 56 + the total current would be (56*.695) ~ 38 amps!

Assume it is not a 48 volt transformer. (Well right away we're in trouble with "assume".) Lets say its a 44 volt transformer. At 130 volts, it will put out 47.7V which after the 1 volt diode drop (2 diodes) becomes ~64.6 (45.7 * 1.414) = ~ 64.6 volts. With Vf = 3.3 the 17 LEDs still drop 56.1 volts, leaving 8.5 v across the 20 ohm resistor for 425 mA current. Goodbye to one or more of powersupply, LED, resistor. The magic smoke will get out.

If you go with with a higher value resistor to get .025 mA, that's 8.5/.025 = 340 ohms (best case transformer, worst case Vf and line voltage). What happens when the voltage is 120, and the Vf is 3.5 if you use a 340 ohm resistor? .5/340 = ~1.5 mA and the LEDs are too dim.

He *must* regulate. *If* he posts that the 60V supply is really a regulated 60 volts, not just a transformer, bridge and capacitor bank, then a resistor current limiter would be appropriate. In that case the regulator is already there, in the supply. He has *got* to have a regulator in there, somewhere. It's an awful lot of work to put 960 LEDs in there - it certainly warrants spending $13.00 on regulation.

Bad idea. He has *got* to regulate. That means a linear, which will waste power as heat, or a switcher, which is not under discussion here. If he is the same person who has been e-mailing me since 6/13 under a couple of different id's, then he has a switcher solution that is cheap ($10.00 for the switcher) and solid, but requires work: 3 LEDs per string. He also has been given a solution in e-mail that works without a transformer and is easier (40 LEDs per string) - but is dangerous due to high (up to ~182) voltage.

The waste heat from the regulator is of no concern, other than cost. It requires about 13 dollars for the LM350K and heatsink. It is not required to be near the LEDs, so it won't cook them.

Yes, he does. It can be very well behaved, but if there is no regulator in the supply, line voltage variation can kill the LEDs, as shown. Varying the worst case the other way (line voltage = 105, Vf = 3.7) will just make them dimmer than nominal.

Exactly.

That's a non-starter. The differential across the LM317 would be huge. 40 leds at 3.7 = 148 volts, which means

what

be

Yes/no. A total of ~79 watts will be produced. Some of it will be heat due to inefficiency, the rest (most, I would think) will convert to heat both in the LED and external (far away) to it. Light energy becomes heat energy as it is absorbed in a surface; even a transparent surface absorbs some and converts it to heat. So he'll have 15% or whatever due to inefficiency, plus an additional amount due to absorbtion. The point is not how much heat exactly - I don't know - but how do you get rid of whatever heat is there. If there is extremeley limited convection, I don't know how to do it. If it's the same person, I told him in e-mail that I was not qualified to tell him how to deal with the heat in the LEDs. But I doubt that he'll be able to cram 960 leds into as small an area as possible without a significant heat problem. In the e-mail he spoke of replacing a projector bulb with these LEDs. If he can spread his LEDs out so that decent convection between the LEDs possible, that worry becomes *much* smaller than what I envisioned.

Ed

If you're going to tear up a post and say four paragraphs to the extent of saying that mains power isn't well behaved, you'd do better to quote the context. I said that in ONE paragraph.

Yes, IF the initial supply isn't well behaved, as in regulated to start with. I said this. You chose to omit that from your quote, but that doesn't make it unsaid.

That means a linear,

Now I get it. :) You're heading me off at the pass, you've offered a private solution and you would rather tear my posts up than let them challenge your efforts, it seems. As I've also mentioned in my posts switching regulators for efficiency and 40-LED chains at high voltage I can see why it might look like competition to you. :) Incidentally, if you also concede that making longer strings for higher voltages is good, why did you say only a few sentences ago that my suggestion of making the LED's drop most of the voltage was a bad idea?

Yes, yes, I DID mention that the mains supply WOULD would vary far more than the 0.5V he'd allowed across his limiting resistor, but you chose to cut that out of your quote before proceeding to administer your professorial thrashing.

Ah, so you did quote it. :) Hell, you even agree. >:) So why bother with those

CURRENT REGULATOR! I guessed someone would see my loose phrasing and make a more serious mistake, but I didn't expect it to be someone who posted like a maths professor. Look at the data sheet. The LM317 can be configged as a current regulator, even though it is a voltage regulator. And as such it does NOT see a high voltage. It doesn't care about the voltage, all it sees is current (actually, the small voltage across a resistor), and the high voltage is mostly dropped across LED's, hence the increased efficiency. And yes, for the record, the LM317 itself IS limited to 37V, and if you actually read my posts you'll know that I also know this. You'd also know by now, even from this quote of my post, that I was insisting on regulation in an ideal solution.

I'll relax my defence on the heat issue, as I have no idea what actual form this thing wil take. A fan won't hurt, for sure. As worst case you can assume heat out locally equals heat in, if you budget for that you won't go wrong, but in practise this thing would have to be DIRE if that little light got out, and it certainly wouldn't be competing with a projector lamp. Yor figure of 79W is the total power drawn by the LED's given 25 mA and 3.3 V dropped for each one. Again, if 15% of that energy got out as light, that's about 70 W to be removed locally as heat, as I said.