LED array current regulation

I'm trying to decide how to drive ~100 UV leds. The Vf is 3.2-3.4 V typ, 3.8 V max. I have a 36.5 VDC power supply (measured @ 0.1A). I need If of 20 mA (no control needed, just on or off) through each LED. Simple and cheap are preferred, provided the results are "good enough". The application is PCB photolithography.

So far, I've looked into a few options:

  1. Resistors. 10 LEDs in a string, 175 ohm resistor. Cheap and easy, but slight variations on Vf would cause big variations in If. Bigger resistors and shorter strings perhaps? 470 ohm w/ 8 LED strings? Still only 10% regulation, not sure if that's OK for LEDs (I don't have much info on the specific LEDs I got, other than what I've listed above).

  1. Infineon BCR402UE6182. This is a discrete single-channel current regulator, but doesn't seem to guarantee a low enough If.

  2. MAX6971 (or 2xMAX6970). Only needs a resistor and clock source to drive up to 16 strings at 20 mA, but the lead time is 4-5 weeks.

  1. MAX6957. In stock at Digikey, but requires an MCU to program it and common-base transistors to isolate it from the high voltages.

  2. Redoing the power supply for 18VDC to accomodate some other solution.

Suggestions? Alternatives?

Reply to
DJ Delorie
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An LM317 and a 62 ohm resistor makes a 20mA constant current series element. 9 LEDs in a chain leaves enough voltage for it.

Digikey have 100mA TO92 LM317s at $0.43.

Reply to
nospam

Here's one way:

4 strings of 9 LEDs & 8 strings of 8. Each string gets an LM317 and a 62 ohm resistor to give you very close to 20 mA. +36.5 ---+--Vin[LM317]Vout---+ | Adj | | | [62R] [.1uF] | | | +----------+---[LEDstring]--+ | | Gnd -----+------------------------------------+

If all the LM317's are physically close, you can use a single .1 uF cap near them.

Ed

Reply to
ehsjr

36.5V - 1.25 - 3.8V * 9 = 1.05V, which isn't a heck of a lot of overhead for an LM317. If that 36.5V turns out to be nominal instead of actual, then you're really in the suds.

Paranoia suggests making it strings of 8, and then only after double checking the minimum _ever_ voltage from the supply.

Alternately you could use transistors at the base of the string, with current regulation via op-amp. See below.

Rs senses the transistor emitter current, which is pretty darn close to the collector (and hence LED) current. Set Vref/Rs to equal your desired LED current. If you make sure to drop a volt or so through Rs and temperature match your transistors you can probably gang the bases of all the transistors to one op amp. This buys you a whole volt or so of overhead, but if you like pinching pennies it has the attraction that the opamp and transistor are probably less than a 317, much less one op-amp and 13 transistors vs. 13 LM317.

| | controlled | | current | | | V |\\ | Vref o---------|+\\ |/ | >------| .---|-/ |>

| |/ | | | '---------------o | | .-. | | Rs | | '-' | | === GND (created by AACircuit v1.28.6 beta 04/19/05

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Reply to
Tim Wescott

How about creating an N-plus-one-transistor current mirror, with N transistors (pick your favorite TO92 NPN jellybean) driving the LED strings, and transistor N+1 being fed the reference current through a resistor to the 36.5-volt rail. Emitters to ground, bases tied together, collectors to the LED strings. With 9 LEDs per string you'd need a total of 12 transistors (pick 'em out of the same manufacturing lot and they'll probably be matched closely enough for your needs) and one resistor to set the reference current (you can drive the 100th LED via this resistor and the reference-setting transistor).

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Reply to
Dave Platt

DJ Delorie wrote in news: snipped-for-privacy@delorie.com:

How I'd do it if you really want identical current: all LEDs in series and use boost regulator. Of course, the switcher will develop over 300 volts, so this approach has its disadvantages. You could also consider using, say, 4 strings of 25 LEDs and 4 boost regulators. There are lots of control chips designed to provide constant current boost output. OTOH, if you reduce the number of emitters in a series string so that there is more voltage across the resistor, you'll find that resistive current limiting actually works pretty well. Emitter follower drive, with a current-setting resistor in the emitter, also provides good current regulation for low cost if you can drive with a constant voltage. Paul Mathews

Reply to
Paul Mathews

Paul Mathews a écrit :

Or use this to share the boost current between the leds strings:

.-----------. | BOOST | | |-----+--------+--------. | | | | | | | V -> V -> V ->

'-----------' - - - ^ | | | | . . . Leds strings | . . . | | | | | V -> V -> V ->

| - - - R1 | | | | ___ | | | +--|___|-. | | | | | | | | | | | |/ |/ |/ | | +--------+--------+----------+ | |> |> |> | | | | | | '-----------+ | | | .-. .-. .-. .-. | | | | | | | | R2 | | | | | | | | '-' '-' '-' '-' | | | | === === === === GND GND GND GND

--
Thanks,
Fred.
Reply to
Fred Bartoli

That's just right for my purposes. Thanks!

Reply to
DJ Delorie

True. But 3.8 is max vf - typical is 3.2 to 3.4. At 3.4*9 + 1.25 in the 317, you still have 4.65 overhead - plenty for the LM317. And, as you suggest, it's easy enough to add another LM317 and drive 13 strings.

Right. If we can't accept the 36.5 figure, then all bets are off. He appears to be trying to get by with the supply he has on hand. He measured 36.5 at 100 mA. If it drops too low when more current (240 mA, 12 strings) is drawn, he needs a different supply or he needs to reduce the current to what the supply can provide. Fewer strings means less total current - and less available overhead for the current regulator. He said he wanted results that are "good enough" - he'll need to decide what that means.

Regarding the information below: Perhaps less costly, but you need to provide power for the op-amp and match the transistors for the single op-amp design and provide for the Vref so it's a bit more complex than shown below. Still, it might be worth the complexity if the extra overhead turns out to be necessary.

Ed

Reply to
ehsjr

I mistakenly read the 3.2-3.4v as the diode spread. As someone else pointed out 9 in a chain is marginal if you get unlucky with the LEDs. The LM317 will need about 3.5v across it to regulate.

Reply to
nospam

The LT3080 looks interesting for this sort of thing:

Last LED ------------ Many LEDs ---+--->!-----! IN OUT !--------- ! ! SET !--- ! ----------! Vcontrol ! ! ! ------------ \\ \\ / / \\ \\ ! ! GND GND

Reply to
MooseFET

That was listed as "nominal Vf" and I measured one at 3.3v @20mA. Anyway, I can prototype it and see what happens, before committing to anything.

Reply to
DJ Delorie

That is slick. I will remember this one.

Reply to
JosephKK

To follow up on this - I got my order of LM317s and built a supply using them - 12 independent 20mA supplies. The bulk voltage drops to

35.5 VDC with an 0.22 amp load (9x11 LED matrix), and a string of 9 LEDs drops 29.4 V, or 3.27 V per LED, leaving 1.25 V for the 62 ohm resistor and 4.85 V across the LM317s. Even at 3.4 V per LED that still leaves 3.6 V across the LM317s. I think this will do just fine for my purposes. Thanks!

Photos:

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Reply to
DJ Delorie

V

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ap and easy,

in If.

470 ohm w/ 8 LED

f that's OK for LEDs

er than what

nt

f.

to

-5 weeks.

t

voltages.

A simple BJT constant current source does it for about 10c for each chain of leds.

Reply to
cbarn24050

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