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Matching 700mA constant current driver to 20mA LEDs?

Hello all:

Ultimately I'm trying to drive upwards of 100 3mm white LEDs in an art project with dimming and a connection to 120v AC power (this is for a chandelier).

I've been playing around with some constant current IC driver like the Supertex CL2 as well as 5 and 12v power supplies but I've been looking for a dimmable solution with a small form factor. I picked up a Robertson constant current LED driver but it outputs at 700mA, I assume because it's intended for 1W or 3W LEDs needing the higher current.

Is there a circuit design that I can use like a current divider to drive smaller loads with this supply?

My current thinking is as follows:

1) I could simply load 35 parallel strings of LEDs since 700/20 = 35 (of an appropriate voltage drop probably between 9v-15v) and rely on the equivalent resistance of each string to act as a defacto current divider.
2) I could do the same thing but with an in series resistor of some value for each string, the constraint being that increasing the resistor will reduce the number of parallel strings and I may need relatively high wattage resistors if I'm driving 9+volts of LEDs on each string.
3) I could do a smaller parallel current divider, but I'm not confident in my math analyzing the ratios of the resistors to achieve this (for instance if I had only two parallel loads and one was drawing 20mA and the other the balance of 680mA the resistors would have to have the same (inverted) relationship i.e. 34:1...to say nothing of the wattage through the 680mA line, which might be as high as 6-8 watts depending on voltage. Is that right?

Are there any other clever solutions I'm not thinking of? Obviously I could buy a lower current driver (and I may ultimately)  but even the lower current options are at 350mA, so the same problem will exist at a smaller scale.

Thanks everybody for thinking about this, I look forward to seeing your thoughts!

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iceng2 years ago

Strategy in 3 points is to ;

1] Let the driver deliver its design 700ma

2] Design the LED load operate in the 9-24 volt regulation range.

3] Use resistors to guarantee equal current distribution.

DESIGN

Select the target voltage say 18 volts to allow for the LED condition of forward voltage drop as the LED warms up.

Next create a 20ma unit string put three LEDs in series with a resistor.

3 leds x 3.2vf = 9.6v || 18v target - 9.6 v leds = 8.4 volts across the resistor.

Resistor = V / amps = 8.4 / o.02 = 420, but we have to use a 430 ohm

Actual current at 18v || A = V / R = 8.4 / 430 = 19.5 ma

Power = V * Amp = 8.4 x o.02 = o.168 Watt we use 1/4w 430 ohm resistors.

Work out units needed to pull 700ma.

Driver current 700ma / 20ma per unit = 35 units in parallel Exactly !!

Total LEDs = 35 unit strings x 3 LEDs = 105 LEDs

Now determine actual voltage VR = R x Amps = 430 x o.02 = 8.6v

Actual driver voltage at 700ma = Vresistor + Vleds = 8.6 + 9.6 = 18.2 volts

You can drive 105 LEDs pulling 700ma at 18.2 Volts .... by wiring

35 parallel unit strings of 3 LEDs in series with a 1/4w 430 ohm resistor.

.

If you use less then 35 units the driver will push more current into the rest and raise its voltage.

If you use more then 35 units the driver will lower its voltage with a corresponding lower unit current.


michael.pokorny.54 (author)  iceng2 years ago

Thanks. This is what I expected I guess (in that I would need to use 35 parallel strands to split the current. It just seems like a lot to set up just to test the circuit, I was hoping for a smaller test case.

Does a resistor in this case provide some fault tolerance the way it would on a constant voltage source? I mean to smooth out any variance between strands or individual LEDs?

Yes a series resistor does limit current to each group of three LEDs and guarantees equal current sharing between units.

Use two 75W incandescent light bulbs in parallel as a 550ma simulated load see first two pics to test eight LED units.

Or use three 75W incandescent light bulbs in parallel as a full 700ma simulated load see the last two pics to test a single 3LED + R unit.

Source2Lamp.JPGTwoLampSupp.JPGThreeLamp.JPGThreeLampsSupp.JPG
michael.pokorny.54 (author)  iceng2 years ago

thanks for the pictures. I've set up something similar based on your suggestion, namely 3 75 watt bulbs in parallel powered from the driver. Since ultimately the intent is to use this with a dimmer I've also put a lutron dimmer between the mains power and the driver input. Since it's a forward-phase dimmer it is working to adjust the bulb intensity (but between the dimmer and the driver the bulbs (at least when all three are in parallel) barely light up.

I will test it later today with an LED strand as you suggest.

Success! This is my hacked together test rig. On the left is the lutron dimmer in a wallbox, that's plugged into the driver, which is connected to the loose sockets. Since they're low voltage and low power I didn't box them up yet. They're plugged into the parallel incandescents and then the LED rig on the bottom right, which is 4 LEDs in series with a 510ohms resistor. The only unusual behavior that I noticed was that in order to get the system to work I had to start it without the LEDs plugged in, add the incandescent lamps one by one and then plug the LEDs in. Once everything was lit I could use the dimmer to adjust the LED brightness (but couldn't see any difference in the lamps, as they were already barely on). As long as I didn't drop the LEDs below some lower limit I could dim them up and down. Once they dropped off I couldn't get them back without 'resetting' the system. I think that has to do with the current draw of the lamps, so hopefully that behavior will go away when I switch to a fully LED system.

Thanks for all your help everybody!

IMG_2929.JPG

NICE picture !

I'm pleased you used my dummy load trick and it worked for you.

Keep in mind incandescent lamps heave very low resistance until they heat up.

.

.

Are you actually able to dim and cause the constant 700 ma current driver to lower its design drive current below say 500 ma ??

michael.pokorny.54 (author)  iceng2 years ago

I don't have a multimeter that will read that high, so I've not confirmed the output current of the driver under different dimming settings. I don't think this answers your question, but the driver works off of forward-phase dimming, which I think is sort of like PWM.

It's called phase control...

Then when you have your driver and LEDs all set up and decide to dim it,

I would recommend you watch the driver for at least 5 hours to see if it starts to get smoking hot.

Then if it does not overheat, it will be good to go forever more.

iceng iceng2 years ago

BTW a multimeter can be had at Harbor Freight for $8.

michael.pokorny.54 (author)  iceng2 years ago

Yeah, I have a multimeter, it's just fused at only 200mA. I need to buy a better one.

michael.pokorny.54 (author) 2 years ago

This is the specsheet for the driver:

http://www.robertsondirect.com/Assets/Docs/spec_sh...

In a constant current driver shouldn't the voltage and the current be somewhat independent of each other? In as much as the load will draw whatever volts it needs but the amps will stay constant (in this case at 700mA?) The question is does the voltage load (of say a string of LEDs) match the current provided by the driver which will be constant given that the relationship between V and I in an LED isn't fixed?

As an experiment I connected to the large driver 4 parallel strings of 4 LEDs each, each with a separate CL2 (spec http://www.supertex.com/pdf/datasheets/CL2.pdf) to force the current down to 20mA. All 16 LEDs lit up for a minute then started to fail, flicker and overheat, though interestingly not all of them (12 out of 16 failed). I can't think of why. Each strand should have been drawing 12v at 700/4 = 175mA, well within the specs of the CL to clamp down to 12V at 20mA. I confirmed that the CL2's had not failed, so I can't see how the LEDs would have, given that they shoudln't have seen more than 20mA of current.

Thanks!

"In a constant current driver shouldn't the voltage and the current be somewhat independent of each other? In as much as the load will draw whatever volts it needs but the amps will stay constant (in this case at 700mA?) The question is does the voltage load (of say a string of LEDs) match the current provided by the driver which will be constant given that the relationship between V and I in an LED isn't fixed?"

To be clear,

1) Small clarification: Voltage is raised or lowered to continue outputting a setpoint current.

2) As I recall, at least some constant-current sources have a finite control band for the source voltages with which they work. So you may to stay within it's voltage control band (if it has one). Otherwise, it simply adjusts it up or down to keep the current flowing at the same rate and once configured (if it requires and/or allows any) it's a set-and-forget (within reason) process.

Parallel connection *does theoretically allow independent calibration of each LED output, but I don't see any evidence that you're doing that using this bulk parallel drive (the 700mA constant-current source) , so there's no real benefit to doing so imo.

It's none of my business, but I'm wondering why you have to push these LEDs to their max ratings in the first place. Doing so places it pretty much out-of-range for durable use in the most sensible space saving design, which would be a series connection, since any LED in the string that deviates from the norm will likely cause problems for the entire string.

Based on the above observation, I recommend that you consider using only up to ~75-80% of the rated I(max) .

BTW, I agree with Steve's suggestion of stringing them as a defacto-direct AC driven layout, using a single load resistor for each string as a limiter. It's an economic use of space and equipment. PS> Dimmer is as easy as a wall dimmer for a lamp.

Say there is no load at all and you turn the device ON

Would you expect the voltage to climb to tens of thousands of volts and ionize the gas between the terminals until 700ma would flow.

Not really ! .... There is a practical limit upper voltage below where the voltage controls the current flow . Your PDF said a control range of 9-24VDC.

That suggests 3 LEDs in series, now lets target 12VDC and assume each led needs 3V at 20 ma and a series resistor to force load sharing.

Three leds are 9v leaving 3v for the resistor at 20ma.

R=V / A = 3 /o.02 = 150 ohms .....P=V x A = 3 * o.02 = o.06 Watt

You already determined 35 of the 20 ma strings should run off of the 700ma except in this case each string lights 3 leds for a total of 105 leds.

The only assumption I made was the forward LED voltage.

michael.pokorny.54 (author)  iceng2 years ago

You're correct of course about the upper V+ limit. I measured the driver with no load and got, as expected, 23v+ as a reading.

I know current and voltage are related in a normal circuit such that you can use a resistor to 'adjust' the current limit for the LEDs, I guess my question is can I do something similar with a constant current source, and if so, how?

You still need to tell me the led forward voltage when running at design current of 20ma ????

michael.pokorny.54 (author)  iceng2 years ago

Sorry, 3.2vf nominal

A constant-current driver is meant to do just that. Drive a constant current at the rated output as long as it can upwardly (or as necessary, downwardly) control the output voltage to maintain that constant current output.

You could apply a known, fixed resistance to adjust the effective output votlage to a specific value, but it will still attempt to deliver the rated output current no matter, since it's job is to deliver current, not voltage. That is, whether your effective load has a resistance of 100 ohms or 10 ohms, it's going to try to deliver N (rated) amps *constantly*

If you put 32 LEDs in each string, with 32 LEDS in a parallel opposite string, in series with a 1K 0.5 Watt resistor, it will work directly on line. Each LED is mains live.

Better still, one LED one way, with one LED in parallel the other way. X 32

Just tell us the LED forward volts.