LEDs different current/voltage

The main problem is that LED voltage at any particular currrent is not a well-specified number, and also varies with temperature. Thus, if your string is made up of only LEDs, the current may vary considerably. If you're driving it off a car-type battery, which is effectively dead if it's actually at 12V, the current may be more than you want, even if the voltage drop for your particular string is actually 12V at 20 mA. Order does not matter. Better LED driving schemes approximate, or are, current sources. LED current is a fairly well-specified parameter, so controlling that and letting the voltage do what it will is a good paradigm.

Other than it not putting out as much of its potential output (ie, it's running at 40% rated current while the others are running at 66%), no.

That's the spirit. They don't cost much.

No. The purpose of the resistor is to set the current. You need to find a higher voltage supply, or divide the LEDs into two groups and run those groups in parallel, each with its own current setting resistor.

(snip) No.

Source is a AC to DC power brick; output 12V 1.5A I am also experimenting with 12 VAC transformer to drive two of these strings on each cycle.

Makes sense. Thanks for the advice, Lawrence. I ponder whether its possible to give 30mA for the red and 20mA for two others, within the same serial string, using two resistors. My gut says no, since two resistors in series would be added together.

• ---100R---30mA---160R---20mA---20mA--- -

Plan B is to remove the 3.2V LED to free up voltage for the 1.9V red LED, but may have to go to (Plan C) two strings with multiple reds driven at the higher current for showing the shorter wavelength light better.

Thanks! Scott

John,

Thank you for your advice. Power supply is an AC to DC power brick outputting 12V 1.5A. Plan B is to remove the 3.2V LED to free enough voltage for the 1.9V LED and using the approporiate resistor value. Plan C, if I find the red light too dim will be to split the string into two and adding more reds.

Scott in Dunedin

...

You would put one in parallel with the LEDs that are to get lower current. ie, you want 30 mA through the red one, and then 20 mA through the others, with 10mA in parallel going through a resistor. However:

Other than elegance, you have no reason not to run as many parallel strings as you like (well, up to 50-75 of them, at 30-20 mA per string), given 1500 mA of available supply. Use one resistor and 1-3 LEDs per string. If your end product requires some sort of balance, one LED per string allows tuning the current of each (by choice of resistor) independent of the others.

"Elegance" in this case being "minimal use of power for the same result"

- one LED per string means more power wasted as heat in resistors for the same light out. But it's fairly minor if this is not a 24/7/365 product.

I'll have to refresh my memory on resistors in parallel. Thank you so much for your advice, it has put me on the right track.

I intended on using 12 of these strings, so the 1500mA transformer is probably overkill, but it does leave me open for further experimentation.

"Minimal use of power for the same result" is what I am after; I was trying to avoid too many resistors and the subsequent power waste/heat through multiple resistors.

Again thank you very much for your assistance. I'm tempted to ask how to calculate the parallel resistor value to obtain a 10mA reduction, but its not something I want to waste your time on.

Much appreciated! Scott in Dunedin

Check the math on that before you get too worried. It's not much. 1/3 to

Measure the voltage across the LEDs you want to run at 20mA when they are running at 20mA. Calculate a resistor that would draw 10mA at that voltage, and put that resistor in parallel with those LEDs. Put the whole business in series with the resistor and LED running 30mA.

ors

If you don't want to waste power don't uses resistors at all. Just wire multiple (same type) LEDS is series to give a voltage drop of

series and no resistor is needed.

Cheers

A temperature increase of x in an LED causes a current increase of exp(x) if the voltage across it is fixed. So, on a hot day, your current string may start putting out enough heat to heat itself faster than it can cool off. This will eventually cause one of the LEDs to fail.

However, if you put a resistance in series with it, any increases of current are balanced by a decrease in voltage across the LEDs(since the increased current causes the resistor to have more voltage across it). This will decrease the current, and so provide negative feedback, preventing runaway current increases.

A small resistance will be fine. For a 12V, string the LEDs in series, and try out various resistances until you find one that works, starting with 1k or so, and moving downwards in resistance until you get the maximum number of LEDs along with the minimum resistance (down to about 47 ohms or so)

Regards, Bob Monsen

When you say the supply is 12V, do you mean 12.0 V? You've specified diode voltages to the nearest 0.1V, but what really is the supply voltage to this accuracy? Did you measure it, or are you just going by what is written on the supply?

For that matter, did you measure the diode voltages, or are you just going by what is written on the diode packages?

Best to measure all voltages, and use those numbers for calculating your resistor. A "12V" supply rated for 1.5A, when drawing only 20 mA, can very well have a significantly higher voltage.

Regards,

Mark

if

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I take your point on termal runaway but if you keep within the current spec of the LED (measure it) they should not run away. If you put enough in series that you don't need to drop 12V-> 1.7 across 1 resistor you'll be much more efficient. So as a better solution how about ust add up the nominal LED voltages to be less than the PS voltage and select a small resisior to control current. e.g. if they are 1.7 V use 6 LEDS in serries to give 10.4V drop and then drop the remaining 1.6 V at 30 mA across a resistor?

Cheers

Ecnerwal,

Thank you for your reply. I just wanted to let you know the preliminary experiments worked out without any immediate failure, and I very happy with the results so far. Although I do not have my notes handy, I did have to tweak the math a little for the resistors immediately available to me. Thinking of the current instead of voltage gave me a better view of what I was working toward. I still feel they may be underpowered because I determined the minimum voltage required for each LED and worked up from there, maintaining the 30mA and 20mA currents.

Thank you for your assistance. Scott in Dunedin

PLOSSL, Thank you for your reply. Indeed my original experiments used no resistors and simply relied upon the even distribution of voltage. Your (or Varactor's) suggestion of a small regulating resitor on a string of similar spec LEDs was what I was attempting to minimize variations in voltage. However, since I wanted to use a red LED within the short strings (3 or 4 LEDs), controlling the current instead of the voltage was the way to go, since it is not a similar spec LED.

Bob Monsen, Resistors as a thermal regulator it turns out, is an added benefit - excellent point. My earlier experiments did seem to run hot during the day. And many of my LEDs have been fried from overvoltage from a non-regulated power supply pushing just a little too much.

Thanks for your help! Scott in Dunedin

Mark,

Thanks for your reply. My 12V 1.5A DC power supply outputs 11.8V to 12.2V with a 20mA draw (during the times I've sampled it, though it's probably even more variable). My 12V AC power supply (88Watt)outputs between 11.8V and 12.4V depending on the load. I've been using 12.5V in my calculations to be on the safe side.

The LED voltages were simply specs from the manufacturer, however experimentation has revealed maximum voltages 0.5V higher than spec (at some risk of early failure), and minimum voltages much lower (1.2V less than spec's typical voltage drop) than I expected. I will keep your suggestion in mind as it is more accurate to calculate with actual values than with specs. Specs are convenient as a starting point though.

Thanks! Scott in Dunedin