How supply power for changing number of LED?

The brightness of an LED depends on the current through it; the voltage across it varies little over its useful range. so, if you want the brightness to remain constant, you need a constant *current* source, not a constant *voltage* source.

In the usual circuit, you have a constant voltage and a resistor. Since the LED has a "fixed" voltage across it, the resistor does too, and you can use the value of the resistor to determine the current through the circuit. However, if you change the number of LEDs, you change the voltage across the resistor, and thus the current.

If you want the current to remain the same regardless of how many LEDs there are, what you need is a constant current power supply. You can use an LM317 with a single 62 ohm resistor as a 20 mA constant current supply, but you need to get one beefy enough to handle the *minimum* number of LEDS (largest power dissipated by the regulator) and the

*maximum* (highest needed input voltage) at the same time. 50 LEDs would require about 100 volts, but 20 could be doable... that's about 40v input, and a big LM317 can handle up to 57v input.

So, example... 20 LEDs at 2v, 20 mA = 40v output. Add 2v for the regulator itself, and 1.25v across the 62 ohm resistor, say 44v input. Now, 5 LEDs is only 10v, possibly as low as 9v. That's about 34v across the regulator, or about 0.7 watts In a TO-220 package, it'll run 50 C/W or 35 degrees C hotter than ambient, well under the 125C limit.

Once she finalizes the design and you know how many LEDs are in the string, you can use the same circuit but with a smaller input voltage to reduce heat loss. If she needs more than 20 LEDs, just make two or more circuits. I've got a project with 11 LM317s in it running 11 strings of 9 LEDs each off a 37v supply, works great.

Alternately, if the input voltage is fixed already, figure out how many LEDs you can put in series before you run out of voltage drop, and put strings of those in parallel with each other (each string has its own suitable current-limiting resistor). For example, if you have a 12v supply and need 18 LEDs, you could have three strings of five LEDs (10v drop) each with a 100 ohm resistor, plus a string of (example) 3 LEDs (6v drop) with a 300 ohm resistor. This is probably the easiest solution if you don't *have* to have all the LEDs in series.

You haven't stated one critical fact: are the LEDs in series or in parallel? If in parallel, is there a single dropping resistor, or one per LED?

The critical facts are this: LEDs have a specific forward voltage drop (depends strongly on the LED color, and somewhat less strongly on the amount of current flowing through the LED). LED brightness is a direct function of the amount of current flow. If you put LEDs in series, their forward voltage drops add together.

Therefore: the trick is to choose the value of the dropping resistor, so that the voltage across the resistor (which will be TotalSupplyVoltage minus the LEDForwardVoltageDrop) results in the amount of current you need to get the amount of brightness you want.

Several scenarios exist:

- LEDs are in series, with one dropping resistor. If you add an extra LED to the string, you'll increase the total LED-string voltage drop, decrease the voltage across the resistor by the same amount, and you'll need to decrease the value of the resistor in order to maintain the desired current. Trying to drive one or two LEDs from a high voltage is wasteful of power... most of the voltage is across the dropping resistor, and most of the power is dissipated in the resistor as heat. Trying to drive a long string of LEDs from a voltage which is just barely above their voltage drop is efficient, but tricky. Small variations in the supply voltage can result in a proportionally- large change in voltage across the small-value dropping resistor, resulting in a large change in current and brightness.

- LEDs are in parallel, with one dropping resistor per LED. No, don't change the resistor value... just add one new LED and one new resistor in series with it.

- LEDs are in parallel, and then hooked to a single dropping resistor. You'd need to change (reduce) the value of the resistor when adding LEDs. This is generally *not* a good hookup strategy, because the LEDs would have to be exactly matched to get equal current sharing, and they usually aren't. Some will usually have a slightly smaller voltage drop than others, and will "hog" the current and be brighter. In extreme cases, one can hog most of the current, get VERY bright for a brief moment, and then burn out... leaving the next one to hog the current, flash, phut... in this mode they can be known as NEDs (noise-emitting diodes).

- LEDs are fed in series-parallel - e.g. they might be arranged in strings of 5 or 6 LEDs, with each string being fed through a single dropping resistor. This is probably the most robust and efficient and flexible arrangement - if your friend wants more LEDs, she would just add as many strings of 5 (each with its own dropping resistor) as the design requires.

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Dave Platt                                    AE6EO
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Uh, series...

"Charles" wrote in news:hdi5br$lvr$ snipped-for-privacy@news.eternal-september.org:

The resistance is fairly critical but constant current control takes care of that by changing it as a result of maintaining current. Another way to make it less critical is to use hyper-bright LED's at very low currents,they should work ok at less than 5 mA and likely withstand up to 45 mA without serious degrading of life expectancy.

Whatever route you choose, start with basic calculations: Find out the Vf (forward voltage rating) for your LED's (which will be connected in series for this calculation) and don't mix LED types if you want this to be easy.

Rs=(VS-Vf)/If

Rs is series resistance, VS is DC supply voltage. Vf is total Vf of string, and If is forward current in string. Calculate power dissipated in a series resistor, or a regulator, by (VS-Vf)*If, to avoid a fire risk. Try the LM317 regulator as a constant current source (see its datasheet for this). LM317's are cheap, and have safety limiting built in so you get good protection.

Lastly, if you're using a lot of LED's the voltage grows too high to be safe. It depends what you're making but if this is to be safe around children as a toy or display, you really need low voltage and be sure that currents are low, so maybe use a 15V supply, buy a few of the smaller 500 mA version LM317 regulators and keep total LED Vf within 12V. This way you also get energy efficiency.

With small strings of 6 or so LED's in each, the control might be easy enough with simple resistors if you have a power supply that won't sag its output voltage much as you add extra strings in parallel as load. You might find a handful of 500 mA LM317's on eBay or elsewhere, cheaply enough that it pays to buy them, and build a MUCH cheaper supply than you'd have to if you wanted it to have fine control of high currents! As you also get good protection for each LED string it's a better choice than series resistors if safety is critical.

50 LED's = high volts.

A constant current supply say 10 ma. is not going to do a lot of damage, but what if it fails.

CAUTION

You may need strings like 24 volts maximum or about 14-16 Led's max in series.

If you gave the application, you could be better informed.

greg

Ans: Yes. The resistor sets the current depending on the voltage drop and the particular LED arrangement. Are you asking a design question: i.e., how to pick the resistor value for a particular unchanging combination of LEDs? Or, do you want to change the number of LEDs on or off at any given time in some short of an array like in a sign? In the second case, each LED is switched independently and requires it's own resistor and transistor or FET switch. It is possible to arrange the LEDs in a matrix, one possibility being where the number of resistors is the square root of the number of LEDs; e.g., 7 resistors for 49 LEDs. The value of the resistors does not change regardless of the pattern of lit and un-lit LEDs. The switching is more complex requiring time division, however, another job for a PIC processor.

What a great idea - now when she puts 50 in series she can drive it all with a very dangerous 100 volt or more power supply!

Another way to make

45 ma? as long as her life expectancy for the LEDs was in the 10's of hours

??? crap. snipped the rest I can't be bothered.

David Eather wrote in news:v- mdnaMwisf_UWDXnZ2dnUVZ snipped-for-privacy@supernews.com:

Snivelling little asswipe! If you'd read the rest you'd KNOW how gormless your first point was! Troll away, if it makes you feel better. Worked for me.

David Eather wrote in news:v- mdnaMwisf_UWDXnZ2dnUVZ snipped-for-privacy@supernews.com:

Rubbish. Should still be in thousands if it's not too hot. Heat has a far stronger effect on lifetime, so if you ignore that (which you evidently did), you could run them in an enclosed space at recommended 20 mA and they'd still die young(ish). Most makers over the last 20 years specified a max current of

50 mA, recommended 20. 45 mA is NOT out of line with maker's intentions, no matter what you might think about that.

David Eather wrote in news:v- mdnaMwisf_UWDXnZ2dnUVZ snipped-for-privacy@supernews.com:

You also ignore that I was stating how wide a margin you get, I was NOT suggesting that they be driven at the high end of that margin. Idiot!