how to connect up bulk leds ?

As LED's are driven with current rather than voltage, it's generally best to connect multiple LED's in series, together with a current-limiting resistor. The number of LED's you can connect in series depends on the available supply voltage and the type op LED's used.

A white LED has a forward voltage drop of approximately 3.6 volts, so if you connect two in series, you need a supply voltage of at least 7.2 volts; with three in series the required supply voltage is 3 x 3.6 = 10.8 volts minimum and so on. Also, you have to add a few extra volts for the current-limiting resistor.

So if you have a 12V supply, and you want to operate each LED at 20mA of current, these are the calculations:

- You can connect three white LED's in series, resulting in a 10.8 volt forward voltage drop, with 1.2 volts remaining (12 - 10.8).

- To get 20mA of current with a 1.2 volts voltage requires a resistor of 1.2 / 0.02 = 60 ohms. So you cold use a 68 ohm series resistor.

With 90 LED's in total, this means that you have to make 30 strings, each composed of three white LED's and a 68 ohm resistor connected in series. These strings are connected in parallel to the 12V supply, resulting in a total current of 30 x 0.02 = 0.6A.

For 24 volts, you can double the amount of LED's in each string:

6 x 3.6 = 21.6 volts. To get 20mA you divide once again the remaining voltage by the desired current: (24 - 21.6) / 0.02 = 120 ohms series resistor. Now you need only 15 strings of 6 LED's and one resistor each. The total current drawn is 15 x 0.02 = 0.3A.

I would strongly recommend not raising the voltage any higher. And note that this is all based on a DC supply. If you have an AC supply, you could double the amount of LED's per string, with two LED's instead of one per

3.6V voltage step, connected anti-parallel, and approximately half the resistor value to double the current (as each LED is switched for on only half the time). Also note that LED's don't take reverse voltages very well.

Richard Rasker

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http://www.linetec.nl/

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The catch with this approach is that the "3.6V" drop per white LED is strictly nominal. The manufacturer's data sheet will also specify upper and lower limits to the LED voltage drop at 20mA (or whatever is approprate for the specific LED which may not always be 20mA) at room temperature - usually (but - not always - 20C).

If you get a batch of LEDs with a lower forward drop, 68R may give you rather more current than you expect.

There is also the problem that the forward voltage across an LED is temperature dependent and will fall by 2mV per degree Celcius increase in junction temperature. This means that you can get more current flowing through the LED as it warms up, and - potentially, higher power dissipation in the LED (if the voltage drop across the resistor is less than the voltage drop across the LEDs).

For sufficiently long strings of LED's and sufficiently small resistors, the process can run away, melting your LEDs. A properly designed constant current driver can be immune to this problem.

-- Bill Sloman, Nijmegen

[snip]

Indeed, with a very small voltage overhead (say less than 5%) and the resulting small resistors, these things must be taken into account. But as soon as you have some 10% or more of the total supply voltage across the current-limiting resistor, things will be fine in the vast majority of cases. But if you get a batch of unknown LED's, it can't hurt all the same to take a handful of LED's and measure the actual forward voltage drop at the desired current, and recalculate the series resistor accordingly.

About temperature: with 25mA through a white LED, the internal power dissipation (~0.1 watts) is hardly capable of raising the temperature more than a few degrees. The increase in forward voltage drop versus increase in current (iow: the dynamic resistance) is usually much larger than the influence of temperature, and prevents runaway effects -- e.g. for a typical white LED, I found a forward voltage drop of 3.6 volts at 13mA; at

20mA, the voltage drop had increased to 4.2 volts. Only when heating up the LED to approximately 70 degrees Celsius was I able to nibble off 0.1 volt.

In over 25 years of using LED's, I have never blown even one using series resistors (I *did* destroy some by accidentally hooking 'em up in reverse, though). And several people I know even say they hooked up white and blue LED's directly to 5V power supplies without failure -- although I definitely wouldn't recommend this.

Richard Rasker

--
http://www.linetec.nl/

Perhaps like this:

--line AC---Rectifier---cap---Pot---LED1...LED?---Com

An Idea for an Experimental Method

0) Set Pot to 0ohm 1) Make an LED string that adds up to the peak voltage for a safe peak current. 2) Power it up and adjust the pot to compensate for self heating. Quickly... 2) Raise temperature of all the LEDs in an oven for the greatest ambient temp. 40C ?? 3) Adjust the pot resistance to maintain a safe peak current.

Even trying to dodge a crapload of math it's still tricky :P

D from BC British Columbia Canada.

He might use a smaller series resistance and then add a variable pot and "extra bright" switch inline with all the other members(or even a 3-way switch). This will give him the option to increase the brightness if he needs(But he would need to know it won't last as long). The pot will allow for brightness controll. he could then, if he wanted, add a fan and/or temp sensor with slow shut off or reduced brightness.

It would really suck to hook all those LED's up and turn on the power and 2 mins later there all dead.

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The Agilent HLMP-CW18 5mm white LED has a thermal resistance from junction to ambient of 240C/watt, which means that the junction is running 24C above ambient when dissipating 100mW.

The numbers are usually available if you can be bothered to look. If they aren't, you'd better look for a more reputable manufacturer.

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For that particular LED. The HLMP-CW18 has a typical forward voltage of 3.4V at 20mA and a maximum of 4.0V, so your parts would seem to have been some kind of cheap rubbish.

Very wise of you.

-- Bill Sloman, Nijmegen