# 1 watt luxeon of 12 volts dc

• posted

hi , i am using 2 1 watt luxeon red leds for brake lights , can i run these of 12 volts via a resistor, or do i need a driver circuit ? what value resister would i have to use , (if it can be done this way ./.)

thanks, mark k

```--
---------------------------------------------------------------------
"Are you still wasting your time with spam?...```
• posted

What colour of luxeon?

You need to find the forward voltage of them. If it's for example 3.5 volts, and you are putting them in series (a good idea) then the total voltage across them is some 7V. Take this number away from 12V, and you get the voltage that you need on the resistor. This is some 5V.

If the current is .3A, then you need 5V/.3A = 15 ohms. And the power needed for that resistor is .3A*5V = 1.5W.

• posted

According to the datasheet at

luxeon reds have a forward voltage of between 2.25 and 3 volts, depending on the current you want to put through them.

If you want 300mA, then they will be at about 2.8V each, according to the current to forward voltage graph. So, put them in series, that'll give you 5.6V. Then, you need a resistor to drop the voltage sufficiently. That is 300mA over 6.4V, which works out to about 22 ohms. However, the power dissipated in the resistor will then be

P = 6.4^2 / 22 = almost 2 W. You'll need a big resistor.

You can use a network of resistors if you can't get 2W ones; if you use

10 220 ohm resistors in parallel, you can use standard quarter watt resistors. The amount of heat generated will be the same, but it'll be spread over the whole set. If you mount them so there is a bit of space between them, it should be OK.

Another option would be to use a switching power supply, like the 'roman black' step-down supply, shown here:

Its simple to build, and you can easily make it put out 5.6V (which will cover your two LEDs) by increasing the zener to a 6.2V one. The nice thing about this is that its much more efficient; this may not be a concern for you, though.

Regards, Bob Monsen

• posted

Do not apply fixed/regulated voltage to an LED - the current will be unpredictable. The LED has a tolerance in its voltage drop (assuming the amount of current specified in the datasheet for characterizing voltage drop). The actual voltage drop *at specified current* has some significant tolerance, and varies with temperature. Since the voltage drop does not vary much with current, a tolerance in voltage drop characteristics can mean current being far from what you want if you apply a fixed voltage. If you apply a fixed voltage, the amount of current usually varies greatly with temperature. Such current variation with temperature can reinforce the temperature change, leading to the current changing even more. Sometimes this even results in "thermal runaway", which is the situation where increasing temperature and increasing current reinforce themselves - sometimes in a spectacular "runaway spiral", sometimes to a lesser extent but to an extent great enough to seriously overheat the LED. The LED may fail in minutes or may merely have a seriously shortened life.

What to do: Use a switching current regulator (or any current regulator), or use a voltage source significantly more than the voltage of the LED (or series string thereof) and use a "dropping resistor". Maybe set the "roman black" power supply to 6.9 volts, and drop the 1.4 volt difference with a 4.7 ohm resistor.

The necessity of limiting LED current has been discussed here before enough times. I mention a bit more of this (and calculating dropping resistors) in a web file of mine,

- Don Klipstein ( snipped-for-privacy@misty.com,

• posted

snip resistor method>>

Much better would be a fixed current output switching regulator such as the LT1618. For a bike it is important not to waste any energy in form of heat. You can even have the options of current adjustment and automatic on/off switch. Only a few external components needed, no power resistor. See design notes in last EDN.

```--
ciao Ban
Bordighera, Italy```
• posted

I copied the schematic which is a bit difficult to explain, because a boost converter is used as a buck. Also that original is faulty, as the SW is not connected, I guessed it goes to the other end of the coil. Efficiency is claimed as being 70% with a 3.6V/700mA Lumiled. Hope I got it right. +9...16V o-+--------------------------------+ | | | ___ 700mA ___ | +----+--|___|--+--->|--UUU-+-->|-+ | | 0.07R | LED 47uH | PMEG2010 | | | | schottky | | .-------------. | | | | ISN | | | | | | | | +--|ISP LT1618 | | | | | | | | +--|Vin SW|----+ | | | | | | | o---|SHDN\ FB|-----------+ | | | | | | | | | |Iadj GND Vc | | | '-------------' | | | | | ___ | --- | | +-|___|---+ | --- | | | 10k | | |4u7 | | --- --- | | | | --- --- | | | | |220p |2n2 | +---------+---+----+---------+----+ | === GND created by Andy´s ASCII-Circuit v1.24.140803 Beta

```--