[sdiy] voltage and LEDs

[Oren Leavitt]

db wrote:
> 
> I now see lm317 regulators can be set up as constant current source..
> Thanks to many of you.
> 
> How about if my existing circuit used an lm317 to power the LED?
> http://img25.imageshack.us/img25/4098/leddriver.jpg
> 
> I'd set the regulator to output 3v or so to match the LED selected.
> 
> I'm guessing I need something bigger than a 2n5088, and maybe better to 
> run this all on 9v, rather than the 12 in my schematic.
> 
>


You'll want to use a power transistor, maybe a power darlington 
transistor for the trigger switch.
The 2N5088 won't handle that high of current.
Using the 317 as a current source the LED goes to the "adj" pin not "Vout"

..Or you can use the 317 as a voltage regulator, with some small R 
between Vout and the LED, and put a small transistor switch in the 
resistor divider that sets the Vout. Use that to electronically turn the 
317 and LED on/off.
The datasheet has an example "5V Logic Regulator with Electronic 
Shutdown" to give you an idea.
http://www.national.com/ds/LM/LM117.pdf

- Oren

> At 2/11/2010 07:41 AM, George Mattson wrote:
> 
>> The forward voltage is what the diode is going to use up across the PN
>> junction. It just "goes away"
>> Actually, that's the voltage required to get the current to cross the
>> barrier region. Consider it the PN road bump.
>>
>> Using 5 V, subtract the 3.9V Vf from 5V to get the difference of 1.1V.
>> 1000mA max current is 1 Amp. Using ohms law divide 1.1V by 1 Amp to 
>> get 1.1
>> ohms to limit the current.
>>
>> That's the minimum resistance necessary to prevent burning up the LED in
>> regards to the specs. Anything less will just dim it down. Usually, 
>> you can
>> get by with half the current. Anything above a certain limit doesn't 
>> let the
>> LED get any brighter, just dissipates as heat at the junction and 
>> shortens
>> the life of the semiconductor.
>>
>> But, 1 Amp is also the high limit of the 7805
>>
>> You'd either need a huge heat sink and a fan to keep it cool. Or, use 
>> two in
>> parallel and let them split the load.
>>
>> The 1.1 ohm resistor at 1 Amp is going to dissipate 1.1 Watts of power
>> (P=IE)
>>
>> I'd suggest using a 2 to 5 W resistor and mounting them above the 
>> board to
>> let them get some air to cool down.
>>
>>
>> My $0.02
>>
>> George Mattson
>>
>>
>> I want to drive a 3watt led
>> # 1000mA - Max Drive Current
>> # 3.90Vf - Forward Voltage
>>
>> Can just use a 7805 to get 5v, and then drop it with two diode in 
>> series to
>> get 3.6v?
>> Or do I still need resistors in series with these LEDS?
>>
>>
>> dennis.barton:skylab2000[socal]
>>
>>
>>
>> : 9.0.733 / Virus Database: 271.1.1/2680 - Release Date: 02/10/10 
>> 11:38:00
> 
> 
> 
> dennis.barton:skylab2000[socal]
> 
> 
> 
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