[sdiy] voltage and LEDs

[Needham, Alan]

Please bear with me on this ...
If an LM317 were used as a constant current source, the output of the
regulator goes via a series resistor to the LED.
The VReg leg of the regulator connects to the junction of the resistor
and the LED
So far, exactly as per the standard circuit; but...

If a resistor (maybe 100k) connected between the VReg pin and the 12v
supply it would have no effect on the circuit, passing too small a
current to affect the biasing; UNTIL...

If the connection between the VReg pin and the resistor/LED junction
were now broken, the regulator would switch off (almost) and the LED
would extinguish. This is a low current path that could be controlled
with a small signal FET. Let the LM317 do all the heavy work!

-----Original Message-----
From: synth-diy-bounces at dropmix.xs4all.nl
[mailto:synth-diy-bounces at dropmix.xs4all.nl] On Behalf Of Oren Leavitt
Sent: 16 February 2010 09:32
To: db
Cc: synth-diy at dropmix.xs4all.nl
Subject: Re: [sdiy] voltage and LEDs



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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