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580-600mA LED driver for my IR-Laserdiodes ?






hi,

thanks for answering

with the lm317 driver, i cant use 5V for Input ... i only have 5V (max. 3A) and i would not use a step up converter - brrrrrrr :yh:
space/heat is no issue (for the driver i have ~ 30x40mm) - a clean output and protecting the laserdiode is most importantly

Greetings
 
hi,

thanks for answering

with the lm317 driver, i cant use 5V for Input ... i only have 5V (max. 3A) and i would not use a step up converter - brrrrrrr :yh:
space/heat is no issue (for the driver i have ~ 30x40mm) - a clean output and protecting the laserdiode is most importantly

Greetings


Why can't you use 5V input? What voltage does your IR diode need? Normally they are around 2.0-2.2V or so - an LM317 driven at 5V will work fine for that.
 
lm317 minimum dropout varies a bit with operating current and junction temperature, but is commonly quoted as 3.5V. This is not enough to drive a 2V load from 5V.

If the 5V source is voltage regulated (most are), you may want to consider a simple power resistor to limit the current and drive your laser diode. >3W 5ohm in series with the diode would be fine.
 
lm317 minimum dropout varies a bit with operating current and junction temperature, but is commonly quoted as 3.5V. This is not enough to drive a 2V load from 5V.

If the 5V source is voltage regulated (most are), you may want to consider a simple power resistor to limit the current and drive your laser diode. >3W 5ohm in series with the diode would be fine.

Datasheet says <2.25V dropout at 1A current and 0°C junction temperature - lower if you go up to 25°C or reduce the current. Unless I'm missing something - I see no reason why it won't work on 5V (See exception below).

That said - I'm assuming OPs diode needs 2.2V - it may need more, in which case there may be issues with the LM317.
 
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It actually depends on the manufacturer a bit, but the original LM317 was guaranteed to work with at least 3.0 volts between input and output pin. For a laser driver add another 1.25 volt between output and adjust.

That is a minimum drop of 4.25 for the whole thing to guarantee proper operation.

In reality you can often get away with much less dropout, especially at lower currents (the 3 volt figure is for 1.5 amp output). Driving a 2.2 volt load from a 5.0 volt supply at room temperature and something like 200 mA could actually work.

The important thing is that it is not guaranteed to work, and the circuit may stop working with temperature changes, or not work at all if you get a 317 from a different supplier or batch, even if that component it very well withing in spec.
 
With two 6.8 ohms in paralel you should be getting 1.25/3.4~370 mA output current.

Anything below that means the unit is not in regulation due to lack of input voltage.

You must add the 1.25 volts on top of the stated drop-out voltage: that voltage is when working in constant voltage mode. If you power a blue diode from it, the input voltage requirement would be in the order of 10 volts, and you should put a heatsink on that lm317 too!
 
it does not work :-( .... the output current is 0,2A (with testload or Laser liode)

i will power a IR-Diode (2.2V, 700mA). How is the circuit to work with 5V input ?
what dirver can i use ?

thx
 
Pretty much any buck regulator for those power and current levels should be fine.

Just a question though: You are using IR diodes, which often have their case connected to their positive pin. Is everything eletrically isolated (i.e. both driver neutral and input make no connection to whatever the diodes are mounted in?
 
where can i get 600mA (output) laser drivers for red/IR diodes ?
with TTL would be fine
 
Or, instead of using the LM317 you could use the AMS1117. Since it has a reference voltage of 1.25V and a dropout voltage of around 1V it would give you about a 2.7V output to the diode.
Additionally, the AMC7135 are dedicated constant current IC's that supply 350mA and have even lower voltage dropout. Although, since they only come in at 350mA per IC than you would have to do 700mA to the diode.
 
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