Voltage tolerance?

[Greenhorn]

I recently ordered these laser diodes:

http://cgi.ebay.com/ws/eBayISAPI.dll?ViewItem&ssPageName=STRK:MEWNX:IT&item=280279354713

They're SLD1236VL diodes. The data sheet says that they run on 2.5 volts. If I used an LM317 which uses 1.25 volts, would these diodes be safe with 3 AA or AAA batteries? That would be 4.5~ volts, 0.75 volts higher than the diode and LM317 use up. Will that extra voltage burn the diode out or shorten its life? Or maybe this is a simpler question - what is the range of voltages that the diode will take? 2.5v up to . . .?

 

[pullbangdead]

With linear current drivers like the LM317, you set the current being pushed into the diode, so the diode takes whatever voltage it wants, as long as there is more than enough available to the driver. If you need a certain amount for the diode and a certain amount for the driver, any access voltage above that will be absorbed in th driver and distributed as heat. Since the driver only sends whatever current and corresponding voltage the diode wants, the extra energy will not go to the diode. It will go to the driver, and that extra energy shouldn't ever be enough to actually hurt the driver.

Just make sure the voltage available from your source (the batteries) is greater than the required voltage of the driver (total operating voltage, which may not necessarily be the reference voltage for a given driver) and the required voltage of the diode combined.

 

[Greenhorn]

With linear current drivers like the LM317, you set the current being pushed into the diode, so the diode takes whatever voltage it wants, as long as there is more than enough available to the driver.  If you need a certain amount for the diode and a certain amount for the driver, any access voltage above that will be absorbed in th driver and distributed as heat.  Since the driver only sends whatever current and corresponding voltage the diode wants, the extra energy will not go to the diode.  It will go to the driver, and that extra energy shouldn't ever be enough to actually hurt the driver.

Just make sure the voltage available from your source (the batteries) is greater than the required voltage of the driver (total operating voltage, which may not necessarily be the reference voltage for a given driver) and the required voltage of the diode combined.

So, theoretically, assuming the LM317 doesn't burn out, I could run a 2.5v diode on any number of batteries I wanted?

But if I fed it, say, 10 volts, and the diode needed 2.5 volts, wouldn't that mean that the voltage drop of the LM317 would be 7.5v? It's always 1.25v, right? The 1.25v drop is how you determine current flow, right? I = V/R? So the extra voltage it gets rid of to feed the diode isn't the same as voltage drop, and doesn't affect current flow?

 

[hydrogenman15]

[quote author=pullbangdead link=1225781402/0#1 date=1225785031]With linear current drivers like the LM317, you set the current being pushed into the diode, so the diode takes whatever voltage it wants, as long as there is more than enough available to the driver.  If you need a certain amount for the diode and a certain amount for the driver, any access voltage above that will be absorbed in th driver and distributed as heat.  Since the driver only sends whatever current and corresponding voltage the diode wants, the extra energy will not go to the diode.  It will go to the driver, and that extra energy shouldn't ever be enough to actually hurt the driver.

Just make sure the voltage available from your source (the batteries) is greater than the required voltage of the driver (total operating voltage, which may not necessarily be the reference voltage for a given driver) and the required voltage of the diode combined.

So, theoretically, assuming the LM317 doesn't burn out, I could run a 2.5v diode on any number of batteries I wanted?

But if I fed it, say, 10 volts, and the diode needed 2.5 volts, wouldn't that mean that the voltage drop of the LM317 would be 7.5v?  It's always 1.25v, right?  The 1.25v drop is how you determine current flow, right?  I = V/R?  So the extra voltage it gets rid of to feed the diode isn't the same as voltage drop, and doesn't affect current flow?[/quote]

I am pretty sure 1.25 volts is the reference voltage and not the voltage drop.

--hydro15

 

[rog8811]

I am pretty sure 1.25 volts is the reference voltage and not the voltage drop

Correct, the LM317 needs 3v overhead so you are looking at a minimum voltage of 5.5v, so 6v is OK.

Regards rog8811

 

[Greenhorn]

I am pretty sure 1.25 volts is the reference voltage and not the voltage drop

Correct, the LM317 needs 3v overhead so you are looking at a minimum voltage of 5.5v, so 6v is OK.

Regards rog8811

So the LM317 takes 3 volts to run, but uses 1.25 of those volts to control the current? (In I = V/R, the V = 1.25?)

 

[hydrogenman15]

[quote author=rog8811 link=1225781402/0#4 date=1225818050]

I am pretty sure 1.25 volts is the reference voltage and not the voltage drop

Correct, the LM317 needs 3v overhead so you are looking at a minimum voltage of 5.5v, so 6v is OK.

Regards rog8811

So the LM317 takes 3 volts to run, but uses 1.25 of those volts to control the current? (In I = V/R, the V = 1.25?)[/quote]

The LM317T needs at least 6v's and the voltage output will conform to the diode your using.
Give it a dummy load and make sure it works properly after that SHORT THE LEADS of the output connections when you turn it off and slowly turn up the current until the diode starts lasing.

Also, if you want want you can use resistors and math so don't have worry about turning up the
current.
Divide 1.25 by the current you want to drive it.

I would drive it at 250ma's just to be safe.

1.25/.250 = 5 ohms (just use two 10ohm resistors)

--hydro15