Gazoo said:
[quote author=IgorT link=1185701612/510#524 date=1194819746][quote author=Gazoo link=1185701612/510#523 date=1194819557]You will want to use at least a 1 watt resistor.
That is true, my bad...
Thanks Gazoo![/quote]
I say at least because it depends on which diode you are driving and so on. But lets use the GB diode for example. The voltage across the diode will be ~2.85 volts with 250ma's of current. 2.85 X .250 = .7125 watts. Therefore for higher currents I would use a 2 watt resistor..

Since we are building the circuit we might as well use a 2 watt for future diodes and what not. Really this is only necessary for experimentation like for a lab laser since it is easy to calculate the resistor needed if using the LM317 as a current regulator. For the open can diode.... I have run 600ma's into it and my resistors get hot. I need to see what the voltage is on the diode with that current..
Added..I checked the voltage across my open can diode running with 600ma's and it is 3.47 volts. Fortunately I am using 3 watt resistors and a 3 watt rheostat. I have two 1 ohm resistors connected in series with the rheostat so current is limited to ~600ma's when the rheostat is set to no resistance. I did this to prevent accidentally blowing the diode..

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hey Gazoo, you have your laws a bit twisted there.
the watt rating of the resistor is how much power it can safely dissipate. using your example, if you have 0.25A running through a 1 Ohm resistor, the voltage drop, V=.25/1 =.25V, power = volts * amps so P=.25*.25 = .0625W
you used the voltage drop over the diode, the resistor is used as a shunt, so its not actually the load, the diode is the load. the example you gave would be the power dissipated by the laser
so you actually could get away with a 1/4W resistor for up to 500mA current, although it will probably get warm. I would still use a 1W. but now you don't need to worry about drawing too much current for the resistor!
this also means that having a 1ohm resistor will only cost your circuit .25V, so it won't really make any difference to the operation of the circuit
Geez i just can't help myself can i?
