1. How do you determine the forward voltage drop of the LD? what do you mean by "drops" quote "assume the red LD drops 2.3V"
You measure it, at the operating current, or get this figure from a datasheet. Like all semiconductor diodes, LEDs and laser diodes have an forward operating voltage, and in a series circuit this is the voltage "dropped" across the diode. If the voltage is below this value, the device can not conduct (turn on). And 5V > 2.3V
If the source voltage is above this "turn on" value, the current will (theoretically) increase without bounds, unless something limits it. Hence a resistor or other means of controlling the current.
Note that as the current is increased, the forward voltage drop of the diode will increase somewhat, so it is important to consult the datasheet (if available) or measure the voltage drop at the operating current you intend to run the diode at when designing your current control circuit.
This probably explains why these diodes survive being connected directly across 2 NiMH cells in series; this is ~2.4V which is apparently within the operating range of these diodes. So from Gazoo's data, we could derive that the forward voltage of his diode is equal to his battery (source) voltage, at 200mA (measured).
Operating that laser at 500mA is asking for trouble and will eventually damage the laser's mirror facet (ablation from excess laser energy), substrate (dissipation limit of semiconductor), or both.
2. lets assume we are using No driver or anything but a straight battery. Voltage at this point isn't relevant as I need to understand how to adapt this to anything. (just for fun lets say it's a cr123 3.0v. lithiums in this case)
3. What is a means of determining the current for a diode? Is there any method? I know there is a trial by fire method where you continue to step up and up until it pops. Would a plain potentiometer work in this?
You can use a pot to adjust the current, but beware the dissipation spec of the pot (more later)
4. Once you have determined the current you want and the source of current (battery) You can figure the voltage or current reduction by subtracting the draw the LD uses minus the voltage of the battery. got it in your example above 18 ohms. What though is the resistor dissapation?
Resistor dissipation is the POWER in watts that the resistor must dissipate as heat. If the resistor is undersized, it will be damaged and may even catch on fire.
Note that the laser diode itself also dissipates heat, in fact most of the energy input to the diode is wasted as heat, only a small percentage is outputted as light. This is why we need to heatsink these laser diodes...
Understand also that batteries have
some (albeit very low) internal resistance that limits the discharge / short-circuit current, though this is typically several AMPS for AA-size cells. Also, an ammeter has a low-value resistor inside, which becomes part of the series circuit when connected (the "ammeter" actually measures the VOLTAGE dropped across this internal resistor in order to measure the current through the meter).
If I get some time, I'll put together a tutorial with some simple schematics, etc. but I suggest scouring Sam's Laser FAQ (link
here)
oo:
