Reducing 3v supply to run 2.4v laser diode
First, the goal is not to reduce the voltage for the laser diode (just like an LED). The diode itself will determine the forward voltage drop across it (2.4Vin this case). This is not a choice. you don't get to choose to run the diode at a higher or lower voltage. The diode will do whatever it can to make the voltage across it 2.4V when you apply a voltage at >2.4V across it. This includes drawing thousands of amps such that the power supply sags down to 2.4V (if the diode doesn't explode first).

Therefore, the goal is not to reduce the voltage for the diode. The diode does this on its own. Instead, the goal to limit the current through the diode so that the diode doesn't explode from overcurrent when the diode forces this voltage across itself.

As for the "two resistor method", I have no idea what you are talking about. Maybe you are talking about a resistive voltage divider? In which case that will NOT work because the resistive divider assumes the load is "infinite" impedance which a laser diode is definitely not. In other words, a resistive divider reduces a voltage and expects to provide ZERO current. A laser diode definitely requires something other than zero current. Trying to power a laser diode off a resistive divider will cause a severe voltage sag on the output due to the current resulting in a circuit that doesn't work (unless your chosen resistances are so low that the resistors are running red hot and wasting more power than the laser diode is actually using).

Current Source:

You can wire up an LM317 (the battery is the load, so in this case that is the laser diode):



From LM317 datasheet.

Current Limiting-Resistor:

Instead, you can just use one resistor. The same way you would size a resistor and known voltage to provide current to an LED:







simulate this circuit &ndash; Schematic created using CircuitLab

Between the resistor and diode, they must drop the entire voltage of the supply. Since the diode's voltage drop is fixed, then that means the remainder of the supply must drop across the resistor. Since the current flowing through the resistor must be the same as the current flowing through the diode, if you size the resistor so that a particular current flows through it when this remainder voltage appears across it, then this is also the diode's current.

\$ V_{supply} = V_{diode} + V_{resistor} = V_{supply} = V_{diode} + (R \times I) \$