As in any oscillator, the fundamental condition for a laser to lase is for the round trip gain to equal the losses. For a laser this means that as the optical wave makes a round trip through the cavity formed by two mirrors, the gain from the excited medium must equal the losses from imperfect mirror reflection, diffraction, etc.
For a diode laser, the gain is roughly proportional to the carrier (electrons and holes) density in the cavity.
So, to turn on the laser, you need to raise the carrier density until it provides enough gain to overcome the cavity's optical losses. Because the gain region is storing charge, electrically it has the behavior of a capacitor, and when the capacitor is charged to a certain level, the laser will begin to lase. But also, the carriers are constantly recombining through different mechanisms so a certain current is needed to maintain a given charge. This means that below threshold the laser's electrical behavior is essentially like a parallel RC circuit, with a nonlinear R.
Now, if you have a laser operating at some low (below-threshold) current level, and you then step up that current to turn on the laser, initially most of the added current will go to charging the "capacitor", and the laser won't lase until the charge reaches the threshold level. If the prior current level was lower, the initial charge on the capacitor would be lower, and so it would take longer for the new current to charge up the capacitor to threshold. This is why the laser turns on faster if the bias is higher.
Note: Above threshold, there is a gain-pinning effect (the gain is locked equal to the optical round-trip losses, and any extra current just goes to increasing the laser output power) so the laser no longer electrically behaves like a capacitor. Above threshold the laser is electrically very close to a short circuit (dependent on parasitic effects).
One common way of doing AM modulation is to multiplying a baseband signal with a square wave that is at the carrier frequency, and then run the chopped baseband signal through a band-pass-filter tuned to the carrier frequency.
The reason this is done is that multiplying by a square wave is pretty easy to do with as little as one high frequency switching transistor.
If the signal plus DC bias is connected through a resistor to the source of a FET, and a square wave at the carrier frequency is connected to the gate, then you will get multiplication by a square wave. When the FET is off the output signal is equal to the input (multiplied by 1), when the FET is on the output is 0 (multiplied by 0).
A square wave at the carrier frequency can be decomposed into an infinite series of sinewaves that are multiples of the carrier frequency. So if we multiply the input by a square wave at the carrier then the output is equal to the input multiplied by a sine-wave at the carrier (which is exactly what we want for AM), plus the input multiplied by sine-waves at the harmonics of the carrier (which is something we don't want).
If the output of the chopper is run through a band pass filter which passes only the signals near the carrier, then the signals at the harmonics are rejected and we are left with only what we want (the AM modulated signal).
See...
http://homepages.udayton.edu/~hardierc/ECE401/ECE401L/ECE_401L_Lab2.pdf
http://www.analog.com/library/analogDialogue/archives/43-09/EDCh%204%20rf%20if.pdf
Best Answer
As I understand it, you want to drive a laser diode with a 100 kHz sine wave with a bias of about 40 mA. The 100 kHz signal is is available from a 0-5 Volt digital output. Here is a circuit that should at least be a reasonable topology:
You will have to fill in the right values yourself. The ones I show represent only a very rough stab at it.
R2, C3, R3, C4 are a two pole low pass filter. This will eliminate much of the harmonic content of the 100 kHz square wave to make it closer to a sine wave. In this scheme more filtering also reduces the amplitude, so you will have to change R5 if you adjust the aggressiveness of the filter. R1 and R4 set the DC bias. In this example is should be roughly 40 mA. C2 AC couples the signal from the filter to the opamp while leaving the average bias alone. The feedback circuit makes sure that this composite voltage (AC signal plus DC bias) appears accross R5. This regulates the diode current since the current thru R5 is the same as the current thru the diode.
Note that this never explicitly sets the diode voltage. Rather, it regulates the current thru it. That's a much better way to drive a LED or laser diode than trying to fix its voltage. This circuit will automatically make the voltage as needed to get the desired current.
Again, you will likely have to adjust the values. You never said how much harmonic content you can tolerate, for one.
Added in response to your two questions:
The FET acts like a variable resistance. It is varied by the opamp to make sure that the input signal (opamp pin 3) appears accross the output current sense resistor (R5). Since the current thru R5 and D1 are the same, and the voltage accross a resistor is proportionaly to the current thru it, the voltage at the top end of R5 is proportional to the diode current.
This circuit would still work somewhat without the opamp, but not as accurately. The opamp drives the gate of Q1 to whatever it takes to get the desired diode current. The gate to source voltage of Q1 will vary with current. This voltage offset is automatically compensated for by the opamp since it is inside the feedback loop. The exact gate-source voltage required for a particular current is poorly specified and will vary from device to device and over temperature. By regulating the voltage on R5 instead of the voltage on the gate of Q1, the gate-source voltage is automatically compensated for by the opamp and becomes irrelevant. Actually, it's only irrelevant as long as the required gate voltage is within the range the opamp can put out. The IRLML2502 has a low enough gate-source turn on voltage so that this is guaranteed to be true. We don't know exactly what that voltage is, but we do know it will be well within the 5V output range of the opamp.
Using the built-in photodiode to regulate the laser's actual light output is even better than regulating its current. In that case you have to add a circuit that creates a voltage proportional to the laser light output, then feed that into the opamp negative input instead of the of R5 voltage. The opamp will then drive Q1 to whatever it takes to achieve the light output proportional to the input signal.