Ehm ..... be careful with these premises, sorry.
Is true that a LD, same as leds, works in current and not in voltage, but is wrong to think that the LD will automatically not pull excessive voltage over what's required to operate .....
Also, the current that flow in a load is a derivate from the resistance (or, more exactly, the equivalent serie resistance), of the load, and from the voltage that you apply at the sides of this load .....
Let me do an example, if you have a load with a resistance of 1 ohm, you will have a current flowing through it that is 1A for each V that you apply to the load ..... if your load is 0.1 ohm resistance, the current is 10A for each V, and so on .....
Now, imagine a laser diode as "a low resistance with a voltage threshold", that basically mean, the resistance is high under a certain voltage value, and after this (the FV) it become very low for the part of the voltage that is over the FV (a bit like a zener diode) ..... the current drivers that we uses, acts regulating the voltage at the LD terminals in the way that the current that flow through it remains stable, and this is done placing a sense resistor in the current path (basical ohm law, the current in the circuit is the same for all the elements in serie), and measuring the voltage drop caused on this sense resistor from the current flowing, and using this as feedback for the regulation circuit ..... when for some reason (junction heating, as example), the current start to change, the driver modify the voltage for keep it constant.
But, with the low serial resistance of the LD in working state, a small increase of the voltage is usually enough for give you a big increase of the current, for this reason the drivers works in current regulation, cause you cannot get a voltage regulation stable enough for grant you that the current don't kill your diode (and, if you give it enough voltage, it take all the current that it can, until it blow up).
In the example made from rhd, your 90V battery will not blow up your diode ONLY if your batteries cannot give more than the maximum current that the chip can hold (about the 10mA in your example, yes, IF the batteries can give ONLY a maximum of 10mA, for their internal resistance) ..... is the same as for some of the small keychain flashlights that uses 3xAG13 button batteries and a led diode without a resistor ..... the led works at 3V, and the batteries gives 4.5V ..... it don't burn ONLY cause those batteries cannot give it more than those few mA that the led can hold ..... but try to use 3 normal batteries, and you will burn out your led in seconds.
Same for a LD, if you use a battery that can give only less than the maximum current that the LD can hold, then also a higher voltage don't burn it, but use a battery that CAN give mor current, and you kill it.