Soft start for 2 volt 4 amp laser diode with fuse instead of CC regulator

Can anyone recommend a soft start circuit for a 2 volt 4 amp 808nm C-mount laser diode without a CC regulator which is fast blow fused? I don't want to use a regulator but still need soft start... If I blow the diode due to over current, so be it.

A very large capacitor and some resistance
would work. For the "resistor" you could
use some thin long straight wire. You
could also get fancy with inductances, but
it's best to keep those away from a diode
unless you know what you're doing.

I can't suffer the loss of much extra resistance in line with the diode due to wanting to use a 2 volt lead acid battery. Since the diode needs at least two volts for full output, even a tenth of a volt drop reduces the current quite a bit. I can build my own inductors, even have a digital inductance meter, just wanted to get an idea of how it is normally done so as not to reinvent the wheel, finding something already designed. If I need to home brew or make my own, I have a full test bench of equipment, just need to get an idea of the voltage or current over time to best use for soft starts with these diodes.

If I can't get enough current due to line losses in the soft start , I will have to abandon the idea of a direct drive soft start fused diode. I have a 5 amp CC driver, but to use it, I need at least 3.3 volts and then there goes my single cell 2.0 VDC lead acid battery idea. I suppose a boost drive could be used with two of them in series and suffer the losses from the regulator, but at this current, I haven't found one yet that is the right size.

Why do this? I have this really nice little Gates cylindrical starved electrolyte lead acid battery which would fit oh so nicely in a pointer tube to keep the unit relatively small, although heavy. Reducing the losses of a regulator also increases battery life, if I can keep from killing the diode from thermal runaway.