Low drop-out linear regulators are great when the input voltage is close to the desired output voltage. Where you start to run into trouble is where the input voltage is significantly higher than the output, as more and more power is dissipated in the regulator itself.

For example, let's say you have 5 LEDs, each with a current of 20mA, so the regulator will need to pass 100mA to keep the LEDs happy. We'll say just for this example that the LEDs and their resistors need 3.0v across them to reach full brightness. Also, let's say that your input voltage starts at 4v, then increases to 8v.

In the beginning, you have 100mA * 3v = 300mW power total being dissipated in the LEDs, and about 100mA * (4v-3v) = 100mW power being dissipated in the regulator.

Then the input voltage goes to 8v.
You still have 100mA * 3v = 300mW power being dissipated in the LEDs, but now you have 100mA * (8v-3v) = 500mW power being dissipated in the regulator! 60% of the power is being used to heat up the regulator. This is also getting close to the maximum power dissipation allowed (625mW) for a TO92 package; if the input voltage gets much higher (9.25v) the regulator will either have to shut down or melt.

This is something that you simply have to be aware of with linear regulators; they get hot when they are required to have a large voltage across themselves while passing current.

As always, good information Sgt.
So we attach the T0-92 pkg. to a heatsink with thermal glue and place a 6.8V Zener across the source to avoid too much voltage on the T0-92 so that when the hurricane comes he won't blow up the circuit.

Doable?

 

This unit looks to have the right specs, and is a TO220 package....better for heat-sinking, no? http://www.nteinc.com/specs/1900to1999/pdf/nte1904.pdf
Another thought would be to have a system that sends the excess to a load?

I'm trying to wrap my mind around this idea, and it seems that either way the regulator passes the same current. If a load is added to the circuit on the ground connection, it also serves to raise the voltage drop-out, right?

How about if I were to use a regulator set at 5v feeding a regulator set at 3v? I am imagining that this way the two units could share the heat load. (I also relize that this would double the drop-out voltage ( .45+.45= .9v). It still comes out to be less than 1v lost, so that isn't too bad.

 

A TO220 pkg would be easier to attach a heatsink and in my opinion, with the circuit constraints, better than adding another regulator. The NTE1904 can provide up to 1A, probably without a heatsink....depending on ambient conditions. Can the Generator you have give that much current when at 8V? I doubt it, but with a heatsink there shouldn't be an issue. Maybe a 3/4W inline fuse cold protect the circuit from some huge wind gust.