Hey Eric. Good question. I'll try to answer as thoroughly as possible as I think that a number of other members might either have a question about this, or may in the future.
First with respect to regulators, the important spec is heat dissipation. In order to figure out whether this will work for you, you first need to determine how much current your circuit is going to consume. You can use the general figure of ~20mA per LED series string. Then you are going to need to know how much voltage needs to be dropped. If you have a 15v source, and you want to drop to 5v, then the regulator will drop 10 volts. Multiply that by current consumed and that will tell you the watts dissipated. I didn't look up the datasheet of the regulator you had in mind so I'll leave that up to you. Once you figure out the amount of dissipation required, then you might need to figure out whether or not a heatsink is required. As you can see, the higher the disparity between source and regulated voltage, the higher the dissipation. As for how much work it will be doing... it depends how much current is being drawn. In other words, if the LED's are off, then even though the regulated rail is active, the amount of current consumed at that point is so low that dissipation at idle is negligible.
However, before you decide to go that solution, you may find the following suggestions more useful and practical for your problem. The reason you are having such sensitive LEDs is because you have too many of them strung in series. I'm sure that you did this to minimize the amount of resistors required since not only does this make for a simpler circuit, it also requires smaller resistors since less current will be drawn and less dropping is required. However with little or no resistance in that LED subcircuit, it becomes very voltage sensitive. A better solution is to break up the LED string so that you have, maybe 2 strings instead of 1. This will require you to increase the dropping resistor for each string but the result is that the LEDs will become less sensitive to voltage transients. There are literally tons of led resistor calculator websites out there so I won't waste any keystrokes addressing that... just google and then punch in the figures and get the resistance/watts value you need.
The other thing that you can do is add a single stiffening cap in the led subcircuit. This shouldn't require much capacitance but you might want to start with maybe 47 or 100 and see how that works out for you. If the source voltage is regulated, then you'd want to be careful about using too much capacitance since the initial draw can be high but if the source is unregulated, you can go crazy. However, I don't recommend that you stuff 10,000uf because although that will work dandy, those LED's will probably remain lit for a few minutes after you shut down the boombox until the stored potential is drawn down.
Ideally, you will use a combination of splitting the LED strings, add a proper resistor AND then adding a stiffening cap. This allows you to have a solution with only passive components which in the long run, is more reliable and simpler solution than using active components path. I would suggest you give it a try and see if the results is satisfactory first before going the regulator route. By if you do decide to go the regulator route, then why not use an LM7812 instead of LM7805? With the 12, the regulator will only need to drop 3 volts instead of 10. The rest of the dropping can be handled by the dropping resistor. The whole point is to have a stable voltage to work from. Dropping the voltage all the way down to 5 from 15 (or 18+ on AC) means that you'll need to split the LED strings which means more current consumption from the parallel strings. Remember the first paragraph above about regulator heat dissipation calculations? Well now you have higher dropping and higher current requirements... a double whammy.
caution said:
In fact, the light behaves like a voltmeter because it's brightness reflects system voltage. As that is the nature of unregulated supplies, that will always be the case unless you can draw from either a different source (like Eric says using a design like the C100 or the M9998) or you can tap from a regulated rail and replace the bulb with one whose voltage requirement matches the regulated rail voltage.
Question - does having a large disparity between a regulator's input and output run hotter, even under no load? I'm wondering if using a regulator that's just over the normal turn-on voltage for an LED, like 5V, would help provide a stable output for a larger range. An LM7805 works down to 7V. But, if it runs hot even when not doing anything I'm not sure that would work.
I need to re-wire the LEDs illuminating my C100-F needles, they're all in series and expect full system voltage at the moment, so any blip makes them dim. I'm curious if I hung a 7805 off of each LED, it would run cooler than running all of them off of one 7805.