In article , "Tim" letter T letter
H @lakeoriongroup.com says...

Not sure why this one is elluding me, but , I have a graphics display with
an led backlight drawing about 450ma @ 5v. While a large 7805 is rated for
an amp it's getting pretty hot even with a nice sized heat sink.

What voltage are you feeding IN to the 7805?

Example: Say you feed 20VDC to the input. That means the regulator
gets to waste 15V of that as heat. Although current in a series circuit
is constant, one must take into account the current used by the
regulator itself to operate. So, in addition to your 450mA for the
backlight, you're going to be drawing up to 8mA more for the regulator.

Given (Ohm's Law) P=IE: You have up to 458mA times 15V giving 6.87
watts worth of power dissipated as heat. That's a lot of heat, so yes,
your regulator getting warm is very normal. It gets worse when the input
voltage goes up, or when the current draw increases.

The 7805 was a fine part in its time, but there are other more
modern (switching) regulators that do a much more efficient job.
Although implementing such may take a few more parts, I would look
seriously at using something from, say, Linear Technology.

Any thoughts on another way to drive this backlight? I was thinking of
running it at 12v (it's going in a radio) and a dropping resistor. What say?

Even worse. You'll need a plenty big power resistor, it'll waste a
lot of power as heat, and you'll have practically no immunity from
spikes and surges. Try the switching regulator idea first.


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