Okay, I bought a JVC KD-HDR1 car stereo unit with HD radio. Among other things, it lets me listen to BBC off the air anytime I like (compliments of a local NPR affiliate's digital sub-channels). After hooking it up for a while at home, I thought I would install it on my boat. The sound quality improvement of FM stations is noticeable.

This unit is now available for $200 which I think is a steal. The one thing that concerned me was that the JVC runs *incredibly* hot. You could fry an egg on it. Other buyers of this unit have noted that.

The incredible heat of the JVC made me wonder if it might be gulping amps; this is a particular concern on a sailboat. I don't really want to run the engine to charge batteries to keep the stereo going.

So I got my multimeter and did a comparison of the existing modest Kenwood receiver and the JVC. With both of them warmed up and volume set low, the Kenwood showed 4.0A draw while the JVC was pulling just 0.8A!

I know there are a lot of other variables here that I might try to solve for, but I am just puzzled. The JVC seems to sound fine when I crank it up, so it is putting out *some* energy that isn't heat, but it just seems like it would have to pull more amperage to heat the case up to such a remarkable degree. What basic principle of solid state electronics am I overlooking?

(If I decide that I am happy with the JVC I may buy another for the car for my "phase 2" Empeg reinstall with the display extender.)

Dismantle the thing and locate the component that's producing all the heat?

You don't have to look to solid-state electronics for a reason to question, only the first law of thermodynamics. The device is consuming electrical power. It must therefore be generating an equivalent amount of energy. Nothing in the device is moving, I would assume, so there's no kinetic energy. There's probably a few LEDs or something producing some light, but not much. Hopefully there's no non-visible EM radiation, or at least very little. That leaves heat. But you'd assume that that would be the same for both stereos.

Are you sure you measured the current properly? You have to break the circuit and put your ammeter in it in-line. Was the JVC hot when you measured it? How long does it take to get hot?

Oo hey here's a thought. Does the new stereo have a built-in amplifier that is perhaps one of those fancy high-power amplifiers? Maybe it's a different kind of amplification technology that generates heat (and the corresponding increase in current draw) even when it's not outputting tons of sound.

Better heat dissipation, to air or to the mounting, on the Kenwood?

Uneven current? Perhaps the Kenwood draws peaks of 4A with a very low duty cycle (i.e. averaging a very low current -- probably a switch-mode PSU would do this), the JVC's current is steady, and your ammeter isn't averaging the Kenwood current in the way you might hope?

Peter