squeezebox power supply

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I tried to upgrade the power supply of my squeezebox by using a diy kit. Tested on the multimeter, it gave me 9.7V, while the squeezebox wall wart is specified at 9V.

My squeezebox worked for 3 minutes playing a piece, then went off.
Several times, I was able to restart it and play for about 3 min, but in the end it stopped working altogether.
Everytime after it stopped, the LM317 voltage controller was getting very hot.

Any idea about the cause and what to do about it?

Thanks

Z.
 
So the question is why does it get so hot?

(Sorry, you may know all this already)

The heat comes from the power to be dissipated, and that is
(Vin - Vout) x I

If your Vin is much more than 9V (your Vout)
and your heatsink is too small, the heat cannot go away from the chip, and the chip shuts down (lucky and smart that it has thermal protection - a transistor would just melt).


Just as a gross guess:
you are at about 500mA;
those TO-220 heatsinks that are like one inch on each dimension, have maybe something like 5°C/W rating;
you said it's "very hot", and you usually cannot stand to touch above 60°C
(when the inside of the chip is quite hotter)

so let's say that this heatsink is at its limits with a raise from ambient temp of about 40°C,
which is 8W as calculated here:
40°C divided by 5°C/W = 8W

At 500mA current, you get 8W for 16V
If your voltage in-out difference is much less than 16V then you should be fine.
(This is large, you'd be feeding 25V input.)

Now if your heatsink is smaller, and has only 10°C/W rating,
then you get that hot at 4W that only leaves you 8V (still large...)


I hope I didn't make too many mistakes...

A much nicer description of the thermal situation is for example in the datasheet from National
http://www.national.com/ds/LM/LM117.pdf


Tell us what transformer you are using, and the dimensions of the heatsink, and number of fins.

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