On Wed, 19 Jun 2002, Chris Loiacono wrote:

>I have a circuit with a 7805 regulator that was set-up with a 12.6V CT
>transformer and two diodes on the input side. This gave it an input voltage
>of 8.9V (or, 6.3V x 1.414 = 8.9).
>
>The 12.6V transformer output is rated with the primary at 230V. Now I am
>considering using it on 277VAC. This should bring the input to the 7805 up
>to about 120% of the 8.9V, 0r about 10.7V.
>
>At 8.9V no heatsink was required with a .5A load. How can I pre-figure how
>much power will be dissipated when on the 277V primary, so I can select a
>heatsink, or perhaps decide that the whole idea is either OK or NG???
>
>I checked the National datasheet for the 7805, and still came up short.
>Is it as simple as saying that I will have 5.7V dropout at .5A, thus will
>have 2.85W to deal with?

Maybe you can remove 1 diode and have more ripple. This will reduce the
power on the regulator. Simulate your circuit with spice. I am amazed that
you ran 9V into it (dissipating 2W) and did not need a heatsink. You will
need a small heatsink for 3W but you can remove the diode as I said or you
can add a resistor to spread the load. 3.3ohms/2W in series will go a long
way I think. You can put it before the filter cap too but then you will
need another value.

Peter

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