At 02:53 PM 6/19/02 -0400, you wrote:

Power dissipation is just:

P= (input voltage - output voltage) * (output current + regulator current)

Regulator current is typically around 5mA for the 78xx05.

You should work out heat sink size based on worst case input voltage,
output voltage tolerance, and ambient temperature.

Best regards,

>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?
>
>TIA,
>Chris
>
>--
>http://www.piclist.com#nomail Going offline? Don't AutoReply us!
>email listserv@mitvma.mit.edu with SET PICList DIGEST in the body

Spehro Pefhany --"it's the network..."            "The Journey is the reward"
speff@interlog.com             Info for manufacturers: http://www.trexon.com
Embedded software/hardware/analog  Info for designers:  http://www.speff.com
9/11 United we Stand

--
http://www.piclist.com#nomail Going offline? Don't AutoReply us!
email listserv@mitvma.mit.edu with SET PICList DIGEST in the body