So in my case (a robot lawnmower I am building) I have a 12v 5A battery.
I was going to use a 7805 regulator to power the PIC side of things, but
that would mean dissipating 7v * 5A = 35 Watts.  Correct?  I guess what I
am asking is is there a rule of thumb (threshold?) for when you use a
switching regulator as opposed to a linear regulator (and heatsink of
course).  In my case I am thinking that I can live with the simple
7805/heatsink solution rather than go with the slightly more complex
switching regulator followed by a 7805 (as was suggested in another
recent post).

This has been a very interesting and informative thread and thanks to all
for the great info.
Adam

On Sat, 15 May 1999 13:22:52 PDT William Chops Westfield
<billw@CISCO.COM> writes:
>     >With a linear regulator you'll have about 100
>     >Watts of heat to dissipate to get 25 Watts of
>     >useful power.
>
>     I'm curious: How did you come up with this? Is there a rule of
> thumb
>     to use for linear regulator efficiency?
>
> P = IV, although it's not quite as intuitive with an active device
> like a voltage regulator as it is with a simple resistor.
>
> If your input is 25V and your output is 5V, you have a 20V voltage
> drop across the regulator.  If you're delivering 5 amps, that's
> going
> through the regulator as well.  So you have a "black box" device
> with
> 20V across it and 5A through it.  P = IV = 5*20 = 100 Watts.
> (depressing, eh?)
>
> BillW
>

Adam Bryant (age 0x23)
abryant@peaktech.com (work)
adamdb@juno.com (home)
Parker, CO, USA
Robotics, RC Airplanes, anything using a PIC

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