Adam,

In the example below the 7805 reg will have to dispose of
(12v - 5v) * current passing thought it. The key point here is the
current passing thought it - not what is available at the supply.
Therefore if you have a PIC and a bunch of logic, LEDs, blar, blar
all on the 5v supply and these are drawing, say, 200mA then the reg
will have to dispose of 1.4 watts of power as heat,
(12v - 5v) * 0.2A. The switch from linear to switching you have to
base on the max power dissipation of the reg in use and the max temp.
you wish the unit to run at (and the reg will tolerate!), and how much
room you have for heat sinking ;)

<flame shields up>
IMHO: Switching regs are not always the golden egg, they do tend to be
a bit noisier than the linear option, they can be filtered but for a
low power supply this all becomes a bit costly.. oh, why do I feel am
going to be very sorry I said that? :)

        Paul Fletcher.

Adam Bryant wrote:
>
> 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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