> But if the amps are too high then sometimes it can make a resistor
> get quite hot, I noticed.  (I wondered which would make it hotter,
> by increasing the amps or increasing the volts. )

Hi Lindy,

consider a common practical example, a 5V regulator like the 7805
which can deliver 5V at 1A, in the TO-220 package. As it's a linear
regulator you can almost call it a resistor

Say you have 8V available as the 7805's input and are drawing 1A.
The difference between the input and output is 3V and the 7805 is
passing 1A, so the wattage dissipated by the 7805 is V * I =3D 3W.
Only just about acceptable without a heatsink. Another way to look
at it is 8V * 1A (8W) in, and 5V * 1A (5W) used by the circuit

Now do the maths for a 25V input. The difference between input and
output is now 20V and the wattage now 20W. This is the limit if you
can find an infinite heatsink.

There are two options - reduce the current or reduce the input-output
differential voltage.

If the load actually uses 1A you can't reduce the current, so the other
option is to lower the 7805's input voltage by adding a series resistor
between the 25V and the 7805 input and limit the 7805 wattage to
3W.

Already known is that 8V input causes 3W of heat, so the resistor
you add needs to drop 25-8 =3D 17V and pass 1A. By V =3D I * R this
gives 17 ohms, and the wattage rating of that resistor needs to be
P =3D I * I * R (or I * V) =3D 17W.

Plus the 3W of the 7805 =3D the 20W that had to be shared

You can see how much energy is lost as heat with linear regulators.
Of the 25 watts going in, 80% is being wasted. For a 12V battery
it would be about 60%. No wonder everybody's into switch-mode
controllers. A good one can be 95% efficient, depending on load.

Joe
--=20
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