"Larry W4CSC" wrote in message
...

(-) on pump---|---.33 ohm 10 watt fuse resistor----ground
|
100 ohm 1/4w resistor
|
|
2N2222 base


+12V-------470 ohms 1/2w------anode LED cathode----2N2222 collector

Ground the 2N2222 emitter (-12V)

At a little more than 2A, the voltage drop on the left side of the .33
ohm "shunt" resistor the pump's current is going through exceeds the
6V required to turn on the base of the 2N2222 (or any common NPN
transistor, about 10 cents). This turns on the 2N2222 from Emitter
(ground) to collector and lights up the LED. When the transistor is
saturated, like it would be at 3A, easily, the 470 ohm resistor limits
the current through the LED (light emitting diode) to around 20
milliamps which is quite bright. If it's too bright, make the
resistor bigger (more ohms). Any pump current over 2A will make the
LED light up.....at almost no current drain at all.

If you put the parts on a small piece of perfboard (or even a little
piece of plastic with holes drilled so the wires go through to hold
the parts in place), you can mount it right on the back of the led
holder. The LED will be running after everyone on the newsgroup has
died of old age..... Should cost under $5 at Radio Shaq. If you want
it to come on at a lower current, raise the .33 ohm to .47.

If you use a .33 ohm fusistor (boxy ceramic resistor, usually buff
colored) you can eliminate the motor fuse. 10W at .33 ohms is 5.5A
which lowers the motor voltage by 1.8V at 5.5A....way more than it
should draw, which is acceptable. It'll blow about 7-8A more quickly.

Make sure the wires to the big fuse resistor can handle 10A to be
safe....same size as the motor wires are now.....

One caveat: if the .33 resistor or fusistor blows, you get 12V through 100
Ohm at the base of the transistor. This will produce a current of 120mA at
the base, which will kill the transistor. A base resistor of 2k2 to 4k7 will
do the trick.

Meindert