Reverse protection

[toneless]

Hi!
Is this the right schematic...?
I only measure 0.38v loss with 1N4002 but it's ok because my
power supply has over 9v output.

  1N4001
+ ----|D----

- -----------
              |
             ---
              -
I'll have to try R.G's mosfet protection,too
Thanks
Nick

[gez]

Anode attaches to V+ of supply and cathode goes to the positive connection of your circuit.

[Ge_Whiz]

Useful tip when using 7809 - type regulators; the output of these regulators is relative to the ground pin, and can be changed. If you place a cheap silicon diode (1N4001 series) between the 'ground' pin and actual ground, cathode (stripe) to ground, anode to regulator, the output of the regulator will rise by approximately 0.7 V. This will then account exactly for any voltage drop in a series protection diode in the supply line to a circuit. Hence you can protect against reversal of the supply, with no voltage loss.

I have a job lot of 7808 (8V) regulators which I stand on two series diodes to boost to 9.4V for general use.