I have a diode cutout for a 3 brush generator. This device is potted into a metal case with two wires exiting, one from each end. One end of the metal case has been stamped with a + sign. Any ideas on how this is installed?
The purpose of the cutout diode is to let current run only one way - out of the generator and into the battery. If the engine has stopped or is running at idle speed, then the generator isn't making enough voltage to charge the battery, and without a cutout current would run back into the generator, eventually draining the battery.
Install the cutout with an arbitrary orientation. Start the engine, and tach it up. Does the ammeter indicate a charge? If so, you are connected correctly. If not, reverse the wires and confirm by restarting the engine.
Another test. With the generator loose from the engine but with the battery wire still connected, ground the frame of the generator to the car. If the generator runs like an electric motor, it means that the cutout is connected backwards or is not functioning. The battery is being allowed to drain into the generator.
I have to point out that if you hook up the diode cutout backwards (you have a 50-50 chance of doing that) and then start the car, the generator will be running with no load on it which can harm the generator when you rev up the motor but if that voltage goes high enough then the diode reverse voltage rating could be exceeded in which case the diode will short and be destroyed. Thus you risk the generator and/or the diode with a live test. It would be far less risk to hook up the thing at both ends with a battery cable disconnected. Then watch the ammeter when you touch the cable back on the battery. If you see a discharge of 6-8 amps (this assumes a 6V battery is being used)) then you have the diode cutout in backwards. Reverse the leads and repeat the test. No need to spin up the generator and thus hurt it or the cutout.