Ok, probably not a stupid question, just a stupid questioner... I've got a Delco 12SI alternator in my Studebaker (converted to 12V). Now before you jump on an answer, maybe go test yours before answering because this seems really strange. Most of you probably have a 10SI or 12SI in your SBCs, so tell me what's going on. The BAT terminal of the alternator has 12V to it, directly from the battery, so it is basically potentially hot at all times. Now, the BAT terminal is not completely electrically isolated from the case on the alternator (high ohms, but not infinite). So... if the alternator case (which is attached of course to the engine and everything else) is at 12V and the neg. terminal of the battery is attached to the engine or ch***is, isn't this a complete circuit and thus a drain on the battery when everything is off? Initially, when I discovered this, I thought my alternator was faulty and went to the parts store and measured the resistance across a new alternator (BAT terminal to case). It was the same as mine. Now when I measure my wife's jap grocery-getter, it doesn't do this. This means that the ch***is to the neg. terminal of the battery has a 12V difference and it seems to me it should drain the battery - what mistake am I making?
An internal diode, if you have a fluke 87 test from red to case and nothing; black to case and you'll read contiuty. It is biased so that ground can read thru to the positive terminal of the windings, but voltage cannot flow.
Diodes will always have a high resistance in one direction and a relatively low resistance in the other direction. The high resistance is a very minimal drain on your battery.
Continuity yes, current flow no = no drain. That said, with the regulator and the field energized you will get the amp draw the rotor requires. Most are around 3 amps. high draw rotors coincide with higher amp output. This is speaking in general terms. I get confused, so if I'm off my rocker, someone else chime in to set me straight.
The Jalopy Journal
