An alternator diode problem.
My track loader is not charging the battery.
Simple rectifier bridge, diode, is mounted in the fuse panel. I’m trying to test. With a cheap $20 multi meter. Rather than measuring voltage drop it measures resistance in Kohms when set to the diode test.
Reads 468. Manual says 1.5 mA is the impressed voltage used.
Should I simply multiply 468 by 1.5 mA To get the voltage drop? Confused by the manual which states that readout is in
K ohms Not ohms! If so that gives 700 V. Does not make sense. Your insight is appreciated.
From what I understand a good fluke meter would simply read out the actual voltage.

Thanks

 

Manual says 1.5 mA is the impressed voltage used

Fact: 1.5 mA is not a voltage
Fact: 468 * 1.5 mA = 468 * 1.5e-3 = 0.700

Some meters read in mV. I forget what my $6.00 Harbor Freight meter reads in,
The diode test generally reads OL in one direction and a voltage (around 0.7) in the other.
Ohms range diode tests generally don;t tell you a lot.
A shorted diode will show up in ohms. Short in both directions.
Diodes normally fail shorted.

Charging system: Check battery voltage under various rpm: ~13.8 Vdc
Check ripple (AC voltage) while charging at various RPM: Probably less than 1V (can't remember).

 

Hmmm
I guess I was using ohms law. Voltage equals current times resistance. I figured this manufactureer of the multi meter wants me to use the resistance they claim is the output at the diode selection times their standard 1.5 mA Impressed current. (I wrongly stated “voltage” in my post, sorry)
So the current of 1.5 mA times 468,000 ohms = 700 volts right?

 

So the current of 1.5 mA times 468,000 ohms = 700 volts right?

See fact 2, post #2

 

So the current of 1.5 mA times 468,000 ohms = 700 volts right?

Slightly different, but the same:

1.5e-3 * 468E3; Exponents add and 1.5*468

or the harder way, 1.5 mA * 1 A/1000 mA * 468 K ohms * 1000 ohms/k Ohm

The 1000's cancel, th mA cancels, K ohms cancel; Unit is Amp*Ohms = Volt

Aside: mA * Kohms is also Volts

1.5 * 468