To be applied only with the diode out of the circuit.
(+) band +18V (two 9V battery) -------- R1 ------ diode ----+ | GND -----------------------------------------------+
Itest = 18V - 12V / R1 = 6 / R1 [Very approximate equation]
R1 = 100K Itest = 60uA R1 = 10K Itest = 600uA R1 = 1K Itest = 6mA
V_across_diode + Plotting the diode | X forward characteristic | | X | X +-------------------- Itest
Now, reverse the diode so it is reverse biased. No current should flow. But we know that the diode does have some reverse leakage.
2uA * 100K = 0.2V , Vdiode = 17.8V, 0.2V across R1. The only effect from changing R1, is to change the drop voltage value. So across R1, we get the "reverse leakage" while a piddly 18V is applied. When there is 12kV across the diode, that's when the real leakage level is realized.That's an example of building a test circuit, when your meter is poorly suited to the task. The meter assumes on "diode" range, that the diode is a single junction, with a 0.7 to 1.0V Vf value.
A TV repairman, spins in his swivel chair, reaches for the Tektronics curve tracer, throws the diode on it, and traces the diode characteristic to 100V or so applied voltage. Which would be acceptable for this task. Curve tracers are good to that sort of test magnitude, somewhere around 100-150V or so. And likely with fine controls, because if you were curve tracing one of the mosfets with dual gates, the ones that are "static sensitive", the gate on those blows out at 40V, and you'd be very careful to adjust the sweep parameters.
In this little experiment, we're not pushing the diode to avalanche, as that requires too much test voltage. And would blow out our multimeter :-) We're manually tracing just a portion of the curve. The nine volt batteries in the example, can't supply 350mA to "test at rated spec".
Paul