From: Patrick Jankowiak (eccm@swbell.net)
Date: Thu Apr 03 2003 - 16:29:41 PST
Well I guess what I meant was the kind of capacitance meter than
derives the capacitance by passing a small AC signal through it and
measuring its impedance (AC resistance). The capacitance is given by:
C[farads] = 1 / ( 2 * pi * F[Hz] * X [ohms] )
Since the system contains diodes and other voltage sensitive devices,
DC bias should be applied from a low resistance, high impedance source
so that on the negative excursion of the tester's AC waveform, any
diodes (such as in the alternator) do not become forward biased.
Because the system is a 24VDC system, the tester must not output more
than 24VAC peak to peak, and a DC bias of half of the peak to peak
testing voltage would be in order. This insures the voltage from the
tester will not exceed 24VDC and will not go negative.
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