Thanks Sean.  I hooked up a really good high impedance  digital volt
meter to a cap.  Charged it to 10 volts, noted it was dropping  about .01
volts in 10 seconds.  It is a 100 uF cap, BTW. The leakage rate was the same
with or without the meter connected. ( I cannot measure the input impedance
of the voltmeter with a 2000 megohm ohmmeter, so it is pretty high.  I think
the input is actually gated so it is open most of the time. )

Disconnected the meter, left the cap on my bench, returned the next morning.
The cap still held 8.9 volts, and the rate of leakage was now about .001
volts in 14 seconds.  Pretty impressive.

Now let's see:  Q = CV  if the cap is 100e-6 farads, the change in voltage
(at first) was .01 volts per 10 sec, that's  10e-6 coulombs of charge in 10
seconds.  If an amp is a coulomb per second, the leakage would be 10e-6/10
or 100 nanoamps.  Does the math sound right?

On the all night test, to lose 1.1 volts in 18 hours would translate to a
leakage current of
100e-6 * 1.1 =  .00011 coulombs in 64800 seconds or 1.69 e-9 amps (about 1.7
nanoamps).


----- Original Message -----
From: "Sean H. Breheny" <shb7@cornell.edu>
To: "Lawrence Lile" <llile@toastmaster.com>
Sent: Wednesday, September 06, 2000 6:42 PM
Subject: Re: [OT] [EE]Low Leakage Capacitors


> Hi Lawrence,
>
> I don't directly know the answer to your question, but you should be able
> to hook up a few of these caps to a good CMOS op-amp (like the LMC6484),
> hooked up as a transimpedance amp and look at the output voltage to
> determine the leakage current. If you wire the current input junction as a
> free-standing connection suspended in the air, you should easily be able
to
> detect leakage as small as 0.01 nA or so (good enough for a 0.01 uF cap in
> your case). You could also, of course, parallel several caps to increase
> the leakage and then just divide the result, too.
>
> Sean
>
> At 03:50 PM 9/6/00 -0500, you wrote:
> >I am trying to analyze some low leakage capacitors.  The leakage
> >specifications are always quoted in numbers like "0.002 CV"  I know that
the
> >"C" is the value of the capacitor.  Does anyone know if the "V"
represents
> >the RATED voltage of the capacitor, or the APPLIED voltage of the
capacitor?
> >
> >To ask the question another way, do low leakage capacitors leak less at
> >lower voltage, and is this phenomenon linear with voltage?  I am using a
16V
> >rated capacitor and applying about 1.2 volts.
> >
> >
> >-- Lawrence Lile
> >Lile's Integrated 'Lectronic Engineering Services
> >http://members.socket.net/~llile/index.htm
> >
> >--
> >http://www.piclist.com hint: The PICList is archived three different
> >ways.  See http://www.piclist.com/#archives for details.
>

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