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Shunt resistance to drop capacitor Q

What parallel resistance is needed across a 100pF capacitor to drop Q factor to 3 at 1Mhz? (Assuming the unloaded capacitor Q is much higher).

Reason I ask is that I have acquired a Boonton 72 capacitance meter thanks to Phil Hobbs's and John Larkin's recommendations in this group and now want to check calibration. Manual lists a procedure using a High-Q/Low-Q 100pF test standard where Hi is 500 and Lo is 3 and then gives an example circuit in Fig 4.3 where Q is dropped by a shunt 1.62kohm.

Trouble is I calculate that 1.62k gives Q=1 and 4.75k needed for Q=3 so have I got something wrong or what please?

piglet

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piglet
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if, Q = 2*pi*freq*R*C.. then I get ~4.7k ohm too. George H.

Reply to
George Herold

Thanks George, glad I remembered it right. Looks like Boonton got things mixed.

piglet

Reply to
piglet

Yes, but isn't that for a series circuit? Where Q=Rp/Xp=Xs/Rs. The resistance is usually ESR (series), right?

Reply to
Tom Miller

The Boonton uses a phase-sensitive detector so theoretically ignores the resistive conductivity. I think there are trimmers to get that right.

It's a really nice instrument.

--

John Larkin         Highland Technology, Inc 
picosecond timing   precision measurement  

jlarkin att highlandtechnology dott com 
http://www.highlandtechnology.com
Reply to
John Larkin

Yes I am more used to the series loss resistance version too. I make it

530 ohms necessary in series to get Q of 3 at 100pF 1MHz or 4775 ohms parallel.

piglet

Reply to
piglet

Yes, that is the trim adjustment I am thinking of verifying.

Yes, really glad I got it - Thanks for the recommendation :}

piglet

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piglet

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