Measuring (pin, trace and wire) capacitance?

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I regularly find myself being worried about whether a design meets requirements for electrical specification or not. One of the parameters is how much a connected device can load a bus, say in the order of 10-400 pF, so that it can be calculated that a total capacitance requirement of all the devices and wires on bus is not exceeded (like 400pF total on open-collector I2C bus with resistive pull-ups).

So do you have a homebrew method that only requires a (good 1 GHz) oscilloscope and some signal generators (HP AWG or even AVR square wave on IO pin) to measure this sub-nanofarad capacitances down to say 10pF with 10pF accuracy so it is known that for example a 50 pF limit per device is not exceeded?

I know there are fancy $5000 capacitance meters that work on multiple frequencies and DC bias voltages but it is so rarely needed and expensive so I won't consider this for my semi-hobby projects.

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You can implement this instrument with the aid of a microprocssor and a resistor.

use the analogue comparator to monitor the bus voltage

Make one pin output pin and connect the resistor ( selected on test)between the output pin and comparator input.

Configure a timer counter to count up ( or down)
Set output pin low
Reset the counter
wait a while to ensure bus capacitance is discharged ( will depend on choice of resistor)

set output pin high and start the counter

wait till comparator trips and stop counter.

Counter content is proportional to the bus capacitance.

Use a 390 pf 10% capacitor to calibrate the unit.

A similar scheme counting at 20 MHz ( and using a current source instead of resistor good for 50 fempto farads( fempto = 10E-15)

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Perfect for hobby purposes or low budget:
http://www.aplomb.nl/TechStuff/E...

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If you can, find a copy of "Everyday Practical Electronics", March 2010. They published a similar L-C meter which I built; it works well.

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St app AN2927, explain what ignoramus said...

Without calculus, but calculing a Resistor, and using an oscilloscope, you could calculate the capacitance.. I mean without calculus because I don't if you can within this low caps..

Quote:
A similar scheme counting at 20 MHz ( and using a current source instead of resistor good for 50 fempto farads( fempto = 10E-15)

Fempto hhuaauuu.... Does it?

Regards,

Bruno Muswieck

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Bruno it does.

I used to run a bare board testing facility using flying probe technology whihc relies on chirge rise time measurement ( elctrostatics).

The measuring system could resolve one plated through hole capacitance worth. That is two doughnuts and the barrel of a typical PTH design.

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It was your work?
How a current source enable you to do that?

Regards,

Bruno Muswieck

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Bruno, charging a capacitance through resistor results in exponential curve, while charging with constant current results in straight line. Q=I*T and Q=C*U.

I shall take your advice into consideration, but I am not sure how they fit when I have pull-up resistance on the lines.

I did read Howard Johnson's book (High-Speed Digital Desing:Handbook of black magic or something like that) where they just used a pulse generator and scope, but my signal generator just does not output so fast pulses.

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I know Japeal, but you see the difference between charging with current and voltage? I mean to measure femt faradays with an AVR?

I'll perform some tests next week, but I think that the pull-ups are part of the show... In part..

I heard about Mr Howard, I will buy into someday...

Regards,

Bruno Muswieck

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Bruno,

Consider a capacitor charging with a source.

The voltage will rise somehow ( either linearly or exponentially).

If you say use a sufficiently slow rate of charging then the capacitor will take longer to charge up.

All you then need to do is et up hihg impedance ( jfet source follower) to buffer the capacitor voltage which is then sent to window comparator to start/stop/reset a counter for an analogue of capacitance.

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You're right Ignoramus...
With voltage pulse, you'll need a high value resistor, so this could let a high value impedance for the AVR pin.. So current source will pass over this problem.. ;)

Regards,

Bruno Muswieck

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