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Measuring Capacitance



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I started a new thread name to make the topic clear to others (Fr. Tom
McGahee)
This is directly Tesla Coil related.

> Date: Thu, 30 Jan 1997 20:56:02 -0500
> From: Steve Falco <sfalco-at-worldnet.att-dot-net>
> To: chip-at-poodle.pupman-dot-com
> 
> There are a number of ways to measure capacitance.  The last one I give
> is probably the easiest, but read them all and see what appeals to you:
> 
> 1) You can measure the RC time constant.  I.e., charge your cap through
> a resistor, and note how long it takes the voltage to build from 0 volts
> to 63% of the battery voltage.  Then T = R x C where T is the time in
> seconds, R is resistance in ohms, and C is capacitance in Farads.  A
> circuit might look like this:
> 
>       ----------- Switch ------ Resistor --------------------
>       |                                        |            |
>    Battery                                 capacitor      voltmeter
>       |                                        |            |
>       -------------------------------------------------------
> 
> 
> Start with the switch open, and the voltmeter showing 0 volts (short out
> the cap for a few seconds to discharge it.  Then close the switch, start
> your stopwatch, and wait till the voltmeter hits battery voltage x 0.63.
> 
> Sadly, this works pretty well for large capacitors like electrolytics,
> because you can do the timing with a watch.  But for small capacitors
> such as are used in tesla coils, you need to use a storage oscilloscope
> or electronic timer/counter to measure the time 'cause it will be in
> milliseconds or even microseconds.
> 
> And don't think you can get around this by making the resistor large,
> like maybe a few hundred megohms, because then the leakage in your
> voltmeter or o'scope will bias the readings.  (The resistor must be no
> more than say 10% of the impedence of your test equipment.)
> 

You can wire a high impedance op amp up as a simple voltage follower,
though, and this would allow large resistor values. But is really isn't the
best way...

> 2) You can also build a simple op-amp or transistor oscillator (sorry,
> this exceeds my ascii art capabilities but you can find a circuit in a
> hobby electronics book :), and measure the frequency of oscillation with
> your home-made capacitor, then substitute known capacitors in the
> circuit until you get about the same frequency.  Then your capacitor and
> the known capacitor are roughly the same value.  (2% caps are pretty
> cheap in the lower voltages and make good references.)  This can be
> pretty accurate, and works for the values used in Tesla coils, but you
> either need a frequency counter or a good ear to estimate the
> frequencies.
> 
> 3) The "Guide to Electronic Measurements and Laboratory Practice"
> (Stanley Wolf, Prentice-Hall, 1973) says the most accurate way is with a
> bridge circuit: 
> 
> 
>          --------------------------------
>         |                  |             |
>         |                 R1             R2
>         |                  |             |
>         |                  |             |
>    AC generator            |--- meter ---|
>         |                  |             |
>         |                  |             |
>         |                 R3            R?
>         |                  |             |
>         |                 C3            C?
>         |                  |             |
>          --------------------------------
> 
> The equations are not too bad:
> 
>         R? = (R2 x R3) / R1
>         C? = C3 x (R1 / R2)
> 
> where R1, R2, R3, C3 are permanent parts of the bridge meter, and R? and
> C? are your unknown capacitor.  (That R? is confusing, but it represents
> the loss or leakage in your capacitor.  If you have a good capacitor, it
> will be pretty close to 0.)
> 
> Anyway, we make R1 and R3 adjustable potentiometers, and play around
> with them until the meter says we have balanced the bridge.  Then we
> disconnect the potentiometers and measure the values they were set to
> with our ohm-meter, do a little math, and we know our capacitor value.
> 

For best results try to get a C3 with a value very near what you THINK the
value of <C?> is going to be. To further improve the accuracy of the
circuit, you can choose R1 and R2 to be resistors whose resistance is close
to the capacitive reactance of C3. These improvements affect the practical
resolution of the answer for <C?>. You will get a greater final accuracy,
because the ranges have been matched.

Also, you can leave out <R3> and <R?> in most cases. Use of R3 can cause
the experimenter to miscalculate <C?> when using just a meter. (meter
should be high impedance or it will only be about 10% accurate). If you
really want to measure the dissipation/leakage, then you need a measurement
of the relative PHASE difference between the two legs of the bridge. With a
differential oscilloscope you adjust R1 and R3 until the scope shows only a
dot in the xy mode. At that point you have all the data to calculate both
capacitance and leakage. (This assumes that the reference capacitor has
little or no leakage of its OWN, of course!)

It is much less hassle to just leave <R3> out and ignore the leakage value.


> 4) Easiest way: you can make a voltage divider:
> 
> 
>                    |------------------------------
>                    |                              |
>                    |                            resistor
>                    |                              |
>         120 volt AC (wall plug)                   | 
>                    |                              |
>                    |                            capacitor
>                    |                              |
>                    |------------------------------
> 
> Measure the voltage across the resistor and the capacitor.  Don't get
> zapped - you are playing with line voltages here and there is not much
> between you and eternity.
> 
>     C = (Vr / R) / (2 x pi x f x Vc)
> 
> where Vr is the voltage across the resistor, R is the resistor in ohms,
> pi is 3.141, f is 60 Hertz (or 50 in some countries - you know who you
> are), and Vc is the voltage across the capacitor.
> 
> So there you are: measure two voltages, do a little math, and get a
> result accurate to maybe 10%.  You need a fairly high impedence meter to
> read the voltages but use what you've got.  Also, I'd pick a resistor
> such that Vr is somewhere around 50 volts for reasonable accuracy.  I.e
> you don't want Vr to be 1 volt or 119 volts - get it somewhere in the
> middle by chosing a different R.
> 
> Happy (and safe) measuring!
> 
>         Steve Falco
>         sfalco-at-worldnet.att-dot-net

There is one more way that uses even simpler math. It uses two capacitors
instead of a capacitor and a resistor.

5)
 
 
                    |------------------------------
                    |                              |
                    |                            Known Capacitor           
                                      
         120 volt AC (wall plug)                   | 
                    |                              |
                    |                            Unknown Capacitor
                    |                              |
                    |------------------------------

This method makes use of the fact that the smaller the capacitor value, the
greater the proportion of total voltage a given capacitor will drop across
itself.
It helps if you choose a known value that is close to the one you want to
measure!

CK=Known Capacitor  CU=Unknown Capacitor  EK= Voltage across Known
Capacitor
EU=Voltage across Unknown Capacitor

Formula is based on the fact that EK*CK=EU*CU  Rearranging it in terms of
CU we get:

	CU=(EK*CK)/EU

Always use a high impedance meter, such as a digital unit or a tube VTVM to
keep errors low.

When done making the measurements, remember to always DISCHARGE each of the
capacitors separately (NOT in series), to ensure that each gets fully
discharged.

By the way, methods #4 and #5 can both be checked for error very simply:
After measuring the voltages across each component, add these values
together and compare them to the actual measured voltage across BOTH
components. If they don't add up within a few percent, then your meter is
loading down the circuit. If this is the case, then switch to using a
signal generator with a sine wave output, and increase the frequency to
about 5Khz or more (don't exceeed the frequency rating of your meter!).
What this does is increase the current flow in the circuit so that the
small amount drawn by the meter is small in comparison. That will reduce
the percentage error considerably. It's a shame that most textbooks do not
tell people up front what the sources of error are so they can correct
them.

6) I can think of at least one more method, which is based on the fact that
the larger a capacitor is, the greater the amount of current it will
conduct *at a given frequency*. This method assumes that the applied
frequency is well below the limit of your digital meter, and that your
digital meter has an AC current range. Many don't!

For a given frequency  C1/I1=C2/I2      so    C1=C2/(I1*I2)

The frequency chosen is arbitrary, but I like something between 1KHz and
about 1/2 the frequency limit of my meter.

Again, try to use as a reference a value that is Close to what you are
measuring. That keeps the resolution of the answer higher. Assume you want
to measure a capacitor that you think is pretty close to .01 mfd.  First
connect a sine wave generator in series with the known capacitor and an AC
current meter. Adjust the output of the generator to about ten volts, and
then put the meter on the scale that gives you the most resolution of the
current value. You may have to play with the frequency to find one that
works well, but stay below the frequency limit of your meter! Then, leaving
ALL things exactly the same, exchange capacitors and use the reference
capacitor. Record the new reading and apply the math to find the unknown
capacitance.

I don't recommend D'arsonval meters because they are not linear with
frequency. A digital meter works best. The value of the internal current
shunt will add less than .1% error to your final calculated value, provided
you made your measurement with the scale that yielded the largest number of
significant digits on the readout. In other words 1.234 is better than
01.23  or 001.2


Fr. Tom McGahee