103
\$\begingroup\$

When I pet my cat, and then touch her on the nose, I get a little shock. Sometimes, when she walks up to something, her nose sparks and she jumps back and puffs out. I was wondering how I might go about measuring the capacitance of my cat.

So how many micro-farads does my cat have? I don't think I can just attach the black thing on the multimeter to her tail and then touch the red side to her nose as in this wikihow article. Neither the wiki article on Body capacitance nor this stack exchange question on the same topic tell me anything about measuring.

I have an I2C capsense chip for my Arduino, but that just seems to throw out randomish numbers between 200 and a couple of thousand, and I'm not sure what to do with those numbers even if there was any repeatability to them.

Would it be possible to create a strap on display for my cat that would show "current charge" for my cat on a bright orange LED grid? Or do I necessarily need to have a reference voltage (my understanding of electricity is that voltage is always relative, does this apply for static electricity as well?)

Thanks in advance,

Tim

EDIT: While Russell McMahon's answer in theory seems to work, I don't think his method is as easy to implement as George Herold's. Both answers do seem to answer the immediate question as posed in the title. However, neither is entirely complete. They both hinge on the requirement of having a fully charged cat. But how do we know how many times to pat our cats before they are fully charged.

It is vital to also be able to measure the charge in real time, as per JRE's response in order to set a foundation for Herold's or McMahon's methods. Using JRE's technique, we can charge the cat until the charge stops rising, THEN measure the cat's capacitance.

Ideally, if we are to verify the potential for petting power as the purrfect post-fosil fuel energy source we will need reliable real time measurement of the cat's stored milliwatt hours as well as purrcentage charge and charge stored purr pat.

\$\endgroup\$
  • 54
    \$\begingroup\$ Do let us know if you end up with an accurate value for the catpacitance. \$\endgroup\$ – Nick Johnson Feb 2 '15 at 12:24
  • 20
    \$\begingroup\$ You cat might appreciate a collar with a long conductive cloth that drags on the floor. I've grounded my children, but never grounded my cat. \$\endgroup\$ – dotancohen Feb 2 '15 at 12:43
  • 5
    \$\begingroup\$ Do you want to measure the capacitance, or do you just want to know if kitty is charged? Charge can be detected with an electroscope (amasci.com/emotor/chargdet.html). I had trouble finding the MPF102, but any N-Channel small signal JFET should work (I used a 2N5464.) \$\endgroup\$ – JRE Feb 2 '15 at 13:07
  • 4
    \$\begingroup\$ You say cat, and all the engineers perk up. \$\endgroup\$ – Matt Young Feb 2 '15 at 15:56
  • 10
    \$\begingroup\$ As the physicist said, we can calculate the catpacitance as follows: first let us assume the cat is a uniform sphere... \$\endgroup\$ – Hugh Bothwell Feb 2 '15 at 23:09

13 Answers 13

11
+300
\$\begingroup\$

Regarding the bounty quest for reliability in a cheap way: to convince yourself that this is a rather difficult task to do reliably (with laboratory precision), have a look at what it entails to measure "it" for a human, e.g. in a paper that studied it for ESD-related purposes, Numerical Calculation of Human-Body Capacitance by Surface Charge Method by Osamu Fujiwara and Takanori Ikawa, doi:10.1002/ecja.10025. Quoting from the abstract:

However, the body capacitance is strongly dependent on the relationship between the ground plane and the body posture. It is therefore not clear what factors govern the body capacitance. In this paper, the static capacitance of a body standing on a ground plane is calculated by means of the surface charge method. [...] It is found that the capacitance increases as the backs of the soles of the shoes approach the ground plane, that the body capacitance at the same height (10 mm) as the soles of the shoes is 120 to 130 pF, and that it is about 60 pF if the location of the soles is sufficiently high. The computational findings are confirmed by measurement of the body capacitance.

And if you're curious about their measurement method, here are the details for that from the paper:

enter image description here

Figure 7(a) shows the method of measurement of the human-body capacitance and Fig. 7(b) shows its equivalent circuit. The person tested (height 168 cm, weight 68 kg) with a body shape close to the human-body model is standing with bare feet on a Styrofoam plate or a perforated acrylic plate (depth 30 cm, width 11 cm) on a metal plate in a Faraday shield. The perforated acrylic plate has 201 holes made with a drill with a diameter of 4.5 mm at random locations over the plate and with an area ratio of about 9%. In this way, the relative permittivity is effectively decreased. Under this condition, a power supply is used for charging to VB0 (= 10 V) via an analog switch (Toshiba TC4066BP). When the power supply is turned off by the analog switch, the body potential vB(t) is amplified by a low-input-impedance amplifier (with an input resistance Ri = 10.2 MOhm, input capacitance Ci = 13.6 pF) and is directed to a computer via an A/D converter. The sampling frequency of the A/D converter is 200 kHz and the quantization level is 12 bits. In the potential measurement, the metal plate is used as the ground to which the grounding connections of the measurement devices are connected. From the equivalent circuit in Fig. 7(b), the body potential vB(t) is given by

$$ \frac{v_B(t)}{V_{B0}} \simeq exp \Big[ - \frac{t}{(C_i+C_B)R_i}\Big]$$

Hence, from the potential decay characteristic, the body capacitance CB can be derived.

This is basically the same time constant method suggested by George Herold (which I upvoted a while back), but at boffin standards. Nobody measures body capacitance with regularity (even for humans), so I don't know why you expect there to be a cheap way to do it reliably... Never mind that it would probably vary quite a bit as the cat changes body position.

Also, if you hope to just do simulate it on a computer... their numerical model likely won't wont be much good for a cat because:

In addition, clothes and hair are not included in the numerical model.

For a somewhat older (but right now freely available) paper, which discusses the problems with getting accurate body capacitance measurements, see N. Jonassen's Human body capacitance: static or dynamic concept?. Reading that, one point that was salient was that the soles of the shoes are actually a major contributor to the human body model capacitance (while hair and clothing can be basically ignored). Alas, that's probably the opposite of what you can expect for the dominant element to be in a cat (in its natural state) as far as capacitance is concerned. Unfortunately bounty points on SE are rather unlikely to be a sufficient "grant" for boffins to tackle this rather different cat body model in their labs...

\$\endgroup\$
  • \$\begingroup\$ Looking up old studies is a good idea. Here is a jn.nutrition.org/content/134/8/2110S.full study which measures capacitance for some purpose or other. \$\endgroup\$ – timthelion Feb 8 '15 at 19:58
  • \$\begingroup\$ @timthelion: that nutrition paper is only briefly mentioning that they measure skin capacitance of dogs. \$\endgroup\$ – Fizz Feb 8 '15 at 20:22
  • \$\begingroup\$ If you're somehow interested in specific tissues rather than whole body... the US Air Force has produced a large review/study on 30 something species (including felines): dtic.mil/get-tr-doc/pdf?AD=ADA309764 Since a tissue is (assumed) homogeneous, they give the conductivity and permittivity. \$\endgroup\$ – Fizz Feb 8 '15 at 20:44
  • \$\begingroup\$ I chose your answer, though I think that Mr. Herold's, and MAC's also deserve a mention. I chose it for it's quality presentation and because it was "interesting". \$\endgroup\$ – timthelion Feb 11 '15 at 21:54
42
\$\begingroup\$

"Touch Not The Cat, Bot a Glove"

  • enter image description here

DTTAH / ACNR / IANAL / YMMV *

Equipment:
High impedance voltmeter / oscilloscope with HV probe.
High voltage low capacitance capacitors (1 10 100 1000 pF) x 2 of each.

Pretest - charge capacitors to some semi known high voltage and measure with voltmeter to determine measurement ability.

For purrfect results there should be minimal paws between first and second iterations of 2.3.4.

  1. Select cap - say 100 pF.
  2. Discharge cap (short)
  3. Connect one end of cap to ground - one end of cap to cat.
    .... ( How "to cat" is achieved is left as an exercise for the reader.)
    .... (Cap and cat are now at same purrtential)
    Disconnect cap from cat
  4. Measure Vcap
  5. repeat 2. 3. 4.
  6. Compare readings.
  7. Repeat with higher and lower caps. Aim is range where V1 / V2 is usefully high - say about 2:1.

Processing.

When cap connects to cat cap is charged. Cat and cap share charge in proportion to capacitances. Overall voltage drops to reflect increase in system capacitance from addin cap to Ccat. If Vcat before and after transfer was known you could calculate Ccat.
But Vcat 'a bit hard' to determine.
Repeating process gives a second point and 2 simultaneous equations can be solved to give Ccat.

If Ccap << Ccat the delta V is small and results are ill conditioned. If Ccap >> Ccat the delta V is large and results are ill conditioned.

If Ccap ~~~= Ccat the porridge is just right and the bed is just right.
If Ccap = Ccat then voltage will halve on second reading.
V = Vcat_original / 2

Otherwise ratio change is related to inverse proportion to capacitances.
V2 = V1 x Ccat/(Ccat + Ccap) or

Say V1/V2 = 0.75 Ccat = 3 x Ccap.

E&OE ....


DTTAH ...... Don't try this at home
ACNR ........ All care, no responsibility
IANAL ....... I am not a lawyer
YMMV ....... Your mileage WILL vary
E&OE ........ Errors & Omissions excepted.

\$\endgroup\$
  • 4
    \$\begingroup\$ Congratulations, I had doubts anyone would come up with a reasonable answer, that would scare off the asker. \$\endgroup\$ – Dzarda Feb 2 '15 at 11:48
  • 43
    \$\begingroup\$ We can get a lower bound on the capacitance. If we assume an isolated spherical cat of radius R in a vacuum, the capacitance will be 4\$\pi \epsilon_0 R\$ F. If the cat is 0.1m radius, then it will be about 11pF. \$\endgroup\$ – Spehro Pefhany Feb 2 '15 at 12:21
  • 4
    \$\begingroup\$ First off, what does ACNR stand for? Secondly, it is not clear to me whether in step 3 the cat should be charged or uncharged. If charged, how do I ensure that the cat is fully charged? \$\endgroup\$ – timthelion Feb 2 '15 at 13:07
  • 7
    \$\begingroup\$ Be sure to use CAT-5 cable \$\endgroup\$ – Jim Garrison Feb 2 '15 at 22:03
  • 8
    \$\begingroup\$ @SpehroPefhany Behold, a spherical cat. Now to isolate it. cdn.techgyd.com/… \$\endgroup\$ – FullmetalEngineer Feb 2 '15 at 23:26
28
\$\begingroup\$

As a bit of a riff on Spehro's, Capacitance ~ radius. You can measure your own body capacitance with your 'scope. Hook up the x10 probe, set you triggering for single shot, Scuff your feet or rub your sweater (jumper in the UK) and touch the end of the probe. You'll see your discharge through the 10 Meg Ohm of the 'scope probe. Find the 1/e point. Here's a 'scope shot for me. (You have to play around a bit to get the right amount of scuffing.) I get about 2.5 ms ~ 250 pF. You could try the same thing with the cat.

/ enter image description here

Oh for the cat (or more accurate numbers) you should subtract the probe capacitance.
(about 16pF for my x10 probe.)

Edit for comments: This is an example of an RC decay. RC is the time constant of the circuit. See the wiki article here. A quick estimate of the time constant is the take the time when the voltage has dropped to 1/e of it's starting value. (1/e is about 1/3) In the 'scope shot above this time is about 2.5 ms = RC (R = 10 Meg ohm)

\$\endgroup\$
  • 4
    \$\begingroup\$ This looks like a reasonable method. However, I don't have a scope at home. I'll have to ask at the hackerspace if cats are allowed... \$\endgroup\$ – timthelion Feb 2 '15 at 15:36
  • 1
    \$\begingroup\$ I always get a positive going pulse. I tried rubbing with other materials.. but nothing going negative. Your results may differ. (just a warning) \$\endgroup\$ – George Herold Feb 2 '15 at 15:48
  • 2
    \$\begingroup\$ @Kynit, Thanks, my first answer at ten, I got a badge or something... figures it would involve a cat. It was J. Larkin on sci.electronics.design who taught me this trick. (I was seeing some sort of weird effect from touching my circuit.) I remember because he was disappointed that I had a longer decay time... it meant I was bigger. 5'11" 165 lbs. What's your decay time? \$\endgroup\$ – George Herold Feb 5 '15 at 0:40
11
\$\begingroup\$

Problem Meowtivation and Purrpose

How does one measure weight in space? Certainly not with a scale, because there's no gravity. One must use a special apparatus to deduce it indirectly - through oscillation.

Similarly, you are trying to measure a value of a cat, whereby you cannot directly measure the capacitance. Luckily, there are a few things we know from physics that occur in capacitors that we can use to deduce our feline Faradicity.

Geomeowtry

Let's start by examining the geometry of this problem. We can't exactly state that the cat is a capacitor in the traditional sense, though it can certainly store charge. Practically, you have described a combination floor-paw-cat system, whereby the cat's paws form a dielectric between it and the floor (or bed, or sheets, or whatever). The cat is just one half of the setup, but I digress.

We will thus avoid taking such drastic measures as frying the cat with 10,000 V from head-to-tail (we already know we can model a cat as a resistor). Instead, we will do something fairly harmless: stick the cat on an insulating mat (just for safety) and pull 10,000 V from cat-to-ground.

What happens when a body stores charge?

  • More charge = more energy. More energy = more mass.
  • More charge = more ions. More ions = more force somewhere.

Looks like we have two different ways we can make a simple measurement.

Meowthed 1: More charge, more mass

Let's do some napkin-derivation from this brilliant revelation from Einstein.

$$ \begin{split} E &= mc^2 \\ m &= \frac{E}{c^2} \quad\text{a little rearrangement} \\ \partial m &= \frac{\partial E}{c^2} \quad\text{convert into differential form} \end{split} $$

Okay, whatever, where am I going with this? Do you see it? We can now relate a change in mass with a change in energy! That nefarious E term isn't so scary, it's equivalent to the amount of energy stored in the catpacitor.

$$ E_{joules} = C \cdot V \quad \text{(coloumb volts)} \\ 1 C = 1 F \cdot 1 V \\ \therefore E_{joules} = F \cdot V^2 $$

Now we're getting there. Let's combine! $$ \begin{split} \partial m &= \frac{\partial[ E ]}{c^2}, \quad E = F \cdot V^2 \\ \partial m &= \frac{\partial[ F \cdot V^2 ] }{c^2} \\ &= \frac{F}{c^2} \partial [ V^2 ] \\ F &= \frac{ \partial m \cdot c^2 }{\partial [ V^2 ] } \end{split} $$

There you have it, my friend - a formula for the capacitance of a cat that you can measure with a household scale and a voltage source - maybe about a thousand 9V batteries in series. Let's give it a try. Assuming cats are similar to humans, we can estimate the capacitance at around 100 pF. Let's see what to expect at 10,000 V one megavolt.

$$ 100 \text{pF} = \frac{ \partial m \cdot c^2 }{ [ 10^6 \text{V} ]^2 }, \quad \partial m \Rightarrow 1.11 \text{fg} $$

Well, if you must complain about something, it is true that we might miss the change in mass from the breathing of the cat or the normal shedding of fur/skin. Also, we might arc across the insulating mat at one million volts, but hey - you wanted something easy to measure, and what is easier than weighing a cat?

Meowthed 2: More charge, more force

We need two levels of indirection for this one because force can be tricky to measure when it's small (see above). Although we could use another scale with the cat on it, let's rely on something simple - the fact that cats always.land.on.their.feet.

This does require some equipment, namely some big magnets. Take our test platform from meowthed one (the cat, the mat, and the ground plane) and drop them together through the magnets.

$$ \vec F = q(\vec E + \vec v \times \vec B) $$

We can start by eliminating the electric field because we haven't specifically created one. Then, note that the charge we are dealing with comes from the capacitance of the cat.

$$ \begin{split} C &= \frac{q}{V} \quad q = CV \\ \vec F &= CV ( \vec v \times \vec B ) \\ C &= \Big ( \frac{1}{V} \Big ) \Big ( \frac{ \vec F }{ \vec v \times \vec B } \Big ) \end{split} $$

Because it's trivial to derive and I've already basically laid out the whole problem for you, I'm going to leave it to the reader to have the satisfaction of this derivation.

If you start the cat in a vertical orientation, it will naturally spin as it falls to correct its orientation so as to land on its feet. Measure the height and length of your cat and determine how high you need to drop it when uncharged so that it has spun exactly ninety degrees when it hits the ground. Repeat and refine until the cat can no longer keep up - it can't spin fast enough. Be very careful here because strange effects come into play when you bring a cat to this limit.

Knowing that the cat is trying its hardest to correct its orientation, you can now charge it up and drop away - bomb bay open. Now, presuming that the cat is energized and forming a capacitor with the ground plane, the charges in its body should have separated: some to its paws and the others to the top of its furry back. As it descends, these charges will each experience a force through the magnetic field according to Lorentz' derivation above and will produce a torque on the cat's body causing it to spin relative to the mat it's on.

Continue to increase the voltage across the cat until the exerted torque matches the efforts of your furry friend to right itself. When the cat can no longer spin at all, you have all the required variables.

\$V\$ is the voltage at your final drop. \$\vec F\$ is derived from the torque on the cat based off of how quickly it was able to spin before applying the voltage. Resort to high-school level physics and your particular cat's geometry to derive this value (N.B., this need only be done once and can be saved for future tabulations). \$\vec v\$ is completely dependent upon gravity and the point in time during the fall when the measurements were made. \$\vec B\$ is the known magnetic field strength based off of the magnets you use.

If this seems too complicated for you, simply drop the cat from a sufficiently high point so that it reaches terminal velocity before starting your observations.

Finally you get the value of \$C\$ with nothing but some fidgeting, a voltage source, and keen eyes.

Conclusion

Obviously this is a simple problem most physics students have done, if indeed they have ever done real physics. The pictures are missing but it's late and I can't spend all of my time helping you out on such basic trivia. There are far more ways to do this measurement, so put your thinking cap on and let us know how it goes!

\$\endgroup\$
10
\$\begingroup\$

You can measure the charge on the cat using an electrocope.

I built one like the one referenced, but couldn't get an MPF102. The 2N5464 worked fine instead. Build the circuit as described, enclose it in a metal box (ground the negative side of the battery to the box) and add an antenna as described in the article. If the LED lights, then you've got a loaded kitty.

Also note, YOU might be loaded instead of the cat. Zap happens when there's a difference in charge levels, so if you are charged more than the cat you will also get zapped. Ground yourself before grabbing the kitty - if you still get zapped then the cat was charged.

Electroscope from William J. Beaty amasci.com/emotor/chargdet.html

\$\endgroup\$
6
\$\begingroup\$

I find it interesting that I didn't see any embedded solutions.

You can create an RC circuit as described above where your cat is the C. Connect your cat C and a resistor R in series from ground to an IO pin on your microcontroller. Set the IO line to high output long enough to be sure you've fully charged your cat. Then switch the IO line to input and count how long it takes your cat to discharge to ground. The input on the IO line will go to zero when this happens.

Charge and discharge you cat repeatedly on a hardware timer interupt to compute an average over time. Your cat's capacitance can be computed from the resistor value and the time it took to discharge from the fully charged voltage to the threshold voltage on your microcontroller's IO line. It's up to you to make a good cat probe and backpack with LCD screen output of the value continuously computed by the microcontroller.

This method:

  1. Requires no extra measuring equipment.
  2. Measures continuously.
  3. Can be output to an LCD screen for live monitoring.
  4. Is small enough to ride on your cat comfortably.
  5. Is fully automated.
\$\endgroup\$
3
\$\begingroup\$

The interfacing of the cat with the measuring circuit is the biggest problem, as there are no international standards on how a measurement probe should be attached to a cat. For the purpose of circumventing the problem, we will look at the cat on a macroscopic level.

Here is a simple(it's not really simple) solution. Things you need:

  • two metal plates that have a bigger area than your cat
  • insulation material for the plates
  • a impulse or continuous source

What to do:

  1. Build a big capacitor with the two metal plates, while insulating their surface. Measure it's response to the voltage input. Increase the voltage until you get something measureable.
  2. Put your cat in the capacitor and measure again. The cat will change the capacitance.

Effectively you will get one capacitor in the first run, and the cat in series with that capacitor in the second run. I would guess you should look at the change in capacitance for different cat positions, different diets, sleeping cat vs. awake cat and so on as to get a relevant model of the capacitance dependent on different cat parameters.

Not knowing the frequency response of the cat, you should try both DC, and pulsed inputs. The cat should be frequency dependent. Especially in regard to the frequency of the flipping of water dipoles, as this is a good part of the cat.

I'll draw pictures soon, now I just wanted to share the idea.

\$\endgroup\$
  • \$\begingroup\$ With the cat placed between the capacitor plates and a high enough voltage I think you're actually going to wind up measuring a furry form of die-electric cat-pacitance. \$\endgroup\$ – Nedd Feb 6 '15 at 21:57
  • \$\begingroup\$ Correct. It won't measure the cat-pacitance of the cat in the sense you wanted it to. I thought it would be helpful to get some more information about the material, as it could help in creating a cat-model. With a cat-model we won't need to use animals in our efforts to advance science! \$\endgroup\$ – WalyKu Feb 8 '15 at 12:21
2
\$\begingroup\$

It is not the capacitance you should ask for. Capacitance of your cat is irrevelant with that shock you had. It is the static charge accumulation. Capacitance is about energy transaction ability by the dielectric material, not the potential of charge bucks. (look: http://en.wikipedia.org/wiki/Static_electricity )

You can have an idea of its level using the fact of electrons repelling each other. You can build an electroscope (helpful: http://www.exploratorium.edu/snacks/electroscope/index.html )

Also, you can use a digital voltmeter instead. Turn it into AC milivolts function. One probe is grounded. Swing the other probe (~2-3 hz) perpendecular to her trunk, without contacting.

\$\endgroup\$
  • \$\begingroup\$ That animal with a trunk would have a lot more capacitance. \$\endgroup\$ – Nedd Feb 6 '15 at 3:11
  • 1
    \$\begingroup\$ I used that word as a synonym of torso. I don't know this language perfectly, sometimes I use dictionary and it can lead to weird synonyms of popular words. I think I will advance faster in English by reading and writing here :) \$\endgroup\$ – Ayhan Feb 7 '15 at 18:01
  • \$\begingroup\$ trunk is a great word.... I think it's more literary however, and not so commonly used these days. \$\endgroup\$ – nielsbot Feb 8 '15 at 17:26
  • \$\begingroup\$ Are you a politician? They, too, tend to say: "The question you should be asking is ..." :-) \$\endgroup\$ – Russell McMahon Feb 9 '15 at 8:58
1
\$\begingroup\$

If the question was feline resistance, would it not be odd if nobody suggested getting up from your computer chair and finding which drawer has your DVM?

Or a BK rlc-meter set to "c" and 1KHz.

One lead touches cat nose. What about the other? Well, Cat-pacitance varies with approach to conductive masses, or distance to ground. So, other lead goes to YOU, and hover a hand near the cat. Should see ~20 pF, much more if cat is in your lap.

\$\endgroup\$
0
\$\begingroup\$

The best way to skin this Catpacitor

While a few of the other methods posted here seem workable this test can be done with out the need for an oscilloscope or several specific values of high voltage capacitors. Though a high impedance/high voltage meter is a necessity.

A better way to handle this would be to treat it similar to a simple current divider circuit then derive the basic equation to calculate Ccat directly.

enter image description here

As in this first circuit determining Rx is not so difficult if the other values are known. At time = 0 the current flows only through Rx. At time = 1 the switch changes and the current divides between the two paths (through Rx and Rref). The amount of current in each path is determined by the resistor values. The main idea is that the total input current remains the same at both switch positions, but the voltage on each resistor will change due to the different currents in each branch. Using Ohms law Rx can be calculated by measuring the voltage changes and knowing the value of Rref.

enter image description here

In this next circuit pair we introduce the Cat with an unknown capacitance (Ccat) to ground. At time = 0 there is fixed total charge in the circuit (Qtotal0). This is the initial charge on the Cat (Qcat0). The initial charge on the Cat produces an initial voltage (Vcat0).

At time = 1 the switch changes and the total charge divides between the Cat and a reference capacitor (Cref). The voltage on the Cat changes (Vcat1), due to the charge transfer. The charge transferred to the reference capacitor produces a voltage (Vcref1) which is equal to the voltage on the Cat (so Vcat1 = Vcref1).

It is important to note that even though some charge has been transferred the total charge now in the circuit (Qtotal1) is still equal to the initial charge, (so Qtotal0 = Qtotal1).

In similar fashion to Ohms law the voltage on a capacitor can be found with the equation V=Q/C. Manipulating this equation the capacitor’s charge can be found by Q=VC. With the capacitor charge equation and knowing the value of Cref it only requires two high voltage measurements (at time = 0 and time = 1) to determine the capacitance of the Cat, as shown below.

.

Terms used:

Capacitor Charge Equation: Q = VC, with Q in Columbs, V in Volts, C in Farads

Qtotal0 = Total initial Charge at time = 0

Qcat0 = Charge on Cat at time = 0

Qtotal1 = Total Charge at time = 1

Qcat1 = Charge on Cat at time = 1

Qcref1= Charge on Cref at time = 1

Vcat0 = Voltage on Cat at time = 0

Vcat1 = Voltage on Cat at time = 1

Vcref1 = Voltage on Cref at time = 1

Ccat = Capacitance of Cat

Cref = Reference capacitor

.

Calculation:

@ time = 0, measure voltage on Cat (Vcat0).

Per capacitor charge equation (Q=VC)

Qtotal0 = Vcat0 Ccat

@ time = 1, change switch, measure voltage on Cat (Vcat1), this is also the voltage on Cref (Vcref1).

Knowing that the total charge in the circuit has not changed:

Qtotal0 = Qtotal1

Qtotal1 is made up of the charge on the Cat and the charge on Cref, so:

Qtotal0 = (Qcat1 + Qcref1)

Rewrite these charges as their equivalent form per Q=VC:

(Vcat0 Ccat) = (Vcat1 Ccat) + (Vcref1 Cref)

Recall that with this switch position Vcat1 = Vcref1, substitute into the equation:

(Vcat0 Ccat) = (Vcat1 Ccat) + (Vcat1 Cref)

Bring Ccat terms to one side:

(Vcat0 Ccat) - (Vcat1 Ccat) = (Vcat1 Cref)

Factor out Ccat:

Ccat (Vcat0 - Vcat1 ) = (Vcat1 Cref)

Isolate Ccat:

Ccat = (Vcat1 Cref) / (Vcat0 - Vcat1 )

Done...

.

Now for an example:

Use a standard value of 100pf for Cref, measure Vcat at time = 0 and at time = 1, (use 9kV and 4kV)

Ccat = (Vcat1 Cref) / (Vcat0 - Vcat1 )

Ccat = (4kV 100pf) / (9kV - 4kV)

Ccat = 400pf / 5

Ccat = 80pf

.

Using a single relatively low value (high voltage) capacitor your Cat capacitance can be calculated, with less chance of any cat-tastrophic consequences.

Also note that this has been a purely theoretical endeavor, no animals were harmed in the process. Your results may vary. Not responsible for any damages due to unprotected, uncontrolled, unsolicited, or careless use of the above information. - .- Enjoy

\$\endgroup\$
  • \$\begingroup\$ Dear sir, It would appear that in your circuit diagram, the cat only has two legs, however my cat has four. \$\endgroup\$ – timthelion Feb 4 '15 at 14:14
  • 1
    \$\begingroup\$ Unfortunately I did state that my derivation was only theoretical. For more of a story book answer you could assume it is done with a Puss-in-boots, (two non-conductive rubber boots). With two of four paws insulated only a two paw simulation is required. \$\endgroup\$ – Nedd Feb 5 '15 at 0:47
  • \$\begingroup\$ Well, I think that the bigger problem, when it comes to reality, is that all paws are common. So in that sense, a cat is not a cap. \$\endgroup\$ – timthelion Feb 5 '15 at 0:58
  • 2
    \$\begingroup\$ If all paws are common I think you should stick with a common Cat-thode connection. \$\endgroup\$ – Nedd Feb 6 '15 at 3:31
0
\$\begingroup\$

You can do this is a two steps measure with a capacitance and a FET amplifier. You can read the result on a scope or on a multimeter.

schematic

simulate this circuit – Schematic created using CircuitLab

First: Short C1 to eliminate all charges.

Second: Connect Input to Cat

You will read on Output the voltage generated on C1 by the cat's charge. Therefore Cat's charge is Vout/C1.

Make sure that Power+ and Power- are sufficient to avoid saturation of LF356. The series resistor is there to avoid sparking your cat when you connect it to the capacitor.

If you want to have continuous reading while you "load" your cat, then you need a capacitive divider. The following schematics could do it.

schematic

simulate this circuit

The output can be read continuously.

\$\endgroup\$
0
\$\begingroup\$

Measuring steady-state cat capacitance by a simple method.

Warning: Don't try this at home. This has not been tested by the author.

Warning: Transmitted power is increased by both voltage, frequency and capacitance. Use a low frequency (e.g. 75 Hz), a low rms voltage (e.g. 1 Vrms), and a low coupling capacitance (e.g 15 pF).

Like others have said before. The ESD discharge doesn't only depend on the capacitance, but also on the charge, both yours and the cat's. This answer proposes a method of measuring the cat capacitance in a steady-state AC scenario.

If the small signal model of the cat looks like this:

schematic

simulate this circuit – Schematic created using CircuitLab

Then you can measure the catpacitance Ccat with a low frequency sine oscillator and an rms or peak voltmeter like in this setup (note ESR above is Rcat below):

schematic

simulate this circuit

Method:

  • First connect the rms multimeter directly to the sine oscillator to assure you have a 1 Vrms sine, and that you get a steady reading.
  • Keep the cat disconnected and apply a sine through C1 with only the probe connected. Then find the probe parasitic Cpar by calculating: \$C_{par} = C_1(V_i/V_o - 1)\$ , where \$V_i\$ is the 1 Vrms input signal and \$V_o\$ is the measured Vrms output voltage.
  • Now connect the cat and apply a sine with the probe connected. Then find Ccat by calculating: \$C_{cat} = C_1V_i/V_o-C_1-C_{par}\$ , where \$V_i\$ is the 1 Vrms input signal and \$V_o\$ is the measured Vrms output voltage.

Getting a good connection to the cat is probably tricky. Something along the lines of an ESD wrist-band, laced with some safe conducting gel or similar, might be necessary. The nose might be a suitable place to get a good connection, but be nice ...

For good sensitivity, the best choice for the value of C1 is when C1 is close to Cpar+Ccat.

The equivalent series resistance shouldn't cause problems but to be sure, repeat the measurement with a frequency one octave down until you get the same result. If you get varying results, but they are repeatable at the same frequency, try to make a bode-plot of the magnitude response, and then ask again with some measurement data. Finally Ccat and Rcat is wetware, so they are probably not too stable over time.

Your cat is also an antenna, therefore she likely picks up 50/60Hz from the mains. This source can maybe be discounted if measuring a small rms output value when connecting the cat while keeping the oscillator output grounded. (If the antenna signal is large you may be able to find Ccat by another method => only varying C1.) In the method above, if the antenna signal is large you should try to pick a frequency for your measurement so the signals won't depend too much on the phase relationship.

To get measurements over time, with a detailed time-resolution it's necessary to measure/record the output voltage with a scope. The same equations apply when measuring peak amplitude as for rms. Therefore, amplifying the output and then using a halfwave rectifier should provide an easy circuit to allow a continuous reading of the catpacitance.

Finally, if you want to measure the the capacitance as a function of the voltage during the discharge, then you should record the discharge with an oscilloscope as also shown by George Herold in his answer. For this setup use a square wave oscillator, and add a largish discharge resistor. Then calculate the capacitance based on the rate of change for the voltage, and the known current (known from the voltage and the discharge resistance).

$$ C_{cat} = \frac{V(t)/R_{discharge}}{\frac{d}{dt}V(t)} - C_1 - C_{par} $$

\$\endgroup\$
0
\$\begingroup\$

To measure the catpacitance of your feline, you first need a capacitor with a known capacitance charged to a known voltage. The amount of charge in Columbs in this capacitor will be CV where C is the capacitance in farads and V is the voltage in volts.

Now fully discharge your cat. One way might be to have then walk across some grounded aluminum foil.

Attach a treat to one of the capacitor terminals, the other to a piece of foil. When your cat takes the treat, the known capacitor will discharge, and the cat will charge until their voltage potential exactly balances.

Now re-measure the voltage on your known capacitor. The change in voltage multiplied by the capacitance will tell you how much charge went into your cat. That charge, divided by the remaining voltage, is the capacitance of your cat.

\$\endgroup\$

protected by Community Nov 10 '16 at 7:14

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?

Not the answer you're looking for? Browse other questions tagged or ask your own question.