The capacitance of a short circuit - Page 1
Forum section: electronics / repair

author: T3sl4co1l
author_rank: Super Contributor
timestamp: May 05, 2019, 09:31:21 am
content: As an idle thought while probing a board for a shorted solder joint -- what capacitance does that correspond to?Infinite capacitance is zero ohms at any frequency, so it's not a dumb question. Impractical, yes, but that's what's fun!I was measuring fractional mV most anywhere on the supply (which is an internal plane, all the harder to track..). Over about ten minutes of probing, I found the culprit, a ceramic capacitor with a drool of solder across it.Sensitivity of the voltmeter was 10uV, and no apparent time dependency was seen. This sets a maximum ramp rate. The current was about 100mA, from a 78L05 (which fortunately, behaves nicely into a short circuit).Therefore, the capacitance is at least: (100 mA) * (10 min) * (60 s/min) / (10 uV) = 0.1 / 10 * 600 MAs/V = 6MF.So, probably the short is more than 6 megafarads.(On a more reasonable note -- the shorted node measured about 380uV, or 3.8mohm.)Tim

author: bd139
author_rank: Super Contributor
timestamp: May 05, 2019, 09:42:01 am
content: Interesting perspective. I do a lot of “lazy” analysis of LC circuits at extremes of DC and infinite frequency so that actually fits that approach as well.

author: magic
author_rank: Super Contributor
timestamp: May 05, 2019, 12:02:39 pm
content: Quote from: T3sl4co1l on May 05, 2019, 09:31:21 amWhat's the impedance of an infinite capacitor at DC?

author: KrudyZ
author_rank: Frequent Contributor
timestamp: May 05, 2019, 03:41:41 pm
content: So what we learn from this is that shorts get more capacitive the longer it takes to find them.

author: duak
author_rank: Super Contributor
timestamp: May 05, 2019, 04:34:07 pm
content: Tim, you used 10 minutes as your charge time. If you repeated the experiment with a pulse generator, would it not decrease? Also, wouldn't the discharge time also decrease?On a different note, what is the inductance of a superconducting magnet? There can't be a voltage present except during charging and discharging.

author: AndyC_772
author_rank: Super Contributor
timestamp: May 05, 2019, 08:28:15 pm
content: Capacitance is defined by Q = CVFor a short circuit, V = 0Pass any current through the short, Q = I*tTherefore, I*t = C*0, so C = (I*t)/0, ie. division by zero error

author: T3sl4co1l
author_rank: Super Contributor
timestamp: May 06, 2019, 12:17:25 am
content: Quote from: AndyC_772 on May 05, 2019, 08:28:15 pmBut I wasn't measuring zero, I was measuring fractional mV. Therefore, C > 6kF?Tim

author: T3sl4co1l
author_rank: Super Contributor
timestamp: May 06, 2019, 12:32:10 am
content: Quote from: duak on May 05, 2019, 04:34:07 pmIf I repeated the experiment, I'd still have the same minimum figure. A pulse generator would only give a weaker minimum.Superconducting magnets are usually in the henries range, AFAIK. You can't measure it just sitting there, of course -- H and F are dynamical units (ohm-seconds and seconds per ohm, respectively).If you know how much energy was put into it, and at what current, you can calculate the inductance, charge time (for a given input power or supply voltage) and whatever. To a certain extent, you can get around this by probing instead (say with a magnetic field probe, measurements of the geometry, and measurement of the enclosed current with a non-contact ammeter).Note that my above measurement looked like it was "just sitting there", but no measurement is perfectly accurate, and following the assumption that it was, in fact, changing imperceptibly within that error range, gives an estimate for the minimum capacitance. It could always be more, indeed it could be infinite as we expect a short to be (i.e., not a capacitance at all, in any meaningful sense, just a resistance).So, this is also a lesson about the ways we can interpret the error bars of our instruments. Are the LSBs changing? Are they noise? How can we tell? If so, what estimate can we give for a min or max condition, that would still be consistent with our measurement?Incidentally, even if the voltage were observed to be zero, that doesn't cause an error -- we interpret division by zero as a single point at infinity. The reals are a continuous ring, connected at \$\pm \infty\$. When we take 1/x, we are in effect mapping that region (i.e., from -infty to +infty) to the region around 0. This seems even more strange on the complex domain, where the perimeter at infinity (in all directions) is a single connected point, by the same mapping function.Less abstractly -- suppose we have a resistor of nearly-infinite value, in parallel with a resistor of nearly-negative-infinite value. What's the parallel equivalent? An open circuit is still a frickin' open circuit!Tim

author: duak
author_rank: Super Contributor
timestamp: May 06, 2019, 05:57:30 am
content: Tim, I read your response and thought about what you were originally asking. Would it be fair to say " what value of capacitance would it take to hold down the rate of rise of voltage to such-and-such given a current of so-and-so?" If so, then I agree with you.I initially took your question a bit differently and more practically - but still not seriously.) A brief digression first: back in the 70s, Mr Peterson was my Physics 11 & 12 teacher. He would say pithy things like "acceleration at the earth's surface is 9.8 m/s. How do we know? Because we measure it!" I applied his dictum to your situation. Hopefully, with a good enough test setup using a pulse generator I would have seen the capacitor in parallel with some small but non zero resistance. I couldn't envisage the capacitor changing value by externally shorting it out so I assumed it stayed the same.On your question about the resistor and anti resistor - I'm not from Missouri but I worked with someone who was and I'm damn sure he would have asked you to show him an anti-resistor first! However, If I consider their admittances to be equal but opposite in sign, wouldn't they cancel and present zero net admittance or infinite resistance?)Cheers,

author: T3sl4co1l
author_rank: Super Contributor
timestamp: May 06, 2019, 06:16:59 am
content: Yes, quite!That would be a more accurate phrasing.Considering more complex waveforms, or more frequencies (same idea), you might not see the parallel equivalent circuit -- since after all, either the resistance is so low, nothing else matters (seemingly a good physical description of the actual situation, a short across a 0.1uF cap) -- but you might see the series equivalent circuit, which in this case reasonably has the ~mΩ ESR plus a very large (probably infinite) capacitance.Indeed, negative resistances are hard to come by, and complex resistors seemingly even harder still -- but their equivalents do indeed arise from active circuits, and you can construct a negative resistor with an op-amp for example.Complex resistors -- are never constant, but always frequency-dependent. Why? Because causality: a constant-reactance element would predict the future, to a certain extent.Given that restriction, it turns out they're just as common, i.e., capacitors and inductors.Taking that one phase (pardon the pun) further, if you consider the magnetic circuit of an inductor: a capacitor in the magnetic circuit, is an element which satisfies the magnetic equivalent of I = C * dV/dt, or rather, MMF = C' * dPhi/dt. The circuit equivalent (as seen by the winding looped around that magnetic circuit) is, da da da da, a resistor.So, flux capacitors are very common and not very interesting -- they're just resistors.Tim