Hello
In a way it's like measuring slew rate in a circuit, but I want to do it with capacitors.
Do your formula could do that and what equipments I need or diy made ?
Thank
Bye
Gaetan
The formula for capacitor charging is well known, one of the first things you learn if you study electronics.
Vc = V(1 – e-t/RC)
Vc is the voltage on the capacitor, V is the applied voltage, t is time from start of charging and RC is the time constant of the circuit.. capacitance times resistance.
The rate of charge is exponential, the more charge the capacitor acquires, the more slowly it charges.
Vc = V(1 – e-t/RC)
Vc is the voltage on the capacitor, V is the applied voltage, t is time from start of charging and RC is the time constant of the circuit.. capacitance times resistance.
The rate of charge is exponential, the more charge the capacitor acquires, the more slowly it charges.
Even simpler, and particularly applicable to slew-rates (meaning constant, not varying in an exponential fashion): dv/dt = i/c (derived from charge definitions)
This means that if you supply a constant charge (or discharge) current to a capacitor, it will result in a constant slew-rate.
That is exactly how most precision relaxation oscillators work, like TV sweep generators and also function generators.
That said, I do not see the purpose of this experiment...
This means that if you supply a constant charge (or discharge) current to a capacitor, it will result in a constant slew-rate.
That is exactly how most precision relaxation oscillators work, like TV sweep generators and also function generators.
That said, I do not see the purpose of this experiment...
Now to actually answer the OP's question… you need:
The (1) siggen must have sufficient low-impedance output to drive the test capacitor - and - resistor combination without being taxed so greatly as to not produce nice square waves.
The first question is … are you trying to figure out a (1a) completely unknown capacitor, or to confirm the more-exact (1b) actual capacitance of a part?
(1a) if completely unknown, this becomes a complicated topic of trial, observation, rescaling, observation, and so on to finally get a solution. Frankly, I'ven't the time for it today.
(1b) if trying to more precisely confirm a marked (i.e. "known") capacitor, then the task is simpler. Using the concept of "RC", choose a frequency 1/(³RC) or so on the signal generator. Put the precision-known resistor in series with the capacitor. Use the oscilloscope to observe both the sig-generator output and the capacitor response.
Without too much trouble, scaling the cap-response for more accurate screen cursor readings, you'll be able to determine how much the capacitor charges and discharges relative to the signal driving it.
From there - and with an evening of probably vexing math - you'll then derive the capacitance.
GoatGuy
(1) a square-wave signal generator
(2) an oscilloscope and
(3) a calculator or spreadsheet
(4) a known precision resistor
(2) an oscilloscope and
(3) a calculator or spreadsheet
(4) a known precision resistor
The (1) siggen must have sufficient low-impedance output to drive the test capacitor - and - resistor combination without being taxed so greatly as to not produce nice square waves.
The first question is … are you trying to figure out a (1a) completely unknown capacitor, or to confirm the more-exact (1b) actual capacitance of a part?
(1a) if completely unknown, this becomes a complicated topic of trial, observation, rescaling, observation, and so on to finally get a solution. Frankly, I'ven't the time for it today.
(1b) if trying to more precisely confirm a marked (i.e. "known") capacitor, then the task is simpler. Using the concept of "RC", choose a frequency 1/(³RC) or so on the signal generator. Put the precision-known resistor in series with the capacitor. Use the oscilloscope to observe both the sig-generator output and the capacitor response.
Without too much trouble, scaling the cap-response for more accurate screen cursor readings, you'll be able to determine how much the capacitor charges and discharges relative to the signal driving it.
From there - and with an evening of probably vexing math - you'll then derive the capacitance.
GoatGuy
The quick answer is what others have said: a capacitor doesn't have a charge/discharge speed, only a capacitor in a circuit has a speed so build the circuit and measure the speed - or simply calculate it.
However, some capacitors used in very fast pulse circuits may behave in non-ideal ways (e.g. the ESR may seem to be higher at higher frequencies). I have no idea what this has to do with audio, which by definition will always be slow.
However, some capacitors used in very fast pulse circuits may behave in non-ideal ways (e.g. the ESR may seem to be higher at higher frequencies). I have no idea what this has to do with audio, which by definition will always be slow.
You do sometimes get some dielectric absorption and related shenanigans, but that is a well understood effect and there are good ways to measure it, slew rate is not one of them.
As to charge/discharge speeds, they are kind of a thing in high energy pulse applications, where special 'photoflash' electrolytic caps are used because they can supply far larger current pulses to the arc tubes then a standard electrolytic can (At least without damage).
You also get 'pulse capacitors' which are usually film types having specifically low ESL construction, manufacturers are folks like General Atomics and application is typically pulse forming networks in radar and laser systems, voltage ratings are typically in the tens of kV region.
None of these things however are particularly relevant to dog slow audio band stuff.
Regards, Dan.
As to charge/discharge speeds, they are kind of a thing in high energy pulse applications, where special 'photoflash' electrolytic caps are used because they can supply far larger current pulses to the arc tubes then a standard electrolytic can (At least without damage).
You also get 'pulse capacitors' which are usually film types having specifically low ESL construction, manufacturers are folks like General Atomics and application is typically pulse forming networks in radar and laser systems, voltage ratings are typically in the tens of kV region.
None of these things however are particularly relevant to dog slow audio band stuff.
Regards, Dan.
I got to thinking… what practical method could be remembered indefinitely, and lo… Math to the Rescue!
Follow if you will… if the defining formula for the accumulated charge on a capacitor-in-series-with-a-resistor is:
And we want to have some useful "measure this, at this point" (which is kind of arbitrary), then let's choose the point. ½ V. Say you've got only one precision resistor (e.g. 1 kΩ ±100 ppm). Hey, they're expensive. And you have a totally variable frequency signal generator, with equally impressive precision frequency determination.
The method couldn't be simpler. Use an oscilloscope plus a low frequency to generate a more-or-less square-wave that more or less fills the oscilloscope screen. Then, ramp up the frequency until the waveform is exactly ½ as high peak-to-peak. Then calculate…
Which ALMOST solves the problem. But square-wave generators tell you frequency, not period. So… there are 2 sides of the square per cycle. Its square!
So substitute that in
And there you go. Read off the frequency, divide 0.7213 by that, and whatever value your precision resistor is, and Voilá, you have capacitance. Granted, in farads, but hey… it doesn't take a post-grad mathematician to multiply by a million to get microfarads, or a billion for nanofarads.
There you are.
Math
GoatGuy
Follow if you will… if the defining formula for the accumulated charge on a capacitor-in-series-with-a-resistor is:
VC = V (1 - e(-t/RC))
And we want to have some useful "measure this, at this point" (which is kind of arbitrary), then let's choose the point. ½ V. Say you've got only one precision resistor (e.g. 1 kΩ ±100 ppm). Hey, they're expensive. And you have a totally variable frequency signal generator, with equally impressive precision frequency determination.
The method couldn't be simpler. Use an oscilloscope plus a low frequency to generate a more-or-less square-wave that more or less fills the oscilloscope screen. Then, ramp up the frequency until the waveform is exactly ½ as high peak-to-peak. Then calculate…
VC = V (1 - e(-t/RC))
VC/V = ½ = (1 - e(-t/RC))
–½ = -e(-t/RC) (change sign)
½ = e(-t/RC) (now take ln() of both sides)
ln(½) = -t/RC (and remember that ln(1/x) = -ln(x)
-ln(2) - -t/RC (negate signs…)
RC ln(2) = t (now solve for C)
C = t/(R ln(2)) (which is memorable enough, but numerically…)
C = 1.4427 t/R
VC/V = ½ = (1 - e(-t/RC))
–½ = -e(-t/RC) (change sign)
½ = e(-t/RC) (now take ln() of both sides)
ln(½) = -t/RC (and remember that ln(1/x) = -ln(x)
-ln(2) - -t/RC (negate signs…)
RC ln(2) = t (now solve for C)
C = t/(R ln(2)) (which is memorable enough, but numerically…)
C = 1.4427 t/R
Which ALMOST solves the problem. But square-wave generators tell you frequency, not period. So… there are 2 sides of the square per cycle. Its square!
t = 1/(2 F)
So substitute that in
C = 1.4427 ÷ 2FR
C = 0.7213 / FR (C in farads)
C = 721,300 / FR (C in μF)
C = 0.7213 / FR (C in farads)
C = 721,300 / FR (C in μF)
And there you go. Read off the frequency, divide 0.7213 by that, and whatever value your precision resistor is, and Voilá, you have capacitance. Granted, in farads, but hey… it doesn't take a post-grad mathematician to multiply by a million to get microfarads, or a billion for nanofarads.
There you are.
Math
GoatGuy
At the risk of making you laugh… A resistor measured to decent precision … makes it a … precision resistor, you know? Ya, ya, I've invested a lot of years doing precision metrology with a good friend who also worked at Lincoln Labs (a big Metrology center). Fun stuff. I still have my “precision 100 Ω standard” resistor someplace. The size of a pipe-fitters lunchbox!!!
Temperature compensated (tho' only certified for 0.1 ppm at 20℃±5, but hey… cost the government over $4,000 new in 1954. Cost me $3 in 1995 when it was retired and put on the scrap-for-auction pile. That resistor in today's money, was over the price of a CAR. I got it for a deluxe hamburger.
Anyway, you're definitely right.
AND it is worth noting that almost any dedent 4½ digit handheld, mainstream brandname multimeter today is delivering rock-solid 0.25% resistor value ratings. Apparently you can buy them for like $99 on EBay. Fluke 8060A 4 1/2 Digit True RMS Multimeter Working! | eBay
Just saying…
GoatGuy
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