Perhaps I misunderstood you, but there seems to be a flaw in what you said.
How does constant current negate ESR? The voltage across a cap will be Vc = Integral(i/C, dt) + i.R, where R is the ESR of the cap.
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It seemed like a more difficult reverse of the previously discussed usage of a constant current current to charge the cap and measuring the time at 2 points (& therefore slope). C = I/(dv/dt) is used in both cases.I'm a little surprised that I didn't get any comments on post #20, regarding C=I/(dv/dt). Maybe I just have an unhealthy ego.
Yeah, I guess the only time it really stands out is when C is a function ofvoltage.It seemed like a more difficult reverse of the previously discussed usage of a constant current current to charge the cap and measuring the time at 2 points (& therefore slope). C = I/(dv/dt) is used in both cases.
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That sounds like an interesting method too, a little hard to apply in real life perhaps...
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Sounds similar to the circuit I posted in post #8.Maybe not. If the comparator setpoints are held at specific levels (or actually it just needs a specific voltage hysteresis compared to Vdd) then the cap voltage will rise at one Tc and fall at one Tc, so total period of oscillation will be 2Tc.
So in implementation all it needs is a frequency meter and to know the value of the charge resistor, and solve for Tc = RC;
Tc = 0.5/freq
C = Tc/R
I'm a little surprised that I didn't get any comments on post #20, regarding C=I/(dv/dt). Maybe I just have an unhealthy ego.
You don't need to measure any voltages if you use precision resistors.Still it would be nice to have a simple procedure to measure a cap within 1% accuracy using nothing more than a multimeter and a 555 or schmidt inverter or something. The comparator oscillator way is nice but it really requires a frequency meter and known setpoint voltage (or a scope to see the amplitude).
Roff, yes your comparator circuit does the same thing, but it's better as your resistors set the switching thresholds to a known voltage value (provided the CMOS output goes from exactly 0v to 5v).
MrAl, ESR elimination is nice but in the spirit of the OP's question I think he was after a way to quantify a capacitor using simple test equipment and a simple circuit, so once he measures the absolute capacitance value of a test cap he can use it to calibrate a home made capacitance meter? I've been in the same place although I found some 1% caps in a junkbox.
Still it would be nice to have a simple procedure to measure a cap within 1% accuracy using nothing more than a multimeter and a 555 or schmidt inverter or something. The comparator oscillator way is nice but it really requires a frequency meter and known setpoint voltage (or a scope to see the amplitude).
Maybe there is a way using mains AC freq from a transformer (AC volts measured with multimeter) into a cap and resistor? Mains freq won't be a great sine as it is usually deformed flat on peaks but maybe this could get close to a 1% reding of the caps value as at least the mains freq is usually within 0.1% freq accuracy?
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Also of some interest is that the series resistance required to get 1/2 of the sine voltage across the cap is equal to:
R=sqrt(3)/(2*pi*f*C)
for what it is worth.
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What values of C did you test? I would think your AC voltmeter input capacitance would be at least tens of picofarads. Of course, that goes in parallel with the R when you measure it also, so it tends to be self-compensating.I did some cap "absolute" testing tonight with my 1kHz sine generator. A cap and a variable resistor in series, the resistor adjusted so AC volts are the same on both components so R = Xc.
C was determined by solving for capacitive reactance at 1000 Hz using;
C(uF) = 1000000 / (6283.18 * R)
It worked pretty well, no problem getting the C value to better than 1% accuracy.
I did some cap "absolute" testing tonight with my 1kHz sine generator. A cap and a variable resistor in series, the resistor adjusted so AC volts are the same on both components so R = Xc.
C was determined by solving for capacitive reactance at 1000 Hz using;
C(uF) = 1000000 / (6283.18 * R)
It worked pretty well, no problem getting the C value to better than 1% accuracy.
Whilst I could order some relatively expensive 1% (even seen some 0.3% silver mica's) before I do so, as a little 'test' I was thinking about a very accurate way to measure absolute capacitance... one that doesn't require a precision capacitor in the first place
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