I have a capacitor that I’d like to measure the capacitance value of, but I don’t have an LCR or other such meter to measure with. (it’s labelled as .01 microfarads, but it’s +/- 20% so I’d like to find out more exactly). Is there a circuit I can construct that will allow me to measure it in some other way?
Thanks!
What equipment do you have, signal generator, scope, DMM, what?
You need to measure the impedance of the thing (voltage across it divided by current through it) at some non-zero frequency.
z = Ve[sup](σ + jω)[/sup] / Ie[sup](σ + jω)[/sup]
That frequency can be purely imaginary (σ = 0, jω ≠ 0), or purely real (σ ≠ 0, jω = 0) or some combination of both.
Either way, you need some way to generate and measure the frequency. Easy if you’ve got the gear, difficult otherwise.
Why not go the local electronics shop, and ask for a demonstration of their LCR meters. Use your cap for the demo.
There’s no easy way to do it as your measuring tools have to be exact. A meter is built to exact specs so it will be precise enough. You could build an LC circuit and see at what frequency it resonates but then you have just enlarged and shifted the problem as you need to precisely measure the L and the F. Tell us what tools you got and we can tell you what you can do but, one way or another, you are going to need measuring something with precision.
I should’ve expanded the LHS of my equation above.
If we assume that the cap is an ideal cap with no series or parallel resistance or inductance, then:
z = 1 / (σ + jω) C
Ideas:
Measure the RC time constant (asuming you can ge R and time precicely)
Measure the impedance and some frequency
There are probably chips whose behaviour is capacitance dependant (555? op-amp? PLL?), but I don’t know the details offhand.
Brian
If you have a half way decent digital voltmeter, then you could try this:
The impedance of a 0.01µF capacitor at 60Hz is 265258 ohms.
Find a resistor of 250000 Ohms. Use your meter to measure the resistance, and make note of it. This is ‘R.’
Get a small plug in AC adapter that puts out AC. These can be bought cheap at Radio Shack - you may even have one laying around the house.
Connect the capacitor and the resistor in series. Connect this to the output of the transformer.
Measure the voltage across the capacitor. This is ‘V1.’
Measure the voltage across the resistor. This is ‘V2’
The effective impedance (AC resistance) of the capacitor is calculated as follows:
X=V1/(V2/R)
The capacitance is calculated from this by:
C=1/(X2pi*F)
Where F is 60 Hertz and pi is 3.14159265
This will get you an value that is about as accurate as your meter.
Measure voltage with the AC voltage setting on the meter.
Your transformer should probably put out something low like 10 volts. This isn’t critical, you just don’t want to take a chance of frying yourself on 110Volts.
Umm. This all assumes that you are in the US. If you are not, then find out what frequency the power line uses where you are. It’ll probably be 50Hz if it isn’t 60Hz. If it is fifty, then the impedance of 0.01µF would be 318309 Ohms, and you will need to change F to 50 in the last equation.
sailor and Desmostylus: as far as measuring tools go, I only have a digital multimeter that meassures current, voltage, and resistance. I guess I should have mentioned that from the outset.
Mort Furd: is the precise value of 250K ohms important, or can I use something in that general range like 200K or 300K?
Thanks, everyone, for all your suggestions!
The resistor value is not important, just that it be known accurately and that it be relatively close to the impedance you expect to measure. Making the values close makes it easier to accurately measure the voltages - you will be using the same voltage scale, which eliminates one source of inaccuracy (different scales may be calibrated differently.) It also keeps you from having to measure extremely small values if you pick a really bad match.
200K-300K should be fine.
Be sure to measure accurately - good contact and use all availabe decimals.
You won’t be extremely accurate, but you should be with in a few percent - which is better than ±20 percent
BTW:
This is the basic principle that I once used to build an impedance curve generator for coils.
We needed to know the impedance of a coil across the audio bandwidth of a portable radio (300-3000 Hz) in order to match the coil properly to the audio amp. Dropping below the nominal output impedance of the amp was a big no-no. The internal speaker would then make noise when the external speaker (the coil) was being used. Since this was for stealth type stuff, this was a BAD THING.
Of course, the equipment was quite different. I used an arbitrary signal generator and fed it my own freqeuncy sweep, and used a two input DAC card to make the measurements and used an FFT to break things down by frequency - but the basic principle is the same.
I just hope I didn’t make some stupid arithmetic error in what I posted above. I fuck up the simple stuff more often than I really like.
Arrgh!
I almost for got:
Make sure not to go over the capacitor’s rated voltage. Granted that staying at low currents by using the large resistor will probably keep you safe, there’s no point in taking chances - especially if you have an electrolytic capacitor. Over voltage can weaken them and cause them to fail at some inconvenient time.
Find a transformer with an output voltage of half of the rated voltage for the cap and you should be just fine.
And the final results were…
0.0098 microfarads for one capacitor, and 0.0102 microfarads for the other (they came in a package of 2), so everything checks out.
Thanks again for all the help!
Glad it worked out.
Just in case someone needs to do it, the equation for figuring the effective impedance of a capacitor at a specific frequency is:
X=1/(2piF*C)
Where F is in Hertz and C is in Farads (microfarads/1000000)
This gets you the value you need to match (approximately) for the resistor.
Also, don’t try this trick for values of C above about 25µF. That’s getting into a range where you’ve got enough current flow to possible toast the resistor.
Do you have access to an oscope?
If so, why not just stick a resistor of known value, apply a DC source, and see how long it takes the graph to rise to 63%. From this, you can just use T = RC, and solve for C.
That’d work just fine for a large, unmarked capacitor when you just need a ballpark figure.
For the 0.01µF part, you’d need a 100Meg resistor to get a charge time of one second - not real practical.