How does a Mutimeter measure capacitors ?

Is there any chance of leaving them charged ??

Just wondering ??

 

Do you mean a multimeter with a dedicated cpacitance range or do you mean how to use an ordinary multimeter to measure caps?

 

Most multimeters cannot measure capacitance directly. Also, a lot of them cannot measure the capacitance indirectly since they cannot determine frequency.
A simple method is to use the capacitor in an astable RC circuit so that you can measure the frequency of the circuit. Since you can measure the resistor fairly accurately, you can calculate the capacitance.
Unfortunately you cannot leave the capacitors charged since they always leak and the charge disappears.

PS: I mean any charge placed on the capacitors will leak so the charge disappears.

 

Is there any chance of leaving them charged ??

Yes. The measurement cannot take place without applying a voltage, and there is always a chance that voltage will remain in the cap at the instant it is disconnected.

 

There are several methods of measuring capacitance.

There are also several types of capacitor.

In terms of measurement the important distinction is between polarised and non polarised capacitors.
The former require a polarising voltage to work properly, the latter do not.

You will not leave a capacitor charged with alternating voltage alone so a test circuit the only applies ac will not normally leave a capacitor charged.

However for those capacitors that require a polarising voltage these methods may well lead to incorrect results or no measurement at all.

Some capacitance meters offer the opportunity to polarise the capacitor under test (CUT) with a DC voltage.
The best of these will discharge or offer discharge at the end of the measurement, but cheaper ones may not.

Possible test (measurement) methods include

Using the capacitor in a timing circuit,
A self balancing bridge,
Measurement of the time to charge to a particular DC voltage with a constant current generator,
Measurement of the ac impedance with using the feedback loop of an operational amplifier,
Meaasurement of the ac voltage across the CUT in series with a high resistance.

There are instruments available that use each of these methods.

 

You will not leave a capacitor charged with alternating voltage alone so a test circuit the only applies ac will not normally leave a capacitor charged.

Not true! The cap will have whatever voltage was applied at the instant of disconnection. I had a friend in grade school that would touch a capacitor - probably a big, high voltage ceramic from an old TV - to the wall outlet and then zap his friends with it. It would deliver a jolt most of the time.

But the AC voltage applied by a meter would be much smaller.

 

Many thanks for the replies ..... it's all becoming clearer ........ I'll have to find a 555 aND GET MY CALCULATOR OUT !

 

I had a friend in grade school that would touch a capacitor - probably a big, high voltage ceramic from an old TV - to the wall outlet and then zap his friends with it. It would deliver a jolt most of the time.

Do you not consider this pretty irresponsible behaviour? He was your friend?

What responsible technician would measure capacitance using the wall outlet voltage?

And if the measuring equipment did use high voltage for some reason what technician would it would incorporate a discharge path.
What responsible technician would ignore this and just pull the cap?

But the AC voltage applied by a meter would be much smaller.

Indeed it would.

Consider one commercial method.

A 1V signal at 1000Hz is applied to the CUT and a low value resistor in series.
The voltage across the resistor is read with an AC millivoltmeter calibrated in capacitance.

Now the reactance of 10 microfarads at 1kHz is 16Ω so a 2Ω resistor will have about 1 x 2/18 = 100 millivolts full scale across it, so the max possible voltage across the CUT is about 1.5 volts. This is about the same as I can measure from random interference and noise in my workshop.

 

The key part of the statement was "in grade school". Not many grade schoolers would be classified as "responsible technicians".

I think wayneh's point was just taking exception to the assertion, "You will not leave a capacitor charged with alternating voltage alone".

 

Yes, I should have said measurment by alternating voltage alone should not leave a capacitor significantly charged.

But that does not detract from the list of measurement methods, perhaps you would like to add to this?

 

Thanks very much for the replies and info ........

I came across this ......

http://www.youtube.com/watch?v=VylC8JFoiBo&feature=c4-overview&list=UUOTPsWDzNAosVd6vc3pCPHQ

which I am going to have a crack at when my DSO arrives this week.

 

They do sell capacitance meters.

 

A 1V signal at 1000Hz is applied to the CUT and a low value resistor in series.
The voltage across the resistor is read with an AC millivoltmeter calibrated in capacitance.

Now the reactance of 10 microfarads at 1kHz is 16Ω so a 2Ω resistor will have about 1 x 2/18 = 100 millivolts full scale across it, so the max possible voltage across the CUT is about 1.5 volts.

If 1 volt is applied to a 10μF capacitor in series with a 2Ω resistor, how can about 1.5 volts ever appear across the capacitor? How can more than the applied voltage appear across the capacitor?

 

1volt RMS has a peak of nearly 1.5 volts.

 

1volt RMS has a peak of nearly 1.5 volts.

You are being sloppy. When you said 1 volt was applied you didn't say that you meant RMS, then later when you described the voltage across the capacitor, you didn't say that you had changed to peak volts. If you are going to change from RMS to peak in midstream you should inform your readers.

Furthermore, you give an example calculation that makes it appear as if a capacitor having a reactance of 16 ohms in series with a resistor of 2 ohms forms a voltage divider with 2/18 of the applied voltage appearing across the resistor. But actually the voltage across the resistor would be 1 x 1/12.4

 

You are being sloppy. When you said 1 volt was applied you didn't say that you meant RMS, then later when you described the voltage across the capacitor, you didn't say that you had changed to peak volts. If you are going to change from RMS to peak in midstream you should inform your readers.

I agree. But I think this is also somewhat uncharacteristic of studiot, so it was probably akin to a typo.

Furthermore, you give an example calculation that makes it appear as if a capacitor having a reactance of 16 ohms in series with a resistor of 2 ohms forms a voltage divider with 2/18 of the applied voltage appearing across the resistor. But actually the voltage across the resistor would be 1 x 1/12.4

How do you get the 12.4?

That would be 2/24.8 implying that the total impedance has a magnitude of 24.8Ω. But the most it can be is (16+2)Ω=18Ω and the least it can be is (16-2)Ω=14Ω. Furthermore, since we know that one is pure real and the other is pure reactive, we know that the lower limit is the larger of the two, so 16Ω.

Since 16Ω >> 2Ω, the magnitude of the impedance is going to be just a bit over 16Ω, so the divider would be close to 1/8.

This is assuming the 16Ω is correct (and it sounds right since 1/(2τ) is 0.159 and everything else is a simple decimal point shift). Oh, hell, might as well do it. 1000*10u = 1/100, so since that's in the denominator it makes it 15.9Ω. Yep, I agree.

 

How do you get the 12.4?

That would be 2/24.8 implying that the total impedance has a magnitude of 24.8Ω. But the most it can be is (16+2)Ω=18Ω and the least it can be is (16-2)Ω=14Ω. Furthermore, since we know that one is pure real and the other is pure reactive, we know that the lower limit is the larger of the two, so 16Ω.

Since 16Ω >> 2Ω, the magnitude of the impedance is going to be just a bit over 16Ω, so the divider would be close to 1/8.

This is assuming the 16Ω is correct (and it sounds right since 1/(2τ) is 0.159 and everything else is a simple decimal point shift). Oh, hell, might as well do it. 1000*10u = 1/100, so since that's in the denominator it makes it 15.9Ω. Yep, I agree.

Yes, I pressed the wrong button on the calculator.

With voltages and currents in RMS I proceed as follows.

The impedance of a series combination of a 16 ohm capacitive reactance and a resistance of 2 ohms is Z = 2 - j16. The magnitude of this impedance is 16.1245 ohms. If 1 volt RMS is applied to this impedance, the magnitude of the current will be 1/16.1245 = .0620174 amps. The voltage across the 2 ohm resistor will be .124 =1/8.06 = 2/16.12.

Studiot apparently forgot about the quadrature relationship between the capacitor and resistor voltages. He added the 2 ohm and 16 ohm as though they were resistors, and the divider ratio would be 2/18.