properties of a capacitor

[Mark S.]

Excuse me for any incorrect terminology as the details of what is actually going on in my experiments are still hazey . Hopefully i will be able to get my question across clear enough.

Okay here goes.

In my beginners book, i have been going through an exercise where you place a resistor in series with a capacitor and measure the voltage across the capacitor at regular intervals.

From a set of results you can draw a graph which shows an exponential curve. I am happy with this and it all adds up as far as the product of resistance and capacitance following some law.

Then my book says do a similar test but this time with no resistor and charge the capacitor fully first then measure the voltage drop over regular intervals.

Apparently there is supposed to be another exponential curve showing that the voltage across the capacitor drops more quickly at first then slows down.

With a 22 micro farad capacitor the voltage should drop to 0.7 within 1 minute.

When i did the test there was still a fair ammount of voltage well after a minute.

Any ideas what i may be doing wrong or does this experiment just not work with any accuracy?

Thanks Marks.

 

[JimB]

There is nothing wrong.

What should happen is that the capacitor will discharge exponentially.
What you have to consider is what is causing the capacitor to discharge.
When you have a perfect capacitor on its own there is no discharge path, and the capacitor voltage will remain as-is for ever.
However, no capacitor is perfect and there will be some internal leakage resistance and the capacitor will self discharge, but for a capacitor in good condition it will take a very long time.
Also, the meter you are using will have some resistance (should be very high) and will also discharge the capacitor slowly, (but faster than the capacitor on its own).

OK?

JimB

 

[Mark S.]

Let me see if i understand this fully.

Measuring the voltage across a capacitor while it is being charged you can be reasonably accurate working out the time it takes to reach .63 of the total voltage. (resistance * capacitor)

but

You cannot work out the length of time it takes for a capacitor to discharge because there are too many unpredictable factors.

Is this correct?

Thanks

Mark S.

 

[audioguru]

Sure you can calculate the length of time for a capacitor to discharge, but the discharge resistance must be much lower than the unpredictable resistances.

A film capacitor hardly has any leakage current. A 1uF film capacitor will discharge to 37 percent of its original voltage by a resistance of 1M in 1 second. You can measure the dropping voltage accurately with an opamp that has an extremely low input current that uses FET transistors on its inputs.

 

[JimB]

You cannot work out the length of time it takes for a capacitor to discharge because there are too many unpredictable factors.
Is this correct?

The parasitic resistances in a real-world capacitor can be ignored for many applications where they become insignificant compared with other real components in the circuit.
By charging a capacitor and measuring the time to discharge, you are in effect measuring the leakage resistance of that capacitor.
But, depending on the type of capacitor, the leakage resistance may be so high that by making a voltage measurement on that capacitor, you are affecting the result because the resistance of your meter is much lower than the leakage resistance of the capacitor.

JimB

 

[Hero999]

Charge the capacitor, then immediately discharge it with a resisitor, then you should have not problem. If you havng around for too long before discharging it it's self discharge.

The discharging charictarisitc should be exactly the same as the charging characteristic providing you use the same resistor. This is because the parasitic leakage resistance is exactly the same under both conditions, and the same goes for your multimeter providing you use the same one for both of your tests.

 

[hawk2eye]

ideal vs real capacitor

an ideal capacitor behaves as shown in your elementary text.
a real capacitor behaves differently. The dielectric:
air, plastic film, paper, or paper impregnated with an electrolyte (a polarized electrolytic) each behave differently when discharged.

The most amazing phenomena can happen with electrolytic capacitors:
1. The electrolytic capacitor can be charge say to +5v through a 1meg resistor.
A DVM is used to measure Vc =+5v, the voltage across the capacitor

2. The 1meg resistor is disconnected from the battery or power supply.
DVM: Vc =+5v

3. place a jumper wire across the electrolytic for say 15 sec.
DVM: Vc =0v

4. Disconnect the short circuit caused by the jumper wire.
the DVM will slowly indicate that Vc is charging from 0v back towards +5v , even though the battery is NOT connected!

This phenomena will not repeat with most dielectrics.
To be fair to use 1uf capacitor when comparing dielectrics.

The explanation has to do with the structure of the dielectric. An electrolytic dielectric flexes and moves while being charged.
The capacitor stored energy is NOT completely depleted by the short circuit.
In fact the electrolytic dielectric will restore itself to its uncharged state.
Only then will the stored energy be released, causing
it to be recharged from 0v towards +5v.

Most other dielectrics will behave similarly, but will cause much, much less recharge voltages say. +0.05 mv to 5 mv

hawk2eye

 

[Hero999]

If I remember rightly it's due to trapped charges within the dielectric, it's similar to the hysteresis in an iron core.

 

[Mark S.]

I think in know what i was doing wrong in reguards to my initial post.( I hope so. Wont have chance to check yet.)

Whilst measuring the voltage across the capacitor as it was discharging ,i had my multimeter clipped to it at all times . Am i right in saying this would slow down the rate that the capacitor would discharge . I think i should have only checked the voltage at intervals ,not throughout the full experiment.

Does this sound right?

Mark S.

 

[Nigel Goodwin]

I think in know what i was doing wrong in reguards to my initial post.( I hope so. Wont have chance to check yet.)

Whilst measuring the voltage across the capacitor as it was discharging ,i had my multimeter clipped to it at all times . Am i right in saying this would slow down the rate that the capacitor would discharge . I think i should have only checked the voltage at intervals ,not throughout the full experiment.

Does this sound right?

No, just the opposite, it would increase the rate of discharge.

What load did you have across the capacitor to discharge it?.

With sensible values the discharge is just as accurately predicable as the charging.

 

[hawk2eye]

ideal vs real capacitor

I agree with Nigel Goodwin.
+++++++++++++++++++++++++++++++++
A multimeter voltage measurement causes a
current to flow from the charged capacitor thru the
multimeter, thus causing the capacitor to discharge.
....................................................................................
I would add that a 22uf electrolytic capacitor is a "real" capacitor not an ideal capacitor shown in the text.

A "real" capacitor is more complicated than the ideal
capacitor. The equivalent circuit for a "real" capacitor
has a internal leakage resistor.

The leakage resistance will cause an electrolytic to
discharge:
1. charge the 22uf electrolytic to +5v
2. disconnect the power supply
3. measure the Vc =+5v ,
quickly DISCONNECT the multimeter.

4. Wait 60 sec. The internal leakage resistance will partially discharge the electrolytic.
5. measure, record the new voltage Vc =?

6. repeat step 4, 5 say 10 times. After 10 x 60sec the voltage will be much smaller than +5v due to the
internal leakage resistance.

hawk2eye