# Metal for capacitor

1. Sep 8, 2009

### jalebi

I am building a simple capacitor for experimentation and i have a question, or two.

At the moment i am using 30cm x 30cm (1 sqft) aluminum plates with a minuscule gap in between them as a capacitor. At the moment, even at the 200micro amp setting on my ammeter I get a very low reading (about 1). What other readily available metals could i potentially use to get a reading of about the same or higher than what i am getting with the aluminum. i want to keep the setup the same, but just change the conductor.

thanks guys

2. Sep 8, 2009

### jalebi

sorry, i forgot to mention that I am no expert, so simple answers would be much appreciated

3. Sep 8, 2009

### Born2bwire

Are you reading DC or AC current? You could use any good conductor which most metals will qualify for, copper, iron, aluminum...

4. Sep 8, 2009

### Bob S

To make a capacitor out of two 30 cm by 30 cm aluminum plates, put 1 sheet of wax paper, 32 cm by 32 cm, between them, and clamp them together with 1 cm of wax paper hanging out on all four sides.. Make sure the aluminum plates are clean and smooth beforehand.

5. Sep 9, 2009

### jalebi

thanks so far guys.

I'll try using the wax paper. Would the clear film paper used for projections work as well?

I have another question: how much affect would humidity have on the results? I'm asking because the capacitor never seems to work in the morning when it is more humid, but works fine (albeit barely) in the afternoon when the air is drier.

thanks once again

6. Sep 9, 2009

### jalebi

Sorry, Born2wire, i forgot to answer your question.

Anyway, I am using a power supply outputting DC power at 25 volts. This is connected into a switch (side A). On side B of the switch is the ammeter which is reading through a resistor. Common to either side of the switch is the aluminum plate capacitor.

I have the switch on A to charge the capacitor. When the voltage is constant (i have a voltmeter attached) - i.e. the capacitor is fully charged - i swith onto B. the capacitor discharges and the ammeter makes a reading.

7. Sep 9, 2009

### jalebi

anybody?
i'm really desperate

8. Sep 9, 2009

### Born2bwire

I do not think an ammeter in this case is a very good measure to see if your capacitor is working. Why don't you just charge up the cap, disconnect the voltage source and then measure the voltage using a voltmeter? A decent cap will slowly dissipate the voltage in an open circuit. I cannot say offhand how much current you should expect to see or if it is even measurable since I do not know all the physical dimensions of your capacitor and value of your resistor.

9. Sep 9, 2009

### jalebi

When I do the experiment with a proper capacitor, it works flawlessly, so it's only the homemade one that's not working well.

I chose to use an ammeter instead of a voltmeter because a voltmeter picks up voltage almost too easily/it's too jumpy and reads a voltage even when there isn't one, etc. An ammeter does not have these problems.

Each aluminum plate in the capacitor is 25cm x 25cm in size. they are separated by 2/5 of a millimetre when air is used as a dielectric. I am using a resistor board with resistances of 33, 100, 330, 1000, 3.3k, 10k, 33k, 100k, 1m, 3m, and 10m ohms. it doesn't work on any.

10. Sep 9, 2009

### uart

The problem is simply that your capacitor is too small (as in too lower capacitance value). It will probably be somewhere around 2 nF, so even with the largest resistor listed the time constant will be sub millisecond and hence too brief to register adequately on your ammeter.

(BTW what type of ammeter is it?)

11. Sep 9, 2009

### jalebi

I got up to 1.5 micrometres yesterday, though. Maybe that was just luck.

The ammeter is actually a multimeter. Here is a picture of it http://www.ruststopnorthamerica.com/images/yellow_multimeter.gif [Broken]

Is there anything I can do so that the capacitance can be measured? Could I use one of those really sensitive ammeters? The ones that measure even minute changes?

BTW could you show me how you made those calculations?

since this is part of an experiment and I need to have some data, what do you suggest I do?

Last edited by a moderator: May 4, 2017
12. Sep 9, 2009

### uart

The forumla for a parallel plate capacitor is very straight forward. It is :

$$C = \frac{\epsilon_r \epsilon_0 A}{d}$$

Where "C" is in Farrads, "A" is the area in square meters and "d" is the plate seperation in meters.

$\epsilon_0$ is a constant equal to approx $8.85 \times 10^{-12}$ and $\epsilon_r$ is the relative dielectric constant which is equal to 1 for air and between 2 and 6 for most paper or common plastics.

The time constant for the discharge is simply equal to the product of the resistance times the capacitance. If this time is too small then it's unlikely that your ammeter will register it very well.

BTW. Have you given 'cling -wrap" type plastic a try. It's very thin and might give you a higher capacitance.

13. Sep 9, 2009

### jalebi

I tried using a variety of plastics, paper, and also film (projection film). I'm going to try wax paper and cling film as well, following your suggestion. Thank you

14. Sep 9, 2009

### jalebi

sorry, just out of curiosity, would measuring the output voltage (i.e. voltage of discharge) provide me with any useful information?

also, i used a different setup before. i now know it is wrong, but i wanted to know what it was measuring.

I had one lead from the power supply connect to Capacitor Plate A, and another lead connect to Cap Plate B through a rheostat. I also had an ammeter connect to both the plates. I would turn the supply on, let it charge for a minute, then turn it off. The ammeter would give a reading. what did it read?

______________________________________,,,,,,,,_____________________
A,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,, capacitor,,,,,,,,,,,,,,,,,,,,,,,,,,,, PS
---------------------------------------------,,,,,,,,,-------rheostat----------

ignore the commas

EDIT: also, do any of you know of a program to draw simple schematic drawings. i have autocad 2010 at my disposal but i can't seem get the hand of drawing simple schematics. all i want is the ability to draw a circuit with voltmeter, ammeters, etc. nothing fancy like autocad 2010 allows

Last edited: Sep 9, 2009
15. Sep 9, 2009

### Born2bwire

1.5 microamps is probably just noise on that multimeter. You probably measured nothing because you kept the multimeter connected as an ammeter when you had the power supply on, all the current went through the multimeter. You should use the voltmeter and measure the voltage across the capacitor when it is set to an open circuit after being charged. This should be a stable reading for you if the capacitor is working.

16. Sep 10, 2009

### jalebi

Thanks. yeah, the ammeter was a part of the circuit the first time around. But i got 1.5 microamps using a switch (i.e. ammeter was not connected to supply).

Could you explain the voltmeter part again? would i use a switch or not?

17. Sep 10, 2009

### Born2bwire

LIke I said, 1.5 microamps is probably just line noise with you multimeter. You should read the documentation for your instrument to learn the precision and noise levels of the device.

Just connect the voltmeter in parallel to your capacitor, hook up the capacitor, in series with a resistor just for current limiting purposes, and charge up your capacitor until the voltmeter reads approximately the voltage of the voltage source. Then, turn off and disconnect the power supply. The voltage reading should remain steady, only decreasing due to leakage currents between the two plates. If it drops off immediately, then either your capacitor is not storing a charge or there is a short between the plates. However, a short could be noticed by the fact that the capacitor could not be charged to the source voltage. As long as it keeps charging, a current runs and this causes a voltage drop across the resistor in the circuit. So this will make the measured voltage across the capacitor lower than the source. The voltage will gradually approach the source voltage. However, if an appreciable leakage current occurs, then the continued current draw through the resistor should keep the voltage from approaching the source voltage. If the plate is not storing any charge (or if it does store charge and you wait long enough), then the capacitor will behave like an open circuit and you will measure the source voltage.

A capacitor stores enough charge to counteract the applied voltage. This has no correlation to the actual amount of charge stored, this is a property of the physical attributes of your capacitor. However, the current when dissipated is dependent upon the amount of charge that was built up. If the charge is very low, then getting a measurable discharge can be difficult. This is why I think voltage would be a more stable measurement since you could have a very low dissipation current or a very low time constant that would prevent you from reading a measurable dissipation current.

18. Sep 10, 2009

### jalebi

Thanks a lot born2bwire

I have another question, though. Using that voltage, could i find out how much energy was stored?

What else could I calculate with that data?

Thank you

19. Sep 10, 2009

### Born2bwire

Not without being able to take a time domain measurement. What you could do is place a known resistance between the terminals of the charged cap and then measure the voltage over time. The quantity V^2/R is the instantaneous power and so you can integrate the measured curve over time to find the energy dissipated.

Otherwise, you will need to know the capacitance of the capacitor. You could use a multimeter to measure the capacitance directly, otherwise you need to calculate it from the physical attributes, or you would again have to do some kind of time domain measurement and calculate it using the time constant or energy.

20. Sep 10, 2009

### jalebi

I have the input voltage, the maximium discharge voltage, and the time it took to discharge. What can I figure out with these (is it enough to calculate experimental capacitance)?

I know how to calculate theoretical capacitance using the dielectric constant, but i want to find the experimental capacitance.

thanks once again