Find Best Metal for DIY Capacitor Experiments

[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

 

[jalebi]

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

 

[Born2bwire]

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

 

[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.

 

[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

 

[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.

 

[jalebi]

anybody?
i'm really desperate

 

[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.

 

[jalebi]

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.

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.

 

[uart]

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.

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?)

 

[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

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?

 

[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 separation 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.

 

[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

 

[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 commasEDIT: 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

 

[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.

 

[jalebi]

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.

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?

 

[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.

 

[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

 

[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.

 

[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

 

[Born2bwire]

That's not enough. The theoretical discharge time is infinite. What you need are at least two points. The initial voltage and then some voltage at a known time t after discharging has started. Ideally you would want multiple points but two would be enough for you to solve for the time constant. Once you have the time constant you can calculate the capacitance with the knowledge of the resistance.

 

[jalebi]

Yeah, so if I know the voltage was X at t=0 and 0 at t=y, would that be enough? i.e. at one point voltage is x and time is zero, at the other point voltage is 0 time is y.

 

[Born2bwire]

No, the theoretical voltage never reaches zero, that data point is not valid for the model.

http://en.wikipedia.org/wiki/Capacitor#DC_circuits

That gives the equation for charging the capacitor but it i easy enough to find the discharge equation given the boundary conditions.

 

[jalebi]

Excuse my ignorance/lack of knowledge, but are you saying that with the data i collected, there is no way to calculate the experimental capacitance?

 

[Born2bwire]

Yes. You need at least two non-zero (well, non-steady-state to be exact) voltage measurments and their associated points in time.

 

[jalebi]

Can't I have one non-zero point, and assume the other point is 0.01V at y seconds. would that work? is there no other way to calculate capacitance

 

[Born2bwire]

Why 0.01 V? Why not 0.001 V or 0.000001 V? This makes a difference and since the voltage has dropped below the sensitivity of your meter you cannot make any assumption about what it is.

 

[Dadface]

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?)

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?

Hello jalebi if your capacitance has a value of the order of nF then unless you have a large resistance your time constant (RC) is going to be extremely small and as explained by uart, difficult to register.Try a long enough time constant, say five to ten minutes.It is easy to calculate the resistance (R) needed but it will be very big.You can use a good quality high resistance voltmeter as your resistor so check what is available and look at the manufactures data to find what the voltmeter resistance is.You could have a circuit where you charge your capacitor up and then discharge it through the voltmeter.Switches will make it neater.Measure the voltage at say half minute intervals as the capacitor discharges and display your results graphically as V against t or better still lnV against t(this will give a linear plot).You can use an ammeter but you will need a high value series resistance.

 

[jalebi]

Thanks guys. Dadface, that's exactly what I've done. however, i only measured the initial voltage of discharge and the time taken to fully discharge (at least up to the limit of my voltmeter). Since, I am working under a time limit, I don't really have time to repeat experiments. Is it possible, in any way possible, to calculate even a rough experimental capcitance?

 

[Dadface]

Thanks guys. Dadface, that's exactly what I've done. however, i only measured the initial voltage of discharge and the time taken to fully discharge (at least up to the limit of my voltmeter). Since, I am working under a time limit, I don't really have time to repeat experiments. Is it possible, in any way possible, to calculate even a rough experimental capcitance?

Thats very difficult jalebi because as you probably know the discharge is exponential.The current and voltage fall to 1/e(1/3 approximately) of their original values for every time interval equal to the time constant.For example if the original voltage is 9V and the time constant 5 minutes then after 5 minutes the voltage drops to about 3V and in the next 5 minutes it drops to about 1V and so on.According to the maths the capacitor never fully discharges but for practical purposes we can say that after about 4 to 6 time constants the voltage/current is so small that we can ignore it.If you want to get something from your results then I would suggest that you take the discharge time as being between 4 and 5 time constants.From this you can calculate a minimum and maximum value of the time constant( RC) and knowing R you can calculate C,of course this gives you a very rough value but it is better than nothing.If you can find some extra lab time I suggest that you do repeats but take measurements at regular intervals .Good luck with it.

 

[jalebi]

Thanks, Dadface. Something rough is better than nothing, i guess.

 

[jalebi]

guys, I have some more questions. these are not really to do with capacitors but with the dielectrics I used.

1) with regards to the type of "film transparencies" used with projectors (e.g. http://content.etilize.com/Large/10639757.jpg), are they polythene or acetate?

2) Is cling film (aka plastic wrap or saran wrap) polythene or something else?

 

[jalebi]

anyone?

 

[alxm]

Why would you use plastic? Plastic is a lousy dielectric. (the cited number of around 6 sounds right. Water is 80). The problem with water is its conductivity - you don't want the capacitor discharging across your dielectric medium.

Perhaps a combination of water, and a thin sheet of plastic to stop any currents?

 

[uart]

Why would you use plastic? Plastic is a lousy dielectric. (the cited number of around 6 sounds right. Water is 80). The problem with water is its conductivity - you don't want the capacitor discharging across your dielectric medium.

Perhaps a combination of water, and a thin sheet of plastic to stop any currents?

Multiple choice question for alxm.

A large (valued) capacitor in series with a small (valued) capacitor results in an overall capacitance which is :

a) Larger than the original small capacitor.

b) Smaller than the original small capacitor.

When you answer that I'm sure you'll figure out why two dielectrics in series won't help.

 

[alxm]

When you answer that I'm sure you'll figure out why two dielectrics in series won't help.

I didn't say it would. Putting a film of plastic there will most definitely hurt the dielectricity and capacitance.

You're totally ignoring my point, which is that water conducts electricity. A capacitor with a lower dielectric is better than a capacitor with a high dielectric constant which is leaking current - because the latter isn't going to act like a capacitor to begin with!

 

[uart]

anyone?

Sorry you didn't collect enough useful data to make any sort of experimental estimate (other than zero) for the capacitance.

Even if you repeat the experiment it is unlikely that you'll manage to get a discharge time constant much more than 100ms (your ohm meter will probably have a 10M input resistance and you're unlikely to get much more than 10nF with the present set up). Now 100ms is going to be too quick to get any meaningful reading from a DVM, the meter time constant may even be longer. So you need to either build a much larger capacitor or give up on that approach.

Alternative approaches would be :

1. Get an ocsilloscope and do it properly.

2. Get an AC plug-pack (some people call them "wall warts") and measure the AC current through the capacitor. I'd try to get a higher voltage unit, like 21VAC or 24VAC (not so common but some devices use them) or at least the more common 15 VAC pack.

Procede as follows

- Set up capacitor with 1k series resistance (to prevent accidental short of pack)

- Connect a DC supply (12 to 24 volts, whatever you have at hand) and attempt to measure the DC current. It should be approximately zero. This part is the control.

- Replace the DC supply with the AC plugpack and re-measure the current (making sure to switch your ammeter range over from DC mA to AC mA).

You'll need a meter that can measure down to 10's of uA AC (a 20mA AC range with a 3 digit display might just do it do) or better yet you could use the 1k resistor as a shunt (and measure it's voltage) if your meter has a 200mV-AC volt range.

The value of the capacitance can be estimated from the measured current as follows :

C = \frac{I}{120 \pi V} *

* Assuming freq=60Hz - use 100 pi if your mains is 50Hz.

 

[jalebi]

Why would you use plastic? Plastic is a lousy dielectric. (the cited number of around 6 sounds right. Water is 80). The problem with water is its conductivity - you don't want the capacitor discharging across your dielectric medium.

Perhaps a combination of water, and a thin sheet of plastic to stop any currents?

I used plastic because i was investigating different dielectrics. I was not designing an optimized capacitor - just carrying out an experiment to learn something.

thanks for all the help uart.

i don't mean to sound rude or unappreciative, but neither of you has answered my question on the materials of the dielectrics. do you guys know what the specific names of those materials are (see about two posts of mine ago)?

 

[alxm]

i don't mean to sound rude or unappreciative, but neither of you has answered my question on the materials of the dielectrics. do you guys know what the specific names of those materials are (see about two posts of mine ago)?

Wikipedia has a pretty comprehensive list:
http://en.wikipedia.org/wiki/Types_of_capacitor#Types_of_dielectric

 

[Dadface]

Hello jalebi.I'm not 100 percent sure but I think the projector transparency is cellulose acetate and cling film is low density polythene.

 

[gary350]

I have built several high voltage capacitors. You did not say what voltage or UF rating you want? I use aluminum flashing available at any lumber yard, hardware store in the roofing section. I buy aluminum 10" wide rolls 20 ft long. I also buy polyethylene clear plastic the thickess I can get is usually 6 mil thick. For high voltage I want the polyethylene to be over lap the edges of the aluminum by 1" all the way around. If I make flat plate caps of rolled caps I still want 1" of over lap to keep sparks from jumping around the plastic from 1 alumimum plate to the other. After cutting the aluminum I file and sand the edges very smooth and cut off the shape corners. Round off the corners real nice and smooth. As for the physical size the rolled caps are always 1/2 the size of the flat plate caps. The flat plate caps have twice the current capablity of the rolled caps. The finished cap is filled with High Voltage oil, or caster oil, or mineral oil, or BBQ grill charcoal lighter fluid, or liquid lamp oil.

 

[jalebi]

Hello jalebi.I'm not 100 percent sure but I think the projector transparency is cellulose acetate and cling film is low density polythene.

thanks again Dadface. you've been too much of help. thanks to the rest of you guys as well - what a helpful forum this is