# simple capacitor test circuit

Discussion in 'Electronic Design' started by [email protected], Mar 18, 2013.

1. ### Guest

Hello, I am new to discussion groups. I have a home built capacitor (it should have a fairly large capacitance) and I would like to find out what its value is as accurately as possible with a standard multimeter. I wonder if it is possible to build a simple circuit with a led light indicator or something that with a little math and good capacitors can tell me ruffly how large it is. Please help. I only need to get a couple of measurements for my project.

2. ### amdxGuest

and I would like to find out what its value is as accurately as possible
with a standard multimeter.

I wonder if it is possible to build a simple circuit with a led light
indicator or something that with a

little math and good capacitors can tell me ruffly how large it is.
Charge it to a voltage, say 30 volts, then put a resistor across it's
terminals and see how long it takes to discharge to 10 volts.
Compare that to known capacitors, add capacitance until the times are
equal. Or, do a search on RC time constant and you can calculate, using
the resistor (R) you pick and the time. Discharge to 7.5 volts is 3
time constants.
These pages discus RC time constants.

The discharge time will depend on the capacitance of your capacitor
and the resistor you choose. If you think your capacitor is not very
leaky, I would make the time for discharge 20 or 30 seconds to improve

What is the input impedance of your meter, or what kind of meter do you
have.

How did you build your capacitor?
Mikek

3. ### Guest

Thanks for all the feedback! I was actually thinking of some kind of simpletiming circuit with a led light indicator but I think I can figure something out with those methods.

I actually have two capacitors to measure. Each are made with 2 sheets of stainless steel foil with 2mm thick corrugated plastic as a spacer on the ends of the foil (the center is not supported by the spacers). They are woundaround each other on a PVC pipe in a double coil so they have two plates but they have about 3/4 more plate surface area. I hope that makes sense. The dielectric will be tap water so I plan on them being very leaky.. The smaller coil plates (foil) are 12cm by 22cm (264 square cm). the larger one are double the length at 12cm by 44cm (528 square cm).

4. ### amdxGuest

Hmm, I'm not sure that will be as big as I was envisioning.

Let us know.

Mikek

5. ### JamieGuest

Don't have any idea how large is large on your end?

But, you can use a resistor to charge the cap and time it for
large types.

Time_TO_Charge_To_63.2% = R * C;

So, if you connect lets say a 1 volt supply through a 1k ohm
R to the cap and have the other end of the cap back to the
opposite terminal of the supply and then, measure the time it
takes to get to 0.632 volts. So Time_Calculated / R = C .

To start the test, you just short the cap and as soon as you release
the short, start the timer.

I am just assuming you have a supper cap!...

Just a thought....

Jamie

6. ### tmGuest

Hey, it's even worse than you were thinking .

7. ### Guest

I remember seeing that calculation somewhere.. Thanks for reminding me about it. I am not sure what you mean by pulse forming and I am not quite sure what I am intending on. All I know is I will be using a pulsed (I have to figure out an ideal frequency to step charge the capacitor) DC power supply at about 3KV or higher with a bifilar wound coil and one of these as a capacitor and maybe a choke coil in there.. The goal is to continuously resonate the tank circuit.

8. ### Tim WilliamsGuest

So like, Tesla coil stuff?

You may find the capacitor is lossier than the coil, and consuming all the
power you're trying to push into it means it won't resonate very well.
Water has a noticable dielectric loss tangent (think of it this way: it's
a *really bad* capacitor all the way up at microwave frequencies, in fact
it's an excellent resistor!).

Tim

10. ### Jasen BettsGuest

perhaps build ac bridge and compare them with a known capacitor.

11. ### Jasen BettsGuest

for some examples of that approach google "555 capacitance meter"
sheets of stainless steel foil with 2mm thick corrugated plastic as a
spacer on the ends of the foil (the center is not supported by the
spacers). They are wound around each other on a PVC pipe in a double
coil so they have two plates but they have about 3/4 more plate
surface area. I hope that makes sense. The dielectric will be tap
water so I plan on them being very leaky.. The smaller coil plates
(foil) are 12cm by 22cm (264 square cm). the larger one are double the
length at 12cm by 44cm (528 square cm).

tap water dielectric won't withstand voltages over 1.48V
so the 555 approach isn't going to work, perhaps you can put them in
series with a smaller capacitor that's good to 10V or so and measure
the pair.

12. ### Phil AllisonGuest

<

I actually have two capacitors to measure. Each are made with 2 sheets of
stainless steel foil with 2mm thick corrugated plastic as a spacer on the
ends of the foil (the center is not supported by the spacers). They are
wound around each other on a PVC pipe in a double coil so they have two
plates but they have about 3/4 more plate surface area. I hope that makes
sense. The dielectric will be tap water so I plan on them being very leaky.

** Very leaky is a massive understatement.

A pair of parallel SS plates the size of your palms will quickly boil tap
water if connected to a 230VAC supply.

That is about 5 amps of " leakage ".

Or 45 ohms of resistance.

The OP mentions using a 3kV DC supply.

He is having you on.

..... Phil

13. ### Robert BaerGuest

And...use the meter resistance (most good DVMs are 10 Megohms) for
the R in series (for charging) or in parallel (for discharging).

14. ### Robert BaerGuest

He lives in the South Pole...

15. ### amdxGuest

Hey diablosdemon, you missed one of my questions.

What is the input impedance of your meter, or what kind of meter do you
have.
Mikek

16. ### John SGuest

Just using the 528 mm^2, 2 mm separation, and 80 for a dielectric
constant, it calculates out to about 187 pF.

Not very large in capacitance and, with tap water for a shunt
resistance, you probably won't even be able to measure the capacitance.

17. ### Jeroen BellemanGuest

He said 12 by 44 cm, not mm. So that should be about 19nF. With
tap water, it would still be more of a resistor than a capacitor.

To the OP: What are you trying to do with all this?

Jeroen Belleman

18. ### Phil AllisonGuest

"John S"

** At last, someone with a tad of common sense.

.... Phil

19. ### Guest

Something is getting lost in the translation, or the OP does not understand the plates need to be insulated from the water:
http://www.electronicsarea.com/pic_display.asp?id=7&title=Capacitive water level sensor

Maybe he thinks the SS passivation is his insulation.

20. ### John SGuest

My mistake. Thanks Jeroen. And, I believe you are correct that it will
still not be easy to measure (if at all).