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simple capacitor test circuit

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.
 
A

amdx

Jan 1, 1970
0
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.
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
your measurement accuracy.

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

How did you build your capacitor?
Mikek
 
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).
 
A

amdx

Jan 1, 1970
0
Thanks for all the feedback! I was actually thinking of some kind of simple timing 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 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).

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

Let us know.

Mikek
 
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Jamie

Jan 1, 1970
0
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.

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
 
T

tm

Jan 1, 1970
0
John Larkin said:
Why not? People used to build homemade electrolytic rectifiers. Think
about
using anodized aluminum, or aluminum foil, to make your own electrolytic
cap.

He did say that he expects a "large" capacitance, whatever that means. No
harm
in asking.




--

Hey, it's even worse than you were thinking :).
 
Compute the nominal capacitance using the plate capacitor formula and a

dielectric constant of 80.



Are you doing high-voltage pulse forming? There is a big literature.



Joe Gwinn

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.
 
T

Tim Williams

Jan 1, 1970
0
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.

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
 
J

Jasen Betts

Jan 1, 1970
0
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.

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

Jasen Betts

Jan 1, 1970
0
Thanks for all the feedback! I was actually thinking of some kind of
simple timing circuit with a led light indicator but I think I can
figure something out with those methods.

for some examples of that approach google "555 capacitance meter"
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.. 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.
 
P

Phil Allison

Jan 1, 1970
0
<[email protected]

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.

Fucking Google Groups gmailers.


..... Phil
 
R

Robert Baer

Jan 1, 1970
0
amdx said:
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.

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
your measurement accuracy.

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

How did you build your capacitor?
Mikek
And...use the meter resistance (most good DVMs are 10 Megohms) for
the R in series (for charging) or in parallel (for discharging).
 
A

amdx

Jan 1, 1970
0
And...use the meter resistance (most good DVMs are 10 Megohms) for
the R in series (for charging) or in parallel (for discharging).
That's why I ask for info about his meter.
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
 
J

John S

Jan 1, 1970
0
Thanks for all the feedback! I was actually thinking of some kind of
simple timing 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 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).

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.
 
J

Jeroen Belleman

Jan 1, 1970
0
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.

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
 
P

Phil Allison

Jan 1, 1970
0
"John S"
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.


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




.... Phil
 
<[email protected]



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.



Fucking Google Groups gmailers.





.... Phil

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.
 
J

John S

Jan 1, 1970
0
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.


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