How can I measure capacitance without expensive equipment?

[zhirong_ab]

Hello,

I am new to this forum (and no physics expert). I apologize if this topic is in the wrong place, but I would really appreciate help on the abovementioned topic.

I am investigating the effect of temperature on the capacitance of capacitors for a school project. Since C = (permittivity) x A/d, I deduced that there would be some sort of relationship as the permittivity of the dielectric is affected by temperature (range of about 10-100 deg C), affecting the value of C.
Given that Q = CV, at a constant voltage, Q will change by an unknown extent.

In order to measure this change, I plan to charge the capacitor at voltage V in a charging circuit, after which the capacitor will be discharged in a separate circuit, and the discharged voltage measured.

However, it occurs to me that the change in Q brought about by change in temperature is independent of voltage, but is instead due to change in C.
As such, the discharged voltage may not vary from the voltage at which the capacitor is charged at all, at any temperature.

Although I am confident that there is a relationship between capacitance and temperature, I am presently unsure of how to proceed. Searching the web has yielded methods involving instruments that I am thoroughly clueless about.

Are there any means of calculating capacitance based on easily obtainable values, with minimal use of expensive equipment? I would really appreciate help on this matter.

Thanks


*If whatever I am saying does not seem to make sense, please do not hesitate to tell me. Capacitance is a rather new topic to me, so I will certainly welcome comments any corrections.

 

[Studiot]

You are right to pause for thought.

Unless you employ some special circuitry that can place a specific quantity of charge on a capacitor, the voltage across it is determined by the external circuitry, not the capacitor.

Capacitors have three main uses in circuits.
1) Blocking DC, whilst passing AC.
2) Providing a charge reservoir to maintain a voltage at some point in the circuit.
3) Providing a time constant.

It is easy to construct a test circuit, based on the last use, to offer the measurement facility you require.
Better still a limited range capacitance sensitive circuit (meter ) can be constructed for pennies with readily available components.

 

[berkeman]

Hello,

I am new to this forum (and no physics expert). I apologize if this topic is in the wrong place, but I would really appreciate help on the abovementioned topic.

I am investigating the effect of temperature on the capacitance of capacitors for a school project. Since C = (permittivity) x A/d, I deduced that there would be some sort of relationship as the permittivity of the dielectric is affected by temperature (range of about 10-100 deg C), affecting the value of C.
Given that Q = CV, at a constant voltage, Q will change by an unknown extent.

In order to measure this change, I plan to charge the capacitor at voltage V in a charging circuit, after which the capacitor will be discharged in a separate circuit, and the discharged voltage measured.

However, it occurs to me that the change in Q brought about by change in temperature is independent of voltage, but is instead due to change in C.
As such, the discharged voltage may not vary from the voltage at which the capacitor is charged at all, at any temperature.

Although I am confident that there is a relationship between capacitance and temperature, I am presently unsure of how to proceed. Searching the web has yielded methods involving instruments that I am thoroughly clueless about.

Are there any means of calculating capacitance based on easily obtainable values, with minimal use of expensive equipment? I would really appreciate help on this matter.

Thanks


*If whatever I am saying does not seem to make sense, please do not hesitate to tell me. Capacitance is a rather new topic to me, so I will certainly welcome comments any corrections.

The temperature coefficient of capacitance for different dielectric capacitors does not just depend on mechanical growth/shrinkage of the capacitor due to temperature. The various dielectrics have temperature coefficients (tempcos) of their own.

A good starting point is to look over datasheets of various construction capacitors -- the different ceramic ones (NPO, X7R, Y5V, Z5U, etc.), tantalum caps, electrolytic caps, metal polyesther caps, etc. The tempcos and tolerances of various dielectric capacitors are some of the main things you look at when selecting what type of cap to use for an application.

 

[zhirong_ab]

Thank you for your swift responses.

It is easy to construct a test circuit, based on the last use, to offer the measurement facility you require.
Better still a limited range capacitance sensitive circuit (meter ) can be constructed for pennies with readily available components.

By "test circuit", do you mean something like figure 14 in this:
http://webpages.ursinus.edu/lriley/courses/p112/labs/node7.html"

...in which I may use equations 19 & 20 to derive the value of C?

I am no university student and have scarce knowledge of electrical engineering so I'd like to keep the circuitry as simple as possible.

A good starting point is to look over datasheets of various construction capacitors -- the different ceramic ones (NPO, X7R, Y5V, Z5U, etc.), tantalum caps, electrolytic caps, metal polyesther caps, etc. The tempcos and tolerances of various dielectric capacitors are some of the main things you look at when selecting what type of cap to use for an application.

I have yet to decide on a specific capacitor to use, but I'm currently looking at electrolytic types which have significant tolerance. I need something that will show a distinguishable trend against temperature. I intend to submerge the capacitor in a water bath to keep temperature constant, so I need something that suits that purpose. Any suggestions?

Oh and I can't seem to understand temperature constant. How is it expressed, and how does it apply to capacitors with non-linear changes in capacitance with temperature? And how does it differ from tolerance?

I apologize if my cluelessness is any cause for annoyance, but I really need help on this matter.

Thanks.

 

[berkeman]

I have yet to decide on a specific capacitor to use, but I'm currently looking at electrolytic types which have significant tolerance. I need something that will show a distinguishable trend against temperature. I intend to submerge the capacitor in a water bath to keep temperature constant, so I need something that suits that purpose. Any suggestions?

Oh and I can't seem to understand temperature constant. How is it expressed, and how does it apply to capacitors with non-linear changes in capacitance with temperature? And how does it differ from tolerance?

Be careful using a liquid bath to adjust temperature -- the liquid can change the capacitance reading if it bridges the capacitor's terminals.

Tolerance is variation in capacitance at room temperature. Tempco expresses expected nominal variation with temperature. If it's linear, you can have one number for tempco. More typically for caps, though, it is non-linear, and just shown as a graph on the cap's datasheet.

 

[Studiot]

By "test circuit", do you mean something like figure 14 in this:

No I mean using the test capacitor as the timing element in a simple pulse oscillator, constructed from a few gates in a couple of digital ICs. The pulses can be converted (added together) to voltage in another capacitor to be read on a voltmeter with a linear scale.

This approach has the advantage of learning a few uses of capacitors and typical circuit configurations to boot.

I don't know if I am allowed provide a circuit, as this is a project, but there are many hobby cap meters out there based on this principle. I don't recommend the ones based on the 555 IC, which I find particularly unreliable.

I don't recommend bathing you caps in water either. Use oil and you will be OK.

 

[zhirong_ab]

I mean using the test capacitor as the timing element in a simple pulse oscillator, constructed from a few gates in a couple of digital ICs. The pulses can be converted (added together) to voltage in another capacitor to be read on a voltmeter with a linear scale.

I'm sorry, but could you explain it in simpler terms? What is this "pulse oscillator"?
I don't expect you to give me everything, but I really have no experience in circuitry and so I'd appreciate it if you could help me understand a bit more.

Just wondering, could I in any case use a multimeter or datalogger to find capacitance?

I'm thinking of using either polycarbonate (polyester) capacitors, due to their lower tolerance, which would otherwise contribute to random error in my experiment. Alternatively, would class I ceramic capacitors be better?

Okay, I admit, I am in quite a hurry.

I apologize for asking so much; maybe this lack of understanding comes from the my poor knowledge of the matter. I have been trying to read up, but I'm starting to think that this is way out of my league, because I just don't know enough and I don't know where to start.

 

[Studiot]

Talk to your project supervisor about what help is permissible and what resources are available.
It may be that you don't need to design the detail of the measuring circuit as the real question seems to be about temperature variation. If that is true I can post some simple circuits.

One way to measure capacitance (and temperature) is to use a bridge. This can be knocked up from a few cheap components. The measurement system is similar for both properties. the output of a bridge may be connected to a multimeter or datalogger or you could just take readings.
The only fly in the ointment is that your bridge will need an ac signal, from an (audio) signal generator if your science lab has one, or you could knock up an oscillator yourself.

The other way that I mentioned will provide direct conversion from capacitance to voltage uses two logic integrated circuits, a 74121 and a 7413 very reliably. This output voltage can again be read on a multimeter or a datalogger or you can build in a meter.

It is an afternoon's work to make these instruments so show this to your project supervisor and let us know how you want to proceed; we will help if we can.

One final question: No simple instrument will cover the very wide range of capacitance available so it is important to decide what base capacitances you will be boiling in your oilbath. Don't make these too large as the variation will be small, and you do not want to be looking to measure a small variation in a large quantity - the most difficult meausrement task in Physics.

 

[zhirong_ab]

According to my supervisor, seeking advice from experts is okay as long as they do not do the project for me, i.e. tell me exactly how to do my experiment.

It's not like you are doing the experiment for me, anyway. Ultimately, I'll be the one doing the data processing/collection myself, as well as the analysis.
Although I try my best to read up, this is out of my syllabus and given my limited knowledge I need some guidance when it comes to experimental design, and required areas of knowledge.

It is in this particular aspect where I need all the help I can get from experts. For example, searching for "Ways to find Capacitance" would never have yielded the idea of using something like a bridge, which I would never have thought of. If possible, could you recommend a few areas in which I could read up?

Thank you for your feedback anyway. Following your suggestion on using a bridge, I thought of setting up something like this:
http://ecelab.com/schering-bridge.htm"

However, there are some things I do not understand. Firstly, why do the ratios of the impedances have to be equal for the current across the ammeter to be zero? Is it because the current is spilt based on the different ratios between the left and right paths, and so for no current to pass across, the ratios have to be maintained?
How do I calculate the impedance of the entire circuit?

I also fail to see the difference between the Schering Bridge, and the Wein bridge in here:http://www.allaboutcircuits.com/vol_2/chpt_12/5.html"

I thought of using an ammeter to measure current, but telling when it reads zero might be difficult, so would a multimeter be better?

Your advice is greatly appreciated.

 

[Studiot]

Electrical bridge circuits have the advantage that you may have to learn about simple ones anyway, so using one will mesh in nicely with your other studies.

Start with a Wheatstone Bridge. Have you met one of these yet?
If not, what on Earth are you studying?

Now a W bridge is the basis or prototype for all others. However it only works for resistance since it uses a DC battery as a supply.

When you come to study AC circuitry ( both mains power and signals ) you will learn that there is an AC equivalent of resistance called impedance.
If you power a bridge with an AC signal you are now comparing impedance, rather than resistance, but the principle is the same.

Have you covered a potential divider yet?

Please answer my questions as they have a direct bearing on what and how I post information. I have asked several, you have not answered. They are all designed to help. So the quicker the answers the quicker the progress.

In particular I suggested you discuss alternative with your supervisor as he will know what resources are available and what is worth pusuing i relation to your course.

To use a bridge you measure temperature you will need a DC supply (battery) a DC voltmeter or microammeter. You will need some standard resistors and a temperature sensitive resistor, called a thermistor.

To use a bridge to measure capacitance you will need an AC supply ( audio frequency sign gen is best) and an something to measure output. This could be an audio amp plus speaker or an audio voltmeter. This same setup couldalso be used instead of the DC for the temperature bridge. You will also need some standard capacitors.

Alternatively to measure temperature you can use an alcohol or mercury thermometer.
Then to measure capacitance my other scheme (which I still like) uses a simple logic oscillator to feed logic pulses to a monostable multivibrator. The capacitor under test is used as the timing element for the monostable so the larger the capacitance the larger the pulse of charge output by the monostable ( because it is switched on longer ).
All you need to do then is to collect these pulses on another capacitor and measure its voltage.

 

[Studiot]

To help understand bridges you need to start with potential a divider.

This is a very simple circuit, which consists of two circuit elements connected in series with a supply. An output is taken as the voltage across one of them.
I have shown this in Fig1.
For purely resistive circuits a DC supply will suffice.
For circuits with capacitance an AC supply is required.

The output is some fraction of the input, depending upon the relative values of the two components as shown in the example calculations beneath Fig1.
Note how the calculations compare for DC and AC.

Now if we connect two such dividers to the same supply, something interesting happens.

I have shown this in Fig2.

If the ratios are the same for both dividers then the output voltage is the same fraction of the supply for both.
In other words they are at the same voltage or there is no voltage diffrence between them.

So if we connect a detector or a wire between these two points no current will flow.

This arrangement is called a bridge and the condition where no current flows is called balance.
Note that there is no requirement for the resistances (impedances) to be the same, just that the ratio is the same.

So how do we use this in practice?

Well if we make the left hand divider from a radio type volume control we can adjust the ratio to achieve balance as shown in a detector connected between the unknown resistor or capacitor. The setting on the control knob then tells us the value.

Alternatively, for small deviations from balance, the difference in output is directly proportional to the change in capacitance or resistance, so we can read this on a meter.

I have shown this in Figs 3 and 4.

 

[zhirong_ab]

Apologies for the late reply.

As a matter of fact, my pre-U school syllabus has not covered anything about circuits in depth yet.At the moment, I have only learned the basics such as parallel and series resistors in circuits, as well as the concepts of current, voltage and resistance.
However, I have been trying to read up further into the topic. We have touched upon potential dividers though.

I thank you greatly for your previously posted explanation as it has certainly helped to clarify my understanding of bridge circuits.
I recently attempted the RC circuit method, where I tried to use a oscilloscope to calculate capacitance through time constant. However, this method failed as the capacitor had capacitance of 0.1microFarad, making the charging/discharging time too short for measurement.

At the moment, what I have in mind is something similar to fig. 4 in your attachment, but instead I will use a variable resistor to control the ratio. I intend to submerge the capacitor in a beaker of oil, which will be placed in a water bath, thus allowing me to quantify temperature via usage of a thermometer.
A low tolerance leads to lower uncertainty in the capacitance value, right?

Correct me if I am wrong, but according to my calculations, if I were to use a 0.1microFarad capacitor at AC current frequency of 10kHz, then:

(Refer to diagram)
When balanced:

Rv/R1 = (1/2piFC)/R2 = 1/2piFCR2

F refers to the frequency of the signal. Note that the variable resistor has a range of 1-10 ohms.
When R1 = 0.33 ohms, R2 = 10 ohms

Subbing in the values:

Rv/0.33 = 1/(2pi x 10000 x 10^-7 x 10)
Rv = 0.33/(2pi x 10^-3 x 10)\
Rv = about 5.3, which falls into the range of my variable resistor.

Is there anything wrong with this setup?

 

[Studiot]

In my opinion 10KHz is too high a frequency.
Bridges like this usually operate in the range 400 to 1000 Hz.

Also you bridges values seem very low. Have you calculated the current draw?

 

[zhirong_ab]

I'm sorry, but what is current draw?

I tried lowering my frequency to about 150, then used 0.1ohm for R1 and 200 for R2, but the multimeter couldn't seem to detect changes in the current significantly. Does this have something to do with current draw?

 

[Studiot]

but the multimeter couldn't seem to detect changes in the current significantly.

I am not at all suprised. To see why look at Fig5.

With a 10 ohm pot at minimum (=0) in series with your 0.1 ohm resistor across the supply you have effectively shorted your supply.

Let us say the supply generates 1 volt. In principle it would need to supply 10 amps.

You need a resistance in the range 5k to 50k for the pot.
It is simpler to connect it as I have shown - you don't need a very fancy bridge.
This also has the advantage of a reasonable fixed current draw so you don't have any worries about this. If you make the pot of too high a value you will have detector problems as the detector needs to have an input resistance of 10 times the pot.

As regards the frequency, yes 150 Hz will work, but try testing your multimeter with just the sig gen attached. most multimeters will go up to several hundred Hz or even 1KHz.
If it is a digital meter then connection polarity is not a problem, but if it is a meter with a needle (analog) then you need to realize that depending upon whether you test capacitance goes up or down will determine which way the meter deflects. Bridges often used a centre zero meter for this reason so the needle could go either way.

Finally I suggest you use a second capacitor, of basically the same value as the test one, rather than a resistor, in the other half of the bridge. This will make setting the intial balance easier and place it somewhere near the centre of the pot range.

The beauty of projects is that there is no fixed 'right answer' like questions from a textbook.
You are gaining valuable experience and hopefully learning by doing this.
Keep it up.

 

[zhirong_ab]

I get that 10 amps = bad, but I'm not too sure of the concept behind it. Is it because of the high impedance provided by the capacitor? I do hope you can make things clearer for me.

Anyway, I figured out that the 0.1uF capacitor I am using isn't sensitive enough to changes in temperature, not to mention that the multimeter can't display measurements for something in that range with enough d.p to be considered accurate.
Hence, with regards to the bridge, I was thinking of using a capacitor with larger capacitance (hence lower impendance), but of a different type.

Anyway, I've been dipping other capacitor types like aluminium electrolytics into hot oil, noticing the significant changes in capacitance. I wonder if there are other types of capacitors that are similarly temperature sensitive, particularly those with inverse temperature-capacitance relationships. Any suggestions?

 

[Studiot]

I get that 10 amps = bad, but I'm not too sure of the concept behind it.

I really think you should get some basic electric theory under you belt, if you do not know Ohm's law.

Talk to your supervisor about

Ohms law
Series and Parallel circuits
Electrical Power / Energy.

 

[zhirong_ab]

Sorry, let me rephrase that.

I understand that the current will be split between the two sides of the bridge, and that the sum of the p.d across the Rv and R1 will be split according to their impedance. A low resistance would lead to large current. However...

With a 10 ohm pot at minimum (=0) in series with your 0.1 ohm resistor across the supply you have effectively shorted your supply

the detector needs to have an input resistance of 10 times the pot.

I don't quite get this. Are you saying the large current is affecting the detector's ability to measurem the current passing through? I thought the size of current passing across the multimeter is a result of the potential difference between the sides of the bridge is proportional to the extent of imbalance of p.d. across the bridge?

 

[Studiot]

I thought the size of current passing across the multimeter is a result of the potential difference between the sides of the bridge is proportional to the extent of imbalance of p.d. across the bridge?

Yes this is true. But that assumes you have a potential difference to start with.

Look again a fig5.

We don't have a bridge here. Just a 10 ohm pot connected inseries with a 0.1 ohm resistor across the supply. You haven't specified the supply, but I have suggested 1volt.

When the pot resistance is at its maximum the total resistance across the supply is 10.01 ohms so 1/10.01 amps ie 0.1 amps flows - so long as your supply can manage this.
When the pot is at its minimum resistance then the load across the supply is 0.1 ohms so 1/0.1 = 10 amps flows again within the limits of the supply.

Now I am assuming we are talking about a standard signal generator ( remember I asked about your resources?).

I don't know of many signal generators capable of supplying 0.1 amps, let alone 10. Somewhere between 0.001 and 0.01 amps would be nearer the mark.

This is one of the reasons why I suggested Fig6. Here there is a constant resistance that does not overload the signal generator.

Now move on to Fig7.

If we choose to use an ammeter (galvanometer) as the meter we must balance the bridge. That is adjust the meter reading for a minimum.
This is because the (ideal) impedance of a current meter is zero. Practically it is very low.
So even a small imbalance in the bridge ( = difference in voltage) will lead to a large current reading.
And you must rebalance the bridge to take each reading.

By the way does you multimeter offer AC current ranges?
Not all do.


So I suggested an alternative whereby we use a voltmeter not a current meter.
An ideal voltmeter has infinite impedance so very little current will flow. However for small changes in the component values in the bridge the voltage reading will be proportional to the change, which is what you want to measure.
So you will not have to keep rebalancing the bridge in order to take each mesurement.

 

[zhirong_ab]

I've acquired a variable resistor and am currently trying to measure capacitance values of 4.7uf to 470uf. Certainly, the bridge does work when I try to measure 470uF with a 1-10 ohm pot.

However, I can't seem to adjust it to reach complete zero. 0.003 mA is sufficiently small right?

My signal generator can generate AC currents up to 100kHz, and up to 6 volts of power.

Just checking, if I want to use the bridge to measure 47 uf, is a pot of about 1-100 ohm sufficient?

 

[Studiot]

Don't worry about getting down to full zero, you will never achieve this. If you did and then
you had a more sensitive meter you would still not be at zero.

All you need is a clear minimum.

My signal generator can generate AC currents up to 100kHz, and up to 6 volts of power.

You still need to clear up the difference between power, current and voltage.

100Khz is too high, remember the test I suggested with your meter to get the best frequency?

Most signal generators in this range expect a load of at least 600 ohms.
6 volts across 600 ohms is 1/100 amps or 0.01 amps, which is about what I said.

But it sounds like you are making good progress. What does your supervisor think?