# Troubleshooting My DIY Capacitive Transducer

• phyky
In summary: It is not necessary to use an amplifier anymore if you have a dielectric permittivity of 1.0 or less. If it changes, it will take a long time for the capacitance to change.
phyky
i built a parallel plate capacitor with brass plates to make water level transducer. A= 12cmx2cm, d=2mm. but the capacitance i measure is μF which far more bigger then theory calculation when it immersed into water. anyone can help?pls

Unless it is very pure, dissolved salts in water make it a reasonable conductor.

So, if your capacitor is touching the water, it will conduct current via the water and this will make measurement of capacitance very difficult.

You could still use it, though, as the resistance should vary according to the water level, so you can measure resistance with a multimeter (or some other way) and compare that reading with the height of the water.

i use pipe water. n 1 more thing is the signal conditioning circuit i used is charge amp with 5Vdc. i used R2=1MΩ C2=4.7nF n RClowpass+buffer.but the ouput vlotage i get is increase for a long time n not stable if i increase water level.why?

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You may be getting polarization of the water at the metal surfaces.

This can be seen as bubbles on the metal surface due to DC current passing through the solution.

You can avoid this by just applying power when you want to take a reading.

it do has bubbles.applying power?u mean drive oscillator ac?

I'm not sure what that means.

The bubbles on the surface of an electrode reduce the surface area for current to flow. So, this may be one reason that your reading is varying.

These bubbles do not form if you use AC and you can also reduce them if you use less DC current or use DC current in short bursts. Stirring tends to sweep the bubbles away too.

If you are familiar with microprocessors, you can use them to generate a DC voltage with one pin and take a voltage reading with another pin and store the value. Then remove the DC voltage.
So, you might take a reading in less than a second and display it for a minute or so, then take another reading.

This is extremely effective at avoiding polarization.

applying power
That means connecting the capacitor to its amplifier only for the time you are taking a new reading.

Could you apply a plastic coating to the surfaces that will be immersed? Is it for only short-term use, or how will you stop the gap becoming fouled with dirt or algae?

vk6kro: so use ac supply or stiring away the bubbles will get stable voltage output? i want connect it to microcontroller to make it a water level sensor. so if the calibration is not stable, it is less precision
NascentOxygen : it use only for short term.

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If you have a microcontroller, you don't need to worry about the bubbles.

Just put a resistor in series with the sensor and connect them between an output of the micro and ground.

Take a wire from the junction of the two and connect it to an Analog to Digital input of the micro.

So,
make the output pin high (ie 5 volts)
wait a second
read the A to D input
make the output pin low (0 volts)
Display the A to D value.
Wait a minute.
Do it again.

So, the electrode is only on for 1/60th of the time. This gives the bubbles little time to form and plenty of time to go away. So, you should get consistent readings.

i used pic18f14k50.sorry, i m not very understand what u mean. i cannot imagine very well. can u show me the circuit diagram?

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Like this:

http://dl.dropbox.com/u/4222062/Water%20sensor%20to%20uP.PNG

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no need charge amplifier anymore?

if the dielectric permittivity change, is it take a long time for capacitor to change its capacitance?

phyky said:
if the dielectric permittivity change, is it take a long time for capacitor to change its capacitance?
The capacitance will change immediately, but if you have it fed from a very high impedance source then there will be an R-C time delay while the charge on the plates decays/changes/settles.

In this case, the sensor is behaving like a resistor because of the conductivity of the water.

You may be able to make a waterproof layer of paint or epoxy by dipping the plates of the capacitor into the paint or epoxy.
If you did this 3 or 4 times and give the paint time to dry, or place it in a warm oven (45°C) to dry between coats.

It is important that there must be no conduction between the plates.

It might suffice to cover each plate with its own sandwich "baggie" before immersing in the water. Some of these are polythene (?) bags and are smooth; if open at the top the water will probably push the bag up against the plate surface. Their permittivity will change the calibration slightly, but this may fulfill OP's insulation needs. Not sure how "short term" the needs are; maybe just one day for a demonstration?

thanks for advices. i did measure the resistance of sensor, and it decrease as the water level increase, and it take a bit time.
so if i use dc, charge amplifier will not amplify the voltage? bcoz most of book show A=-Cs/Cf, A=gain, Cs=capacitive capacitance, Cf=feedback capacitance.but wikipedia say it can only amplify unless exciting input voltage(which is ac?), or the circuit acts as a charge-to-voltage converter. is it true?
http://en.wikipedia.org/wiki/Charge_amplifier

Your capacitor is not working as a capacitor, because it is in water which has a relatively low resistance.

So, you can just accept this and use it as a variable resistor.

This means that you can just put it in a DC circuit and amplify with a conventional opamp if you like.

I notice in your first post that the spacing is only 2 mm so you don't have room to add paint or plastic bags.

I think you could get heat shrink tubing to squash down flat and fit over a 2 cm wide electrode.

If you did this, you could make it "U" shaped, so that it went from one electrode to the other under water and then you wouldn't have to worry about sealing it under water.
Heat it up with a hot air gun and it would be a tight fit on the electrodes.

but the capacitace is also increase for my measurement, it can't be a capacitor?

It is still a capacitor but with a low value of resistance in parallel with it.

So, you will have a lot of trouble trying to measure the capacitance while the resistance is there and varying.

no wonder my measurement is not stable.
so according to my circuit, i can't use Vo/Vin=Cs/Cf, how i calculate output voltage?

Your measurement is not stable since the DC current through the water is decomposing it and you are getting bubbles of gas forming on the electrodes.

To measure the resistance of the water between the electrodes, you could use the "OHMS" function on a multimeter or you could use an opamp circuit like this:

Many opamps would work OK in this circuit.

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so i need take it as resistive transducer. the charge amp will not working if i use charge amp? but the output voltage did increase for my measurement

Yes, it would, but it is just working on the resistance of the water.

that mean it just working on feedback resistor, Rf not Cf?

Thank u

phyky said:
that mean it just working on feedback resistor, Rf not Cf?

Yes, that is right. It is just measuring DC resistance.

But did you understand the comment about using heat shrink tubing?

i know heat shrink tubing but how to make it "U" shaped?

1 last question for charge amp, is charge amp really amplify output signal for dc or just a charge voltage converter?confuse to looked at the wiki as above..

is my μC can take ac voltage output to do ADC or i nid a rectifier to convert to dc?

Your A to D converter must get rectified and filtered DC and it must usually only have DC in the range 0 to +5 volts.

Check your Micro's instructions to check on this.

But why would it have AC output if it is an opamp using only DC and having a DC input?

just curious if pic can connect ac.thanks

sorry, i just realize that dc ll not pass through capacitor make my charge amp won't working for my capacitive transducer.so i change to ac supply now with 5V amplitude.but the output of charge amplifier will oso change to ac so how could i connect it to my microC to do ADC? do i need a ac-dc converter?

phyky said:
sorry, i just realize that dc ll not pass through capacitor make my charge amp won't working for my capacitive transducer.so i change to ac supply now with 5V amplitude.but the output of charge amplifier will oso change to ac so how could i connect it to my microC to do ADC? do i need a ac-dc converter?
How rapidly can your PIC take repeated samples of the analog signal? It may be sufficient to add a DC offset in your "charge amplifier" so the voltage fed to the PIC alternates from 0 to +10V rather than from -5V to +5V, and make allowance for this in your PIC coding. (Should 10V exceeds the PIC specs, obviously attenuate it, say by 50%.)

vk6kro said:
Your measurement is not stable since the DC current through the water is decomposing it and you are getting bubbles of gas forming on the electrodes.

To measure the resistance of the water between the electrodes, you could use the "OHMS" function on a multimeter or you could use an opamp circuit like this:http://hampatong.com/wp-content/uploads/2010/04/Wheatstone-bridge-single-op-amp.png

Many opamps would work OK in this circuit.

i try did it as resistive water level sensor,it did quite well. but what is the reason of water level increase and resistance decrease?is it because the conductivity of water?because i never heard of resistive water level sensor. can u explain it for me?

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