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Measuring Liquid conductivity

Discussion in 'General Electronics Discussion' started by andre, Oct 6, 2010.

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  1. andre

    andre

    5
    0
    Oct 6, 2010
    Hi all,

    I am working on a project that aims at measuring the conductivity of sea water. For this purpose I will be building a simple circuit made of a two electrodes dipped into the water and a resistor, then I will be using an instrumentation differential amplifier to measure the voltage sensed across the resistor, from which water conductivity can be determined. I need an opinion about what kind of Instrumentation amplifier I can use that would do a good job ( meaning has high rejection mode ratio, works well for high frequencies).

    Please let me know of any recommendation

    Andre
     
  2. Militoy

    Militoy

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    0
    Aug 24, 2010
    What frequency and waveshape are you planning on using for your measurement circuit? You seem to have selected a project with some real challenges. The patent documents describing commercial versions of this type circuit all discuss the difficulty of getting an accurate and repeatable measurement, because of almost instant ionization of seawater molecules - and polarization of those charged particles in the test sample. Along with the continuous changes in measurements caused by galvanic and corrosive effects of the seawater on the electrodes, the measurement seems to be a constantly "moving target". Most inventors appear to have settled on a square wave signal - and a comparison of the measured sample to a sample of known salinity - with conductivity ratio between the two determining salinity in the unknown sample.
     
  3. andre

    andre

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    Oct 6, 2010
    Thanks Militoy for your feedback, right now as a first stage goal I just need to read the voltages and monitor how those voltages change over time in open seawater.

    what I am going to use seawater as a battery: have two electrodes dipped inside and a resistor connected in series with the electrodes, and I would just measure the amplified differential voltage at the output of the instrumentation amplifier. I will not be using any sample liquid or sample seawater, the electrodes will be dipped directly into the open seawater. I have no idea if this will work out or not, that's why I wanna experiment with it and see if it will give a correct data. What do you think?
     
  4. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,396
    2,777
    Jan 21, 2010
    That won't measure conductivity in any meaningful sense. All you'll be doing is using the sea as an electrolyte in a battery -- presumably the electrodes will be different materials. In theory the voltage is determined by the electrode materials.

    If you want to measure conductivity you need to apply a voltage across 2 electrodes (made of the same material) and measure current, or apply a current and measure the voltage. is you just stick in a series resistor and use some power source, you'll have to measure both voltage and current.

    As Militor (;)) suggests, DC will result in your electrodes being eaten away and other effects you probably don't want.
     
  5. andre

    andre

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    Oct 6, 2010
    Thank your steve, I will probably apply some vollage across the electrodes, but not much just enough to get some current flowing through the very close spaced electrodes. I know applying higher voltage will result in having my electrodes being eaten away over time due the reaction that takes place due to salty water. But still do you guys think an Instrumentation IC will be need to get an amplified voltage that is proportional to the conductivity of sea water.
     
  6. Militoy

    Militoy

    180
    0
    Aug 24, 2010
    A straightforward differential amplifier should be stable enough to get a reading - the measurement will be somewhat subjective though - as the electrodes will immediately begin to react with the seawater, and change the nature of the circuit. Even if you have a supply of constantly renewing seawater (like in a moving boat), the degradation of the electrodes will likely result in changes in the reading that are much greater than any drift in the diff amp.
     
  7. Trent

    Trent

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    Apr 17, 2010
    Here's a link to a circuit that may help you with a little modification.
    http://www.high-voltage-lab.com/230/plants-watering-watcher

    It uses a square wave to reduce electrode oxidation.

    I've tried doing this on a simpler scale and I could not get any repeatable results because of oxidization and polarization of my electrodes.

    Hope this helps.
     
  8. KMoffett

    KMoffett

    718
    71
    Jan 21, 2009
    That circuit won't reduce corrosion. Well,maybe by half. It's pulsating DC, not AC.

    Ken
     
    Last edited: Jan 30, 2011
  9. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,396
    2,777
    Jan 21, 2010
    No, that will produce AC. The probes alternate between 0V and 5V. Is it a 50% duty cycle though?
     
  10. KMoffett

    KMoffett

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    71
    Jan 21, 2009
    No! 0v to +5v to 0v to +5v, regardless of duty cycle, is pulsating DC. The current is always going in the same direction, it does not "alternate" directions. However, if you pass it through a capacitor, the DC component will be removed, and then it will be AC. ;) And, the duty cycle in the schematic was 10%on/90%off.

    Ken
     
  11. Resqueline

    Resqueline

    2,848
    2
    Jul 31, 2009
    No-no!! While one probe is at 0V the other is at 5V, and vice versa - so the current does indeed alternate.
    The water is not in contact with the power supply and so does not "know" any DC level present except for the one effectively created by a duty cycle different from 50%.
    D1 & R2 is responsible for the 10% duty cycle and ensures that there will be corrosion.. The reason for that setup eludes me.. Granted, a capacitor would remove any DC.
     
  12. KMoffett

    KMoffett

    718
    71
    Jan 21, 2009
    Yes...yes...yes. Looked at the circuit again. I concede. ;)

    Ken
     
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