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Could a voltage signal system also work by measuring current?

  1. Apr 3, 2013 #1
    00185.png

    The circuit above uses a voltmeter to measure the voltage allotted to it through a potentiometer and the varying voltage acts as a signal.

    Could this signal system also work by measuring the current allotted to a load based off the position of the potentiometer (i.e. higher voltage = higher current)? What are the differences in measuring voltage and current?

    Thank you in advance.
     
  2. jcsd
  3. Apr 3, 2013 #2
    Yes it could. You would replace the voltmeter with an ammeter and remove the connection between the bottom of the potentiometer and the return. In fact there is a signalling system based on current called a 4-20ma current loop.
    http://www.bapihvac.com/CatalogPDFs/I_App_Notes/Understanding_Current_Loops.pdf [Broken]
     
    Last edited by a moderator: May 6, 2017
  4. Apr 3, 2013 #3
    Why does the connection between the bottom of the potentiometer and the return need to be removed? Wouldn't the current change proportionately with potentiometer adjustment even if that connection was still there?
     
    Last edited by a moderator: May 6, 2017
  5. Apr 3, 2013 #4

    sophiecentaur

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    I don't know either. You would need to replace it with something else or the mAmeter wouldn't work. I think skeptic knows what to do - he has just given you a part answer.
     
  6. Apr 3, 2013 #5
    That's odd because in a circuit simulator, the current read by the ammeter changed with variation in the potentiometer. I suppose I'll wait for someone to clarify.
     
  7. Apr 3, 2013 #6
    Let's say you want to make your circuit into a 4 - 20 mA current loop using the changes I suggested. You could do that by using a 10 volt battery, a 2K pot and adding a 500 ohm resistor between B+ and the top of the potentiometer. With the tap at the top of the pot there would be 500 ohms in the circuit so the current would be 20 mA. With the tap at the bottom of the pot and without the connection between the pot and the return, you'd have 2500 ohms in the circuit for a current of 4 mA. But with the connection between the bottom of the pot and return, how much current through the ammeter would you have then? True, you wouldn't have to move the tap very far from the bottom of the pot before the currents in both circuits would be indistinguishable from each other, but good design isn't just designing something so it will work, it's designing it so it can't fail. With that connection between the bottom of the pot and return, you have an unnecessary failure mode.
     
  8. Apr 3, 2013 #7
    So from what I'm understanding, putting the connection between the potentiometer and the return would result in zero current going into the load when the tap is at the bottom of the potentiometer, which is not a good thing?

    Doesn't the same thing occur when measuring voltage? In the diagram in my original post, the voltmeter would measure zero volts if the tap was at the bottom of the potentiometer.

    Thanks for the help, and my apologies, I'm just a bit confused.
     
  9. Apr 3, 2013 #8
    One of the advantages of using a 4 - 20 mA loop is that since 4 mA is the minimum current, you can easily tell if the circuit is open. Another advantage is that being a low impedance the circuit it is more immune to noise even over long distances. In addition, with that connection, you're always wasting 4 mA which is not a good thing if you're using batteries.

    Yes, when measuring voltage, the voltage does go to zero and unless you use a pot of resistance close to that of your voltmeter, the voltage response will be linear right down to zero V. In measuring current with that connection in place and with a low resistance pot, the last little distance close to zero it will be very nonlinear.

    It's your choice which method you use and if it were me, I'd probably use the voltage method - but you asked about the current method.
     
  10. Apr 3, 2013 #9

    jim hardy

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    I tend to write choppy sentences when explaining things, so be patient with me please.
    I'll try to keep to one thought per sentence.

    I think this thought is more significant to your question :
    whenever that connection IS there, you have division of current between lower portion of your measuring potentiometer and your meter.

    As you pointed out, that's true regardless whether it's an ammeter or a voltmeter.

    So the answer must lie in the difference between ammeters and voltmeters..


    An ideal voltmeter draws zero current and a real one comes pretty close to ideal.
    An ideal ammeter has zero voltage drop, a real one has small voltage drop and so do the real wires going to it..
    So the division of current is negligible when you use a voltmeter,
    But substantial when you use an ammeter. Worse, it varies with position of float.

    So, if you use an ammeter you should remove that connection so ALL the current has to flow through your ammeter. That eliminates division of current.

    That's a basic rule for measuring - Voltmeters go in parallel, ammeters go in series.

    Was that any help ?

    Skeptic's 4 to 20 milliamp scheme is in very wide use for the reasons he mentioned.
     
  11. Apr 4, 2013 #10
    That was actually extremely helpful. It's pretty clear to me now. Much thanks to you sir! And thanks to skeptic as well!
     
  12. Mar 22, 2014 #11
    Voltage to current

    One of the problems with your setup is a potentiometer does not properly convert a position to a current. Some additional electronics is needed to convert the position of the potentiometer to a current. Hence this need to be done linear otherwise your system is useless. See this explanation about the problem you are facing. [noparse]http://www.divize.com/techinfo/4-20ma-potentiometer.html[/noparse] [Broken]

    Regards,

    Heinrich
     
    Last edited by a moderator: May 6, 2017
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