calculating shunt for ammeter, part II?

Discussion in 'General Electronics Chat' started by BillRush5, Mar 26, 2013.

  1. BillRush5

    Thread Starter New Member

    Jan 13, 2013
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    This time I have a 10 mA FS ammeter that I want to lessen the range so that it only measures 1 mA FS. It currently takes 1.51V to reach the 10 mA FS. How would I do the calculation? Thanks.
     
  2. k7elp60

    Senior Member

    Nov 4, 2008
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    Unless the 10mA FS meter has an internal shunt, the only way I know to accomplish what you want to do is use amplifier with a gain of 10 to feed the meter and when 1mA is flowing the meter will show full scale.
    Maybe there is someone else on the forum that has additional information.
     
  3. crutschow

    Expert

    Mar 14, 2008
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    I agree with k7elp60. You would need to change the shunt resistor value or use an amp.
     
  4. BillRush5

    Thread Starter New Member

    Jan 13, 2013
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    Ok, here's what I have guys. I took the meter apart and there is a diode bridge network (I am assuming it is there because this is a DC ammeter), and across the diode network, opposite the +/- meter outputs, were two precision resistors adding up to 19.4 ohms when I disconnected them. The internal resistance of the meter without the resistors present (but with the diode network in place) is 158.8 ohms. So this was originally a 10 mA FS meter with 158.8 ohm internal resistance with a 19.4 ohm shunt across the diode network. Without the shunt in place, the total FS current is 217.5 uA FS (with the shunt of course the current FS is 10 mA).

    So, what needs changed to get a 1 mA FS reading? Thanks....
     
  5. BillB3857

    Senior Member

    Feb 28, 2009
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    edit. Remove the 19.4 ohm resistor. Put meter in series with a pot and adjust the pot for full scale on the meter. Measure the voltage across the meter. Use ohms law to calculate a resistor value, that when in parallel with the 158.8 ohms will give the measured voltage when 10ma goes through the network.
     
    Last edited: Mar 26, 2013
  6. crutschow

    Expert

    Mar 14, 2008
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    That should be 1mA external current. :)
     
  7. BillB3857

    Senior Member

    Feb 28, 2009
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    Fingers got going to fast. Yes, 1ma. As you see, I also did a total re-think about the whole thing. Pretty sure either way would work.
     
  8. timescope

    Member

    Dec 14, 2011
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    It appears to be nonlinear : 19.4 // 158.8 = 17.288 ohms.
    17.288 ohms x 10mA = 172.88 mV
    217.5uA x 158.8 ohms = 34.539 mV
    It might be better to find the value of the shunt empirically using a variable resistor as a shunt, initially set to zero ohms and then adjusted to obtain FSD with 1mA. A fixed resistor could then be substituted.

    Timescope
     
  9. BillRush5

    Thread Starter New Member

    Jan 13, 2013
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    I agree and that was my next thought as well. Thanks again guys for the help.

    Bill
     
  10. richard.cs

    Member

    Mar 3, 2012
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    Do you want to measure ac or dc current? The diodes are there to allow measurement of AC but if you don't need that then you would be better off removing them.
     
  11. timescope

    Member

    Dec 14, 2011
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    No. The diodes are there to protect the moving coil meter from excessive currents.

    Timescope
     
  12. BillB3857

    Senior Member

    Feb 28, 2009
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    That would depend upon how the diodes were connected. If back to back across the actual meter movement, I would agree. If a bridge prior to the movement, I would say to allow AC. What does the label on the meter say?
     
  13. timescope

    Member

    Dec 14, 2011
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    From post #4.

    Timescope
     
  14. richard.cs

    Member

    Mar 3, 2012
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    Timescope

    Re-reading the thread I think you're right about these being protection diodes. I wonder what causes the non-linearity then.
     
  15. BillRush5

    Thread Starter New Member

    Jan 13, 2013
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    Just an update, I gave up trying to change the meter range. I tested it on a couple of different power supplies and got different mA results with each. This may have been because one supply was full wave DC and the other had a single rectifying tube I don't know, but it's just as cheap to continue using my cheap DVM's as current monitors but I am going to be installing battery eliminators in them soon.

    If I get back to this analog mA meter, I will use the variable resistor method and calibrate it to the single electronic piece I need it for and just keep it in line with that particular piece.

    Thanks again for your help and thoughts. Just FYI, this was one of the cheap Hong Kong/ Chinese analog DC mA meters as seen on ePay. They can be completely disassembled with little effort making modifications easily done. Even the scale can easily be changed.
     
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