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To shunt or not to shunt?

Discussion in 'General Electronics Discussion' started by Gadgetastic, Jun 22, 2013.

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  1. Gadgetastic

    Gadgetastic

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    Nov 22, 2011
  2. poor mystic

    poor mystic

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    Apr 8, 2011
    :) Hi
    I believe you want to use this meter to measure 2 or 3 mA.
    No shunt resistor is needed in that application. However you'll only be using 10% to 15% of the meter's full range, which seems to me a bit of a waste.
     
  3. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    A bit of a waste, but the only practical option. A 2 mA (0-1.999mA) meter would not have sufficient range to cover the intended currents.
     
  4. Gadgetastic

    Gadgetastic

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    Nov 22, 2011
    Wiring the Ammeter

    Getting frustrated.

    I know it's likely just something obvious that I'm missing, as is usually the case it seems, but I can't seem to find anything clear enough online that helps, so, here goes.

    I have an ammeter designed for mA. It is panel mounted and digital. There are four wires coming out the back, Black, Red, Yellow, Blue. I hook the black wire up to the 9v battery by way of a 470ohm resistor. The Red wire to the ground. At this point the meter lights up and says 16.66. I have 2 electrode outputs which have 4.5k Ohm resistance to brig them down to 2mA which my multimeter confirms. So i hook the yellow wire to one electrode and teh blue wire to the other, seeming the same thing I did with the multimeter, but the gosh darn panel mounted meter just goes BLANK!

    What is the obvious thing I am doing wrong here?
     
  5. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    Jan 21, 2010
    Don't start new threads for the same thing please.

    Firstly, you need to read the specs (they're on the auction you pointed us to).

    The specs say 5V +/- 5% -- you supplied 9V. There is a significant chance you have caused something to fail (I know this because I applied the incorrect voltage to a similar module and that's exactly what happened to me).

    You need to power it from a source of 5V (a 9V battery and a 78L05 regulator would be sufficient).

    The meter should display 0.00 when you power it up.
     
  6. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

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    Nov 28, 2011
    As well as providing a 5V supply instead of a 9V supply, you need to make the connections as shown in the "Wiring connection" diagram on the auction page. Red and black are for positive and negative connections (respectively) to the 5V power source (there is no 470 ohm resistor), and yellow and blue are for the positive and negative connections (respectively) in the path whose current you want to measure.

    The auction does not state what the requirements are for voltages on the input pins in relation to the power supply; it simply says "please use the independent power supply", which means that the 5V power source must be independent and electrically separate and isolated from the circuit whose current you are measuring.

    If it doesn't do what you expect, I suggest you buy another one, since the one you have may well be damaged, as Steve says.
     
  7. Gadgetastic

    Gadgetastic

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    Nov 22, 2011
    You say it does not need a 470 resistor, but the specs say it requires 20mA max and the 9v puts out a lot more. So don't I need to cut the current with a resistor?

    Sorry about starting a new thread. I wasn't sure if this counted as the same topic.

    Thanks.
     
  8. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

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    Nov 28, 2011
    The specs say that the measurement range is 0~20 mA.

    The power supply current is specified as 100 mA maximum.

    You need to use a 5V regulator connected to your 9V battery, to produce a 5V supply rail for your meter. You do not need a 470 ohm resistor. Have a look at the 78L05 data sheet: http://www.fairchildsemi.com/ds/MC/MC78L05A.pdf

    You need to learn more about voltage, current and resistance. You could try Googling explanation of Ohm's Law.
     
  9. Gadgetastic

    Gadgetastic

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    Nov 22, 2011
    There's a lot I need to learn, which is why I'm a newbie. ;)

    Thanks for the advice. I'll give it a try.
     
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