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    This simple micro ampere meter circuit can help in measuring small currents in five ranges: from 1 µA to 10 mA. The meter is working in this way: the current being measured Ix shifts the input voltage resulting to an output voltage with an inverted polarity.

    The output voltage of the opamp CA3130 is proportional to the measured current Ix.
    By selecting the proper feedback resistors through S1, the output voltage by full meter deflection is 1 volt in all measuring ranges.

    The value of the series resistor R must be selected for the particular meter being used. For example if a 1 mA meter is used, the total resistance (the sum of the resistor R and the coil resistance Ri of the meter) must be 1kΩ. If a 100 µA meter is used, the total resistance must be 100 kΩ. If needed, a potentiometer can be used for the R.

    Micro ampere measurement circuit schematic

    micro ampere meter circuit schematic

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    11 Responses to "Simple Micro Ampere Meter Circuit"

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    1. I want Measuring nanoamperes and nanovoltage of microamper and microvolt for 0-4000mv and 0-3A. also nanoameres and nanovolt is beter for me.
      i want circuit for it. please help me.
      Measuring low currents can be tricky. Clever analog-design techniques and the right parts and equipment can help.

      • اندازه گیری در حد نانو باید در تجهیزات مدار قطعات و ادوات نانویی باشد. یعنی ایسی هایی که با فناوری نانو ساخته شده اند. اما ایسی هایی که در نسل امروز هست به فناوری میکرو ساخته شده اند .

    2. Udayanga says: on October 5, 2012 at 4:12 am

      Can i use above circuit to measure AC Micro ampere current??

    3. The 2 diodes are input voltage transient protection for the sensitive op amp–they do not normally conduct current.

      The input voltage across the two input terminals is zero V, so if there is input current, the op amp must generate an output voltage sufficient to cause an equal but opposite polarity current to flow in the feedback resistor. The output voltage is then measured via the analog meter–or you could use a DMM without the series resistor.

      EO /R feedback = I in

      e.g. Eo = 0.75V

      R feedback = 1M

      I in = 0.75V / 1M = 0.75ua

    4. Abhishek111 Abhishek111 says: on August 11, 2013 at 5:10 pm

      Sir, if you don’t mind, plz send some more information about this ckt, we r not getting the working of that circuit, so plz send depth information. What is the input for inverting and non inverting inputs?

    5. One thing you may have missed is the fact that each of the two batteries must be grounded. Also recommend diodes in series with the batteries to prevent accidental voltage reversal–op amps are very unforgiving if the battery gets reverse–very easy to do.

      The op amp has differential inputs–a positive change in voltage at the non-inverting (+) input causes the output to swing negative–a positive change in voltage at the inverting (-) input causes the voltage to swing negative. With proper negative feedback (output tied to inverting input via resistor etc.), the op amp does everything within its capabilities to make these two inputs the same voltage.

    6. How can I convert the output voltage of 0 to 1 volt to the range of 0 to 5 volts? Will there be any chip needed?
      I am connecting the voltmeter to my arduino and I hope the 0-5v would give a better contrast for my arduino to read.
      I thank you in advance for your assistance.

    7. Salut.
      De ce ai ales o tensiune de lucru de 8V si nu de 9V sau de 7,5V ? Aceasta tensiune este critica? De ce intreb ,pentru ca vreau sa fie portabil si nu exista combinatii de baterii sau acumulatoare sa-mi iasa 8V.Vad ca circuitul nu are punct de masa au 0V-e un lucru bun pentru mine.
      Multumesc se raspuns.

      Hello.
      Why choose an operating voltage of 8V or 9V not 7.5 V? This voltage is critical? Why ask because I want to be portable and no batteries or accumulators combinations out my 8V. I see no point circuit have 0V (+8V 0V -8V )- it’s an advantage for me.
      Thanks be answered.

    8. mojtaba says: on July 3, 2014 at 2:58 pm

      Hello
      Thanks for the useful article.
      I have a few problems to run the circuit. solving the problems got my time. you don;t show the potential of Current in node. I connect it to the ground (The midpoint of +VCC and -VCC) and then the circuit work properly.
      I run it to sens current from 10 to 30 micro ampere and that was successful.
      I then try to run it for 5 mili ampere and wasn’t successful. I think about how to work the circuit…
      I have a question. if the diodes are only for protection, Where the input Current go? The input impedance of the Op-Amp is too large, isn’t it? so the big currents have to become low!!!
      I think that is the reason of working the circuit for 10-30 micro amperes and not working for 1-5 mili amperes.
      please help me to run the circuit for 5 mili ampere and then answer this question : ” if the diodes are only for protection, Where the input Current go?”

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