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Need help in current measurement using CT.

Discussion in 'AVR' started by BHARGAV SHANKHALPARA, Nov 5, 2013.



    Nov 5, 2013
    Hello everyone...

    i am new in current measuring concept, and i make one of control application type project...

    actually i am controlling one ac induction motor using microcontroller (ATTINY88). and i want to add one extra function into it is over-current protection.(when current more than 6 amp, automatically gets turn off by microcontroller) for that i decide to use current transformer for current measurement. so i need to interface it with microcontroller.

    i have current transformer with 1 (one) turn of primary winding and 350 turn of secondary winding. i know that to convert secondary current into voltage i need to connect burden resistor across it. i need to obtain voltage between 0-5v DC from that to apply microcontroller. also for that i need to signal conditioning for that before apply it to microcontroller.

    problem is that...

    i don't now that which value of burden resistor is best for my application and after obtaining well voltage signal condition circuit require which include (rectifier and amplifier).

    after searching about signal conditioning circuit i found this document and circuit.


    in this circuit i understand about all resistor and op-amp but initial two zener diodes and at last RC time constant make question in my mind that why its here and if here than what should be its value.

    please help me in this project...

    thank you...
  2. JimB

    JimB Super Moderator Most Helpful Member

    Sep 11, 2004
    Peterhead, Scotland
    The two diodes across R1 are there to protect the OP-AMP from voltage spikes due to any large transients in the measured current.

    The R-C network on the output is there to give a smooth DC output. The output from the op-amps will be halfwave pulses, not very usefull for reading with an ADC.
    Look at the last paragraph of the application note.
    It clearly states that the RC network should have a time constant which is at least 10 times greater than the period of the measured current.
    So for a 50hz current, the period is 20ms, then make the time constant greater than 200ms.

  3. NorthGuy

    NorthGuy Well-Known Member

    Sep 8, 2013
    Northern Canada
    6 * sqrt(2) = 8.5 peak.

    If maximum primary current is 10, then secondary will be 10*1/350 = 28 mA
    Since you do not need high precision, you can simply get rid of negative half-wave and spread the positive half-wave over your 0-5V range.
    A resistor to spread 28 mA to 5V is 5/o.o28 = 179 Ohm.

    Then you can use a small RC filter to get rid of spikes. This will lower peaks a little bit, so for 6A, instead of 8.5A , you get, say 8A.

    Approximate critical value at microcontroller will be 5*8/10 = 4V. This will need some calibration because of RC filter and CT errors.

    You can use a comparator, which will signal to the microcontroller's pin when voltage goes above 4V. Since this is a peak, you will get a series of pulses, but this is easy to deal with in software.
    • Disagree Disagree x 1

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