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How to interface Current Transformer ( 30:5 ) into 16F877 PIC

Discussion in 'Microcontrollers, Programming and IoT' started by fromdawest, Dec 25, 2012.

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  1. fromdawest

    fromdawest

    7
    0
    Nov 17, 2012
    Hi all,

    Please help me to interface 30:5 A current transformer into a 16F877 PIC .

    I want to get the analog output of the current transformer ( ex.5A ) to be used as the input to the PIC . And according to different current levels ( 0A - 5A ), i want the PIC to drive a small LCD to display a corresponding value .

    Simply, i want to make a simple AMMETER using 16F877 PIC which displays current through a line from 0A to 32A in a small LCD .

    Hope you'll help .

    Thanks in advance .
     
  2. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,448
    2,809
    Jan 21, 2010
    The traditional approach is to place a resistance in series with the current (in this case across the current transformer) and measure the voltage.

    Alternatively, you could use one of the current monitor ICs to measure the current.

    I googled "PIC current sense" and got a plethora of answers.
     
  3. eem2am

    eem2am

    414
    0
    Aug 3, 2009
    welcome to the mystical world of current transformers, -these are steeped in industrial secrecy...........there are few if any application notes telling how current transformers really work.

    The only half decent reference for current transformers is page 3.173 onwards of "switchmode power supply handbook" by billings and morey, 3rd edition.

    Current transformers are not fully explained anywhere, and you will have your brain frazzled contemplating such things as ensuring reset of the current transformer, ensuring you avoid saturation due to magnetizing current............the reflected value of the secondary resistance used, and its relationship with the current transformer's primary inductance......etc etc

    You must ensure non saturation operation over all transients.

    If you are going to build your own current transformer, then may the angels come down and help you.................consider that the primary usually has just one single turn.......so how do you wind this?........ there's a real danger that you could end up with only half a turn unless you somehow ensure that the single turn genuinely forms a "closed" ring around the core.

    ....then consider leakage inductance, and how on earth do you get good pri:sec coupling when you only have one primary turn to play with?.....you obviously can't do sandwich winding....unless you sandwich the primary between two secondary layers.

    ...If you dont get good coupling, then your 1:100 current transformer will not really be 1:100.

    ....and how do you measure leakage with a current transformer?........the extremely low primary inductance will not make this easy for you.

    You are treading the heights of industrial secrecy with current transformers, and google will not be of much help to you with them.

    Good luck............probably trial and error with a core and somw magnet wire is you best bet
     
  4. Harald Kapp

    Harald Kapp Moderator Moderator

    10,800
    2,435
    Nov 17, 2011
    @eem2am: I don't know what experiences you have using current transformers, but you sound frustrated. Ct's have been (and are being) used routinely for many decades to measure primary currents in e.g. high voltage installations.

    The "secret" is: the transformer must be specified for the nominal primary current as well as for a certain allowable amount of overcurrent. It is not unusual to have a ct specified for a rated current plus 50* or even 100* overcurrent.
    In addition you should keep the burden on the secondary as low as possible to minimize saturation of the transformer's core and losses within the transformer.

    In this example the rated secondary current is 5A. Let's assume that a temporary overcurrent of 10*Irated = 50A is allowed. These are typically RMS values. Therefore the peak-to-peak value of the secondary current is 5A*10*sqrt(2)*2=142A (+-71A).

    Let us further assume that the input of the AD-converter (PIC) is 0V...5V.
    We now want to convert +-71A to 0V...5V. A burden of 35mOhm could reduce the level from +-71V to +-2.5V. All you need is a summing amplifier (e.g. OpAmp) to add a fixed voltage of 2.5V to this signal and voilá: +-71A -> 0V...5V.
    You can even do without the amplifier by referencing one end of the transformer's secondary to a fixed 2.5V level.

    However: 0.35mOhm is hard to realize reliably and you will face additional losses due to the wiring from the transformer to the burden resistor. In such an application typically a ct with a much higher transformer ratio would be used (e.g. 32A -> 1mA or 32A -> 10mA). this reduction in secondary current allows for a much better reproducible burden (in the example of a 32A-1mA transformer this translates to Rburden = 5000*35mOhm=175Ohm.) and less losses on the secondary.

    Be aware that in any case you should add protective circuitry (series resistors, overvoltage protection diodes) to prevent damage in case of overcurrents. Also some filtering (low pass, you can use the protection resistors plus a few capacitors for this) to avoid aliasing will be required.

    Harald
     
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