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I am using external CT to measure Current parameter. I am using PIC16F1938 to measure analog . Now here i am looking for circuit and calculation involved to get proper current rating

datasheet

Min Current : 0.8A max Current :30A Primary Current 30A nom., 75A max Turns Ratio 1000:1

I have 10 bit resolution ADC. Kindly suggest me the circuit how we can choose the value. I got article with arduino Energy meter reading with ac-1030

Let me know best circuit to be achieve better accuracy. Some forum recommend to use amplifier circuit . Let me know which is best

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    \$\begingroup\$ Your link [2] is broken. 'Better accuracy' is meaningless. What accuracy do you want to achieve? \$\endgroup\$ – Neil_UK Dec 27 '18 at 10:55
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Some missing specs, will assume PIC DC supply (along with voltage reference for 10-bit ADC) is +5V. Therefore, the maximum peak-to-peak voltage signal at ADC input should be 5V peak-to-peak, with a +2.5V DC offset. ADC reference voltage would be +5V.

If you really must provide a linear current-sense up to 30A from the current transformer, then peak-to-peak current would be 84.8A (the 30A is RMS current). This current is reduced by a factor of 1000 to 84.8 mA peak-to-peak flowing through the current transformer load resistor. A load resistor of 58.92 ohms would produce five volts peak-to-peak for the ADC.

schematic

simulate this circuit – Schematic created using CircuitLab
You might use the next smaller standard value for R1 (like 56 ohms) so that noise and transients won't over-voltage the ADC input. The large value of R2,R3 may help prevent latch-up on really large transients.

As always, when dealing with high-potential AC with that current transformer, be careful!

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  • \$\begingroup\$ If you really must provide a linear current-sense up to 30A from the current transformer, then peak-to-peak current would be 84.8A (the 30A is RMS current). I could not able to understand this part. How it will be 84.8A. As always, when dealing with high-potential AC with that current transformer, be careful! What is protection circuit i need to build. Why didnt u used amplifier circuit here. \$\endgroup\$ – AMPS Dec 28 '18 at 4:28
  • \$\begingroup\$ @AMPS A sine-wave current of 30A swings to a positive peak of 42.4A and a negative peak of -42.4A: there is a \$ 2 \sqrt 2 \$ factor between RMS and PEAK-to-PEAK. The result must fit within the ADC's range of 5V. By choosing R1 value properly, no amplifier is needed. A larger R1 (within reason) makes the current transformer more sensitive. \$\endgroup\$ – glen_geek Dec 28 '18 at 14:34
  • \$\begingroup\$ it is said lower value better accuracy what you recommendation.. \$\endgroup\$ – AMPS Dec 31 '18 at 3:39
  • \$\begingroup\$ Yes, for the CT itself, linearity improves with lower R1. But you pay with less voltage output - hence your ADC resolution suffers. Design this way: decide what maximum current you anticipate. Scale R1 so that current yields 5V peak-to-peak into ADC. The example above calculates R1 for 30A(rms) yielding 5v (p-p). Values of R1 much above 100 ohms result in poorer linearity. \$\endgroup\$ – glen_geek Dec 31 '18 at 14:43

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