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I'm hoping to use this(CR8420-1000-G) current transformer to detect ground faults of 20mA. Using the formulae in the datasheet: enter image description here

Here are my calculations:

enter image description here

They say

For best linearity, choose R such that V < 0.8 VL

I'll go with 0.5:

enter image description here

Now say I'd like to have a 500mV output for 20mA:

enter image description here

enter image description here

35.3KOhm seems way out in left field for the burden resistor. I was expecting something around 500 Ohm.

Did I miss something?

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  • \$\begingroup\$ Can you add definitions to the formula that you used? A link to the data sheet may also be useful. \$\endgroup\$ – Dwayne Reid Dec 30 '19 at 4:10
  • \$\begingroup\$ @DwayneReid The datasheet is linked. To the left of the model number... I'll add the fomrulae \$\endgroup\$ – Matthew Goulart Dec 30 '19 at 4:13
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My off-the-cuff calculation is 0.5 / (0.02 / 1011) which is the same as (1/(0.02 / 1011)) * 0.5 = 25,275 Ohms.

I work in RPN, so the second calculation is how I solved the equation. 0.02 [Enter] 1011 [Divide] [Invert] 0.5 [Times]

Yeah - that resistor value seems high but upon inspection, is reasonable.

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