Sanity check for current transformer burden resistor calculations

Question

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

Here are my calculations:

They say

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

I'll go with 0.5:

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

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

Did I miss something?

Can you add definitions to the formula that you used? A link to the data sheet may also be useful. — Dwayne Reid, Commented Dec 30, 2019 at 4:10
@DwayneReid The datasheet is linked. To the left of the model number... I'll add the fomrulae — Matthew Goulart, Commented Dec 30, 2019 at 4:13

Dwayne Reid · Accepted Answer · 2019-12-30 04:35:41Z

1

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.

answered Dec 30, 2019 at 4:35

Dwayne Reid

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