1
\$\begingroup\$

Reading this article I'm trying to understand this fragment where they are using a 1000:1 CT, with 0.1 V/A for a 100 ohm load.

A simple 12bit ADC will give us 2^12, or 4096 possible values. Applied to a 3300mV range, this gives us a resolution of about 0.8mV. The load pictured above has a total peak-to-peak voltage span of 1219mV, and should be about a 4.5-4.7A load. The 1219mV give us a possible 1523 values for the voltage displayed at our calculated resolution, which translates to about 3mA per ADC value when using our 100 ohm load resistor (remember: it’s RMS!).

I know that 0.8 mV comes from 3.3/4096. I think there's an error with 1523 value, since I get Adc = (1.219*4096)/3.3 = 1513.

My question is: how is the ratio 3 mA per ADC value derived?

\$\endgroup\$
3
\$\begingroup\$

4.6 A gives about 1219 mV.

1219 mV ---> 1513 counts


4.6 A ---> 1513 counts

??? ---> 1 count?


Ans: ~3.02 mA

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy