# SCT013-000 and Capacitor C1

I have read and implemented the SCT013-000 with Arduino following this article.
Article

The explanation is very helpful but it does not completely explain the role of capacitor C1. All it says is this: “Capacitor C1 has a low reactance - a few hundred ohms - and provides a path for the alternating current to bypass the resistor. A value of 10 μF is suitable.”

Why does the current need to bypass the resistor?

Thanks in advance.

Welcome, Dave, to OEM.

Because the Atmel ATMega data sheet recommends that the ADC input is fed from a source of less than 10 kΩ impedance. In the absence of the capacitor, and if you’re using the maximum value shown of 470 kΩ for each resistor, you effectively have 235 kΩ in series. That’s the Thévenin equivalent of the two bias resistors in parallel. Otherwise, the sample and hold capacitor inside the ADC wouldn’t charge fast enough, and the ADC wouldn’t report the correct voltage.

Thanks you so much, Robert.

Fascinating! So, the capacitor allows some of the current to drain off effectively lowering impedance and a cap is chosen for this role because it it neither a short nor an open. Yes?

Not quite. It’s an open circuit to d.c., but close enough to a short circuit as far as the 50 Hz a.c. is concerned.

You’ve got to think of the a.c. part (that’s all we’re really interested in for now) as flowing in a loop. The voltage generated across the burden resistor for a c.t. input, or the bottom resistor of the voltage divider for the a.c. voltage input, drives a current into the ADC input, through that input to ground, through the common ground to the bottom bias resistor, then through that and the capacitor in parallel back to its source. Because the “reactance” of the capacitor is very small in relation to the resistor, almost all of the a.c. part of the current flows in the capacitor and none in the resistor. But in the d.c. world, the capacitor carries no current at all, and plays no part in it.