# Capacitor for current transformer measurement circuit

I want to measure an AC current with a current transformer and an arduino device. I found on a website (http://www.homautomation.org/2013/09/17/current-monitoring-with-non-invasive-sensor-and-arduino/) that measurement circuit:

My question is, why do I need a capacitor here? The explanation from the website is:

The capacitor C1 (10uF) has a low reactance – a few hundred ohms – and provides an alternative path for the alternating current to bypass the resistor.

But why do I need an alternative path for the alternating current?

The explanation is confusing, what RA, RB do is divide the 5 V of the Arduino's supply by a factor 2 to 2.5 V. This is used as a reference voltage because the sensor can generate positive and negative voltages.

Suppose the sensor makes -1 V, by itself the Arduino cannot handle this, you have to be between 0 V and 5 V. But if you "lift" that -1 V by 2,5 you get 2.5 V - 1 V = 1.5 V which is perfectly OK.

But a little bit of AC current will flow, disturbing the 2.5 V reference voltage. This is where the capacitor C1 comes in, it decouples that reference voltage. Indeed, the AC component will now travel (have a path through) the capacitor instead of the resistors. That's what they are saying.

But stating: "C1 decouples the 2.5 V reference voltage" would probably be easier to understand.

• That schematic is drawn in the most confusing way possible...
– jms
Commented Feb 4, 2016 at 14:44
• Indeed, that doesn't help either. Commented Feb 4, 2016 at 14:46

But why do I need an alternative path for the alternating current?

Without the capacitor and driving a perfect ADC it will theoretically work. However, when a capacitor is not used, the source impedance of the CT is made much bigger due to the resistances of the potential divider. Again this isn't a problem until you start to look at the ADC spec for such devices as PICs and MCUs on arduinos.

Typically, the ADC on a device like this won't like the source impedance of the signal it is measuring to be much more than a couple of kohms and if your potential divider is 2x 10k resistors then the net source impedance will be 5k + a few ohms for the burden.

So putting a capacitor across the lower resistor in the potential divider means you have masterminded a way to significantly reduce the source impedance without busting-up the DC offset voltage.

10 uF at 50 Hz has an impedance of 318 ohms and clearly this is low enough for most MCUs with in-built ADCs.