5
\$\begingroup\$

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:

enter image description here

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?

\$\endgroup\$
  • \$\begingroup\$ If you read the comments in your link you'll find the solution "Vincent Demay" \$\endgroup\$ – R Djorane Feb 4 '16 at 13:53
5
\$\begingroup\$

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.

\$\endgroup\$
  • 4
    \$\begingroup\$ That schematic is drawn in the most confusing way possible... \$\endgroup\$ – jms Feb 4 '16 at 14:44
  • \$\begingroup\$ Indeed, that doesn't help either. \$\endgroup\$ – Bimpelrekkie Feb 4 '16 at 14:46
3
\$\begingroup\$

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.

\$\endgroup\$

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.