I just started working on interfacing a current sensor (SCT-013-000) with a micro-controller (ESP32.)

I am referring to this video, which shows this diagram:

Circuit diagram

In this I have some confusion :

  • How it is generating the DC voltage to power the Arduino? Is this due to the capacitor that converts AC to Dc? Is it really safe to use this circuitry in the same way for powering the ESP32 (with different burden resistance value?)
  • Text in the image shows it generating 4.66 volts which is considering 50 mA output from the current transformer (CT.) What will happen if the CT output is low? (In my understanding, the CT output will vary from 0 to 50 mA depending on input current from 0 to 100A.)
  • What is the use of R1, R2 and C?

I have seen several such videos which use the same method of powering a micro-controller from a current transformer.

EDIT1: Adding image for circuit diagram after Andy answer : enter image description here

  • 1
    \$\begingroup\$ That isn't a circuit diagram; it's a wiring diagram. We (probably me more than most) are quite OCD about people calling it the right name on this site. \$\endgroup\$
    – Andy aka
    Oct 7, 2021 at 15:36

1 Answer 1


I have seen several such videos which use the same method of powering micro-controller from ct

You've probably watched videos that are wrong or you have misinterpreted their message incorrectly. As far as I see in your schematic, Arduino 5 V d.c. is a external power supply and not anything to do with the CT nor derived from it. That 5 volts creates a 2.5 volts mid-point (using R1 and R2 and smoothing it with C1) for biasing one leg of the CT's output. The other leg of the CT then feeds into an ADC or other circuit for processing the signal.

How it is generating dc voltage for powering Arduino, in above diagram?

It isn't; it's using the AC signal from the CT and feeding that into another circuit that probably measures it in some way.

Text in image shows generating 4.66 volt which is considering 50 mA output from CT

Just think about it; if the CT output current is 50 mA RMS and it connects to a 33 ohm burden then the voltage produced is only 1.65 volts RMS (4.667 volts peak-to-peak) and cannot power an Arduino.

A current of 141.4 mA peak-to-peak into 33 ohms can produce a waveform voltage of 4.6662 volts peak-to-peak but, that isn't a d.c. voltage of 5 volts no matter how hard you try with diodes and capacitors; you might get a d.c. voltage of maybe 3.5 volts but, that's all.

  • \$\begingroup\$ Thanks for your quick response. Please have a look at schematic in this. \$\endgroup\$ Oct 7, 2021 at 15:18
  • \$\begingroup\$ I didn't see a schematic - I saw a cartoon wiring diagram. Anyway, do you have some reason for me to look at it? Does it contradict anything in my answer? The video in your question DOES NOT use the CT output for powering the arduino <--- let me emphasize that. As a footnote, neither does it measure AC power correctly but, that's another story. \$\endgroup\$
    – Andy aka
    Oct 7, 2021 at 15:22
  • \$\begingroup\$ Yes You are right, I misinterpreted that. Could you explain a bit more like why we are applying 2.5V for biasing one leg of the CT's output. What will be the effect if don't use that R,R,C circuitry ? \$\endgroup\$ Oct 7, 2021 at 15:38
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    \$\begingroup\$ You bias one leg of the CT at 2.5 volts so that the other leg (that has the signal on it) is forced to have a DC offset of 2.5 volts. This means that the peak-to-peak voltage excursions on the signal "leg" fall within the ADC input range of your arduino (powered from 5 volts and 0 volts). If you didn't do this, your signal digital inside your arduino will appear to be half-wave rectified and, quite possibly (in fact likely), your arduino input transistors will fail i.e. arduino damage is virtually guaranteed. \$\endgroup\$
    – Andy aka
    Oct 7, 2021 at 16:14

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