i have this CT current transducer: it outputs 333mV at 5A full scale.

I'm planning to use it in my house/office in order to detect if a load is switched on or off, but i'm worried about how i can protect my arduino pro mini.

The arduino pro mini (based on atmega328p) is running on batteries (2x1.5V) and has some modifications in order to lower the power consumptions (at this time, the arduino itself consumes only 5uA).

Digging around, for my project purpose i've found a lot of solutions but this seems the most pratical:


(The generator and the 33ohm resistor on the left side of the schematic, represents the current sensor)

This is a simple comparator circuit, as you can see from the schematic, the 1.3V power input represents a low power voltage reference based on LM385, that i'm planning to use in order to have a stable reference of my circuit during the battery discharge over time (if you have some alternatives these will be appreciated!)

The treshohold is regulated by hand using a trimmer with the load switched on.

I've not yet tested physically this circuit, but only simulated using this CircuitJS, but i'm not sure about some things:

  1. The 200k voltage divider that aims to "scale" the ac 333mV output but it seems not scaling nothing in the simulated circuit. enter image description here

  2. The aim of the placed diodes is to clamp the voltage if some spikes (that may occour) are injected in the circuit, protecting the arduino pin, i've seen that if i change the AC generator from 50mV to 600v (simulating consistent noise signal spikes) the output signal oscillates between -700mV and 2V...why? enter image description here

I have some doubts about my circuit...i'm too much worried about my battery consumption that i'm making working bad the circuit? maybe i'm overthinking on that?


  • \$\begingroup\$ If you do not want measuring exact current, use optocoupler. \$\endgroup\$
    – user263983
    Jan 18 at 22:12
  • \$\begingroup\$ @user263983, this can be a solution, you can add some details? in how to connect it and how i can use it? \$\endgroup\$
    – VirtApp
    Jan 18 at 22:15
  • \$\begingroup\$ Schematic can be different, depend of electronic elements, you have in hands or can buy. Current from CT need to be rectified, so or bipolar optotransistor, or diodes bridge. Current loop seehttps://en.wikipedia.org/wiki/Current_loop \$\endgroup\$
    – user263983
    Jan 19 at 0:08

1 Answer 1

  1. That 200K resistor pair is not scaling input voltage, it is setting the zero point. Think about this way: what if there is no voltage? Then there is no AC in the circuit, and the capacitor behaves like open circuit. So what is the voltage on the comparator's input? It is 1.3V/2, the value of power input divided in two by dual 200K resistors.

(As a side affect, this forms a secondary divider, with (10K + 4.7K) on one side, and (200K || 200K) on the other side. This divider does decrease input voltage, and this is not desired. To minimize voltage decrease, the resistances in the input path (10K, 4.7K) are much smaller than resistors in zero-setting path (200K, 200K)

  1. Regular diodes open (start conducting electricity) when when anode is (approximately) 0.7V more positive than cathode, and they do not allow voltage to raise to more than that. So as as you increase the voltage, eventually the top diode will open. It's cathode is connected to 1.3V, so the diode will start the conduct when anode is 0.7V bigger, 1.3V + 0.7V = 2.0V which what you see.

The bottom diode has anode connected to 0V, which means it'll open when lower voltage is 0.7 less, which means -0.7V, which is what you see as well.


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