I've bought this AC current switch, which is sensitive to currents from 50 mA to 5 A on the single AC wire that passes inside its hole. When currents from 50 mA start to flow in the wire, it allows its output terminals to "conduct".

I need it to be sensitive to lower currents, in the range of 20 mA, so I opened it and drew the schematics of it.

While drawing it I noticed its output terminals are not isolated from the input, but I could wire an SSR to its output terminals.

Can anyone help me in what I could do to make it more sensitive to a lower current flow on the input (20 mA) in order to allow current flow on the output? I suspect if it's possible, it would be something about changing the values of R3 and/or R4.

VDA3510CTA voltage detector datasheet

Photos and schematics below:

Photos taken from Aliexpress listing here: Link

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  • \$\begingroup\$ What do you mean not isolated from the input? The current transformer would provide isolation. \$\endgroup\$
    – Hearth
    Commented Apr 28 at 0:30
  • \$\begingroup\$ @Hearth indeed! I meant the whole internal circuit is connected to the output circuit… the output is not isolated from the circuit, like an opto… but you’re right… the input load is in fact isolated from the output by the transformer. \$\endgroup\$
    – Rodrigo
    Commented Apr 28 at 13:21

1 Answer 1


You can't easily change the basic sensitivity of the device but if you put multiple turns of the AC cable through the centre hole you can make it trip at lower currents.

For example if you put the load carrying cable through the hole twice it should trip at 25mA. If you put it through three times it would trip at 16.7mA.

As others have commented, this would also reduce the upper limit of allowable current. This is difficult to determine without knowing the number of turns on the secondary of L1.

I would expect the main limit to be excessive power in R3 or ZD2.

If R3 is reduced in value (eg 470 ohms) that would keep the maximum power the same as when used as intended with one turn on the primary. It is not obvious why R3 could not be reduced to zero.

ZD2 power dissipation is less obvious. If L1 had 1,000 turns the current in ZD2 would be about 10mA with two turns on the primary when 5A is flowing in the primary. This results in about 50mW dissipation which should be well within the capabilities of the existing device. However more than about 100mW would probably be problematic.

Additional zener diodes of the same voltage in the ZD1 and ZD3 positions may be one way to increase the allowable current although at some point the maximum dissipation in the enclosure or saturation of L1 would impose a limit. (Saturation of L1 would actually help as it would limit power dissipation automatically).

  • 5
    \$\begingroup\$ It worked!! Solved my problem!! That's perfect! Thanks! \$\endgroup\$
    – Rodrigo
    Commented Apr 26 at 20:22
  • 3
    \$\begingroup\$ The upper current limit will change as well \$\endgroup\$
    – fraxinus
    Commented Apr 27 at 16:57
  • \$\begingroup\$ @fraxinus Good to know… so.. the upper limit will also be divided in half, if the lower limit is divided in half? In my case it’s still ok.. \$\endgroup\$
    – Rodrigo
    Commented Apr 28 at 13:19
  • \$\begingroup\$ @fraxinus But even so it worked for my purposes, would there be any component change to allow it work without changing the upper limit? \$\endgroup\$
    – Rodrigo
    Commented Apr 28 at 13:24
  • \$\begingroup\$ @ Rodrigo * Change the current VDA voltage detector (3.5V) to a model with lower threshold voltage (e.g. 1.5V). These are available in 0.1V steps, beginning at 0.8V. \$\endgroup\$
    – Raonoke
    Commented Apr 28 at 15:42

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