I Need to design and make a CT (current transformer) to energize a 3ph relay with sufficient voltage and current from low load to full load. I'm thinking of a toroidal transformer with very few windings of 10 or less since I'm not running anything except a relay

A CT is preferred as it allows (I hope) me to monitor all 3 phases at once and output to a 3 phase relay connected to a shown 3ph MIG wire feeder.

So what I hope will happen is when power is ON but the stick welder is not energized the wire feeder is NOT ON either. When the machine is on and power is going in to the machine the CT energizes the 3ph relay powering on the wire feed unit in a pass through type configuration.

I guess my main questions are if a CT divides amperage, is voltage multiplied by the same factor?

A CT with that is "measuring" no load outputs no current either? so not energized i.e. 3 phase delta, 3 live wire going somewhere but the load is yet to be turned on...

From what little that I know/understand Its like voltage transformer in reverse (voltage increases at the expense of reduced amperage). the opposite could be said about a CT from what I little know/understand, amperage is reduced while voltage if left open circuited can quite quickly rise to unsafe levels.

Will 3 wire Delta pass through a single CT and function as normal, reduced or just not possible at all. Will I need to have 3 CTs wired in series with a relay as load w/ or w/o a resistor?

One last Question googled and what not still couldn't find a definitive answer with a formula and example of how to find the output voltage of a CT, I'm still not sure, answers were all to do with current/amps etc...


simulate this circuit – Schematic created using CircuitLab

  • \$\begingroup\$ How big is the range from minimum sensed load to maximum? How big of a load is your control logic, if any/no auxiliary power supply? \$\endgroup\$
    – winny
    Nov 26, 2016 at 10:43
  • \$\begingroup\$ @winny Its a very large 3 phase welder. 7.5KW~ (2.5KW~ per phase) at 300A welding output, I usually use it on the low ish end of its output 75A - 150A. anywhere from 0.75KW~ - 1.5KW~ per phase input (@450V 1A~ - 4A~) link \$\endgroup\$ Nov 26, 2016 at 20:40
  • \$\begingroup\$ Sure, but you want to measure this and turn something on/off accordingly? Whats the requirements on that part? How much does that glue logic/relay logic consume? \$\endgroup\$
    – winny
    Nov 26, 2016 at 21:45
  • \$\begingroup\$ Enough to power a 3 phase relay or 3 single phase relays 20W=< \$\endgroup\$ Nov 26, 2016 at 23:50
  • \$\begingroup\$ Can you accept having a dedicated auxiliary power supply for it? \$\endgroup\$
    – winny
    Nov 27, 2016 at 10:37

1 Answer 1


Will 3 wire Delta pass through a single CT and function as normal, reduced or just not possible at all.

No that won't work under any normal load conditions because the instantaneous sum of all three currents is zero for a perfectly balanced load.

Will I need to have 3 CTs wired in series with a relay as load w/ or w/o a resistor?

Just one CT attached on one phase is sufficient because if one phase is taking current then the other two can be presumed to be taking current also.

Will it drive a relay? I don't think it will because the relay coil doesn't represent much of a burden resistor and the toroid core may sature and you'll not get enough to power the relay. However, that is a generalization and you might find one that works BUT you are now starting to have problems deciding which cores will saturate and which won't.

Almost certainly, a more reliable solution is to use a conventional CT with a burden and use the signal to turn on a transistor that then activates the relay.

  • \$\begingroup\$ Any reason why a CT can't be used as a 1:1 or 1:10 ratio instead of 10:1 or 100:1? i.e. instead of dividing a large current in to a small current, keep the same amount of current in primary winding(s) and have many more times current in the secondary? \$\endgroup\$ Nov 26, 2016 at 19:58
  • \$\begingroup\$ Most CTs I've seen have a big single turn power conductor in order to handle the current. To get more current you need to have a fraction of a turn on the secondary and this is impossible so, you end up with say ten turns on the primary and one turn on the secondary and your core saturates. \$\endgroup\$
    – Andy aka
    Nov 27, 2016 at 10:14

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