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I have a 12VAC-controlled contactor (ESB 20-20, rated for 12V @50Hz), and a 220/12 transformer that I intend to connect to the contactor. Unfortunately, rather than 12V, the transformer outputs 14V (which is outside of the 85—110% range of allowed voltage). Which option should I employ?

  • Put a resistive voltage divider between the transformer and the contactor?
  • Use two opposite 12V Zener diodes to clip a half-wave each?
  • Return the transformer to the shop and try to find a more precise one?
  • Throw out both the transformer and the contactor, buy a DC-driven contactor and a stabilized power supply?
  • Stop caring, connect them as they are?

Edit: forgot to add, while 14V is the measure of an unloaded transformer, under the contactor load the transformer voltage drops an insignificant amount, to around 13.8V.

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  • \$\begingroup\$ I will eliminate the first and the last options for you. \$\endgroup\$
    – Eugene Sh.
    Nov 3, 2015 at 19:29
  • \$\begingroup\$ The last one is the best one. \$\endgroup\$
    – Marko Buršič
    Nov 3, 2015 at 19:31
  • \$\begingroup\$ @MarkoBuršič If you wish to switch to the previous one right after that. \$\endgroup\$
    – Eugene Sh.
    Nov 3, 2015 at 19:36
  • \$\begingroup\$ The zeners are a bad idea. Once the AC voltage exceeds their breakdown voltage they will conduct and try to pass whatever current the transformer can deliver as it approaches maximum voltage. \$\endgroup\$
    – Transistor
    Nov 3, 2015 at 19:56
  • \$\begingroup\$ Something like a DC solenoid doesn't need a regulated power supply. It won't care if there is ripple on the supply and may work fine if it wasn't even smoothed and was just full-wave rectified. Remember that the inductance of the coil will tend to smooth out the current somewhat and the inertia of the solenoid armature will keep it in position between mains half-cycles. \$\endgroup\$
    – Transistor
    Nov 3, 2015 at 19:58

3 Answers 3

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If you measure the transformer output voltage without load, this value can be higher than the nominal value. You must measure with the maximum load that the transformer support; this is the voltage that you must take care of.

For the contactor, I think that you can connect it, and measure the voltage on the coil when the device is operating, to verify that the voltage applied is safe.

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  • \$\begingroup\$ The voltage under the load of the control circuit is still higher than tolerance, around 13.8V. I think a single coil is far from the maximum load, but there is nothing else requiring 12VAC anyway. \$\endgroup\$
    – Maxim
    Nov 3, 2015 at 22:51
  • \$\begingroup\$ @Maxim Check the temperature of contactor. I have so much experience with this device, and I think that's no problem, but to be sure, you can operate it and check the coil's temperature. Have You a datasheet? Can you post it? \$\endgroup\$
    – Martin Petrei
    Nov 3, 2015 at 23:09
  • \$\begingroup\$ farnell.com/datasheets/1605549.pdf — it's ESB 20-20, rated for 12V @50Hz. \$\endgroup\$
    – Maxim
    Nov 3, 2015 at 23:25
  • \$\begingroup\$ @Maxim The tolerance shown to the coil voltage is the same as is usually indicated to a power line 220 V. What you can do is to insert in the circuit of the coil, a fuse of 1 A, type gL. \$\endgroup\$
    – Martin Petrei
    Nov 4, 2015 at 0:50
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If the 14 V AC that you're measuring is the open-circuit (unloaded) transformer output you might find that when loaded by the contactor the voltage drops closer to 12 V. Connect them up and switch them on for 10 s while you measure the coil voltage. If it's within spec then you're OK.

If not, then I recommend adding some series resistance. Note that since the coil has inductance and you're running on AC that it won't be a simple Ohm's Law resistance calculation but that would be a good place to start to get a rough idea of the series resistance value to choose.

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  • \$\begingroup\$ It indeed drops closer, but not by much, to around 13.8V. So, instead of a voltage divider I should just put a resistor in series with the solenoid? \$\endgroup\$
    – Maxim
    Nov 3, 2015 at 22:53
  • \$\begingroup\$ That should do the trick. \$\endgroup\$
    – Transistor
    Nov 3, 2015 at 23:31
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The easiest solution is to add either a series resistor OR add a pair of parallel but reverse-connected diodes in series with the contactor coil. That is: two diodes connected together in parallel but back-to-back.

I would use the pair of diodes in series with the coil: this will provide about a voltage drop of about 0.5 Vac (Average) voltage drop. This should bring you right in line with the maximum rating of the contactor coil.

I do NOT like to use a series resistor on the contactor coil because the coil inrush current can be extremely high. A series resistor can lead to problems with the contactor not pulling in or sealing correctly.

Be sure to choose diodes that will handle the contactor inrush current.

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  • \$\begingroup\$ @maxim, coil inrush is high because the resistance and inductance is low. When the contactor pulls in it closes the magnetic circuit (the iron core and armature) and increases the inductance. This increases the impedance (resistance to AC) and that reduces the current in operation. This is good and bad. It means in normal operation that there is very strong pull and adequate retaining force. It's bad if the contactor doesn't pull in properly because the inductance remains low and the coil can overheat and burn out. \$\endgroup\$
    – Transistor
    Nov 3, 2015 at 23:35
  • \$\begingroup\$ Do I understand correctly that, if the data sheet says “pick-up consumption 8 VA, holding consumption 3.2 VA”, then the inrush current is 8/3.2=2.5 times higher than the operating current? \$\endgroup\$
    – Maxim
    Nov 4, 2015 at 13:52

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