Electrical – Meaning of symbols used in ABB contactor circuit

I think that design is ok. If you assume that the inductance of the coil doesn't delay the rise of current significantly and that the power supply doesn't contribute significantly to maintaining the capacitor charge during pull-in, you can just use the time constant of the 0.9 ohm coil resistance and the 0.47F capacitor to calculate that the discharge during the 50 ms pull-in time is less than 10%. That should be enough for the contractor to pull in.

The power supply has electronic protection that trips between 110% and 145% of rated current. There is probably a slower trip for 110%, so you could probably safely use a 2 ohm resistor to limit the charging current. That will allow the capacitor to fully charge in 5 seconds. Recharge after pull-in will be much faster because pull-in only causes discharge to 90% voltage. The 2 ohm resistor will only reduce the voltage by 1% when the contractor is drawing hold-in current. The peak power will be 288 watts during charging. Even though the average power will be much lower, you will probably need a 25 or 50 watt wire wound resistor in order to avoid over stressing the resistance wire.

You should probably fuse the power supply with a slow acting fuse to allow continuous current somewhat more than the holding current but less than full power of the resistor.

I think that it would be less complicated and less expensive to use a transformer, rectifier and smaller capacitor. Think that could be economically designed to supply high pull-in current and low holding current.

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.