After some advice here I'm wondering if this is the correct design of my circuit.
I'm now using a SSR as the control device to the contactor. The diode, I believe, should not only protect the SSR from back EMF, but also the capacitor as well. I know I will also need a resistor or thermistor prior to the capacitor so the capacitor itself does not cause a large inrush current, both when the system is turned on, and again after the switch is opened again.
The value of that resistor is something I'm still somewhat unsure of how to calculate.
For application details, the system will be on for at minimum 2 minutes before the relay is closed to engage the contactor. The contactor will then be on for up to 60 seconds before being disengaged. The capacitor needs to be able to recharge within 60 seconds to be able to close the contactor again if necessary.
If I'm understanding correctly, upon startup, the resistor/thermistor will suppress the capacitor inrush and slowly charge it (charge time depending on resistance value). As soon as my switch is closed, the capacitor will provide the full energy required to close the contactor coil. From that point on, the 24V source will provide enough energy to hold the contactor in, while also slowly charging the capacitor.
Once the switch is re-opened, the diode with suppress the back EMF, while the capacitor recharges.
Does this design seem reasonable? If so, how do I determine the value of the resistor to complete the design?
http://www.kvc.com.my/EnterpriseChannel/SharedResources/Datasheet/0/?ProductId=1000066755&Filename=SCHNEIDER-LX4-FH024.pdf
http://www.mouser.com/ds/2/293/e-nt-15067.pdf
https://www.us.tdk-lambda.com/ftp/Specs/dpp120-240.pdf
Best Answer
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.