Relay latching and getting stuck

[dansku]

Check if my ideas are correct
I found this resistor for example: http://www.mouser.de/search/ProductDetail.aspx?R=0virtualkey0virtualkeyAS25J10R0ET
It's a 1.5W / 10ohm and the pulse


What do you think?

[tatus1969]

What do you think?

Any resistor that has enough wattage to deliver the current to your load should be suitable. In order to calculate this, we need to have a rough figure of the current consumption of your load. The bulk capacitance that it has would also be important to thermally estimate the switch process.

For example: I assume that your load has a maximum current consumption of 100mA at 55V. You want to limit the inrush current to a value that is below the relay's contact rating. As good engineering practice, I would not go to the limits but use half of that rating instead, especially because you are also operating it above its voltage rating.

Your relay has a rated current of 10A for DC, so lets limit the current to 5A.
The resistance value would be 55V / 5A = 11 ohms, your choice of 10 ohms should be good then.
The power dissipation would be 10ohms * 100mA * 100mA = 0.1W. Double this again to be on the safe side. So you could safely choose a 1/4W leaded resistor here. But you should know your maximum load current well here. If you are in doubt, use an NTC instead, because it is protecting itself just in case. But be aware that an NTC can also act as a "resettable fuse". If it "trips" (heats up too much) during the switch, or during overcurrent, then it can stay hot because of enough static current draw of your load, and you need to completely disconnect the load in order to "reset" the NTC.

I also assume that you have 1000uF of capacitance at your load. LTspice (attached) tells me that the resistor has to "digest" 1.55 joules of energy during the switch transition. A quick googling for a rough figure of the thermal capacity of a 1/4 watt resistor gave me this: http://electronics.stackexchange.com/questions/131808/whats-the-thermal-time-constant-capacitance-of-a-1-4w-resistor. They calculate 18mJ / K for that part, the math looks good. If I take that here (although I think that the number is too optimistic because the metal film will heat up more than the ceramic substrate), I get an instantaneous temperature rise of 1.55J / (18 mJ/K) = 86 K. Add max ambient temperature to this, lets say 50°C, then you get the max temp of the resistor = 50°C + 86K = 136°C. That should be okay for any 1/4W part. You may choose a 1/2W resistor again to be on the safe side (needs some more research go get a good number for the heat capacity).

Remember, all these equations base on my guess of 100mA and 1000uF.

Edit: regarding your chosen resistor: we can derive the thermal capacity from its datasheet. The single-pulse graph shows that the device can handle 4000W for 100us. That is probably valid for ambient = 25°C, although they missed to state the conditions. The max operating temp is 155°C, equivalent to a single-pulse temperature rise of 130K. From this, we get the heat capacity = (4000 * 100e-6) / 130 = 3mJ/K. Divide our 1.55J by this number, and you get a temperature rise of 503K, which is by far too much. I would expect that this device will just vaporize 

Edit2: I recommend to choose a wire-wound resistor instead of a metal film type, because these simply have a lot more copper to take the pulsed heat. For example: this one http://www.nikkohm.com/nikkohm_e_pdf/rnp50s_e20140701.pdf is a very large device, and still has only 34mJ/K of heat capacitance (!!) according to a Nikkohm appnote (http://www.wdi.ag/files/manufacturer/nikkohm/Nikkohm_Capability-Brochure-Power-Resistors.pdf). The IRV100 (http://www.nikkohm.com/nikkohm_e_pdf/irvirh_e20150101.pdf) has the same rated power (comes with builtin heatsink) but is wire-wound and has a heat capacity of 2.2J/K. That is 65 times better!

The heat capacity calculated above for the 1/4W metal film resistor is most probably way off. And a 1/2W metal film resistor will neither do it.

Edit3: found some information at Vishay: http://images.vishay.com/books/VSE-DB0007-0805_Leaded%20Fixed%20Film%20Resistors_INTERACTIVE.pdf. On page 32 they specify pulse ratings of a 1/4W resistor (MBA/SMA 0204). Doing the same calculation as for your part, I get (9W * 10ms) / (155°C - 25°C) = 0.07mJ/K. With our 1.55J of energy that one would heat up by 1.55 / 0.07m = 2239K (!!). So definitely not suitable, and also 1/2W and 1W metal film resistors will be inadequate.

You see that this topic involves quite some math and research beforehand. If my guess with 1000uF and 100mA is close to your reality, I suggest to use a wirewound resistor and check its datasheet for the heat capacity. Should be more than 15mJ/K to stay on the safe side.