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Considering this:

schematic

simulate this circuit – Schematic created using CircuitLab

  • V1 is a beefy battery - it will happily provide 1000A or more;
  • RLY1 is a beefy contactor, the LEV200 series, rated for 500+A, 12-900VDC;
  • C1 is the capacitor in question.

When closing the contactor, the inrush current to energise C1 is capable of welding the main contacts. Of course if C1's capacitance is small enough the issue subsides to manageable levels. This amount of capacitance is what I want to compute, if possible. A crude/conservative figure is reasonable given that the datasheet does not mention anything about (non-precharged) capacitive loads.

Ideally there are many methods to avoid the welding even if C1 is large (precharging, adding series inductor and/or NTC resistor), but my question pertains to the above model only, I want to know when these measures are needed.

EDIT: As per Macrus Müller's comment, the source impedance of V1 is very important, it is approx. less than 0.1 ohms.

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  • \$\begingroup\$ Hey, if we act as if all components including the batteries are perfect, than even a 10 pF capacitor will exhibit an infinite inrush current. So, it's important to add the source impedance in series of the battery voltage source, and suddenly, this becomes pretty calculatable with classic methods of network analysis! \$\endgroup\$ Commented Nov 10, 2020 at 10:17
  • \$\begingroup\$ So, I'd recommend to add an explicit resistor in series with your voltage source, and try to estimate its value. Usually, you could also do things like have (a second relay + series resistor) in parallel to your contactor, which you close first, so that C1 charges slowly, and only after C1 is close to 350 V, you close the contactor and turn on any device attached to the outpu. \$\endgroup\$ Commented Nov 10, 2020 at 10:20
  • \$\begingroup\$ Yes, this is the precharging method, the standard approach for this kind of problems. I'll employ it if needed, however the real question is when it becomes needed. \$\endgroup\$
    – anrieff
    Commented Nov 10, 2020 at 10:39
  • \$\begingroup\$ as recommended, no way around estimating the internal source resistance and writing it down as series resistor. Also, at these currents, the ESR of the capacitor becomes pretty important, as well as the resistance of the cabling/traces. Not to mention inductance, too! \$\endgroup\$ Commented Nov 10, 2020 at 10:41
  • \$\begingroup\$ Datasheets give me 84 milliohms, so let's assume less than 0.1 ohms for V1. For C1 ESR and cabling the addition is small, say 20 milliohms. \$\endgroup\$
    – anrieff
    Commented Nov 10, 2020 at 10:45

1 Answer 1

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[Battery] datasheets give me 84 milliohms, so let's assume less than 0.1 ohms for V1. For C1 ESR and cabling the addition is small, say 20 milliohms

so you got \$R_\text{series}\$=0.1 Ω in series with a short on connection, i.e. 350 V · 0.1 Ω = 3.5 kA inrush current, but for very short time (hence the importance of the parasitic inductivity)!

You can calculate the charge current curve for an RC element over time as

\begin{align} I(t) &=\frac{V_1}{R_{\text{series}}} e^{\frac{-t}{R_{\text{series}}C_1}}\\ &= 3.5\,\text{kA}\cdot e^{\frac{-10t}{C_1·Ω}}. \end{align}

Note how the value of the capacitor doesn't change the peak current, but only the duration of high current. So, basically, you're asking the wrong thing here; you don't want to find the larges capacitor for which your contactor doesn't weld, you're looking for the lowest \$R_\text{series}\$ with which it still works!

Realistically, your cabling, contactor and connectors do have some parasitic reactance, so I'd encourage you to play around with models of these in SPICE. For a start, simply put a 10 nH in series with your battery (in addition to your \$R_\text{series}\$) and see how that changes things in a step response.

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  • \$\begingroup\$ I have to say I'm not thoroughly convinced by the reasoning why nonideal behaviour alone saves the contactor from welding during closure. Welding requires some minimum energy in order to occur. This depends on the duration of high current. Or, if you will, this is correlated with the joules that eventually end up in the cap. \$\endgroup\$
    – anrieff
    Commented Nov 10, 2020 at 17:04
  • \$\begingroup\$ well, how exactly does the contactor get hot if it doesn't have a resistance? \$\endgroup\$ Commented Nov 10, 2020 at 17:55
  • \$\begingroup\$ I cannot say for sure. I guess the bounce time is something like 10-30ms and it has resistance during that time. But I think if the cap charges in 1µs then the welding likely cannot occur because too little energy is deposited onto the contacts during that time. \$\endgroup\$
    – anrieff
    Commented Nov 10, 2020 at 21:17
  • \$\begingroup\$ well, that's why I gave you the easy formula, you can calculate that yourself. But yes, these times are dominated by parasitic effects, and you cannot ignore them in this case. \$\endgroup\$ Commented Nov 10, 2020 at 22:02

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