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I am using the en=g5la relay to activate the charge of a capacitor. Does it make sense to use a relay when you have a capacitive load? The general layout is that I have a diode brigde ( 4 diodes rectifier) on one side of the relay side which feed power and I got a 2.5 F capacitor on the other side which serve as energy pool. I switch on/off by using an arduino ouput the relay with a BJT who drive the inductor.

schematic

simulate this circuit – Schematic created using CircuitLab

[EDIT] This circuit worked perfectly in the end.

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  • \$\begingroup\$ @MathiewL: You have enough reputation to know that you need to supply relevant data, links to datasheets and that there's a schematic editor button on the toolbar. Can you fix your question? \$\endgroup\$ – Transistor Mar 10 '16 at 23:15
  • \$\begingroup\$ @transistor I'll do it right now. \$\endgroup\$ – MathieuL Mar 10 '16 at 23:15
  • \$\begingroup\$ @transistor there you go \$\endgroup\$ – MathieuL Mar 10 '16 at 23:22
  • \$\begingroup\$ Why do you want/need to switch the capacitor? \$\endgroup\$ – EM Fields Mar 11 '16 at 0:56
  • \$\begingroup\$ @EMFields With out going more into details of all the project the capacitor power an electromagnet \$\endgroup\$ – MathieuL Mar 11 '16 at 0:59
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It does make sense in the way that both the relay and capacitor would work fine in that set-up. There would be a large current spike when the capacitor is empty and gets connected to the supply, but it shouldn't last long enough to melt the wires. It will, however, cause the supply voltage to dip down, and that could mess with other things if they rely on it (like the Arduino could reset).

But I don't see any reason in your description to use the relay instead of a transistor to control charging, so in that way it doesn't make sense.

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enter image description here

Figure 1. Data sheet extract.

All the information you require is on the datasheet.

You haven't specified the voltage of the supply or it's internal resistance so we can't calculate the peak current. You need to remember that if the capacitor is completely discharged then it looks like a short circuit the instant the contacts are closed. Your 10 A contacts might not be enough.

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  • \$\begingroup\$ I will check but i highly doubt that the power consumption is too high. The signal that is rectified come from a transformer which is drive on the primary by 5 V / 0.8 A. I wil check for sure the voltage when I active the charge but I doubt that this is the problem \$\endgroup\$ – MathieuL Mar 11 '16 at 0:26
  • \$\begingroup\$ But, is it logical to use a relay with a capacitor? I mean if you check the datasheet, it is written load = resistive load. Is that the recommend load or the test load. \$\endgroup\$ – MathieuL Mar 11 '16 at 0:27
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Since the datasheet does not mention testing with anything other than a resistive load, we must assume that it has not been tested.

Because of the size of the capacitor (2.5 Farads), currents could be quite high.

When charging a discharged capacitor, the current could be quite high, perhaps high enough to damage the contacts in the relay. Note that this may ALSO exceed the maximum current rating of the capacitor, so may also damage the capacitor.

The "easiest" fix would be to include a resistor in series with the relay in order to limit the current, though this would waste energy (Joule heating in the resistor). Perhaps the source resistance of the (assumed) AC supply transformer would limit the current is this way. Assuming a 24 V supply, a 5 ohm resistor would be required to limit the current to less than 5 A.

A second fix could be to include an inductor (with parallel diode) in series with the capacitor to limit the current. Assuming a 2.5F capacitor, 24VDC supply, and a 1ohm series resistance, spice simulations suggest that a 50 Henry inductor would be required to limit the current to less than 5 A. Charging would take about 20 seconds.

The best solution would likely be to use a current source to charge the capacitor instead of a voltage source. A MOSFET in saturation provides constant current, so may be appropriate. Also, a linear or switching regulator could be used as a precise constant current source.

Since you have the circuit working, I'd suggest using an oscilloscope to measure the switching currents and voltages, in order to ensure that the relay's and capacitors maximum currents and voltages are not exceeded.

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