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I am using a Panasonic relay (ALQ105) rated for a resistive load of 10 amps 5A at 250V to drive LED bulbs. These LEDs have capacitive circuits in their ballasts and hence draw high inrush current when switched ON. Steady state current is quite low (less than an amp at 220 Vac).

This inrush current (and possible spark during switching) is causing contact welding in the relay. I cut open one and found the contacts stuck together. A gentle prying with my finger nails separated them together. I couldn't find any black deposits (which I suspected would be there due to sparks). The value of capacitance in the circuit is unknown and different in different cases. As such I am looking for a general method to get rid of this issue. I could think of these three possible ways:

1) Use a series resistor and possibly an inductor to limit the high inrush current - Possible negative effects would be steady state power dissipation in resistor. Also, I am not sure about what values of R and L should I choose.

2) Use NTC thermistor like this - http://www.cantherm.com/media/productPDF/MF72_JUNE_2016_1.pdf

This looks a better idea as compared to L-R circuit but I am not fully sure of the calculations that I need to do for selecting the correct part.

3) Switching to solid state devices like triacs. Since there won't be any mechanical contacts, there won't be any welding. I am currently using BTB-16-800-BW triac in other applications. (ST BTB16 Triac datasheet)

This has a steady current carrying capacity of 16A which is much more than what my requirement is. It also has a surge current limit of 160 A. However I am not sure whether 160 A is good enough for the load I am dealing with.

Please help me selecting the best feasible solution for this issue.

[Updated relay contact rating to correct value of 5A at 250VAC]

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    \$\begingroup\$ Maybe it would be worth it to first check with a scope the inrush current duration and peak value. Then you can choose the appropriate device that will be able to switch that, without having to overspecify things. \$\endgroup\$ – dim Sep 6 '16 at 14:06
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    \$\begingroup\$ If you are not sure about the surge current capabilities, then how can we be? \$\endgroup\$ – PlasmaHH Sep 6 '16 at 14:06
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    \$\begingroup\$ As well as reducing the current with some kind of device, you might want to consider using a relay with contacts rated for 'tungsten load'- which are designed (metallurgically) to withstand a large surge without welding. \$\endgroup\$ – Spehro Pefhany Sep 6 '16 at 14:10
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    \$\begingroup\$ In some supplies such an NTC is used to limit the inrush current. What is also used is a series resistor (like you suggested) but which is shorted by a second relay a few moments after the main relay closes. \$\endgroup\$ – Bimpelrekkie Sep 6 '16 at 14:17
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    \$\begingroup\$ Power supplies can draw huge currents on power-up. I once tested a 15 W supply that triggered the protection circuit on a 200 W AC power supply. \$\endgroup\$ – Robert Endl Sep 6 '16 at 17:01
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You might want to use an inductor after all. Besides limiting the inrush current, it will improve your circuit's power factor which right now should be reduced by the capacitive nature of the load, at least judging by your description.

Of course, you'll need to find out how much of capacitance you have to counter to calculate the right inductor value. You'll have to either obtain that information from the datasheet or measure it. If you can't, just pick a value and try it out (I'd expect something in units or tens of mH to work).

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    \$\begingroup\$ Thanks a lot, Dmitry. I'll solve the differential equation for multiple expected C values and calculate L. Once I am done, I'll order a bunch of them and try them out. In parallel I'm also considering NTC because it appears like no brainer. Probably I'll end up using both of them for more robustness. Let's see how the testing goes. \$\endgroup\$ – Whiskeyjack Sep 8 '16 at 6:36
  • \$\begingroup\$ @Whiskeyjack Hint: you don't have to solve diff equations here, just calculate the total complex impedance of your circuit (L in series with (R in parallel with C)) and find the value of L which zeroes the imaginary part. Don't hesitate to ask a new question, should you be stuck with calculus. \$\endgroup\$ – Dmitry Grigoryev Sep 8 '16 at 8:02
  • \$\begingroup\$ Also, I'm not advising you against NTC resistors; however keep in mind they are non-linear devices so you may not be able to calculate the peak current, only try and measure it. \$\endgroup\$ – Dmitry Grigoryev Sep 8 '16 at 8:06
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You might want to also consider the triac and then turning it on at the zero crossing either using either a specialized trigger circuit, or monitoring the line with a microcontroller and sending the turn on signal near the zero crossing.

The reason this will help is because the turn on is at a very low voltage in the AC cycle and then the current is naturally limited as the AC cycle ramps up.

Anyway, just another option that may work for you. -Vince

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You could try use a soft-start module to control the inruch current. These are common in large (audio) power amplifiers with linear (transformer) power supplies. A few examples here.

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If the load is actually a fully discharged capacitor, its initial inrush current will certainly be very high and well above the capacity of the relay contacts. As the other answer suggest, in-rush current limiter is needed.

The most easiest of all is the low ohmic high inductance in series with that contact. High inrush current will thus be limited by the inductor because of its inherent property of the device of opposing large changes in current. This is much better than the resistive current limiters with relayed bypass because the latter has power loss and additional switching components, which makes it little complicated than a simple passive inductor.

Choose the inductor current capacity based on nominal circuit current and calculate the inductance so that the peak current is within the capacity of the relay.

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