I'm looking at the datasheet for TE Connectivity Gemini Series switches and contact rating is given as: 5A @ 125VAC or 28VDC resistive; 0.4VA @ 20V AC or DC. I want to use this switch for DC application and the first part of specs: 5A @ 28VDC seems inconsistent with the second one 0.4VA @ 20VDC, which suggests 0.4W/20V = 0.02A current. Can someone explain discrepancy between 5A in the first part and only 0.02A in the second one?
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\$\begingroup\$ 20 mA sounds like the current required to hold the relay. \$\endgroup\$– winnyCommented Jul 20, 2016 at 19:07
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\$\begingroup\$ @winny It's not a relay. It is a toggle switch. \$\endgroup\$– Paul JurczakCommented Jul 20, 2016 at 20:41
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2 Answers
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Inductive loads will force the voltage to increase on a circuit break, so there is a much greater chance that the switch will arc. The maximum voltage is therefore much lower under such a load. The reduced power is so that the switch itself doesn't melt or ignite should it arc.
answered Jul 20, 2016 at 18:19
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\$\begingroup\$ OK, but going from 5A for resistive load to 0.02A only for inductive load doesn't make much sense. \$\endgroup\$ Commented Jul 20, 2016 at 20:44
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\$\begingroup\$ It's not 0.02A, it's 0.4VA. Since the heat generated is proportional to the power, it's the power that's limited, not the current. \$\endgroup\$ Commented Jul 20, 2016 at 20:48
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\$\begingroup\$ Do you mean 0.4VA is a maximum power loss on the switch? \$\endgroup\$ Commented Jul 20, 2016 at 21:13
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\$\begingroup\$ The arc is likely to be the highest resistance in the circuit, so in a sense, yes. \$\endgroup\$ Commented Jul 20, 2016 at 22:28
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VA is used for reactive power - ie inductive switching.
When an inductive circuit is broken the current MUST instantaneously continue. To do this, if a formal path is not available the voltage will rise until the current does continue to flow. This results in a n inductive "voltage spike". The energy may end up stored in stray capacitance (so that 1/2.L.i^2 is transformed to 1/2.C.V^2) or if there are paths to complete the circuit at voltages below that required for capacitive storage, a spark will occur. Such sparks dissipate part of the energy in the switch contact surfaces causing pitting and general damage and "should be avoided" [tm].
You will accordinly find that the inductive rating of switches is far far far lower than the AC values. DC circuits fall in the "may be inductive" category.
Even with only stray inductance a DC arc may form and self maintain.
In AC circuits if an arc forms it will be extinguished half a cycle later. As this occurs usually during contact opening, when the voltage next rises to the value where the arc occurred the contacts will be further apart (usually much further. In a DC circuit there is no 100 or 120 zero crossings a second to extinguish the arc. As the contact gap widens the conducting channel of ionised gas provides a conduction path to maintain the arc.
Magnetioc blow out units are used in higher current AC & DC switches to extend the arc length or the arc may be directed to a pair of diverging conductors so the spark rises under thermal convective effects up the widening gap,
answered Jul 21, 2016 at 9:25