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Comments on How Relay inrush current limiter works?

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How Relay inrush current limiter works?

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In this circuit below the relay circuit part is acting as an inrush current limiter, but I can't understand how it works since the current is flowing in both conditions where the relay is open or closed!

the circuit

The source

Edit: so it's all about the switching time, relays are known for slow switching time which is supposed to be a negative criteria, I checked the Datasheet of the relay and its 9~10 Ms.

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It's a current limiter, not eliminator. The current is always supposed to flow, but be limited at startup to avoid high inrush current.

All the relay does is short out R103 when activated. The relay starts out off, so R103 is in series with the input power when the device is first plugged in or turned on. The 10 Ω of R103 keeps the current limited. It's not meant to allow the device to fully function, but enough to allow the input capacitors of the power supply, or whatever else might take a large initial slug of current, to get past that point.

Eventually the 12 V supply comes up, which then turns on the relay. That shorts out R103, and the device can draw its full normal operating current without its input voltage sagging. There may be a small current spike right when the relay closes, but presumably that has been calculated to be within tolerable limits.

The device may have additional circuitry inside that doesn't run the main device function until after the relay has been on for a few 100 ms or so. By that time, the internal DC supplies should be up to their intended voltages.

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Why the resistors R104 and R105? (2 comments)
Why the resistors R104 and R105?
Lundin‭ wrote almost 2 years ago

Any idea why they added R104 and R105? Looks very strange. That they are in parallel is probably just some Bill of Material optimization(?) but why limit the coil current way below the spec? (It's 30mA for this relay.)

Olin Lathrop‭ wrote almost 2 years ago · edited almost 2 years ago

@Lundin You are right, those resistors don't make sense. At total of 37.5 kΩ wouldn't let the relay turn on with 12 V as shown. I suspect something wasn't copied correctly. Maybe the C109 and 12 V are supposed to be on the left side of the resistors with high voltage on the right. Even then there wouldn't be anywhere near enough coil current. I don't know what the circuit is really supposed to be, but certainly not what is shown. 7.5 kΩ instead of 75 kΩ might almost work with 240 V line voltage.