Snubber circuit across relay contact.

[XaviPacheco]

Hello,

I attach a fragment of a circuit that I found. Some stuff can't be well appreciated, but what I want to note is that there is an snubber network across the relay contact. As far as I know, snubbers are needed when there is inductive load. But that part is just feeding a rectifier bridge. So I think that's not really needed and can be removed, right? I would like to know your viewpoints.

[Paul Price]

Leave well enough alone. Any abrupt sub-nanosec switching of high currents at peak instantaneous voltage will create a small arcing and RFI and possibly some pitting of the relay contacts.

[T3sl4co1l]

Wow, I just have to say, that's an impressively badly rendered PDF!  Who knows how that happened...

Inductance: it comes in many scales, from small to large.  You don't need a large motor to still have a spark -- though you do get a bigger spark in that case.

If the contacts open while the bridge is drawing current, that current is suddenly stopped.  But current cannot stop any faster than the speed of light.  The "stop" event propagates up the wires, as a wave, back to the source.  It takes less than a meter of wire for this to be noticeable (that is, the spark only takes a couple of nanoseconds to jump -- quite fast indeed!).  Whether this is enough to upset nearby devices, depends on a lot of geometry (how the wires are arranged, how close a device is, etc.), but it is definitely a real effect!

The snubber provides a current path, so that the wave is absorbed at the source (the contacts), rather than bouncing around the wires.  Alternately: the resistor terminates the wires (which act as a transmission line), absorbing the wave.

Tim

[XaviPacheco]

Yeah, it's a badly rendered PDF. It's from a treadmill controller that I found while searching on the web. I wish I could find it in better quality.

[XaviPacheco]

Can anyone help me understand how it works? I mean, for example, AC signal won't go to rectfier until relay contact is closed. When the relay contact is open, AC input charges snubber capacitor through resistor and can't go to rectifier, is that right? Or does it go to rectifier for a brief period of time until capacitor charges?

[apis]

You are overthinking it. You only have to consider the relay switch and the snubber network (R1 and C1). It's like Tim said. There is some inductance in every wire. Imagine the contact is closed and there is a current flowing through it. The inductance in the wires prevents the current from stopping instantaneously when the switch opens. When the switch opens current cause charge to build up at the terminals and eventually the potential can become large enough that a spark forms across the gap which lets the excess charge flow across and equalise the potential difference. But sparks cause noise and wear down the contacts so they are undesirable. Therefore a snubber network is added across the contact so that the current can take the path through R1 and C1 instead of creating an arc across the switch.

I find it helps to think about it in terms of the hydraulic analogy. I.e. imagine a pipe (wire) with water (electrons) and a valve (switch) in the middle. If there is a current flowing through the pipe when you suddenly slam the valve shut, the waters inertia (inductance) will cause a pressure buildup (voltage) across the valve. That pressure could be high enough to damage the valve.

To prevent that add a second pipe in parallel across the valve that is blocked with a flexible membrane (capacitor). The current from the "inertial kick" when you close the valve can now pass through the second pipe causing the membrane to flex a little bit. That means the pressure buildup across the valve is much lower.

[XaviPacheco]

I get your idea. I just was imagining when the contact is initially open and then closes, not viceversa. The idea is when I power up the board, the contact closes and feeds the rectifier. So I wanted to know what exactly happens in the RC network before the contact closes, o what happens everytime the contact is open for a long time even though AC input is present. Of course, the rectifier won't be fed if the contact is open, but what is happening in the RC network at that moment. As yu say, I might be overthinking it, but it something that confuses me.

[apis]

To AC a capacitor looks like a resistance whose value depends on the AC frequency (usually called a reactance or impedance in that case, to separate it from "normal" resistance).

So for a given frequency, the RC network just looks like a resistance in parallel with the switch. The values of R and C are probably chosen such that the resistance is high enough that the current across it during normal operation can be ignored.

I.e. you can think of RC as a high value resistor.

[T3sl4co1l]

Note that the snubber allows some current to pass -- this may not be suitable for all loads.  Example, LED lights will flash when the switch has leakage like this, because the leakage charges their capacitors, which eventually get up to operating voltage, turns on, flash, uses up the stored energy, repeat.

The snubber doesn't do anything for turn-on, because the contacts are a much lower resistance, they are in parallel, and the lower resistance dominates.

The snubber only works for turn-off, because the contacts go high impedance, in parallel with the snubber, and the lower impedance dominates: the snubber.  So the snubber should be dimensioned for the open-circuit condition, which has to do with the wire length and/or inductance around that circuit.

Ideally, you'd use the inverse of this network: an inductor in parallel with a resistor.  This is connected in series with the contacts, to snub the turn-on condition as well.  This is not commonly done, unfortunately.

Tim