Relay contact protection

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Hello,
My first post on this forum.

I've recently built a unit for remotely switching on my Rotel power amp using a power relay.
I added an RC spark suppressor across the relay contacts to protect them.
The suppressor is a combined 100nF capacitor in series with a 100ohm resistance rated at 250VAC.

The problem is that it with the suppressor in place, when the relay is off the amp buzzes so assume that there is some leakage current.

I looked at the schematics for the amp and it appears to have the same kind of suppressor across the power switch so I don't understand why I'm having problems.

Any help would be appreciated.

Thanks in advance
 
The caps reactance does allow a small current to flow all the time, effectively partially powering the amp. Is the cap that is internal to the Rotel a 100nf or smaller ? (just for curiosity)

What can you do about it... not much because nothing is faulty as such. You could try a 10nf rather than a 100nf cap for the relay suppressor if such a thing is available as an all in one suppressor including the resistor. If not a Class X 10nf cap (it must be class X) can be used with a suitable resistor (say 1 watt carbon or metal film).

If the relay is adequately specified (15 or 30 amp rated) then I'm pretty sure you could use it without any suppressor. If its a miniature type that is only just suitable then you could have issues long term with contacts burning)
 
I haven't opened the amp to look at the suppressor. The circuit diagram and parts list just give a reference DE7150F472M but I can't find any specs. It appears to be a Murata spark suppressor.

I was indeed thinking of trying a lower cap value using separate components so I'll take your advice.

Also I saw some circuits that put the suppressor across the load rather than the contacts but would that be really effective?

The amp is rated at 800W max power consumption and the relay contacts are 10A.
 
Just looking at two typical amps now, a Rotel and a Pioneer rated at 150wrms. The Rotel uses a 4700pF (so that's 0.0047uF) and the Pioneer a 0.01uF (10nF). No series resistors.

Across the contacts is the correct location for spark suppression. Across the load is really more line filtering.
 
Class X2 are fine and across the switch contacts is the time honoured place for this when you look at typical service manuals for audio gear. I don't doubt what Dave says is also correct... probably pros and cons to eah approach.
 
"I still can't figure out why putting them across the transformer primary would reduce sparking across the the relay contacts."

What is storing the energy to maintain the arc across the contacts? Obviously the total loop inductance, dominated by the Xformer.

So a cap across the I will kill the dv!

AFAIK
 
I just ordered some X2 0.01uF caps which is the lowest value I could find.

I still can't figure out why putting them across the transformer primary would reduce sparking acoss the the relay contacts.
it's back emf.
When you turn off a load, the inductive part of the load tries to maintain the current flow and generates a back emf in an opposite direction to the ON voltage.
It's the strength of that back emf that "jumps" the gap of the opening contacts.
look at any switch at turn off in the dark and you will see the spark !
A light switch is especially good at this.
Closing a switch rarely creates a big spark and rarely damages the switch contacts. It's the opening that is the damaging problem.

The snubber is a method of reducing the back emf. Make the back emf small enough and the spark does not "jump" the gap.
Making the back emf small enough is achieved by establishing a "path" for the back emf current to flow. If the current is allowed to flow there is no back emf, because the current has not been stopped (instantly).

This is why a relay coil gets a back emf diode.
 
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Hi,
You can use a Metal Oxide Varistor (MOV) in parallel with the transformer. If you used it make sure you have a fuse in the line. Also it will protect the equipment from lightning. I used it in all my equipment for surge and lightning protection.
 
Mooly said:
You do indeed see a spark from switching off a resistive load. Back emf though ?
The spark is created by back emf from the inductance of the wiring. Current through an inductance stores energy in the magnetic field. When the circuit is broken this energy has to go somewhere. The normal route is for the magnetic (kinetic) energy to turn into electrostatic (potential) energy across the capacitance of the switch gap. If there is no breakdown then you just have an LC oscillator, damped by some combination of radiation and resistance. In most cases the switch capacitance is quite small so a very high voltage develops. Air breaks down and some current continues to flow. The process continues until the energy has been dissipated.

Add a capacitor in the circuit and the voltage developed is much smaller (it varies as the inverse square root of C). Also the oscillation frequency is much lower, so most of the energy cannot be lost by radiation but by resistive heating in the circuit. In many cases a transformer wire resistance is plenty enough to do this. If not the C can have a small R added in series with it, but too large an R will add to the voltage at the gap as the full circuit current passes through it at the point of switch off.
 
That's interesting, and its just the wiring inductance doing that ? or wiring plus distribution tranny at the other end.

(That got me thinking... the spark you get from a thermostat on a 3.4kWh storage heater (so a resistive load) is pretty spectacular and the stats on some models do have limited life due to fairly rapid cycling when up to temperature. Those high temperatures (perhaps upwards of 150 to 200C on the stat would preclude the use of a suppressor but one could be hard wired and located at the bottom of the heater)
 
Lets do a 'back-of-envelope' calculation to see what spark voltage you might get across a mains switch. Assume 1A of current at the point of breaking, and fed from 10m of cable.

Wild guess no. 1: the cable has an RF impedance Z somewhere around 100ohms.
Wild guess no. 2: the propagation speed v is 2/3 of c i.e. v= 2x 10^8m/s.

Then we can calculate the cable inductance as Z/v = 0.5uH/m. Cable capacitance is 1/(Zv) = 50pF/m. So we have L=5uH and C=500pF. Note that I am doing some serious 'hand waving' here, as I am equating distributed reactance and lumped reactance - but I did say it was just 'back-of-envelope'!

Stored energy = L I^2 /2 = C V^2 /2, so V = I sqrt(L/C). Well fancy that - we have got back to the normal expression for a transmission line impedance. So our 1A gives us 100V peak. In reality even a resistive load will have some inductance, and the capacitance isn't all at the switch but distributed so the back emf voltage will be higher than I have calculated here.
 
You do indeed see a spark from switching off a resistive load. Back emf though ?
Not necessarily: even in a purely resistive circuit, a spark may appear (an arc in fact): the contacts do not separate instantly, and just before the separation actually begins, the current is concentrated on a tiny area. This causes a localized heating, and when the contact actually begins to separate, the gap is initially very small, and this, aided by the heating, initiates an arc that will be sustained until the separation distance is too large for a given current.

Note that a mains power cord is not actually an inductance: it is a transmission line, with a rather lowish characteristic impedance, like 50 ohm or so. When the current is interrupted, an emf will appear, but it won't be a (theoretically) unlimited one: it will be something like Iload*Zc
 
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