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Help with a chattering relay!

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Alpina540

New Member
My cars interior lights dim on and off when you lock or unlock the doors.

I have a relay tapped into the interior lights to control a fog light.

When the interior lights dim to go off the voltage drops thus the relay chatters.

How can I make the relay stop chattering?

should I add a cap? where? :confused:

Thanks for your help!:)

-Anson
 

PatM

Member
My cars interior lights dim on and off when you lock or unlock the doors.

I have a relay tapped into the interior lights to control a fog light.

When the interior lights dim to go off the voltage drops thus the relay chatters.

How can I make the relay stop chattering?

should I add a cap? where? :confused:

Thanks for your help!:)

-Anson
You should tap into another circuit that doesn't change voltage.
Why not tap into the cigar lighter circuit?
That would have sufficient current capacity and would not vary.
PatM
 

BrownOut

Banned
A capacitor probably wouldn't be the best solution. First, I'd ask myself if there is a better way to control the fog lights, besides using a signal that doesn't drive it correctly. Why do you need to use this signal? If you just can't think of a better way, there might be a way to add hysterisis to your relay, using the self latching feature. Maybe someone has done this before???
 

Alpina540

New Member
You should tap into another circuit that doesn't change voltage.
Why not tap into the cigar lighter circuit?
That would have sufficient current capacity and would not vary.
PatM
See I am using the interior lights power circuit so i can turn my fog lights on and off via my remote key fob. So When I walk to the car at night I can unlock the car the interior lights come on then it triggers the relay and then the fogs come on.
 

BrownOut

Banned
Something like this might work. Work out the resistor values based on the current requirements of the relay. When the relay snaps in, part of the output is fed back to the coil to keep the contacts held in. When de-energized, that feedback is removed, so there is no chance it will re-energize on the decaying signal.
 

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Alpina540

New Member
So R1 goes to ground but what is the purpose of R1 ?

I am a bit of a noob...

wont this defeat the mechanical isolation of the relay?
 
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BrownOut

Banned
There is no ground. R1 goes to the dome light, where you have the relay connected now. R2 ensures that when the relay energized, it remains energized till you want to turn it off.

For example, say the relay needs 30 mA to energize, and coil resistance is 100 ohms. Further, assume you want to turn it on at 8V and off at 1V. As the voltage rises during trun on, it gets to 8V, and you want 30 mA at that point. Then select R1 = 8V/30mA - 100 ~= 170 ohms. While the voltage is falling during trun off, you want R2 to hold the relay till the input falls to 1V. You need 3V across the coil (3V = 100*30mA) So, you want for R2, 12 - 3/IR2, where IR2 is given by the sum of the currents thru R1 and the coil, ie IR2 = 3/170 + 30mA = 47mA. Thus, R2 = 190 ohms.

I've made a number of assumptions about your relay. This method may or may not work for the relay you're using.
 
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Alpina540

New Member
Thanks for all the info let me chew on that and digest it then I will ask my remaining questions once I compose them.

Thanks Brown Out
 

BrownOut

Banned
PS If you decide to try this, the realy might still chatter a little, as the contacts have a little inertia, and tend to 'bounce' a little. You might still need a capacitor to smooth out those effects. Use a capacitor across the relay coil, and an bigger one ( 10X ) across R2. The values will depend on your relay, and you might have to experiment a little to get it right. This solution is going to be tricky, but it's the simplist one I can think of.
 
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Alpina540

New Member
Ok, so I may not have told you the whole story.... I have a few more components in the circuit.

Attached is a sketch of the complete circuit.

My relay coil has 72.3 ohms of resistance. and its a 30 amp model 12v of course.

I would want the light to turn on at 10 volts and off at around 9 volts if possible.

What is in red is what you say I need to add.

I have two SPST switches to activate and deactivate the steady on and interior feed.

Does this give you more to go on?

Thanks

-Anson
 

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BrownOut

Banned
Now you only need to know the minimum current to energize the contacts. Also, the diode should be moved so that it's across the coil, the marked end connected to the ground lead of the coil. That will help the chattering a little.
 

Alpina540

New Member
Well I just tested low voltage relay activation and it activated at just 5.7 volts.

I don't have a current measuring multimeter so I can't get the current measurement.
Could you give me an educated guess?

I have a confession..... I actually don't even have the diode in the circuit yet it needs to be there so that when I have the solid on switch thrown I don't turn on my interior lights.
 

BrownOut

Banned
5.7V/72 ohms = 80mA. Yeah, the circuit won't actually work with the diode as drawn. You already have a switch to disconnect your interior light when your solid on switch in on. Also, you want to break the connection to R2 when the other relay is energized. Using a double-pole, double-throw relay will work for that.
 
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Alpina540

New Member
Yeah thats what I was saying earlier about the R2 defeating the relays mechanical isolation. If I don't use a DPDT relay the interior lights will be energized by the parking lights.

Sorry I forgot to use the simple ohms law 11th commandment to figure 80ma...
 

BrownOut

Banned
I understand a diode only allows current to flow one way but why is it necessary across the relay coil?
It will quash inductive kickback from the relay coil, and also add a little capacitance to help the chattering. Quashing the kickback will also help the chattering.
 

BrownOut

Banned
Just use any of the normal rectifier diodes. 1N4002 - 1N4007 will do.
 
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