Simple relay circuit, killing relays... not what i expected, please help...

[champ222]

Morning all, i need a little bit of help here.

I have this circuit here:

**broken link removed**

What i was trying to acheive was to have the output stay at 12V for a while after the switch was been turned off. also, another goal was for there to be no load on the supply when all was switched off, as this is for use in a car.

The idea was that the switch (actually going to be the remote wire in a car, but was a switch when i was testing this, the main 12v supply would be the permanent 12v in the car) powers the coil of relay 1 which switches the supply to the coil of relay 2 and the capacitor (i want to isolate the main supply from the charged capacitor, i only want the delayed switch off on the output, not the remote lead.

relay 2 is activated and the output goes to 12v

the cap is charged up.

when the switch (remote) switches off, relay 1 cuts the supply to the coil of relay 2 but the capacitor keeps it switched on while it discharges through the coil of relay 2 (approx 1k ohm)

then once the voltage across the capacitor reaches below the drop off voltage for relay 2, it switches off, and the output goes off.

does this make sense?

i tested this, and it seems to damage relay 1, and the relay switch terminals then becomes permanently closed.

I cant understand why this is, i cant see any high currents or voltages that could damage the relay.

the relays in question are reed relays.... see here:

MEDER|SIL12-1A72-71D|SIL REED RELAY 12V | CPC

the diodes you can see in the diagram are "built in" to the relays

any ideas why relay 1 keeps getting damaged? is it the charging of the capcitor thats drawing too much current for the switch contacts of relay 1?

Thanks guys, i'm really confused by this one. i'm sorry if i havent really made myself too clear.

Thanks in advance

 

[Nigel Goodwin]

The high current is caused by the capacitor, which is effectively a complete short when it's discharged. Add a current limiting resitor in series with the contacts of relay 1.

However, this is a VERY clumsy and unreliable way of trying to get a relay to 'hold' for a while.

 

[JimB]

When the capacitor is discharged it effectively presents a short circuit to the supply. When the contact of relay 1 closes a large current will flow through the relay contact to charge the capacitor, after a few milli-seconds the capacitor charging current will drop to zero.

I am sure your problem is due to this current surge.

To overcome the problem, either:

1 use a relay with larger contacts
or
2 put a resistor in series with the relay contact, I suggest a 47ohm resistor.
This will limit the charging current to 12/47 = 250mA, well within the rating of the relay contacts.

JimB

 

[Boncuk]

I guess the high cap charging current welds the relay contacts together.

Nigel had the right idea about reducing the charge current. Use a 33 to 39Ω resistor to limit charge current.

The relays can stand 1A of switched current. Using a 33Ω resistor will reduce the initial charge current to safe 363mA and the relay won't weld.

Boncuk

 

[champ222]

Thanks guys, that makes sense.

I realise that this is a very clumsy way of creating a delay, but a clumsy delay is all i require. it doesnt matter how long the delay is really, i just need SOME delay.

So like this then:

**broken link removed**

Thanks for your help guys

 

[champ222]

another question.... the output of this circuit switches on another circuit, and the first thing in that circuit is a capcitor to ground, so i guess i'll have the same problem again?

its actually this circuit

**broken link removed**

if you look at the power supply bit at the bottom of the first diagram, i was going to connect the power IN to the left side, effectively across C8 and C4 (dont know why there are two C8s in this diagram) then i connect the 12v tag in the diagram to the +VS of the opamps.

does this make sense? is there any point in doing this? ie would i get the same result, ie smoothed and filtered 12v to the opamps +VS, if i connect the input to the 12v tag as well? would this be a better way to do it?

which ever side (left or right side of the 12v rail) i connect the output of my relay to, i guess i need to add a 47ohm resistor to stop welding relay 2?

i am assuming that C8 and C4, and the "other C8" and C7 are there to help smooth out and reduce noise from the 12v supply to the opamps?

what would be the best way about going about this?

I hope that makes sense?

thanks again

Andy

 

[BrownOut]

You could try moving the 10 ohm resistor in the amplifier diagram before the input capacitors. Otherwise, you might be OK connecting this circuit with your relays. The input caps are smaller than the one that was causing your a problem.

 

[champ222]

thats true, the input caps are smaller there.

what effect would it have if i moved the 10ohm resistor to the input, and increased the value of it to 47ohms? would that affect the 12v output going to the opamps?

regarding the suggestion above for the series resistor for the relay, would a standard 47ohm 0.6w resistor be ok?? because if the capacitor acts as a short, there would be 12v across it, and 250mA through it, doesnt that equate to 3 Watts? although it would only be for a short time until the capacitor charges up?

the reason for the tiny relays is because i'm pretty short on space.

Thanks again guys