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Off - On - Off timed circuit.

dezweb52

Aug 28, 2013
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Hello!
I'm hope you guys can help me. I'm a bit stuck with a circuit that I am building. Sorry about the title of the thread, but I'm not sure how else to describe it...

I'm in the middle of building a circuit that will switch a relay on for a short period of time, and then switch it off. However, the switch is going to be held in the same state UNLIKE a hand dryer circuit. The circuit should do the following:

Switch toggles, relay switches and after a short period (1 -2 seconds), turns off.

Switch toggles, relay switches and after a short period, turns off.

I have come up with a circuit and build it in Croc-Clips (Yenka) but it doesn't seem to switch the relay. I cant seem to figure out what is wrong. I guess I'm loosing current somewhere, but can't seem to find out where....

I hope you guys can help, I know its something silly that I have missed, but can't seem to find it!

Thanks in advance!
 

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Korvette88

Aug 28, 2013
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Aug 28, 2013
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Hi
I was looking at your circuit and it looks like when the DPDT switch is in position 1 it activates the relay, and then when the capacitor is charged the relay is deactivated. When you switch to position 2, The first capacitor discharges to itself, and the second relay activates until the second capacitor is charges.

What if you put an LED on the discharge cycle of each capacitor? That way you know its discharging and that the discharge its being used, instead of recharging itself.
Ben
 

dezweb52

Aug 28, 2013
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Aug 28, 2013
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Hi,
The capacitor is being discharged, however I don't think the current is high enough for the relay to switch before it becomes fully charged. If I put an LED into the circuit instead of a relay, I get a brief flash. I just can't work out where I am losing the current....
 

duke37

Jan 9, 2011
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You have a 1k resistance in series with the relay coil. This will limit the current.
What voltage and current is needed by the relay?
 

dezweb52

Aug 28, 2013
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You have a 1k resistance in series with the relay coil. This will limit the current.
What voltage and current is needed by the relay?

Ah, that might be it. The relay is 6V, dont know about the current. Yenka is slightly too dumb to give me that information....

The reasoning for the resistor is to limit the 'flow' rate, or how long it takes to fill the capacitor up.

This is a small personal project, I'm use to using a small microchip for this but I thought I would give it a go with something like this. Although my knowledge of electronics is rather limited, and you know what they say "A little knowledge is sometimes very dangerous!"
 

KrisBlueNZ

Sadly passed away in 2015
Nov 28, 2011
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Start by deleting the resistors, and the series diodes too. The diodes are not needed and will just waste voltage.

You should connect a diode across each relay coil to suppress the back EMF that is generated when the current flow is interrupted. Connect them with the anode to the negative side (the side that returns to the battery negative). This will protect the switch against possible arcing if the circuit is broken before the capacitor is fully charged.

You might want to put a small resistor in the discharge path to protect the switch contacts against damage from sparking. A 6.8 ohm resistor will limit the current to about 1A but will still discharge the capacitor quickly. Use a 1 watt resistor; the instantaneous dissipation will be about 6W but only for a very short time.

The duration of the relay closure pulse depends mostly on the capacitance of the capacitor and the resistance of the relay coil. A typical 6V relay coil has a resistance of 200 ohms. Sensitive relays are available with 720 ohm coils. I'll assume you can get a relay with a 720 ohm coil.

The formula for the time constant of an R-C circuit is t = R C where t is measured in seconds, R is resistance in ohms, and C is capacitance in farads. This gives the amount of time taken for the voltage across the relay coil to fall to 37% of its initial voltage. This is VERY ROUGHLY the voltage at which the relay will drop out.

With those values, R=720 and C=0.00047, the calculated time is 0.34 seconds. So your relay will pull in for about 1/3 of a second. If you want a one second pulse, you'll need to increase the capacitor to about 1500 uF. The exact value you need depends on the characteristics of the relay - that is, what voltage it drops out at, so I suggest you get several capacitors in the range 470~4700 uF and try them out.

BTW, relay coils are not simply resistive; they have inductance, but in this application the inductive reactance is low, because the frequency is low, so they can be approximated as a simple resistance.
 

dezweb52

Aug 28, 2013
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Aug 28, 2013
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Hello.
Thanks for your help! It makes a lot of sense now that you explained it. The diode was there to stop the back EMF from the relay, of course, I put it in the wrong place.....DOH!

I have made the modifications as per your recommendations and its working! :D I'm going to build up the circuit when I can source the parts. I'm guessing if the resistance across the coil is 220 ohms, I can make it up to 720 ohms with a resister in series with it?

I have posted the image here so that if someone else comes looking, hopefully it will be of help to them.

What do you call this type of circuit? Time delayed isn't right as that means it takes a while to turn off after turning on (according to what I found on google anyway!)

Thanks again! So chuffed that its working (in the sim) :D
 

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KrisBlueNZ

Sadly passed away in 2015
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That's looking good.

No, you can't add a resistor in series with the coil. If you do that, the resistor and the coil resistance will form a voltage divider, and you will only get a fraction of the total voltage across the coil, so the relay will not pull in. Make sure you get a relay with a 720 ohm coil. These may be described as "sensitive" relays.

I guess I would call the circuit a "pulse generator" of some kind.

Re your schematic. You should add a "1W" marking to the resistors; the convention is that if a resistor needs to be rated for significant power, e.g. more than around 1/4W or 1/3W, the wattage should be indicated on the diagram.

Also, the capacitor marking "1.5 mF" could be confusing because capacitors are not normally marked in millifarads; people might think that "mF" means "microfarads". Strictly speaking you're totally right, but it's better to follow convention.

You may need to adjust the capacitor values to get the desired pulse durations when you are testing with actual relays.
 
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