0.5 second relay

Thread Starter

orwikcons

Joined May 2, 2008
22
I need to make a device, something like relay, but which will, when plugged to +12V, connect input and output, but for only 0.5 seconds, and then disconnect, although it is all the time connected to +12 V



(My friend told me that he had seen some circuit about this on the Internet, but hadn't been able to find it...)
 
Last edited:

Wendy

Joined Mar 24, 2008
23,415
How precision does it have to be? We had a similar query about 2 weeks ago, and used something like this. If you need precision it will need to be done differently.
 

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Thread Starter

orwikcons

Joined May 2, 2008
22
I don't understand the scheme (I am not electric expert...:rolleyes:). It is not important to be exactly 0.5 seconds, but to be under 1 second. So when I put the switch on, it has to give an impulse.
 

SgtWookie

Joined Jul 17, 2007
22,230
Bill,
That's almost the circuit, but it must have a bleeder resistor across the capacitor. Otherwise, the cap will charge once, but won't discharge - unless the energizing source is a SPDT switch that alternates between Vcc and ground.

orwikcons,
We need to know what the resistance is of the relay's coil in order to calculate the RC time.
 

mik3

Joined Feb 4, 2008
4,843
I don't understand the scheme (I am not electric expert...:rolleyes:). It is not important to be exactly 0.5 seconds, but to be under 1 second. So when I put the switch on, it has to give an impulse.
The capacitor will work fine. Also, use a 5K resistor in parallel with the capacitor to discharge it when the power is off to be sure the next time you will turn it on it will give you the impulse. Play with some capacitors to find the one appropriate for you if you dont know to calculate it.
 

SgtWookie

Joined Jul 17, 2007
22,230
See the attached.

Times are approximate. Relays make and break at different voltages, and have different resistances in their coil windings.

I tried similar to the attached with a relay that had a 400 Ohm coil with 1000uF and 2000uF caps, the latter worked fine for what you want to do. You should use an electrolytic capacitor with at least a 25v rating; somewhat higher is better.
 

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Thread Starter

orwikcons

Joined May 2, 2008
22
I've got the time of 1 second with 1000 uF and 5.1 k Ohms. I cannot tell you the coil impendance now. If I put the capacitor 500 uF, to short the time, do I have to reduce od enhace the resistor impendace?
 

SgtWookie

Joined Jul 17, 2007
22,230
Your relay must have a pretty high-resistance coil, probably around 2k Ohms.

Try doubling the resistance to 10K Ohms first. That should decrease your relay ON-time by a fair bit. If it is not enough, then reduce the size of the capacitor. 470uF is a standard value. If you happen to have two of the 1,000uF capacitors, you could use the two of them in series to get 500uF.
 
Last edited:

Wendy

Joined Mar 24, 2008
23,415
Bill,
That's almost the circuit, but it must have a bleeder resistor across the capacitor. Otherwise, the cap will charge once, but won't discharge - unless the energizing source is a SPDT switch that alternates between Vcc and ground.

orwikcons,
We need to know what the resistance is of the relay's coil in order to calculate the RC time.
I hear you, but if the cap/coil and the circuit are on the same side of the on/off switch then the circuit will provide the bleedoff, which is what I was thinking in both cases.
 

SgtWookie

Joined Jul 17, 2007
22,230
I hear you, but if the cap/coil and the circuit are on the same side of the on/off switch then the circuit will provide the bleedoff, which is what I was thinking in both cases.
I'm afraid I'm not quite following you.

A typical on/off switch has basically two modes; connect the load to the source of power (basically, very low resistance) and high impedance (open circuit). Therefore, the capacitor would have a charge path, but no discharge path except via internal leakage.

If the on/off switch were SPDT, and had the output on the common, the power source on one input and ground on the other, that would take care of charging/discharging the capacitor. Of course, the surge current on the ground side would be mighty high with nothing to limit the current flow.
 

Wendy

Joined Mar 24, 2008
23,415
If the coil/cap were a permanent part of the circuit the rest of the circuit would discharge the cap when power was disconnected. You need a bleedoff resistor if it were floating by itself, on the other side of the power cutoff switch.
 

Wendy

Joined Mar 24, 2008
23,415
Yeah, and we really don't know the application on this one. If it is anything like a CMOS design the bleedoff is a must.
 

Thread Starter

orwikcons

Joined May 2, 2008
22
Thank you for advices. I will set the time by changing the capacities and resistances. But there is another problem. When I switch the circuit on, it does its work, and then I switch it off and again on, it will not give an impulse, probably because the cap didn't have enough time to discharge, if I wait a few seconds, and then switch it on, then it works again. How can we discarge it fast? If I put the LED parallel with resistor and capacitor?
 

Wendy

Joined Mar 24, 2008
23,415
Depends on your application. If this is the norm, then you may have to up the complexity of what you need and include a timer, which will do what you're wanting

What kind of circuit do you need to pulse? It appears to be some kind of reset.
 

Wendy

Joined Mar 24, 2008
23,415
I was thinking about it the shower, the only thing you can do before going to somethng that much more of a PITA is drop the resistance to the absolute minimum that you can get by with. It might be worth using a variable resistor across the cap. and turning it down until the relay engages, then backing it off until the relay disengages. It that doesn't work it gets harder, not too hard, but harder.
 
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