Relay power on- power off 1 second

Thread Starter

Jovyevane

Joined Apr 6, 2018
6
Hello everyone. I tried looking for a similar project, but I couldn't find him. I have an energized circuit with 12V dc. With a switch I open and close my power. I would like to energize a relay for about 1 second when I open the switch and when I close it. My relay is 12vdc.
Can anyone help me with a very simple scheme? Thank you very much
 

Thread Starter

Jovyevane

Joined Apr 6, 2018
6
Thanks for your response Dana. A timer relay is fine when I open the circuit, but when I close it of course the relay is no longer powered. I think I need a configuration with a capacitor that functions like a battery. I have a decent experience in the manufacture of PCBs. You can achieve this effect by using bjt?
 

AnalogKid

Joined Aug 1, 2013
8,678
What is the resistance of the relay coil? With that you can calculate the capacitor size. Warning, it might be pretty large. For example:

Relay coil resistance - 120 ohms
Nominal relay coil current - 100 mA
Dropout voltage - 6 V

Approx capacitor size: 12,500 uF / 25 V

ak
 

danadak

Joined Mar 10, 2018
4,057
You could always use a SPDT break before make switch to generate
triggers to a 555 timer to generate the 1 sec on time and use the 555
timer to drive the relay.

Regards, Dana.
 

Thread Starter

Jovyevane

Joined Apr 6, 2018
6
What is the resistance of the relay coil? With that you can calculate the capacitor size. Warning, it might be pretty large. For example:

Relay coil resistance - 120 ohms
Nominal relay coil current - 100 mA
Dropout voltage - 6 V

Approx capacitor size: 12,500 uF / 25 V

ak
Thank you.The resistance value is 230 ohm
 

eetech00

Joined Jun 8, 2013
2,034
HI

Here's a optional circuit with a 555 Timer.
I show a pushbutton but its just for the simulation and would actually be a switch contact.
Probably could use a digital Lead/trail edge detector for the input but I didn't want to add a chip(s).

Relay1SecOnOffTriggerTimer.png
 

AnalogKid

Joined Aug 1, 2013
8,678
The resistance value is 230 ohm
Does the datasheet for the relay say what its dropout voltage is?
I'm not sure, but it seems 6v
For this discussion, note that the capacitor charges from 0 V to 12 V through the relay coil when the switch is closed, and discharges through the coil from 12 v to 0 V when the switch is opened. I'll discuss everything in terms of charging, but these are symmetrical operations so everything applies equally to the discharge cycle.

A first order approximation is that the capacitor voltage increases linearly rather than exponentially as the capacitor charges. We are interested only in the first 50% (12 V to 6 V) of the charge curve, and that is the more linear portion. To make my life easier, lets round up 230 ohms to 240 ohms (less than 5% change).
Start: 12 V, 24o ohms >> i = 50 mA >> relay pulls in
End: 6 V, 240 ohms >> i = 25 mA >> relay drops out
Therefore, the average capacitor current is 37.5 mA
Charge voltage = 12 V - 6 V = 6 V

Constant current capacitor equation:

EC = it
charge voltage change x capacitance = charge current x time
6 x C = 0.0375 x 1 (sec)
C = 0.0375 / 6 = 0.00625

C = 6800 uF - ish

This should be pretty close. 6800 is over 8% greater than 6250, which offsets the 5% adjustment we made to the coil resistance.

ak
 
Last edited:
Top