Quote: Originally Posted by
dupa2 
cheap....
http://www.newark.com/jsp/displayPro...K=821TD10H-UNI
ok, if you say so. I mean compared to your usual industrial time-delay relays, sure it's cheap. but compared to a cap, a resistor and some elbow grease, it's a lot.
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In order to make a delay circuit, you will need to know your relay's dropout voltage, and coil side resistance.
The relay-coil side of your circuit is much like the example circuit in
this tutorial, with the coil of the relay replacing the resistor. The idea is that with supplied power connected, the cap will charge to supplied voltage, and will stay there until voltage supplied lowers below the voltage supplied by the cap. Once this happens, tthe capacitor will discharge, but it needs a path for current to flow in order to do so. that path in your circuit is through the coil of the relay. The time that it takes to discharge is inversely proportional to the current that can flow I=V/R. if it discharges at a lower current it will last longer. But keep in mind, as it discharges it's voltage drops. The result is that the voltage decays at a non-linear rate. The rate can be found by multiplying the series resistance (coil) times the capacitance of the cap (uF). The result is time in seconds and is called the time constant. Time constant=R*C. For our purposes it can be assumed that it takes 5 time constants for a cap to discharge completely, once power is removed.
Each time constant lowers voltage by about 63 percent of the previous value.
if coil resistance is 200 ohms and dropout voltage is 1.2V and is at 13.8Volts and capacitance is (2 4700uF capacitors in parallel) 9.4mF.
RC= (.0094)(200) = 1.88 seconds
this is the time that it takes for the capacitor to discharge to 37% of the previous voltage which was 13.8. V=13.8*.37=5.106
If you know the voltage you want it to decay to and want to know how long it will take to decay to that value, use this eqn. V(t)=V
0e^-t/RC.
t= (RC(-ln(v(t)/V
0))
I'm attaching a excel file that you can use to find an appropriate capacitor or combination of capacitors in parallel to do the job.
There are also many calculators online for this.
If you find you want to change the total resistance to influence RC, than maybe
add a small resistor in series with the relay, but keep in mind, this will lower the voltage across the coil. Be sure to not reduce voltage so low as to cause the coil not to pick up. Pickup voltage is usually much higher than dropout voltage.
Just click the picture if you want it.
this is designed after another similar circuit a found online. the acc wire runs into a 1 amp fuse(F1) then a 1 amp rectifier diode(D1). then the wire splits and a 18v capacitor(C1) causes the delay depending on the uf and the other wire runs into the relay normally. I hope this makes sense. as I said i am pretty new to wiring things like this. So anyone who knows what there doing and is willing to comment on weather or not this will work I would be very appreciative to you. thanks.
So it seams like i can just put a capacitor before the relay and maybe a fuse before that for saftey and it will give me a delay. Now if i can get the delay I want without using a resistor is there any advantage to using one or am i just as good without it? Right now i view the circuit as:
Linear Mode
