I want to find a way to keep a relay circuit on for at least 50 milliseconds after power is disconnected from it. If I connected a large capacitor across the rails of the relay, in series w/ small resistor, would the cap discharging be able to hold the the relay on?

Thanks in advance for any replies, I appreciate any info.


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Yes, this will work, but (depending from relay coil) maybe a free-wheel (or snubber, antiparalel) diode make enough delay.

 

How about a cheap resistor instead, (series w/cap) to delay the discharge time.

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Sorry, I guess you already answered my question. Since I have a lot of capacitors, not many diodes, I'll probably go with a capacitor in series with a resistor. Thanks, Sebi.


Circuit is for a soft start. A double pole double throw switch is used. When at 1st position, simple resister keeps surge current down, when when ready light lights, operator can power up circuit by putting switch from 'warm-up' to 'on'.

I wanted to keep the initial/first warm-up switch position intact for a while, when switching it to the final on position. I figure, with a relay in parallel with the switch, when I flip it from warm-up to on, if the relay stays on for 50 - 100 milliseconds, it will give time for the final switch to de-bounce and take over, and the circuit will not see any momentary voltage drop, as would be the case without the relay. Although the momentary drop would only probably be less than 10 milliseconds, I am more comfortable to leave the warmup resister in parallel until totally on.

Have not used relays before now. Hopefully it will work out as planned.

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hi sebi,

can you give me the formula to compute the discharge time of a capacitor over a given resistance value. i have seen a formula given here in the forum, to guy who was asking how to discharge.

the formula he gave was something like this;
Discharge time = 5*80K*C

is this the correct interpretation: 5 x 80K x C

where did the figure 5 come from or how was it arrive.

thanx :?

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RC is a time constant... 5RC is when the capacitor is fully charged...

take a look at
http://www.tpub.com/neets/book2/3d.htm

 

Use a 1N4001 diode instead of a resistor. Otherwise, much of your discharge current may flow back through the resistor, instead of through the relay coil where you need it.

 

hello,

hey EXO thanx for the site and your tip on the diode to be used instead of a resistor. this set up has a lot of applications. and also thanx to the guy who originally posted this problem.

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Thanks for all the info.

Mozikluv, you prob. know this already, but to elaborate a little on time constants for others who read the forum, the formula is T=RC where T is 1 time constant, R is the resistance in series with the capacitor, and C is the capacitance of the cap in question.

Also, a capacitor is considered fully charged/discharged after 5 time constants. The rise or fall of Vc (cap voltage) changes by about 63% in one time constant.

The significant time constant factors are:

.2 -- 20%
.5 -- 40%
.7 -- 50%
1 --- 63% (63.2)
2 --- 86% (86.6)
3 --- 96%
4 --- 98%
5 --- 99%

There is a very slight decimal deviation of these percentages, but these the workable, and the ones I was given at school. (ref Txt: Malvino, Grob)

Ron, I think I probably want to use a capacitor discharging to hold the relay open, since I intend to use a capacitor to help filter the diode feeding the relay, I'm thinking the diode will block the cap(s), and they will have to discharge through the relay coil.

To better explain, I drew the circuit, and it is posted Here:

http://www.geocities.com/hamfiles/Dualregulator.htm


I think this will work, but I have to wait a couple of days to buy a relay to try it out. Basically, I wanted a cheap, simple way to soft-start my supply, and I didn't want to have to use another transformer to power a relay, or a couple of Triacs on the dc side. I know I will need some high wattage resistors for R surge, and R relay, but it will be worth not having to use another transformer. The circuit is almost finished. I also might add a Green LED to light up within about 4-5 seconds, when the capacitors are charged after warm-up, kind of a "Ready" light. I don't really care about delaying the relay turning on, I just want it to stay on to short switch 'B' for longer than the few microseconds when switching from B to C, so that the circuit will not turn off. I know that a simple fix would be a custom switch that leaves "B" on when switching to "C", but I think it is out of my price range.

Also, I'm thinking of using position 'A' on the switch in series w/ some resistors to short the rails after the rectifier, and after the regulators, to discharge all of the caps once the power is off.

Sorry about writing a book here, I very much appreciate all of your info and feedback.

Thanks.

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