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Time delay for cheap.

boyd

Oct 7, 2016
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I am using a 3-30vdc relay to start a 230vac machine by using a 12vdc supply directly from a power supply connected into another machine. There are also other operations going on at the same time but this is just one in the circuit.

I wish to hold the relay closed for approx 30 seconds for run-down time after I remove the 12vdc. The relay is solid state so will close contacts at >1vdc and operates using milliamps.

Any ideas for something very easy, simple and cheap? I was wondering if a Capacitor in series with a zener would run a solid state relay for 30 seconds or so. I am trying to avoid buying timer delay relays and wish to keep it cheap and simple simple. Or would a resistor slow down the discharge of a cap? Or.....can you buy slow release capacitors?

Very keen to hear thoughts from members here please.

Regards Boyd
 

(*steve*)

¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
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Jan 21, 2010
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One on the problems with just using a capacitor is that you need to charge it when power is applied and this may require significant current.

Is power completely removed?

Could you use the 230vac that is switched to power (say) a small USB charger to provide power to hold the SSR on?
 

AnalogKid

Jun 10, 2015
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10,000 uF / 25 V

eC=it
voltage drop times capacitance equals current times time
12 V supply - 3 V min operating voltage = 9
C = holdup capacitor in farads
3 mA operating current = 0.003
30 seconds = 30

9 x C = .003 x 30
C = 0.01 F = 10,000 uF

ak
 

boyd

Oct 7, 2016
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Hi. Great thanks. I don't have a lot of understanding about capacitors, how they discharge, and their capacity, so appreciate your reply. I am on the correct track then.
 

boyd

Oct 7, 2016
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Hi. I am just thinking about this. Do I need to install a resistor or zener so the cap doesn't discharge all at once rapidly? I did read your calculations and will study them more. But I am just checking the the cap wont just discharge as fast as the relay will accept the current which may event in a straight couple of seconds and then nothing. Do solid state relays have their own internal resistance?

This relay I am using is 1-30vdc. If I was to put any of the operating voltages on it, would it still regulate on its own the current required to operate or just soak up what ever is put on it? Probably a simple question but I am not familiar with using caps for a time delay.

Regards Boyd
 

(*steve*)

¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
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Jan 21, 2010
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The cap on its own is probably ok because these devices are designed to connect to a voltage source.

My concern is the charging of the capacitor. What is the source of the 12V applied to the SSR and can it withstand what is effectively a brief short every time the delay is offered on? You may have to place a low value resistor in series to limit the maximum charge current.
 

boyd

Oct 7, 2016
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Hi. I am using a 12vdc 1.25A power supply. I would have thought this would be able to do the job? I will be running the machine for at least, a minimum for 60 seconds so if you recommended a current limiting resistor to charge the cap time is on my side? Or do you think I should slow down the current to the capacitor?
 

(*steve*)

¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
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I'd put a 100 ohm resistor in series with your power supply connection to the relay.

This will limit the charge current to about 125mA.
 

AnalogKid

Jun 10, 2015
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Depends on the solid state relay. Many that list a wide control voltage range have a small constant-current circuit in series with the optocoupler LED, so the input current is almost constant no matter what the input voltage. Best to test this theory with a series resistor, but if the datasheet does not say you need one (and it certainly would if needed), then you probably are ok.

How a capacitor discharges depends on what it is discharging into. If connected to a fixed resistance, the voltage of a discharging capacitor decreases exponentially with time. If discharging into a constant current load, then the voltage decreases linearly. This is the case I used for my example.

https://en.wikipedia.org/wiki/Capacitor#Current.E2.80.93voltage_relation

In the first equation, if the current is a constant then the integral equation reduces to the one in my post.

ak
 

boyd

Oct 7, 2016
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Far out. Aliexpress comes through every time. I bought three of these:

https://www.aliexpress.com/item/1pc.../1100514896.html?spm=2114.13010608.0.0.SMMiOt

With Aliexpress I never expect 100% functionality so I usually purchase an additional unit just to be sure I get full operation.

Gee. I have bought just about everything off Aliexpress for under US$200. I was quoted NZ$1200 just for the pneumatic cylinders......

I am liking this forum. My wife and daughters get that "glazed" look in their eyes when I start talking about this ha ha ha.
 

Bluejets

Oct 5, 2014
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Yep, amazing what one can get for a couple of dollars and most seem to work just fine. Many are delivered also with an all up cost of at least 1/3 of what it would cost just for postage here in Aus.
 
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