Can a capacitor hold open a relay ?

Discussion in 'The Projects Forum' started by RiG615, Nov 20, 2008.

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1. RiG615 Thread Starter Active Member

Nov 13, 2008
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can this capacitor(1000μf) hold open this relay?
and for how long? I just want a few seconds say 5

Would it need a resistor?

thnks

2. mik3 Senior Member

Feb 4, 2008
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63
If you are going to use a voltage higher than 12V to power the relay you will need a resistor in series with the relay coil to limit the current through it to a safe value and do not destroy the relay. If you will use a 12V supply connect the relay coil directly to it. In both cases, you will connect the capacitor in parallel with the relay as when the power is switched off the relay will stay energized for a few seconds. The time it will remain energized depends on the capacitors value, the resistance of the relays coil and the pull-out voltage of the relay.

If you measure the resistance of the coil with a multimeter then the time will be approximately equal to:

t=-RC*ln(V/Vm)

where

R=coil resistance
C=capacitance of capacitor
V=pull-out voltage
Vm=initial voltage across the capacitor (working voltage of the relay's coil)

3. RiG615 Thread Starter Active Member

Nov 13, 2008
50
0
Thanks mik3

For "C=capacitance of capacitor" is the units μf (microfarads) of F farads ??

also - i'm using a 9V battery

so if i add a resistor it would slow it down right?
whats the limit to the size of the resistor i could use??
i don't think the formula takes into account the cut off current

if i had a 50kohm resistor then the formula would=
(-50 000) * 0.001 * ln(1.2 / 9) = 100.745151

but the current would be=

9 / 5000 = 0.0018

4. Wendy Moderator

Mar 24, 2008
20,772
2,540
Don't think so, think in terms of a 1/10 of a second. 1 RC time constant = R C = 150Ω X 1000μF = .150 sec. I suspect you're going to have to build a timer connected to the relay. I didn't bother figuring a divider circuit. You could even use a transistor to boost the time, it doesn't need to be that fancy.

********************************

Here is something I sketched out. Never built it, but it should work. You may need to use a Darlington transistor to extend the time. R2 will adjust the voltage down where the relay won't fry, and R1 might not be needed at all.

A 555 version will be much more predictable, with a slightly higher parts count. I'd be glad to sketch that up if you want.

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Last edited: Nov 21, 2008

Feb 4, 2008
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6. Wendy Moderator

Mar 24, 2008
20,772
2,540
The problem is the length of time (5 seconds is a long time in electronics) and the relay (which is pretty low ohmage on the coil, check the specs out). A very rough rule of thumb is the RC time constant (which I showed previous), R X C = Seconds. K in ohms is X1000, and µF is X 0.000001F . The transistors gain increases the apparant resistance on the emitter (the arrow) to increase the RC time constant.

It occurs to me you may not have a clue what I'm talking about. Feel free to ask questions and we'll do our best to answer them.

If you were to make the capacitor 20 to 100 times bigger you would be in the range you're looking for, but 100,000µF is not a small part, in fact it is physically huge.

7. RiG615 Thread Starter Active Member

Nov 13, 2008
50
0
Thanks Bill,

Can you explain whats happening and how your sketch works?
i don't think i fully get it. -thanks

8. Wendy Moderator

Mar 24, 2008
20,772
2,540
In a common collector circuit the apparent resistance from the base to ground (through the emitter) is the gain of the transistor (β) X the emitter resistor. Figuring the resistor is around 180Ω and β is 50, this brings the resistance up to 9000Ω. 1000µF X 9KΩ = 9 seconds.

Last edited: Nov 22, 2008
9. mik3 Senior Member

Feb 4, 2008
4,846
63
Rig,
Just to clear things up,
Do you want the relay to stay on for a while after you switch the power off or you want to switch the power on and the relay to be energized after a while?

10. RiG615 Thread Starter Active Member

Nov 13, 2008
50
0
mik-
I want the relay to stay open for a while when i switch the power off

11. RiG615 Thread Starter Active Member

Nov 13, 2008
50
0
I really just want a delay from when I flip a switch
either on or off

12. RiG615 Thread Starter Active Member

Nov 13, 2008
50
0
-Bill
in your diagram can you explain what the capacitor does when you shut off the supply voltage?

13. Wendy Moderator

Mar 24, 2008
20,772
2,540
It will slowly discharge. If it does it too slowly then we add a resistor across it. The 555 circuit is a lot more predictable, but a bit more complex.

Are you planning on cycling the switch off/on/off/on, or are you thinking in terms of once every hour or so?

14. mik3 Senior Member

Feb 4, 2008
4,846
63
Bill's circuit does exactly the opposite, it will let the relay to be energized for a while after you power it and then it will de-energize it. To make it work the way you want you have to flip the position of the capacitor and the resistor connected to the base.

15. SgtWookie Expert

Jul 17, 2007
22,183
1,728
You're not going to be able to use that relay with a 9v "transistor" battery.
Even when brand new, those batteries only put out about 8.6v. They're generally rated at 150mAh, which translates to 15mA over 10 hours.

The relay's minimum pull-in voltage is 8.4v. You might actually get it to work a couple of times if you used a MOSFET to switch the power, but then you'll need a fresh battery.

You'd be better off using eight AA or D cells.

16. RiG615 Thread Starter Active Member

Nov 13, 2008
50
0
how did you get 9000Ωs?
wouldnt it be 5400Ωs [180*50=5400]
unless im missing something- just curious

17. RiG615 Thread Starter Active Member

Nov 13, 2008
50
0
so if i flip c1 and r1?

18. mik3 Senior Member

Feb 4, 2008
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Yes flip C1 and R1 and to protect the transistor place a 1K resistor between the connection of C1-R1 and the base.

19. RiG615 Thread Starter Active Member

Nov 13, 2008
50
0
What are the factors that determine the time now??

20. mik3 Senior Member

Feb 4, 2008
4,846
63
When you say you want the relay to stay energized for a while after the power goes off you mean the entire power supply or just the control signal of the relay?