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Post by Disco » Mon Dec 01, 2003 8:08 pm

I have two identical 12 volt dc solenoids that draw about .5 amps coil current. I need to connect them so that one switch controls them both (coils in parallel). I need make sure one makes before the other. Will a choke provide the delay I want? If so, what size? This is an outside application, so I want to keep parts count down and circuit to be simple. If not a choke, will some passive componet give me a delay? I am trying to stay away from IC's, transistors, etc. I need something to give me a simple delay on make just like a large capactior connected across one of the coils (isolated from the other coil with a diode) will give a delay on break.

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Chris Smith
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Post by Chris Smith » Mon Dec 01, 2003 10:23 pm

On one positive feed wire to one of the relays/solenoid attach a small resistor, experiment with its value by using a pot if necessary and change it to a fixed value resistor later.<p>The resistor then goes to a capacitor positive lead and onto the relay/solenoid positive lead, while the other lead from the cap [-] goes to ground. Chose a 10 uF cap or bigger to start.<p>Again experiment to find these/your timing values, and when the power is applied to the one circuit, VS the straight through wire on the other relay, the cap and resistor will need time to charge up holding back that relay by the value of the size of the cap and resistor feeding it.<p>This is known a R/C timing value.<p>Starting values for a short duration delay might be 50 ohms or less, and a 10 Uf cap.<p>Keep in mind the wattage needed to feed your soleniods must also cover the resistors wattage or rating, or it will get hot, smoke, and burn out. 1/4 watt wont do? Also the cap will hold that relay on for the same delay time, after you cut off the power.<p>[ December 01, 2003: Message edited by: Chris Smith ]</p>

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Post by MrAl » Tue Dec 02, 2003 2:18 am

Hello there,<p>Another way to get the kind of delay you
are looking for is to simply use another relay.<p>This new relay will drive the delayed relay,
and the first relay will be directly driven.
In this way, the delayed relay is the last to
turn on.
The sequence of operations is as follows:<p>1.
The drive power is turned on which energizes
the coils in relays 1 and 2 only. Relay 3
(the delayed relay) isnt energized yet.<p>2.
The first and second relays turn on after a delay. The third relay is still off.<p>3.
After the delay introduced by the second relay
has expired the third relay coil is energized.<p>4.
After the delay of the third coil it turns on.<p>Turning off:
The third relay would be last to turn off.
If you need all relays to turn off at about
the same time, you'll need to wire the power
through the main switch.<p>
Take care,
Al<p>[ December 02, 2003: Message edited by: MrAl ]</p>
LEDs vs Bulbs, LEDs are winning.

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Post by dacflyer » Thu Dec 04, 2003 9:21 am

how much of a delay do you need?
* just currious and why do you need the delay ? *
is the supply AC or DC ?<p>another idea is to use a inrush limiting resistor
apon powering this slowly gives power to a device,rather than a sudden jolt..
usually a second or 2 delay.
we use then in low voltage / hi current applications,,in lighting helps prolong filiment life..
this inrush limiting resistor has a hi resistance when cold...when you power it up,,,the resistance drops in about 1-2 seconds to a very low value
so it will light a low voltage . hi current lamp in a few seconds, rather a sudden jolt...which is a # cause of lamp failure...but anyway,you can use this to delay your relay....<p>or if nothing else use a 555 timer to create your delay...very easy....

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