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Author Topic: Time delay relay  (Read 8053 times)
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Skeleton Man
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« on: September 15, 2006, 07:18:35 pm »

Hi guys,

I tried wiring up my time delay relay for my pepsi vendor as per the attached diagram and it doesn't work.

Terminal A is AC hot, and terminal 3 is AC Neutral, and I get 110V between them. Between terminals A and 6 I get 55V, and between A and B I get nothing at all (which is why the relay isn't energising).

It looks like I connected something wrong, but I'm not sure where..  any help would be much appreciated..

Chris
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Skeleton Man
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« Reply #1 on: October 08, 2006, 08:01:06 pm »

Well it turns out all I had to do was bypass the resistor.. replaced it with a small length of wire and it now works a charm..
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bubba
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« Reply #2 on: October 09, 2006, 05:54:12 am »

Hmmmm...you may want to find out just why that resistor is there...  Now also, looking at that drawing, that item with the black arrow and line on it above the resistor is a diode. Diodes are like an electrical check valve and if installed backwards, will not work.
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Ken

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« Reply #3 on: October 09, 2006, 03:19:53 pm »


(bubba @ Oct. 09 2006,6:54)
QUOTE
Hmmmm...you may want to find out just why that resistor is there...  Now also, looking at that drawing, that item with the black arrow and line on it above the resistor is a diode. Diodes are like an electrical check valve and if installed backwards, will not work.

My guess is that the resistor is there to drain the cap or something..  all I know is that all the components were fried when I got it, and the resistor measured at 565 ohms..  it could've gone up or down in value or anything though and it was burnt so I couldn't read the stripes..

I am familliar with what a diode is (I drew that diagram myself), and it's oriented correctly.. it's there to feed DC to the capacitor (which is polarised)..

The original schematic for the machine doesn't have any of the components on it.. just a box with "time delay relay"..

I might add that when I tested the original time delay relay it didn't work with the resistor either, but worked ok with a wire in its place..

I am thinking somebody added the resistor later for some reason..  the main thing is that it works correctly now.. the solenoid stays down for about 1 second to allow the can to pass.. (instead of springing right back up immediately)
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bubba
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« Reply #4 on: October 09, 2006, 03:32:33 pm »

'<img'>

Nice job on the drawing '<img'>  What program did you draw it in? I've been looking for something to do basic electrical in... Right now I'm using Visio, which I like, but a bit overkill for what I normally need.
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Ken

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« Reply #5 on: October 09, 2006, 05:49:50 pm »

Actually with this diagram I just added the time delay components..  the relay diagram itself was taken from the data sheet..

I use visio for all my drawings..  I redrew the entire schematic for my machine using it..  (viewable here)
Only thing I do need to add to that is the compressor wiring..  

If anyone knows how to get striped wires (e.g. red/white) in visio let me know..  I couldn't find any way of doing it so I mixed colors to represent the wires.. (e.g. red/white wire = pink, blue/red wire = purple)
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johnieG
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« Reply #6 on: October 09, 2006, 07:18:15 pm »


(Skeleton Man @ Oct. 08 2006,9:01)
QUOTE
Well it turns out all I had to do was bypass the resistor.. replaced it with a small length of wire and it now works a charm..

I hope your Capacitor is A.C rated, or you'll be in for a bit of a suprise in a while...( can you say BANG!?) you need to add a Diode in series with terminal "A" this will produce D.C. to charge the cap & create a much more consistant discharge rate

the way you had it wired, with the Cap' & restistor in series would produce a delay in the relays pulling in ( turning on) but it wouldn't allow the Cap' to "hold" the relay "In" at all.

 remember  R x C = T    [resistance x capacitance = time]

so then ... .050f x 535 ohms = 28 seconds until "C" reaches 2/3'ds of "T" & then  the relay would pull in.

insofar as the discharge  this depends on the resistance of the relays coils in ohms. assuming say 100 ohms , then again R x C applys....about 5 seconds of "hold time" on the relay....


you'll have to excuse me now.. I have to go help Kim Jon Il-brains of north Korea... something about setting off a firecracker down the girls toilet..what a card! Hmmmm '<img'>
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« Reply #7 on: October 09, 2006, 07:55:30 pm »


(johnieG @ Oct. 09 2006,8:18)
QUOTE

(Skeleton Man @ Oct. 08 2006,9:01)
QUOTE
Well it turns out all I had to do was bypass the resistor.. replaced it with a small length of wire and it now works a charm..

I hope your Capacitor is A.C rated, or you'll be in for a bit of a suprise in a while...( can you say BANG!?) you need to add a Diode in series with terminal "A" this will produce D.C. to charge the cap & create a much more consistant discharge rate

the way you had it wired, with the Cap' & restistor in series would produce a delay in the relays pulling in ( turning on) but it wouldn't allow the Cap' to "hold" the relay "In" at all.

 remember  R x C = T   Â
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Skeleton Man
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« Reply #8 on: October 09, 2006, 08:11:53 pm »

Everyone is probably thinking that there's just AC hot on terminal A and AC neutral on terminal B..  hence thinking the diode, etc is not in series..   but it's not wired like that..  check out the picture below and you will see that the diode and resistor are wired in series with the cap..



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