12v DC relay wiring & control circuit, Help

Discussion in 'The Projects Forum' started by gturnbull, Jun 19, 2008.

  1. gturnbull

    Thread Starter Active Member

    Jun 19, 2008
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    Firstly, A hello to everyone, I just found this place and it is really good :)

    I am trying to wire a relay into a control circuit, as the Float switch (reed contact switch) that is controlling the pump is only rated at 1a and the pump is drawing 1.5a, I know you may not think this is a great problem (as I have been told already, by others) but I thought about putting a relay in the circuit to help bring the power down that is going to the float switch (reed contact switch).

    I cant get my head around this, I think it I wire it up like the attached file, but would that bring the Amperage down that is getting pulled through the Float switch or do I need a completely different power source IE a power source to operate the contacts and a power source to operate the pump.

    I have just looped A1 & A2 (relay power) into 21 & 11 (fixed end of contacts)

    I hope I have explained this right and please excuse my rather rubbish drawing I used Paint, I did not know about the PDF converter till after the drawing had been done.

    :thanks:
     
  2. SgtWookie

    Expert

    Jul 17, 2007
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    You are close!

    1) Remove the BROWN wiring from the battery to the magnetic float switch to the relay terminals A1 and 11.
    2) Connect the 12v DC supply + terminal to relay terminal 11 using a RED wire.
    3) Connect the 12v DC supply + terminal to one side of the magnetic float switch using a RED wire.
    4) Connect the other side of the magnetic float switch to relay terminal A1 using a RED wire.

    Make those changes, and re-post so we can check it.

    You don't need to convert it to a PDF file. Save it as a .PNG file and attach it; it will load much faster.
     
  3. gturnbull

    Thread Starter Active Member

    Jun 19, 2008
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    Thanks for your help on this, I have made the changes you suggested and have hopefully attached the revised drawing.

    :thanks:
     
  4. SgtWookie

    Expert

    Jul 17, 2007
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    That will work OK :)

    But, right now you are switching both the + and - supply leads to the pump - you really only need to switch one side.

    You can just about double the life of the relay if you:
    1) Remove the blue wire from between relay terminals A2 and 21.
    2) Disconnect the blue wire on relay terminal 24, and reconnect it to the battery - terminal.
    3) Run another red wire from relay terminal 11 to 21.
    4) Run another brown wire from relay terminal 14 to 24.

    This way, instead of having a single pair of contacts on each power supply wire, you have both in parallel on the positive side sharing the load.
     
  5. gturnbull

    Thread Starter Active Member

    Jun 19, 2008
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    Thanks for your help wookie, you have been great :) will post my little project that it was for later on :)
     
  6. gturnbull

    Thread Starter Active Member

    Jun 19, 2008
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    When I do the above changes I gather I cannot just connect the wire that is currently in 24, in with A2.

    I forgot that I have 2 led's in this circuit as well, I will post a drawing up including the led's, 1 indicastes power to the system, and 1 indicates power the the pump, I will post it later on as I am supposed to be working right now :) I think I know where they go, but will check with you, just to make sure,

    Thanks again
     
  7. gturnbull

    Thread Starter Active Member

    Jun 19, 2008
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    Hi Wookie,

    Could you have a quick look at the diagram attached and see if all is ok.

    Do you have an idea what amperage the float switch shall pull when it is energised, it will be as the diagram pulling in a 12v coil, if you require more information just let me know.

    Thanks for all your help.
     
  8. gturnbull

    Thread Starter Active Member

    Jun 19, 2008
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    I Have just noticed that the attached image is not complete, something wrong with the format I think.

    See new attached drawing (amended for the last time, I promise :))

    Thanks Again
     
  9. SgtWookie

    Expert

    Jul 17, 2007
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    Sorry, I wasn't ignoring you - I didn't see that the thread had been updated, and it's been a busy day!

    OK, it looks good so far - BUT, I do not see current limiting resistors for your LEDs. If they are designed to run from a 12v supply, then you are OK. If not, you will need current limiting resistors in series with the LEDs. You will need to know the Vf (forward voltage) of the LEDs and their current rating.
    To calculate the resistors:
    Rlimit = (SupplyVoltage - VfLED) / LEDCurrentRating
    For example, your supply voltage is 12, and let's say for the moment that your LEDs run on 2.3 volts at 25mA (plug in your numbers to the formula below):
    Rlimit = (12 - 2.3) / 25mA
    Rlimit = 9.7 / 0.025
    Rlimit = 388 Ohms
    Use a resistor that is not less than the result.

    The shorter LED lead is the cathode; this goes to the more negative of the supply.

    Something else - now that you are using semiconductors in the circuit, you must add "snubber" diodes across both the motor and the relay coil to take care of the reverse EMF when the circuits to the inductive loads are opened up. These can be even an old-fashioned 1N4001 through 1N4007 1A rectifier diode that you can pick up at a Radio Shack; they have an assortment for a couple of bucks.

    You connect the snubber diodes so that the cathode (the end with the band) is connected to the positive lead side, and the anode (the band-less end) is on the more negative side. If you omit the snubber diodes, you will let the smoke out of your poor little LEDs.
     
  10. gturnbull

    Thread Starter Active Member

    Jun 19, 2008
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    I did not think you were ignoring me, I know how busy these boards can be, and posts can be easily missed :)

    The are actually indicator lamps, not LED's (sorry me using the incorrect terminology) they are designed to run on 12v dc, so I should not need to do anything further? is that correct.

    Wiring it up according and the float switch should have a very small amount of amps running through it, do you know how many amps it will have? Can you tell me how to calculate it, for future reference.

    Again Thanks
     
  11. SgtWookie

    Expert

    Jul 17, 2007
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    OK. If they are incandescent, you might get away without the snubber diodes, but your bulb life will likely be shorter. The reverse EMF can be quite high.

    With the float switch open, your current will be just that consumed by the Power On indicator lamp. If the lamps have a wattage rating, then Current = Watts/Voltage.
    When the float switch closes, you will need to add in the current of the relay's coil, the 2nd lamp, and the 1.5A that the motor draws.
     
  12. gturnbull

    Thread Starter Active Member

    Jun 19, 2008
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    This is what I am trying to get around as the float swith is only rated at 1.0A,and possibly weld closed due to the high amperage, I assumed that the float swith would receive very low amperage as it is only closing the coil, would I need a completely different power source to do that, ie a source for the pump and a different source for the float switch to operate the coil
     
  13. SgtWookie

    Expert

    Jul 17, 2007
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    No, the only load the float switch will see in this configuration is the current through the relay's coil.

    The "RUN" bulb and motor current come straight from the power supply via the relay. The "Power ON" bulb is powered directly by your supply.
     
  14. gturnbull

    Thread Starter Active Member

    Jun 19, 2008
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    Right got it now :rolleyes: I mis-read your other post :)

    Many thanks for your help on this, you have been great, I doubt I will be able to return the favour, so all I can do is say, THANKS
     
  15. SgtWookie

    Expert

    Jul 17, 2007
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    When the float switch closes and the pump turns on, you'll know that it's time to buy a beer for a Vet. ;) That's all I ask.

    Have a great weekend.
     
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