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A question for you electrical genius's!

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  • A question for you electrical genius's!

    What do I need to do to cut down the power going into a relay that I think is rated at .5amp 24v from a 24v 20amp power supply? What happens now is the relay activates to the closed position but when the signal is sent to open the relay the relay stays closed. The relay is controlled over an Ethernet cable which is what I need. I think the problem is the relay is controlling two solenoid valves that require just under 1amp each to activate. So in my thinking I might need to run the signal from the Ethernet controlled relay into a higher rated relay to do the heavy lifting on the hydraulic valve.........?????

    Or is there some lectrical magic little thingy I could put inline?

    It would be no problem to use a PLC to actually control the solenoid valve and run the Ethernet relay power into the input on the PLC with relays that could easily handle the current.
    Location: The Black Forest in Germany

    How to become a millionaire: Start out with 10 million and take up machining as a hobby!

  • #2
    Is the coil ac or dc?
    Helder Ferreira
    Setubal, Portugal

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    • #3
      DC 24v.
      Location: The Black Forest in Germany

      How to become a millionaire: Start out with 10 million and take up machining as a hobby!

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      • #4
        Strange. Does the relay work well without the load? If so, connect a freewheel diode to the load.
        Helder Ferreira
        Setubal, Portugal

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        • #5
          Originally posted by Noitoen View Post
          Strange. Does the relay work well without the load? If so, connect a freewheel diode to the load.
          Yes the relay works fine with no load.
          Location: The Black Forest in Germany

          How to become a millionaire: Start out with 10 million and take up machining as a hobby!

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          • #6
            Then, the freewheel diode should do the trick.
            Helder Ferreira
            Setubal, Portugal

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            • #7
              Originally posted by Noitoen View Post
              Then, the freewheel diode should do the trick.
              I just googled freewheel diode......the problem is I don't speak Greek! Would you mind to tell me exactly where I would mount the diode and in which direction?
              Location: The Black Forest in Germany

              How to become a millionaire: Start out with 10 million and take up machining as a hobby!

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              • #8
                You should also have a BEMF diode across the relay coil. 1N4007.
                Max.

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                • #9
                  A freewheel diode is connected in parallel with the load (coil) and the power to the load goes, positive to the cathode (stripe) and negative to the anode side. Any general purpose diode will do.
                  Helder Ferreira
                  Setubal, Portugal

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                  • #10
                    A relay is the quintessential isolation between control and load. The coil cannot be affected by load unless there is wiring that you aren't mentioning. What I suspect is happening here is that the relay _is_ opening when the coil is de-energised, but the DC load current is arcing across the relay contacts. Especially likely given the inductive nature of the solenoid load.

                    The recommended diode is the solution for this. Two diodes - one across each solenoid, with the diode's anode (+) to the solenoid minus. This allows the solenoid's current to continue through the diode when the relay is opened, but not conduct with the relay closed. The diodes need to be rated for the solenoid's current - 1A, you said.

                    Speaking of current - you are driving 2 1A solenoids with a 1/2A relay? That's not going to work for long. Unless the 1A is just the initial draw-in current for the solenoid and is brief, before dropping to not more than 1/4A hold current.

                    Before you continue, check the relay contacts for being burned from the arcing.

                    Bob

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                    • #11
                      Originally posted by Bob Engelhardt View Post
                      with the diode's anode (+) to the solenoid minus. Bob
                      The diode Anode is connected to the negative side of the circuit, the Cathode to +ve (normally reverse biased).
                      See free wheel diode http://www.electronics-tutorials.ws/diode/diode_4.html
                      Max.
                      Last edited by MaxHeadRoom; 05-21-2017, 05:24 PM.

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                      • #12
                        Originally posted by MaxHeadRoom View Post
                        The diode Anode is connected to the negative side of the circuit, the Cathode to +ve (normally reverse biased).
                        ...
                        That's kinda' vague - the diode needs to be specifically connected across the solenoid. Not just anywhere to negative and +ve ("+ve" is not a term likely to be used in a circuit controlling solenoids).

                        Since my previous reply, I have realized that the relay coil might also need a diode. If it is being controlled by a relay driver, that driver might have the diode built in. If it doesn't, the driver could be damaged by the inductive kick when the relay is de-energized. Wouldn't hurt to add a diode to the relay coil. Same orientation as for solenoid: diode + to coil -.

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                        • #13
                          Can you isolate the power that activates the coil from the load the relay is controlling? Somewhere in your circuit the relay coil feed is getting voltage from the relay load. It sounds like you have inadvertently created a latching relay.

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                          • #14
                            Problem is that you are using a

                            +solid state device to turn on the relay, and either you are using an AC output - and it is not turning off because you are feeding it with DC, or if it is a DC device the leakage current is high enough to hold the relay closed once it is activated.

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                            • #15
                              The commutating (or free-wheeling) diode is very important to add across an inductive DC load. The relay contacts should be rated higher than the continuous load. It is possible for a DC coil relay to become magnetized enough for the armature to stick closed, so you could try reversing polarity on the coil. A solid state relay with DC output might work better, but may be more sensitive to inductive transients if they are not properly handled. An ordinary 1N400x 1A diode should work well enough, but in some cases a Schottky or other fast-acting diode may be better. Also, the diode may cause a delay in opening, so sometimes a series resistor about the same value as the coil resistance will dissipate the inductive energy more quickly to get better response.
                              http://pauleschoen.com/pix/PM08_P76_P54.png
                              Paul , P S Technology, Inc. and MrTibbs
                              USA Maryland 21030

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