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  • Another OT electrical design question

    We have a 20 watt, 36v DC solenoid valve we use for a particular application that includes timing cycles.

    We have some prototyping to do, and would like to try a new low-power version of this valve that is only available in 24v and uses 5watts.
    We would still need the same timing cycles as the previous valve, so would like to run it off the 36v timed source.

    What's the best way to wire up this 24v valve to run off that 36v output?
    Voltage divider circuit?
    Last edited by T.Hoffman; 05-25-2012, 08:21 PM.

  • #2
    I would first try a series resistor. Measure the current drawn by the new relay, calculate the resistance for 12 VDC across your dropping resistor and get that size and wattage.

    R = E / I (Resistance = Voltage divided by Current)
    R = 12 / I (Make sure you pay attention to the decimal point when you record the current.)

    Power (wattage) = E*I (Voltage times Current)

    You only have to be close on the resistance. +- 10% is not going to make that much difference.

    I had a similar problem once, and had to get a pump running NOW!. I had 24 VDC available to the burned out relay and 40 miles to town on a Sunday. I grabbed a couple of 12 VDC automotive relays from the mechanics and wired them in series. Then I paralled the contacts to insure they would not burn too badly before I could acquire the proper relay tomorrow.

    Pops

    Comment


    • #3
      That would take a 56 ohm resistor, power rating 5 watts would be good. There's about 2.5 watts of heat to dissipate. For a cooler running resistor use a 10 watt.
      I seldom do anything within the scope of logical reason and calculated cost/benefit, etc- I'm following my passion-

      Comment


      • #4
        Originally posted by darryl
        That would take a 56 ohm resistor, power rating 5 watts would be good. There's about 2.5 watts of heat to dissipate. For a cooler running resistor use a 10 watt.
        You are assuming the new relay consumes 5 watts. Which is what he said ("Uses 5 watts.") I was assuming, (I know!) that it was rated at 5 watts, which would mean that it actually draws less than 208mV. That was why I wanted him to get the actual specs or measure the draw.

        I suspect the difference is going to be pretty slight.

        Pops

        Comment


        • #5
          Since 110V AC controls are pretty much standard in the US and 24vDc in the rest of the world How did you wind up with 36 vDC and why would you want to stay with it?

          Comment


          • #6
            Thanks for the info gentlemen.

            Originally posted by tdmidget
            Since 110V AC controls are pretty much standard in the US and 24vDc in the rest of the world How did you wind up with 36 vDC and why would you want to stay with it?
            I have to deal with the cards I'm dealt. It's going into an application that has been in production for many years, using a number of 36v solenoids.

            They're looking for the cheapest solution working with the current arrangement. If a simple valve type change can make a big improvement vs minimal costs, I'm sure there is no motivation to be changing anything else.

            Comment


            • #7
              Originally posted by T.Hoffman
              Thanks for the info gentlemen.


              I have to deal with the cards I'm dealt. It's going into an application that has been in production for many years, using a number of 36v solenoids.

              They're looking for the cheapest solution working with the current arrangement. If a simple valve type change can make a big improvement vs minimal costs, I'm sure there is no motivation to be changing anything else.
              That being the case, I'm sure the cost of a 10 watt resistor is not going to be a breaker. Check sizes and be sure to include mounting for the resistor, rather than just hanging off the end of the connecting wire.

              Pops

              Comment


              • #8
                Originally posted by armedandsafe
                That being the case, I'm sure the cost of a 10 watt resistor is not going to be a breaker. Check sizes and be sure to include mounting for the resistor, rather than just hanging off the end of the connecting wire.
                This is just for prototype experimentation.
                If it turns out that this is a decent idea and worth pursuing, the valve company says with the amounts we need, they would make a 36v version...

                Comment


                • #9
                  For prototype operation you can get by with the resistor inline with the connecting wire. I've discovered over the years of designing machine controls that it is not good in the field when the left-footed maintenance people get to messing around with your pretty machine.

                  One trick I did, which won't be applicable in your case, was build the resister into the relay mount. That one didn't work, because the techs insisted on using Tandy relay mounts as replacements because they were cheaper/easier to find.

                  Pops

                  Comment


                  • #10
                    I would first try a series resistor. Measure the current drawn by the new relay, calculate the resistance for 12 VDC across your dropping resistor and get that size and wattage.

                    R = E / I (Resistance = Voltage divided by Current)
                    R = 12 / I (Make sure you pay attention to the decimal point when you record the current.)

                    Power (wattage) = E*I (Voltage times Current)

                    You only have to be close on the resistance. +- 10% is not going to make that much difference.
                    Turns out the new solenoid only uses about 2.4watts, measured at ~100ma with 24v DC supply.

                    So my resistor should be around a 120ohm 5watt (or 10watt) resistor inline to my new solenoid with the 36v supply? sound right?

                    Comment


                    • #11
                      120ohm is correct but you should only need a 2w resistor at that current?
                      Max.

                      Comment


                      • #12
                        A 12VDC drop at 0.100 amps would be 12/0.1, thus 120 Ohms. Yup, you got it.

                        The 5 Watt will be sufficient, but check the heat given off by that component in operation, with concern to other items near/touching it. The 10 watt ones might be needed for that reason, while not needed for operational reliability.

                        This sounds like a good solution. Keep us posted, eh?

                        Pops

                        Comment


                        • #13
                          The resistor is only going to be dissipating 1.2 watts, but go with a 5 watt to keep the temperature rise down. No need to start melting insulation, etc.
                          I seldom do anything within the scope of logical reason and calculated cost/benefit, etc- I'm following my passion-

                          Comment


                          • #14
                            Max, 2 watt resistor will be fine, for the steady state math, but you have to take into consideration the flash heating at pull-in. You are driving an inductive device and the initial in rush through that resistor is going to be pretty hefty. I ran into that when designing trigger circuits for cannons. In that case, the relay was tripping and releasing many times per second, so the demand was much higher that I think we will see on this application. However, I would still start with the 5 watter. (Yes, I do tend to get anal, at times. )

                            Pops

                            Comment


                            • #15
                              Originally posted by armedandsafe
                              Max, 2 watt resistor will be fine, for the steady state math, but you have to take into consideration the flash heating at pull-in. You are driving an inductive device and the initial in rush through that resistor is going to be pretty hefty.
                              Pops
                              But that would only apply to AC solenoids, with DC the inrush is not seen, only the coil resistance.
                              With AC it is not only the inrush which would be less than one cycle duration there would also be high current until the armature shifted over.
                              AC solenoids have a very low resistance and the current is not limited until the inductive effect takes place, (after 1 cycle and pull in occurs).
                              This is why DC solenoids and relays are superior, if the armature/spool on an AC solenoid does not shift, or someone pushes it over manually, a burn out results.
                              Max.
                              Last edited by MaxHeadRoom; 05-29-2012, 05:08 PM.

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