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  • LED issue of interest, to me anyway.

    I have need of a pilot lamp on a 220v circuit to tell me when the voltage is on. I bought a LED panel lamp with 12-16v rated and forward current 12ma typical- 20ma max. I have a 400v 1A diode and the web calculator said 800 ohm .5 watt for the resistor so I used a 1k resistor.

    The resistor smoked so that didn't work then I tried 20K and the LED was barely visible. Then I tried a 10K and the LED looked good but the resistor was getting to hot.

    So the question is, what wattage and resistance do I need to use to make this work? It's going across the 220v element on a water heater.
    It's only ink and paper

  • #2
    I can provide you with a low power capacitive reactance circuit that will do the same job but without the heat losses associated with you dropper resistor..

    But the simplest way to go would be with a neon indicator... just two wire and a mounting hole..


    Rob

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    • #3
      18,000 Ohms
      3 Watts. It'll get hot.

      This is a bad idea.
      You'll be much better off with a neon lamp.


      Mike
      Mike

      My Dad always said, "If you want people to do things for you on the farm, you have to buy a machine they can sit on that does most of the work."

      Comment


      • #4
        I would try something like this http://www.allelectronics.com/make-a...EDS-RED/1.html or this http://www.allelectronics.com/make-a...ASE-RED/1.html as they are not internally regulated. Probably about 6K to 10k OHMs will make it happy and bright. The first link they are 10 for a dollar, so you can afford to experiment, the second a couple bucks ea. so order two and start high on the resistance. More: http://www.allelectronics.com/mas_as...k_Up_Guide.pdf

        If you can find a 120V or 240V xenon bulb, that would be the quick/easy. 120V would be happy from one leg of the 240 to ground.

        Comment


        • #5
          Mike and Rob hadn't posted when I started my reply. What can I say? I'm slower than a slug going up an icy hill wearing a teflon jacket with oil on it

          A series resister with one of the LEDs I linked won't get hot at any voltage. The LED will die before it can draw enough current to heat up even the smallest resister. The current limit resister keeps the current below the rating of the LED, which is only a handful of milliamps. I have to revise my resistor size recommendation though. Try:

          15K @240 volts = 16mA (safe for most LEDs)

          20K @240 volts = 12mA (most like a red LED running at 12 mA will outlive us all)

          You can clculate any size with OHMs law:
          I=E/R
          I = current
          E = voltage
          R = resistance

          If you leave off the Kilo from the resistance, the answer, by happy mistake, comes out in mA, rather than Amps. For a 25KOHM:

          240 / 25 = 9.6mA

          Comment


          • #6
            A .1 mfd mylar capacitor with a 600 volt rating in series with a diode will give it about 10 ma. No heat will be generated but it will flicker at 60hz. If that is a problem put a in a full wave bridge instead. Pay attention to wire dressing. It is all at mains voltage, but then so are neon lamps.

            Knudsen,

            12 ma at 240 volts is almost 3 watts.
            Last edited by Evan; 02-15-2010, 12:01 PM.
            Free software for calculating bolt circles and similar: Click Here

            Comment


            • #7
              Originally posted by Evan
              A .1 mfd mylar capacitor .
              Not wanting to be pedantic ..

              but a X2 class capacitor would be the preferred option for no sleepless nights...

              Rob
              Last edited by MrSleepy; 02-15-2010, 12:22 PM.

              Comment


              • #8
                I wouldn't worry about it. A 1N4004 diode makes a really effective fuse.
                Free software for calculating bolt circles and similar: Click Here

                Comment


                • #9
                  As it happens I am using a 1N4004 diode on one leg and the resistor on the other leg. Should they be on the same leg and does it matter if the resistor is before or after the diode? I don't see where it makes a hill of beans if the resistor is before or after the LED as long as the voltage drop is there, there's no reverse current and nothing gets hot. I sure don't want it to cause a fire in the hot water heater, the water is hot enough as it is.

                  I had forgotten about the neon panel lights, I used them in the past. I will see if the Shack has them, if not it's to the electronic supply house in Louseyville (really Louisville Ky) to get two of them.

                  My suggestion is DON'T buy a Smart water heater because they aren't really smart and just a bag of worms with a control board full of issues. If all else fails it's off with the control board and on with thermostats for each element. I don't want to replace the water heater as it's only about 5 years old.

                  When I was 9 or so (about 1950-51) I started dabbling in electronics in the tube days. I have since forgotten almost everything I learned and now it's solid state so woe is me. I still have a tube tester, signal generator and oscilloscope though as if they mean anything now days.

                  EDIT: I plan to use shrink tubing on the leads, is that an issue?
                  Last edited by Carld; 02-15-2010, 01:57 PM.
                  It's only ink and paper

                  Comment


                  • #10
                    Originally posted by Evan
                    A .1 mfd mylar capacitor with a 600 volt rating in series with a diode will give it about 10 ma. No heat will be generated but it will flicker at 60hz. If that is a problem put a in a full wave bridge instead. Pay attention to wire dressing. It is all at mains voltage, but then so are neon lamps.

                    Knudsen,

                    12 ma at 240 volts is almost 3 watts.
                    +1 for this solution. His current design will also flicker and I can't see how that would matter (it won't be visible).

                    Comment


                    • #11
                      Originally posted by Carld
                      As it happens I am using a 1N4004 diode on one leg and the resistor on the other leg. Should they be on the same leg and does it matter if the resistor is before or after the diode? I don't see where it makes a hill of beans if the resistor is before or after the LED as long as the voltage drop is there, there's no reverse current and nothing gets hot. I sure don't want it to cause a fire in the hot water heater, the water is hot enough as it is.

                      EDIT: I plan to use shrink tubing on the leads, is that an issue?
                      It's a series circuit, the order is not important, but polarity is as I'm sure you know.

                      Heatshrink is good.

                      If you continue with the resistor plan rather than the capacitor solutions suggested you will be dissipating over 2.5Watts. If you use a bigger 10W resistor you will still be dissipating the same power but it's spread out so temperature will be lower.

                      I really can't see why you want to use an LED for this.


                      Mike
                      Mike

                      My Dad always said, "If you want people to do things for you on the farm, you have to buy a machine they can sit on that does most of the work."

                      Comment


                      • #12
                        So I need to eliminate the resistor and use the capacitor? I thought I was to use both.
                        It's only ink and paper

                        Comment


                        • #13
                          Originally posted by Evan
                          A .1 mfd mylar capacitor with a 600 volt rating in series with a diode will give it about 10 ma.
                          what will be the current the moment you turn on the power?

                          Comment


                          • #14
                            The capacitor acts as an AC resistor because of it's capcitive reactance. No other resistor is required. Instead of dissipating the excess energy it stores it and gives it back to the line on the following half cycle. Therefore the only power consumed is that which the LED circuit needs. Changing the value of the capacitor or the voltage input changes the amount of current that flows. Be sure that you use the type I specified as they are the most reliable in this type of application. They are specified as either mylar or polyester capacitors.

                            [edit to answer BB]
                            Instantaneous current will be limited by the diode. The LED can withstand at least perhaps 10 times rated current for a few milliseconds. To be on the very safe side a 1000 ohm resistor can be put in series with the rest.
                            Last edited by Evan; 02-15-2010, 04:08 PM.
                            Free software for calculating bolt circles and similar: Click Here

                            Comment


                            • #15
                              Originally posted by Evan
                              The capacitor acts as an AC resistor because of it's capcitive reactance. No other resistor is required. Instead of dissipating the excess energy it stores it and gives it back to the line on the following half cycle. Therefore the only power consumed is that which the LED circuit needs. Changing the value of the capacitor or the voltage input changes the amount of current that flows. Be sure that you use the type I specified as they are the most reliable in this type of application. They are specified as either mylar or polyester capacitors.

                              [edit to answer BB]
                              Instantaneous current will be limited by the diode. The LED can withstand at least perhaps 10 times rated current for a few milliseconds. To be on the very safe side a 1000 ohm resistor can be put in series with the rest.
                              This will put 2x line voltage across the LED during the non-conducting half-cycle, no? Perhaps back to back LEDs would be safer.

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