How can I wire a 9-position switch so that each position turns on one more LED than the one before?





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$begingroup$


I have a 9-way switch like this guy:



9-way



And I'm trying to figure out how I can turn on one LED with position 1, 2 with position 2, all the way up to all 9 in position 9.



Obviously I can repeat all the wiring for the LEDs at each position, but that seems silly.



My idea is that with a layout like below, the switch would represent the circled red line (shown in position 3), which would elongate to the right in each successive position until it connects all the lights. How can I do this?



schematic






led switches wiring







share|improve this question









$endgroup$












  • $begingroup$
    What operating voltage does each LED require, how much current does each LED draw, and what is the supply voltage?
    $endgroup$
    – Bruce Abbott
    5 hours ago










  • $begingroup$
    Your logic defines OR input logic for each LED but easier tinyurl.com/y38aomlp
    $endgroup$
    – Sunnyskyguy EE75
    3 hours ago










  • $begingroup$
    @BruceAbbott the idea was to use a 9V battery, but if I have to, I can use a DC power supply that fits a 9V-type switching jack. The LEDs draw about 25-30mA but they do the trick at even half-brightness, so there's some wiggle room there.
    $endgroup$
    – Isaac Lubow
    2 hours ago


















2












$begingroup$


I have a 9-way switch like this guy:



9-way



And I'm trying to figure out how I can turn on one LED with position 1, 2 with position 2, all the way up to all 9 in position 9.



Obviously I can repeat all the wiring for the LEDs at each position, but that seems silly.



My idea is that with a layout like below, the switch would represent the circled red line (shown in position 3), which would elongate to the right in each successive position until it connects all the lights. How can I do this?



schematic






led switches wiring







share|improve this question









$endgroup$












  • $begingroup$
    What operating voltage does each LED require, how much current does each LED draw, and what is the supply voltage?
    $endgroup$
    – Bruce Abbott
    5 hours ago










  • $begingroup$
    Your logic defines OR input logic for each LED but easier tinyurl.com/y38aomlp
    $endgroup$
    – Sunnyskyguy EE75
    3 hours ago










  • $begingroup$
    @BruceAbbott the idea was to use a 9V battery, but if I have to, I can use a DC power supply that fits a 9V-type switching jack. The LEDs draw about 25-30mA but they do the trick at even half-brightness, so there's some wiggle room there.
    $endgroup$
    – Isaac Lubow
    2 hours ago














2












2








2





$begingroup$


I have a 9-way switch like this guy:



9-way



And I'm trying to figure out how I can turn on one LED with position 1, 2 with position 2, all the way up to all 9 in position 9.



Obviously I can repeat all the wiring for the LEDs at each position, but that seems silly.



My idea is that with a layout like below, the switch would represent the circled red line (shown in position 3), which would elongate to the right in each successive position until it connects all the lights. How can I do this?



schematic






led switches wiring







share|improve this question









$endgroup$




I have a 9-way switch like this guy:



9-way



And I'm trying to figure out how I can turn on one LED with position 1, 2 with position 2, all the way up to all 9 in position 9.



Obviously I can repeat all the wiring for the LEDs at each position, but that seems silly.



My idea is that with a layout like below, the switch would represent the circled red line (shown in position 3), which would elongate to the right in each successive position until it connects all the lights. How can I do this?



schematic






led switches wiring




led switches wiring






share|improve this question













share|improve this question











share|improve this question




share|improve this question










asked 5 hours ago









Isaac LubowIsaac Lubow

1164




1164












  • $begingroup$
    What operating voltage does each LED require, how much current does each LED draw, and what is the supply voltage?
    $endgroup$
    – Bruce Abbott
    5 hours ago










  • $begingroup$
    Your logic defines OR input logic for each LED but easier tinyurl.com/y38aomlp
    $endgroup$
    – Sunnyskyguy EE75
    3 hours ago










  • $begingroup$
    @BruceAbbott the idea was to use a 9V battery, but if I have to, I can use a DC power supply that fits a 9V-type switching jack. The LEDs draw about 25-30mA but they do the trick at even half-brightness, so there's some wiggle room there.
    $endgroup$
    – Isaac Lubow
    2 hours ago


















  • $begingroup$
    What operating voltage does each LED require, how much current does each LED draw, and what is the supply voltage?
    $endgroup$
    – Bruce Abbott
    5 hours ago










  • $begingroup$
    Your logic defines OR input logic for each LED but easier tinyurl.com/y38aomlp
    $endgroup$
    – Sunnyskyguy EE75
    3 hours ago










  • $begingroup$
    @BruceAbbott the idea was to use a 9V battery, but if I have to, I can use a DC power supply that fits a 9V-type switching jack. The LEDs draw about 25-30mA but they do the trick at even half-brightness, so there's some wiggle room there.
    $endgroup$
    – Isaac Lubow
    2 hours ago
















$begingroup$
What operating voltage does each LED require, how much current does each LED draw, and what is the supply voltage?
$endgroup$
– Bruce Abbott
5 hours ago




$begingroup$
What operating voltage does each LED require, how much current does each LED draw, and what is the supply voltage?
$endgroup$
– Bruce Abbott
5 hours ago












$begingroup$
Your logic defines OR input logic for each LED but easier tinyurl.com/y38aomlp
$endgroup$
– Sunnyskyguy EE75
3 hours ago




$begingroup$
Your logic defines OR input logic for each LED but easier tinyurl.com/y38aomlp
$endgroup$
– Sunnyskyguy EE75
3 hours ago












$begingroup$
@BruceAbbott the idea was to use a 9V battery, but if I have to, I can use a DC power supply that fits a 9V-type switching jack. The LEDs draw about 25-30mA but they do the trick at even half-brightness, so there's some wiggle room there.
$endgroup$
– Isaac Lubow
2 hours ago




$begingroup$
@BruceAbbott the idea was to use a 9V battery, but if I have to, I can use a DC power supply that fits a 9V-type switching jack. The LEDs draw about 25-30mA but they do the trick at even half-brightness, so there's some wiggle room there.
$endgroup$
– Isaac Lubow
2 hours ago










5 Answers
5






active

oldest

votes


















3












$begingroup$

use a regulated current source to light them, wire them in series and short out the segemnt you want to be dark.





schematic





simulate this circuit – Schematic created using CircuitLab



you can possibly use a buck-boost converter to make the 30V if you don't already have a suitable voltage.



here's a simple way to build one using a LM2596S module
remove the potentiometer, and both large capacitors, connect one of the salvaged
capacitor between +in and +out (positive to +in), fit a 1uF ceramic capacitor where the output capacitor was. connect a 100 ohms resistor from the -output to the centre potentiometer terminal.



modified in this way it will create a negative voltage on the -out terminals and act as a 12.5mA current sink at the centre potentiometer terminal (with source at +out) if power is applied between +in and +out





schematic





simulate this circuit






share|improve this answer











$endgroup$





















    1












    $begingroup$

    If you can afford another 0.5V drop you can use a massive array of diodes. Here is an example with three LEDS which requires 6 diodes.

    (Sorry for the SW, SW2.., circuit lab does not have a rotatory switch symbol)





    schematic





    simulate this circuit – Schematic created using CircuitLab






    share|improve this answer









    $endgroup$





















      0












      $begingroup$

      Here is a low tech solution that requires a lot of parts. Only 4 positions shown, you need 45 diodes for 9 positions.



      Sunyskyguy has a clever solution if you have a high voltage available.



      enter image description here






      share|improve this answer









      $endgroup$













      • $begingroup$
        Can you do that with fewer diodes if you bring each one in before (to left of$ the one above it?
        $endgroup$
        – Bob Jacobsen
        2 hours ago



















      0












      $begingroup$

      If you’re not wedded to the specific switch you have, get a “progressive shorting rotary switch” to replace it. That works just like your drawing.






      share|improve this answer









      $endgroup$





















        0












        $begingroup$

        It might seem like overkill but it would be fewer parts, and possibly less expensive, than some other solutions to use a micro-controller. Many Ardunio boards have over 9 digital output pins - you could drive one LED with each of nine pins. By having the switch pick different points of a voltage divider and feeding it to one analog pin, you could determine the switch position and light up what ever you have decided should be lit up.






        share|improve this answer









        $endgroup$














          Your Answer






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          5 Answers
          5






          active

          oldest

          votes








          5 Answers
          5






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          3












          $begingroup$

          use a regulated current source to light them, wire them in series and short out the segemnt you want to be dark.





          schematic





          simulate this circuit – Schematic created using CircuitLab



          you can possibly use a buck-boost converter to make the 30V if you don't already have a suitable voltage.



          here's a simple way to build one using a LM2596S module
          remove the potentiometer, and both large capacitors, connect one of the salvaged
          capacitor between +in and +out (positive to +in), fit a 1uF ceramic capacitor where the output capacitor was. connect a 100 ohms resistor from the -output to the centre potentiometer terminal.



          modified in this way it will create a negative voltage on the -out terminals and act as a 12.5mA current sink at the centre potentiometer terminal (with source at +out) if power is applied between +in and +out





          schematic





          simulate this circuit






          share|improve this answer











          $endgroup$


















            3












            $begingroup$

            use a regulated current source to light them, wire them in series and short out the segemnt you want to be dark.





            schematic





            simulate this circuit – Schematic created using CircuitLab



            you can possibly use a buck-boost converter to make the 30V if you don't already have a suitable voltage.



            here's a simple way to build one using a LM2596S module
            remove the potentiometer, and both large capacitors, connect one of the salvaged
            capacitor between +in and +out (positive to +in), fit a 1uF ceramic capacitor where the output capacitor was. connect a 100 ohms resistor from the -output to the centre potentiometer terminal.



            modified in this way it will create a negative voltage on the -out terminals and act as a 12.5mA current sink at the centre potentiometer terminal (with source at +out) if power is applied between +in and +out





            schematic





            simulate this circuit






            share|improve this answer











            $endgroup$
















              3












              3








              3





              $begingroup$

              use a regulated current source to light them, wire them in series and short out the segemnt you want to be dark.





              schematic





              simulate this circuit – Schematic created using CircuitLab



              you can possibly use a buck-boost converter to make the 30V if you don't already have a suitable voltage.



              here's a simple way to build one using a LM2596S module
              remove the potentiometer, and both large capacitors, connect one of the salvaged
              capacitor between +in and +out (positive to +in), fit a 1uF ceramic capacitor where the output capacitor was. connect a 100 ohms resistor from the -output to the centre potentiometer terminal.



              modified in this way it will create a negative voltage on the -out terminals and act as a 12.5mA current sink at the centre potentiometer terminal (with source at +out) if power is applied between +in and +out





              schematic





              simulate this circuit






              share|improve this answer











              $endgroup$



              use a regulated current source to light them, wire them in series and short out the segemnt you want to be dark.





              schematic





              simulate this circuit – Schematic created using CircuitLab



              you can possibly use a buck-boost converter to make the 30V if you don't already have a suitable voltage.



              here's a simple way to build one using a LM2596S module
              remove the potentiometer, and both large capacitors, connect one of the salvaged
              capacitor between +in and +out (positive to +in), fit a 1uF ceramic capacitor where the output capacitor was. connect a 100 ohms resistor from the -output to the centre potentiometer terminal.



              modified in this way it will create a negative voltage on the -out terminals and act as a 12.5mA current sink at the centre potentiometer terminal (with source at +out) if power is applied between +in and +out





              schematic





              simulate this circuit







              share|improve this answer














              share|improve this answer



              share|improve this answer








              edited 2 hours ago

























              answered 3 hours ago









              JasenJasen

              11.9k1632




              11.9k1632

























                  1












                  $begingroup$

                  If you can afford another 0.5V drop you can use a massive array of diodes. Here is an example with three LEDS which requires 6 diodes.

                  (Sorry for the SW, SW2.., circuit lab does not have a rotatory switch symbol)





                  schematic





                  simulate this circuit – Schematic created using CircuitLab






                  share|improve this answer









                  $endgroup$


















                    1












                    $begingroup$

                    If you can afford another 0.5V drop you can use a massive array of diodes. Here is an example with three LEDS which requires 6 diodes.

                    (Sorry for the SW, SW2.., circuit lab does not have a rotatory switch symbol)





                    schematic





                    simulate this circuit – Schematic created using CircuitLab






                    share|improve this answer









                    $endgroup$
















                      1












                      1








                      1





                      $begingroup$

                      If you can afford another 0.5V drop you can use a massive array of diodes. Here is an example with three LEDS which requires 6 diodes.

                      (Sorry for the SW, SW2.., circuit lab does not have a rotatory switch symbol)





                      schematic





                      simulate this circuit – Schematic created using CircuitLab






                      share|improve this answer









                      $endgroup$



                      If you can afford another 0.5V drop you can use a massive array of diodes. Here is an example with three LEDS which requires 6 diodes.

                      (Sorry for the SW, SW2.., circuit lab does not have a rotatory switch symbol)





                      schematic





                      simulate this circuit – Schematic created using CircuitLab







                      share|improve this answer












                      share|improve this answer



                      share|improve this answer










                      answered 3 hours ago









                      OldfartOldfart

                      8,9362927




                      8,9362927























                          0












                          $begingroup$

                          Here is a low tech solution that requires a lot of parts. Only 4 positions shown, you need 45 diodes for 9 positions.



                          Sunyskyguy has a clever solution if you have a high voltage available.



                          enter image description here






                          share|improve this answer









                          $endgroup$













                          • $begingroup$
                            Can you do that with fewer diodes if you bring each one in before (to left of$ the one above it?
                            $endgroup$
                            – Bob Jacobsen
                            2 hours ago
















                          0












                          $begingroup$

                          Here is a low tech solution that requires a lot of parts. Only 4 positions shown, you need 45 diodes for 9 positions.



                          Sunyskyguy has a clever solution if you have a high voltage available.



                          enter image description here






                          share|improve this answer









                          $endgroup$













                          • $begingroup$
                            Can you do that with fewer diodes if you bring each one in before (to left of$ the one above it?
                            $endgroup$
                            – Bob Jacobsen
                            2 hours ago














                          0












                          0








                          0





                          $begingroup$

                          Here is a low tech solution that requires a lot of parts. Only 4 positions shown, you need 45 diodes for 9 positions.



                          Sunyskyguy has a clever solution if you have a high voltage available.



                          enter image description here






                          share|improve this answer









                          $endgroup$



                          Here is a low tech solution that requires a lot of parts. Only 4 positions shown, you need 45 diodes for 9 positions.



                          Sunyskyguy has a clever solution if you have a high voltage available.



                          enter image description here







                          share|improve this answer












                          share|improve this answer



                          share|improve this answer










                          answered 3 hours ago









                          Mattman944Mattman944

                          2015




                          2015












                          • $begingroup$
                            Can you do that with fewer diodes if you bring each one in before (to left of$ the one above it?
                            $endgroup$
                            – Bob Jacobsen
                            2 hours ago


















                          • $begingroup$
                            Can you do that with fewer diodes if you bring each one in before (to left of$ the one above it?
                            $endgroup$
                            – Bob Jacobsen
                            2 hours ago
















                          $begingroup$
                          Can you do that with fewer diodes if you bring each one in before (to left of$ the one above it?
                          $endgroup$
                          – Bob Jacobsen
                          2 hours ago




                          $begingroup$
                          Can you do that with fewer diodes if you bring each one in before (to left of$ the one above it?
                          $endgroup$
                          – Bob Jacobsen
                          2 hours ago











                          0












                          $begingroup$

                          If you’re not wedded to the specific switch you have, get a “progressive shorting rotary switch” to replace it. That works just like your drawing.






                          share|improve this answer









                          $endgroup$


















                            0












                            $begingroup$

                            If you’re not wedded to the specific switch you have, get a “progressive shorting rotary switch” to replace it. That works just like your drawing.






                            share|improve this answer









                            $endgroup$
















                              0












                              0








                              0





                              $begingroup$

                              If you’re not wedded to the specific switch you have, get a “progressive shorting rotary switch” to replace it. That works just like your drawing.






                              share|improve this answer









                              $endgroup$



                              If you’re not wedded to the specific switch you have, get a “progressive shorting rotary switch” to replace it. That works just like your drawing.







                              share|improve this answer












                              share|improve this answer



                              share|improve this answer










                              answered 2 hours ago









                              Bob JacobsenBob Jacobsen

                              1,13758




                              1,13758























                                  0












                                  $begingroup$

                                  It might seem like overkill but it would be fewer parts, and possibly less expensive, than some other solutions to use a micro-controller. Many Ardunio boards have over 9 digital output pins - you could drive one LED with each of nine pins. By having the switch pick different points of a voltage divider and feeding it to one analog pin, you could determine the switch position and light up what ever you have decided should be lit up.






                                  share|improve this answer









                                  $endgroup$


















                                    0












                                    $begingroup$

                                    It might seem like overkill but it would be fewer parts, and possibly less expensive, than some other solutions to use a micro-controller. Many Ardunio boards have over 9 digital output pins - you could drive one LED with each of nine pins. By having the switch pick different points of a voltage divider and feeding it to one analog pin, you could determine the switch position and light up what ever you have decided should be lit up.






                                    share|improve this answer









                                    $endgroup$
















                                      0












                                      0








                                      0





                                      $begingroup$

                                      It might seem like overkill but it would be fewer parts, and possibly less expensive, than some other solutions to use a micro-controller. Many Ardunio boards have over 9 digital output pins - you could drive one LED with each of nine pins. By having the switch pick different points of a voltage divider and feeding it to one analog pin, you could determine the switch position and light up what ever you have decided should be lit up.






                                      share|improve this answer









                                      $endgroup$



                                      It might seem like overkill but it would be fewer parts, and possibly less expensive, than some other solutions to use a micro-controller. Many Ardunio boards have over 9 digital output pins - you could drive one LED with each of nine pins. By having the switch pick different points of a voltage divider and feeding it to one analog pin, you could determine the switch position and light up what ever you have decided should be lit up.







                                      share|improve this answer












                                      share|improve this answer



                                      share|improve this answer










                                      answered 1 hour ago









                                      George WhiteGeorge White

                                      34727




                                      34727






























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                                          Península de Txukotka

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                                          Península de Txukotka Tipus Península Ubicació 66° N, 172° O  /  66°N , 172°O  / 66; -172 Mar mar dels Txuktxis La península de Txukotka a l'extrem oriental de Sibèria La península de Txukotka o península dels Txuktxis (en rus Чуко́тский полуо́стров , Txukotski poluóstrov ) és una península que es troba a l'extrem oriental d'Euràsia. Contingut 1 Descripció 2 Població 3 Vegeu també 4 Bibliografia Descripció L'estret de Bering, entre la península de Txukotka i la península de Seward a Alaska Foto de satèl·lit de la zona amb la península de Txukotka a l'esquerra El seu punt més extrem és el cap Dezhnev a la vora del nucli d'Uelen, un dels poquíssims llocs habitats de la zona. Al nord limita amb el mar dels Txuktxis i amb el mar de Bering al sud i amb l'estret de Bering a l'est. A la costa sud es troben la badia de Sant Lawrence i la badia de Kresta. La penínsu...
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