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I have a 9V battery, a rotary switch (one pole, 5 positions), and yellow LEDs (2V drop). How can I wire it so that as I rotate the switch the LEDs turn and stay on until I reach the end of the switch?

Meaning as I turn the switch first I get one led on, then two, three, and so on

I'm a beginner trying to wire a very simple toy for my niece.

So far I tried wiring it so that I connect + to led then led to next led, each one to its switch, and then ground from the switch back to the battery. But had no luck.

I'm not too concerned about some loss in light intensity since this is a homemade toy.

Thank you!

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3 Answers 3

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You can do it as following (example for 3 position switch, the arrow is indicating the switch, because the schematic tool doesn't seem to have the component). Then it is in R1 position, the closed circuit is R1+D1 in series. When in R2 position, then you have R2+D2+D1 in series. The last one is R3+D3+D2+D1. You will have to calculate the proper R1, R2 and R3 values to get a relatively even light intensity in each position (R3 < R2 < R1).

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ WOW I got it!! Thank you so much!!! \$\endgroup\$
    – Afff
    Apr 5, 2022 at 20:18
  • \$\begingroup\$ Upvoted - I naggingly knew there was a simpler way of doing it than mine as I was drawing it... :-) Particularly because the resistor values here can be chosen to keep the brightness more even across switch positions. \$\endgroup\$
    – TonyM
    Apr 5, 2022 at 21:51
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I'm not too concerned about some loss in light intensity since this is a homemade toy

With that in mind, you can use the below circuit.

The BAT81 diodes have a drop of 0.2..0.3 V, so D2..D5 will get a varying voltage depending on the switch position because of the cumulative diode drops.

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ thank you so much for your reply, unfortunately, I only have the yellow LEDs (2V drop). \$\endgroup\$
    – Afff
    Apr 5, 2022 at 20:10
  • \$\begingroup\$ @Afff, understand. The BAT81 diodes aren't LEDs anyway, they're Schottky diodes. \$\endgroup\$
    – TonyM
    Apr 5, 2022 at 21:09
  • \$\begingroup\$ Oh I see, thank you! \$\endgroup\$
    – Afff
    Apr 5, 2022 at 21:30
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You can do it with 6 or 10 diodes (the ones shown in boxes are optional but make the LEDs that are on more even) As the battery dies the LEDs will get a more uneven with the 6 and also will change brightness as more of them are on).

You connect the common on the switch to +9V.

schematic

simulate this circuit – Schematic created using CircuitLab

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