I am working on a hobby layout that will have LED lighting for some of the elements. I would like to individually switch LEDs in series on/off. I used ledcalc.com to generate the circuit diagram below. I am stuck on if it is possible to switch one or more of the LEDs in a given series without affecting the LEDs that follow.

For example, Lamp 1 below has on/off switch while Lamp 2 and Lamp 3 are always on.

Lights would be common 5 mm LEDs if that matters.


simulate this circuit – Schematic created using CircuitLab

  • \$\begingroup\$ Why is R3 so much higher than R1 and R2 if all of the LEDs are identical? \$\endgroup\$
    – Barry
    Commented Dec 23, 2019 at 21:42
  • 1
    \$\begingroup\$ Does this answer your question? LED Matrix Leakage LEDs \$\endgroup\$ Commented Dec 23, 2019 at 22:16
  • \$\begingroup\$ @Barry I'd guess it was an editing miss BUT it is entirely valid to run different LEDs at different brightnesses of the application calls for it. With white LEDs the 5.6 ohms would result in ABOUT 0.5A/LED. The 470 R to ABOUT 6 mA/LED. The 0.5A LEDs may be in a spot light. The 6 mA ones may be eg power-on indicators. Or bling illuminators. Or ... . \$\endgroup\$
    – Russell McMahon
    Commented Dec 23, 2019 at 22:40

2 Answers 2


Placing a switch ACROSS lamp1 will extinguish it when the switch is closed.
The current in the "leg" will increase.
You can design the cct so current is "somewhat low when switch is open and somewhat high when switch is closed so brightness is "much the same" overall.

eg with 3V LEDs. If series R is 100 Ohms.

  • 2 LEDS I = V/R = (12-9)/100 = 30 mA.

  • 2 LEDS I = V/R = (12-6)/100 = 60 mA.

You'll notice a 2:1 difference but it will not APPEAR to be that much.


Alternative -

Use a SPDT switch and operate EITHER the LED or a zener of equal voltage.
LEDS shown are example only.

With eg 3V white LEDs I ~~= V/R = (12 - 9) /100 = 30 mA.
Choose zener to suit LED.
Zener voltage will depend on current.
eg for a 2V7 rated zener, Vz is usually about 2V7 at 10 mA and somewhat more at higher currents.

As jsotola notes - a resistor could be used in place of the zener.
R = V/I = V_LED_replaced / Istring For white LEDs at Vf = 3V and 30 mA LED string current,
R = V/I = 3V / 30 mA = 100 ohms
So 3 LEDs have 100 ohms in series and 2 LEDS have 100 ohms in series.
(Following that line of though 4 LEDS need NO Ohms in series at 12V BUT this is "not really so". The equation relies on Vsupply being exactly 12V and each LED having a Vf (forward voltage) of 3V at 30 mA. If that was exactly and always so then it would work. In practice the spread in LED Vfs and small supply variations would usually lead to "very poor" results.


simulate this circuit – Schematic created using CircuitLab

  • \$\begingroup\$ Thanks for the detailed reply - up-voted your answer but my rep is not high enough to record. \$\endgroup\$
    – jshenn
    Commented Dec 23, 2019 at 21:47
  • \$\begingroup\$ you could also use a resistor instead of the zener \$\endgroup\$
    – jsotola
    Commented Dec 23, 2019 at 21:47
  • \$\begingroup\$ @jshenn, upvoted for you ... you can also "accept" the answer if it solves your problem \$\endgroup\$
    – jsotola
    Commented Dec 23, 2019 at 21:49

Another option is to use 'smart' LEDs, 9 wired in series, with a small microcontroller (such as one of the Arduino models) to read switch contacts and send the appropriate data to the LED string (such as WS2812Bs).


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