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schematic

simulate this circuit – Schematic created using CircuitLab

I have a question regarding the schematic shown above where I have 8 LEDs connected in parallel, each having a switch. Since the supply voltage is 12 V constant, there wouldn't be a problem of lighting all the LEDs at the same time.

However, I assume if I want to light a single LED with this supply, I would probably burn it. My question is how to handle this power problem (how can I adjust the circuit by adding different elements) if I want to light only a single LED or multiple ones with a constant voltage supply.

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    \$\begingroup\$ Are these LED systems designed for 12 V operation? Or just regular LEDs? Can you specify the part number, etc., for the LEDs you have under consideration? \$\endgroup\$
    – jonk
    Commented Sep 29, 2022 at 8:38
  • \$\begingroup\$ @jonk They are regular LEDs having 2V forward Voltage and 20mA forward current \$\endgroup\$
    – Alex
    Commented Sep 29, 2022 at 8:51
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    \$\begingroup\$ Then your circuit is a disaster. No matter how may LEDs you place there if that constant 12 V supply is anything like a car battery, they will all disappear in a puff of smoke. If you don't care about power waste, you can just add one resistor per LED and you will be fine, so long as the power supply itself can deliver the required sum of currents to each LED in series. (Make sure each resistor is at least a 1/4 watt, maybe even 1/2 watt to be safe. 470 Ohm should be okay.) \$\endgroup\$
    – jonk
    Commented Sep 29, 2022 at 8:59
  • \$\begingroup\$ Do you know what a buck converter is ? Are you building this projet yourself ? \$\endgroup\$
    – Rahmany
    Commented Sep 29, 2022 at 9:02
  • \$\begingroup\$ The regular LEDs having 2V forward Voltage and 20mA forward current would need about 17 V when connected in series with a single current limiting resistor. All 8 LEDs plus the resistor in series. A single switch in series would switch all LEDs together on and off. There is no chance to switch each single LED only. \$\endgroup\$
    – Uwe
    Commented Sep 29, 2022 at 9:24

1 Answer 1

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LEDs are current operated.
Operating voltages given are typical at a specified current.
Connection of a 2V typical device at 12V will cause it to die so fast it probably will not even emit any discernible light.

To operate a 2V LED from 12V at known current and no concern about efficiency, a series resistor is added for each LED.
The resistor is dimensioned to drop the excess voltage.

Here:

  • Vsupply = 12V
  • Vled = 2V
  • V resistor = Vsupply - VLed = 10V.
  • I LED = I resistor = 20 mA (say).
  • R = V/I = 10V/20 mA = 500 Ohm.

A standard 470 ohm resistor will be adequate.

Current per LED can be adjusted by altering the resistor values.
As Iled increases Vled increases at a slower rate. See the LED data sheet for a Vf vs If curve.

Power loss per resistor = Vr x Ir = 10V x 20 mA = 200 mW.
Using a 0.5 Watt resistor is recommended.
A 250 mW resistor is nominally OK but operation close to its power rating leds to shorter life and more risk of catastopic failure (usually open circuit).

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  • \$\begingroup\$ Thank you so much for your reply. It helps so much! Cheers. \$\endgroup\$
    – Alex
    Commented Sep 29, 2022 at 10:33

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