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These LED’s I have are incredibly bright. So when using just one, I would typically use a 10-15k resistor in series to dim them down. Now I need to use those same LED’s in a parallel array. I used an LED array calculator to compile a circuit. The values I put through were taken from the data sheets. 20mA is the forward current, but that would be way too bright. The calculator chose 100 Ohm

In order to maintain roughly the same brightness per LED as in a series circuit with a 10-15k resistor, how do I calculate the new resistors? Should I simply use the same resistor values?

enter image description here!

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  • \$\begingroup\$ Do you have an ammeter or a multimeter of any kind?? They are NOT expensive. The TekPower TP7040 is cheap and when in the \$5\:\text{mA}\$ full-scale setting its resistance is only \$20\:\Omega\$, so it won't affect the current measurement with your resistor values. You should just measure the current and then also measure the voltage across the LED at that current. Two easy measurements to make. \$\endgroup\$
    – jonk
    Jan 26, 2020 at 2:44

3 Answers 3

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Your assumption is correct, just use the same resistor values (10k-15k) in your parallel circuit.

In this simple circuit, from the point of view of the individual LED and current limiting resistor, they still see the 5V supply. Now that you have 14 in parallel, just copy your series circuit 14 times, and make sure the power supply can handle the load.

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  • \$\begingroup\$ So if I want to use this calculator to estimate current draw, should I set the forward current value to my desired current for the brightness? For instance, I want about .3mA for my brightness. If I enter .3 into the calculator’s forward current, it outputs 4.2mA. My PSU is simply an arduino nano which can handle 40mA per output pin. \$\endgroup\$
    – BobaJFET
    Jan 26, 2020 at 2:25
  • \$\begingroup\$ @StrugglingStudent117 You are correct! Limiting the current will control the brightness. \$\endgroup\$
    – DSI
    Jan 27, 2020 at 3:15
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Since you have several LED/resistor pairs in parallel, you can consider each LED/resistor independently, so you should use whatever resistor value gives a suitable brightness when testing with a single LED/resistor.

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  1. If you are trying to turn on all the LEDs using a nano GPIO pin it will be severely limited.
  2. If the purpose of the LEDs is to only indicate then limit the current severely to about 1 mA or so.
  3. Nano pin cannot supply 15 LEDs where each LED is sinking about 4 mA.
  4. When the current drawn from a particular Nano pin increases, there will be drop in the output voltage. This will again limit the current through the LED reducing the brightness of each LEDs as and when more LEDs are parallelled.
  5. My proposal is to use a simple N MOSFET as a switch to control that much high current load.
  6. From the Nano datasheet, the output high current is characterised only till 20 mA. At 20 mA the output voltage will be 4.2 V at the output pins.

enter image description here

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  • \$\begingroup\$ To clarify, the purpose is for an LED sign. I’d like the LED’s to be around 0.3 mA so they don’t hurt the eyes. I’ve decided not to use arduino for this after all but instead use a more capable PSU. I will look into N MOSFET. \$\endgroup\$
    – BobaJFET
    Jan 26, 2020 at 4:10
  • \$\begingroup\$ @StrugglingStudent117 as a hint.. the power wasted here due to heat is roughly 40% of the input power. You can also look at LED drivers \$\endgroup\$
    – User323693
    Jan 26, 2020 at 4:35

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