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I want to power two red LEDs from a USB charger or USB battery pack. I want both LEDs to light up at the same time from the same USB cable. Can anyone advise what resistor/s I need to fit in the chain and where, please?

Below are the two different types of LEDs I have available to use:

Red LED Ammo Through Hole, Broadcom HLMP-EG1T-Z20DD
Datasheet & sales page

2.4 V Red LED 5mm Through Hole, Broadcom HLMP-EG1G-Y10DD
Datasheet & sales page

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2 Answers 2

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You can connect both LEDs in series with a resistor. The order of the components does not matter, but the LEDs will only work the correct way round.

schematic

simulate this circuit – Schematic created using CircuitLab

As Russell explained, the LEDs each 'drop' 2.1 volts from the 5 volt supply, leaving a 0.8 volt drop across the resistor. The value of the resistor controls the current around the circuit, so for 20 mA (bright), the resistance should be 0.8 / 0.02 = 40 Ohms. Increasing the resistance will lower the current and therefore the brightness.

It is easiest to experiment to find the brightness you want. Try 47, 100 and 220 Ohm resistors.

Connecting the LEDs in series with a single resistor will last twice as long on a battery pack compared to connecting the LEDs in parallel, but the brightness will be more sensitive to variations in the 5v supply voltage.

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I want to power two red LED's from a USB charger or USB battery pack. I want both LED's to light up at the same time from the same USB cable. Can anyone advise what resistor/s I need to fit in the chain and where, please?

The LEDs should be powered from the USB power leads - giving about 5V DC.
Both LEDs have a Vf (forward voltage drop) of about 2.1 volt typical (datasheet figure 2 page 5) at 20 mA current.

The series resistors shown in the diagram drop the difference in voltage between the 5V supply and the LEd voltage drop.
Here Vresistor = Vsupply - Vled = 5-2.1 = 2.9 V.
This is approximate but an OK starting point.

The resistors shown are sized to give about 10 mA - you can reduce them to drwa less current or increase them to draw less.
Here R = V/I = (5V-2.1V)/0.010 A = 290 Ohms.
330 Ohms is a common standard value of resistance.

Note that at the reduced current Vf will be slightly lower.

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ Thank you for doing this. I would have never figured it out. Excuse my beyond crude drawing but is it correct: imgur.com/a/QYXalcX Also is there a certain wattage the resistor needs to be? \$\endgroup\$
    – Mark
    Jun 10, 2020 at 11:04
  • \$\begingroup\$ @Mark You better than some drawing is correct. Note that the resistor could go in +ve or -ve lead. It's usual to put them in the + lead but not essential. The power levels are such that almost any resistor will be OK. Power = V x I. Here total V = 5V and resistor V is about 3V. Total power = V x I = 5 x 0.010 = 50 milliwatts. Resistor power is about 3 x 0.010 = 30 milliwatts. That's well below the rating of any resistor you are liable to encounter. Even at say 100 mA (a VERY bright LED) resistor power is about 300 mW - a very small wattage resistor would get excessively hot. \$\endgroup\$
    – Russell McMahon
    Jun 10, 2020 at 11:20

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