You'll still need a series resistor to set the right current. You could halve the amount of current consumption by having the LED wired in series instead of in parallel.
The upside to driving them in parallel is that it's more robust to Vf changes due to LED batch variation and ambient temperature.
In the series case, the total Vf is 2 x 2.2 = 4.4, leaving only 0.6 V to drop across the resistor. So, you choose a 30-ohm resistor to allow 20 mA. If you get a batch of LEDs with a low Vf, and/or the ambient temperature rises, the Vf might drop to 1.9 or less. Now, you're driving 40 mA into the LEDs.
If you did them in parallel, you would choose (5-2.2)/.02 = 140 ohms. If the Vf drops to 1.9 now, the LED is only getting 22 mA.
The downside is wasted power in the resistors in the parallel case.
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I think the LEDs are inverted.
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Oops.
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Fixed, thanks!
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The Amber LED could be in series for 5V I/O as they have forward voltage of 2.1-2.2V.
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I thought you always need a resistor on a LED, due to its non-linearity.
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You'll still need a series resistor to set the right current. You could halve the amount of current consumption by having the LED wired in series instead of in parallel.
Are you sure? yes | no
Oh, I see. But I guess I can also achieve the same halving by using larger resistors.
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No you are not as you are *always* using 2X the current because you are supplying 2 branches of LED instead of a single branch.
The power wasted in one of the series resistor *could* have powered the second LED. You throw away about 2.8V in that resistor instead of 0.6V.
Are you sure? yes | no
The upside to driving them in parallel is that it's more robust to Vf changes due to LED batch variation and ambient temperature.
In the series case, the total Vf is 2 x 2.2 = 4.4, leaving only 0.6 V to drop across the resistor. So, you choose a 30-ohm resistor to allow 20 mA. If you get a batch of LEDs with a low Vf, and/or the ambient temperature rises, the Vf might drop to 1.9 or less. Now, you're driving 40 mA into the LEDs.
If you did them in parallel, you would choose (5-2.2)/.02 = 140 ohms. If the Vf drops to 1.9 now, the LED is only getting 22 mA.
The downside is wasted power in the resistors in the parallel case.
Are you sure? yes | no