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Is it possible to give LEDs varying current, without sacrificing efficiency? Basically I have 4 LEDs which could all use varying current, because they are different distances from the receivers. The top LED doesnt need to be as bright as the bottom LED, since it is much closer to the photodiode.

I expected that I would be able to get away with less power, since I could more finely tune the current going to each LED, but it seemed to require more power to work as well as the first circuit did. Is my thinking on this wrong?

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    \$\begingroup\$ Have you done the math? If so, please share. If not, why not? \$\endgroup\$ – brhans Mar 13 at 19:30
  • \$\begingroup\$ Please share your resistor values. \$\endgroup\$ – TimWescott Mar 13 at 19:40
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    \$\begingroup\$ Also share your LEDs forward voltage and nominal current (hint: there's one of your two configurations that don't work, and it's possible for you to find out which yourself, and I hope it's relatively easy for you!) \$\endgroup\$ – Marcus Müller Mar 13 at 19:43
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    \$\begingroup\$ it didnt seem to work is not a useful description of the observed results \$\endgroup\$ – jsotola Mar 13 at 20:29
  • \$\begingroup\$ @brhans The tricky part with the math is I am not certain of what to make the ratio between the resistors. However, basic ohms law seems to suggest that splitting the lines generally requires more power. \$\endgroup\$ – Tapatio Sombrero Mar 13 at 20:52
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Assuming you set your goal to drive the LEDs at their minimal (optimal) currents i1 > i2 > i3 > i4, your initial (a) design will dissipate power as:

Pa = U × i1 + U × i3 = U (i1 + i3)

(Each branch driven at the highest current of the LED pair since shining brighter is OK, but dimmer isn't)

Your secondary (b) design goes as follows:

Pb = U × i1 + U × i2 + U × i3 + U × i4 = U (i1 + i2 + i3 + i4)

From that it's easy to see that the initial design is more efficient.

But...

Seeing that your supply voltage is 3.3 V, you should watch for the combined voltage drop of the LEDs. By having two LEDs in series you add up 2 * Vled which, if close to the supply voltage, leaves little margin (if any) for regulating the current through a resistor.

It's also useful to know that, leaving a small voltage drop across the limiting resistor calls for a small valued resistor. The consequence is that your circuit gets more sensitive to small fluctuations of the supply voltage and also depends more on non linearities of the LEDs themselves.

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  • \$\begingroup\$ The voltage drop is 1.5V which I guess would leave just .3V for my resistor. How do I calculate the margin of current I can regulate? Thank you. \$\endgroup\$ – Tapatio Sombrero Mar 13 at 20:58
  • \$\begingroup\$ Is it 1.5V or does it increase with the forward current? You should check the datasheet if available. Also does your power supply maintain 3.3 V or does it drop to 3.2 under load, or is it 3.45 to begin with? All these things can make a difference. \$\endgroup\$ – RaphaelP Mar 13 at 21:25
  • \$\begingroup\$ For the LED Im using the SFH 4545 and it seems to max out at 1.8V according to the datasheet. The power supply does maintain the 3.3V and I measured the voltage across the diodes and it was 1.2V each with the initial design using 56 ohm resistors. \$\endgroup\$ – Tapatio Sombrero Mar 14 at 0:06
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I thought I could be more efficient by making all 4 parallel [...] but it seemed to require more power

Yes. It's easy to understand when you consider that:

  1. the current through each LED must be the same to get the same brightness.
  2. in the parallel circuit the overall current is four times the LED current
  3. in the first circuit the overall current is only two times the LED current.

No need to look at voltages or resistor values.

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  • \$\begingroup\$ Yes this should be obvious. I guess I confused myself by expecting that by decreasing current on some emitters I could tune it to the point where it is overall more efficient but I guess that doesnt negate the extra current necessary for multiple lines. \$\endgroup\$ – Tapatio Sombrero Mar 14 at 0:09

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