I have a display that has six white LEDs in parallel (common anode) as backlight.

The LEDs' forward voltage is 3.2V, the current per LED is 20mA, so 120mA in total.

The available voltages in the circuit are 3.3V and 5V.

Of course I could use 5V and a serial resistor for every LED, but that is not very efficient, and the device which contains the display is supplied by a battery.

PWM control should also be possible...

Most small LED boost driver ICs need their LEDs in series. Using small DC-DC converter to create a voltage between 3.3V and 5V, and smaller resistors would be better, but PWM should probably be done with a MOSFET, not with the EN pin of the DC-DC converter.

What is the best solution to power the parallel LEDs?

  • \$\begingroup\$ Put them in series and use a current limiting booster. \$\endgroup\$
    – Andy aka
    Apr 23, 2020 at 16:19
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    \$\begingroup\$ ^^^ Spatz stated the LED's are common anode - Can't put them in series. @Spatz - Is your primary concern power efficiency, cost, size of PCB, or ??? There are dozens of ways to do this, but each has tradeoffs. For example, if you're making only one of these as a demo, I'd approach it much differently than if you're making 100,000. Can you share a little more about the nature & goals of your project???? \$\endgroup\$
    – Kyle B
    Apr 23, 2020 at 16:23
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    \$\begingroup\$ I'm going to buck the trend here (and in general) and tell you to wire all your LEDs directly in parallel with each other and use a single resistor for them all (preferably from the 3.3V supply for better efficiency). The LEDs used in a commercial display backlights are going to be so closely matched that all the traditional (and sensible) reasons to either run them in series or use individual resistors do not apply. Open up any commercial LED "light engine" and you'll find LEDs grouped into parallel banks of easily 5 or 6 LEDs, sometimes more - bcos the manufacturer knows they'll be matched. \$\endgroup\$
    – brhans
    Apr 23, 2020 at 17:33
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    \$\begingroup\$ " that is not very efficient, " Say what? You're at 66%, which is not all that bad. Put it another way - Assuming the LEDs are the only load on the battery, going to a 100% efficient limiter will only extend battery life by about 50% compared to using resistors. If, for instance, you're also drawing 30 mA for the rest oft the electronics, going to a 100% efficient driver will gain you 40% in lifetime. With no other current drain and an 80% efficient driver you'll get 20% extra battery life. Are you sure this is an issue? \$\endgroup\$ Apr 23, 2020 at 17:39

2 Answers 2


You're on the right track with the DCDC converter. Use the 5V and step down, and sense the LED current on the low side. If you choose a regulator that has a 0.6V reference, the loss will be low - you'd use a 5 ohm sense, so the loss would be 72mW at 120mA.

If you want even lower loss, use an op-amp with a lower-value sense resistor to boost the sense voltage up to the reference. That's also an opportunity to apply dimming.

On the other hand, using a purpose-built current-sensing LED driver will use a much lower overhead voltage for sensing as it has the sense amp built in. More about this below.

To get dimmability, you can feed in a PWM chop to the sense voltage to vary the current, or if you have the space, a current sink/source DAC can do that (Maxim makes an I2C part that does this.)

If you're not so fussy about the chop frequency, just use the EN pin of the device.

Here's a cheap DCDC that I've used many times from Diodes Inc. It has direct crosses with a lot of vendors including Richtek, TI and others. It's 2A, more than you need, but very small. https://www.diodes.com/assets/Datasheets/AP3402.pdf This would use 5 ohm sense, or if you insert an op-amp, you could use a lower sense value. I've bought AP3402's (or similar) for as little as 7-10 cents in volume.

This is a purpose-built, quite fancy buck DCDC LED driver with high-side sensing, dimming and other features, also from Diodes (Zetex): https://www.diodes.com/assets/Datasheets/ZXLD1320.pdf It's a complete solution. This part's over a dollar. Much more expensive than the AP3402 + op-amp.

And finally, probably the best solution. Since I seem to like Diodes so much, here's a cheap-and-cheerful LED DCDC with 0.1V reference voltage for low-side sense, and it specifically supports PWM chop: https://www.diodes.com/assets/Datasheets/PAM2804.pdf You'd use a 0.8 ohm sense resistor with this. Very cheap too: 16 cents from Digi-Key if you buy a whole reel.

One problem with LED backlights is matching. If the individual LEDs have small variations in Vf, then when they're driven in parallel there will differences in brightness: the ones with the lower Vf will get the lion's share of the current. To help with this, even with a current mode driver like I'm proposing, keep a small series resistance on each LED cathode (maybe 10-20 ohms) to help compensate for the variance. If they are matched, so much the better, then you can replace these with zero ohms and reduce the losses.

A more sophisticated approach would be to use multiple current mirrors. That's probably more complicated than you want to get into.

  • \$\begingroup\$ Thank you, the PAM2804 was exactly what I was looking for! \$\endgroup\$
    – Spatz
    Apr 24, 2020 at 13:49

To maintain even brightness with each LED would require a resistor connected to each LED. To save power, you can reduce voltage using a DC/DC converter. If you brought your voltage from 5 to 3.4V you can connect 10 ohm resistor for each LED. Total power loss in resistors of 24mW. If using a DC/DC converter with 80% efficiency and 5V input and 3.4V output with output delivering 0.12A, then you have power loss in dc/dc converter of 102mW. Lot better than using 5V directly to give 216mW of power dissipation.


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