I am designing a high current LED driver circuit to power a matrix of pulsed IR LEDs. The LEDs will operate at 5A; 4.6V; 0.1ms pulses (~10 pulses at this frequency, wait 300s, repeat.) The device is powered off a 3.3V Li-Po battery.

In order to deliver sufficient power, I need a DC-DC converter as part of my driver circuit. I would have assumed that a boost converter would be a no-brainer for this application, however doing some reading seems to suggest that for such high-current applications, the size of properly rated inductors would start to become prohibitive, and so I am wondering if it is worth looking at charge pumps instead.

I've done some reading on it (https://www.maximintegrated.com/en/design/technical-documents/tutorials/7/725.html) and while the losses would be much higher than a boost converter, would the size and cost savings be significant enough to support the use of a simpler charge pump? Are there any drawbacks from a charge pump that makes it unusable for a battery-powered device?

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    \$\begingroup\$ Isn't this article exactly what you are looking for: maximintegrated.com/en/design/technical-documents/app-notes/3/… ? \$\endgroup\$
    – Eugene Sh.
    Commented Jan 22, 2020 at 19:17
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    \$\begingroup\$ IR LEDs usually don't have 4.6V forward voltage... Can you rewire your IR LEDs to have Vf lower than battery voltage? \$\endgroup\$
    – bobflux
    Commented Jan 22, 2020 at 19:31
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    \$\begingroup\$ @EugeneSh. Hmm that is useful, however that is at much lower current than my application. Still insightful however, will take that into account! \$\endgroup\$ Commented Jan 22, 2020 at 19:45
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    \$\begingroup\$ @peufeu No can do, the LED in question typically operates at a much lower voltage but for my application we need maximum radiant power (Page 2 on this datasheet vishay.com/docs/84249/vsmy99445ds.pdf) \$\endgroup\$ Commented Jan 22, 2020 at 19:46
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    \$\begingroup\$ "losses would be much higher" Seems like a pretty big drawback for a battery-powered device. "the size of properly rated inductors would start to become prohibitive" I think you have this backwards. The biggest drawback of capacitive switching compared to inductive switching is their limited current capability. I've never seen capacitive converters of greater than 60mA, whereas you can easily find inductive converters in the tens of amps, and far higher than that can be found. Remember, higher losses tends to mean everything has to be bigger for the same power to handle the inefficiency & heat \$\endgroup\$
    – DKNguyen
    Commented Jan 22, 2020 at 20:33

1 Answer 1


You do not say how long between pulses in your bracket of 10 x 0.1 ms pulses - this may alter the solution (slightly easier)

Your peak power is P = I x V = 5A x 4.6V = 23 Watts.
This is not an overly demanding boost converter requirement, but even that is not essential.
Your mean power is
ton/tcycle x Pmax = 0.001/300 x 23W = < 0.1 mW = tiny.

As long as you can store enough energy to provide the LED during pulsing you can recharge it at leisure.

If using a capacitor as a direct to LED store then the allowed voltage droop sets the capacitor size.
C = t x i / V
t - pulse time total, i current, V allowed voltage droop

For 0.1V allowed capacitor droop - C = (0.0001 x 10) x 5A / 0.1V = 50 mF (50,000 uF)
If you allow 1V droop then 5000 uF suffices.

If you follow the capacitor with a 5A LDO (low dropout regulator) with say 6V into LDO then a say 10,000 uF capacitor would allow zero output droop.


Charged to 4.6V this 3.5F supercapacitor will drive the LED directly without an LDO being needed. Wiring drop is liable to be more than capacitor droop over 10 pulses.

This $US3.94/1 3.5F capacitor will supply the LED with 'nary a ripple' (OK - maybe 2 mV droop over 10 pulses). It will take a few minutes to charge intially with a 5V x 0.1A supply and thereafter recycle after each pulse train in well under a second. A 5V x 1A inverter will shorten the initial charge accoringly.

DGH355Q5R5 3.5F supercapacitor. Pricing page.
Shortform datasheet
This 1.5F part is notionally outside its Imax spec BUT as that is for 1s your 1 ms requirement may not phase it. Or may.
It's hardly any cheaper - the 3.5F part is more attractive.

  • \$\begingroup\$ So, how should the 3.5F supercap be charged from the 3.3V Li-Po for the 10x0.1s 4.6V 5A output? \$\endgroup\$ Commented Oct 24, 2021 at 21:03
  • \$\begingroup\$ @MicroservicesOnDDD The question assumes a boost converter - with the main discussion being between capacitor or inductor converter. \$\endgroup\$
    – Russell McMahon
    Commented Oct 24, 2021 at 22:47
  • \$\begingroup\$ So you're suggesting a boost converter using a supercapacitor? Like one of the chip solutions, and the flying capacitor is a supercap? \$\endgroup\$ Commented Oct 24, 2021 at 23:46

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