After reading on regulators, and asking few questions here and there, I think I should change my circuit to the following scheme : basic power circuit

These images are from datasheet of the LDO regulators (both MIC29302A).

first graph

dropout voltage


Questions : Does the first curve mean that the output voltage will change with changes in output current? (since the required current will be anything from zero to ~920mA as the LEDs will be turned on/off). if so, Should I adjust the first regulated voltage (marked in red in diagram) to 5.8v or what?

Basically I want constant voltage for the LEDs (the voltage marked in blue in diagram). and need to adjust all other parameters (other regulator voltages?) for this purpose.

  • \$\begingroup\$ The dropout voltage is the voltage across the regulator that below this value, the regulator can no longer regulate - ie the output voltage will decrease. According to the graph, you need at least 5.2V in for 5 V out at 1A. Is there a reason you want to separate the led and the controller supply? Why not just two switching regs? \$\endgroup\$
    – Kartman
    Apr 30, 2021 at 9:04
  • \$\begingroup\$ @Kartman I've read it's the best cheap way to get rid of noise and ripple from switching regulator. \$\endgroup\$ Apr 30, 2021 at 9:25

1 Answer 1


The dropout voltage is the minimum input to output voltage that allows the regulator to still regulate. Knowing the output voltage, that gives you the minimum input voltage for regulation at the current you want.

Note power supply rejection ratio (PSRR) of LDO decreases as they go near dropout, so YMMV. But, PSRR is not specified in the datasheet anyway.

You don't need a linear regulator for your LEDs. Just use the output of the switcher directly. If they are WS2812B they already regulate their current. If they're just LEDs, you can wire them in series and use 12V directly. It's much simpler.

Unless the micro is doing sensitive analog things you also don't need a LDO for the micro, it'll run just fine with a bit of switching noise on VCC.

However, I hope you don't plan to use the counterfeit cheapo "LM2596" modules?...

  • \$\begingroup\$ 'counterfeit cheapo "LM2596" modules' - is there anything but counterfeit modules?? Along with the various other 'knockoff' modules.......... We get what we pay for. \$\endgroup\$
    – Kartman
    Apr 30, 2021 at 11:32
  • \$\begingroup\$ Thanks for clarifications. So I should tune the Vin of the LDO based on the needed output voltage. (And ignore the output current requirement of leds?!!!) The first graph still confuses me. Cuz it suggests that the Vin depends on the output current. The LEDs are not in series and each one is controlled independent of others. (Hence the wide range of current requirement). Is it good solution to just set the Vin of LDO to 6v and not to worry about a little more heat dissipation? \$\endgroup\$ Apr 30, 2021 at 11:36
  • \$\begingroup\$ Between two LM2596S ics, I bought the expensive one and hope it's not fake. (Costs like 8x) \$\endgroup\$ Apr 30, 2021 at 11:39
  • \$\begingroup\$ The MINIMUM Vin depends on the output current. According to the graph 5.2V is the minimum, so 6V is fine. You'll want to do your temperature rise calcs - it might get toasty. \$\endgroup\$
    – Kartman
    Apr 30, 2021 at 12:16
  • \$\begingroup\$ Say you need 5V, at X amps, look LDO datasheet for max dropout at X current, you need at least 5+X volts at the input. Add tolerance for your buck. So yeah you need to factor in current. But what you should especially do is not use a LDO to power LEDs, just use the buck regulator, it's fine. \$\endgroup\$
    – bobflux
    Apr 30, 2021 at 16:33

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