I'm trying to design a boost converter for a backlight which has Vf:10.6-13.6V, If:140mA
I picked AP3031, I did steps required to find component details e.g Inductor Value, Iswmax


So far I understand. Low input in and you get high input voltage out.

According to the application note, if Iled is set to 140mA, the corresponding \$R_{ISET}\$ value is 1.43 ohms since the feedback reference is fixed at 200mV. However, I'm having trouble grasping the exact function of \$R_{ISET}\$/ fixed 200mV and its impact on the output voltage. And how do I achieved dimming if needed?

Block Diagram

Circuit I'm using

Source of both above images: Mouser - AP3031 Data Sheet from BCD Semiconductor

  • 2
    \$\begingroup\$ FYI, running strings of LEDs in parallel with a single current limiting resistor or other current limiting circuit (as is the case in your schematic) risks the strings carrying different currents, and may shorten the lifespan of the LED strings. \$\endgroup\$ Apr 17 at 2:06

1 Answer 1


The LEDs must be driven from a current source not a voltage source. The circuit shown applies a constant current to the LEDs automatically adjusting the voltage s required.

\$R_{iSET}\$ converts (senses) the total LED current to a voltage which is then compared to the 0.2V reference through the FB input pin. If greater than 0.2V the output voltage is reduced. If less than 0.2V, the output voltage is increased. The voltage may vary as you suggest between 10.6 to 13.6 depending on the LEDs. The current through LEDs is therefore fixed at 140mA regardless of the voltage requirement.

The total output voltage is \$V_{LED}+V_{RiSET}\$.

To dim the LEDs adjust \$R_{iSET}\$ to control the brightness. Increase \$R_{iSET}\$ to dim, decrease \$R_{iSET}\$ to brighten.

  • \$\begingroup\$ Hi Russell, Thanks for the quick reply. It was my mistake I didn't full read the operation mode from the datasheet and I mistakenly assumed it functioned solely as a voltage source. I think I understand better now. \$\endgroup\$
    – S_D
    Apr 17 at 2:15
  • \$\begingroup\$ Your welcome. @S_D \$\endgroup\$
    – RussellH
    Apr 17 at 6:28

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