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I use a FP6296XR-G1 (U10 / 10A boost converter) together with a MBR20100CS (D5 / dual Schottky diode in a TO252-2 package) in order to make 13 V out of 2.5-5 V. When I connect BOOin to 5 V from an Arduino, it sucks a lot of current (LEDs go dark) first and then it settles at 11 mA.

Are these 11 mA the internal LDO of the U10, that sucks 4 mA? And to make 13 V with 4 mA (=52 mW) it needs 4.7 V@11 mA?

Or does D5 need a lot of time to stop the flow back, when U10 shorts the inductance to ground? I could not find any t_rr (reverse recovery time) in the datasheet of D5. And I'm just an informatician, so that my knowledge about diodes is quite limited. In "production" another part of the circuit will need up to 12 V@1 A until a bolt lock has been pulled open.

cutout of the schematic (painted by KiCad 7.0)

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  • \$\begingroup\$ Please refer to components by their reference designator to avoid confusion. After all BAT54 diodes are Schottky types too. Please add pdf data sheet links for the IC and the other components of relevance to your question. \$\endgroup\$
    – Andy aka
    Jan 18 at 10:23
  • \$\begingroup\$ @Andyaka it is done... \$\endgroup\$
    – RRIDDICC
    Jan 18 at 12:34
  • \$\begingroup\$ When I connect BOOin to 5 V from an Arduino, it sucks a lot of current (LEDs go dark) first and then it settles at 11 mA That does not sound good. Please describe the 5V source, current, power, etc. \$\endgroup\$ Jan 18 at 22:35
  • \$\begingroup\$ @StainlessSteelRat the 5V source is from a USB port... so there is a current limit at 500mA... and it seems like my mainboard turns off a misbehaving port for a short while... i cannot investigate that so good, because my oscilloscope does not work so good... \$\endgroup\$
    – RRIDDICC
    Jan 19 at 7:52

2 Answers 2

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A 10 A capable, 400 kHz converter operating at very low current and does not feature skip or burst mode will have high static consumption/bad light load efficiency.

What the converter "needs" 11 mA for is mainly switching losses. For a converter made for high current (10 A), the MOSFET (parasitic) capacitances will scale accordingly. Total switching losses will go up with higher output current but just outputting a tiny duty cycle will mean charging and discharging all parasitic capacitances. Combine that with a high fixed switching frequency of 400 kHz, it will mean a tiny amount of energy needed 400,000 times per second. It adds up.

Look for a (synchronous) boost converter with skip or burst mode if you want to save losses. Preferably from a reputable vendor. They will instead of trying to switch a tiny duty cycle all the time simply go big for a a few cycles to charge the output capacitor and then stop switching until your load has discharged it below a threshold, where it will start again. Your switching frequency could still be 400 kHz, but it will only run for say 1 % of the time so your switching losses will be potentially 99 % less.

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  • \$\begingroup\$ can it be that the FB/COMP networks cause that DCM behavior, that Rohat K. describes in the other answer? I mean: maybe the switch never switches, if FB is high enough or COMP denotes a certain trend? \$\endgroup\$
    – RRIDDICC
    Jan 20 at 20:01
  • \$\begingroup\$ @RRIDDICC At light load, you want DCM. At forced CCM, your switch losses you’ll be even higher. With asynchronous rectification, this isn’t an option to you. \$\endgroup\$
    – winny
    Jan 20 at 20:15
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The snubber network across the LX pin will consume some power.

The network shows ~485 Ohms at 400 kHz. If we assume CCM the duty cycle should be ~0.6 for 5V input and 13V output. So the voltage across the snubber network will be a pulse train with a duty cycle of ~0.6V and with a peak of ~13V. This makes the RMS voltage ~10V, therefore the snubber will draw 10/485 = 20 mA already. But for DCM, if there's something like pulse skipping or hiccup/burst mode the average current might be lower.

Also, if the purpose of the network on top-right (Q24, D9, R64, and R65) is self supply (i.e. supply the IC from the output once the output is present) then expect some more current consumption from the input.

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  • \$\begingroup\$ thx... when u write "13V input": do u mean "13V output"? \$\endgroup\$
    – RRIDDICC
    Jan 18 at 12:39
  • \$\begingroup\$ @RRIDDICC yeah, sorry. Fixed now. \$\endgroup\$ Jan 18 at 12:58
  • \$\begingroup\$ so at a duty cycle of 0% it will need 5.85mW and with a switching frequency of 400kHz it will need 52mW... in addition the snubber network needs 0.5*820*10^-12*13.3^2*f [W]... by trial and error and with ChatGPT i got f=137kHz... so it skips some cycles? and the frequency is down from 400kHz to 137kHz? \$\endgroup\$
    – RRIDDICC
    Jan 20 at 20:38

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