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I am working on a soft-start for a capacitive load. I using module QSB40024S24 to create 24V output. At first, I allowed the output of the module to connect directly to a capacitor array (21.4mF in total) in parallel with the load. So you can imagine the inrush current is quite high and it causes a large drop voltage at Vin. So the module ran into the UV protection so that the output capacitor array could not be fully charged. After many tries fail,(ex: soft start with driving Vgs of mosfet or work with NTC mostly fail with my short-circuit test) I found this way: enter image description here

It worked well and ensure that the capacitor would be in full charge. If the load suddenly consumed more current the capacitor will discharge through the Schottky diode. It worked well even with a short circuit test. The only problem I see is you have to choose the power resistor to endure the high current at the initial time when V_cap=0. Does this design still remain any problem that I can not see for a soft-start circuit? Any idea would be very appreciated!!

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  • \$\begingroup\$ It is very inefficient. \$\endgroup\$
    – D.A.S.
    Commented Aug 7, 2021 at 5:12
  • \$\begingroup\$ Why Tony? Can you explain more? \$\endgroup\$
    – Lan Tran
    Commented Aug 7, 2021 at 5:29
  • \$\begingroup\$ @LanTran I've no idea if you can consider an inductor. Back in the day, we used them. You can examine what I write here to see if any of it makes sense in your situation. Usually ... not ... these days, considering the space and cost. But there it is, anyway. Also, your approach seems okay to me. I like it, too. \$\endgroup\$
    – jonk
    Commented Aug 7, 2021 at 6:49
  • \$\begingroup\$ With R1 almost = R2. you can see 50% of the power is lost., but using a DCDC converter with current limit will be more efficient. \$\endgroup\$
    – D.A.S.
    Commented Aug 7, 2021 at 11:50
  • \$\begingroup\$ Power resistors often specify a higher power rating for short pulses at low duty cycle. So, reading resistor datasheets carefully may allow you to economise... @TonyStewartEE75 that 50% efficiency is only momentary as C1 charges. For rare brownouts it's not worth worrying about. \$\endgroup\$
    – user16324
    Commented Aug 7, 2021 at 14:58

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