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I am analysing the an eval board circuit for a battery charger IC which has the following inrush control circuit.

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Question 1: I understand that C22 is used to "slow start" Q6 limit inrush current into C3,C4,C23,C24,C44. Signal CHG_OK goes high, when a power source is detected at J1 (12V). With CHG_OK high, Q10, Q9B and Q9A are turned on. I don't understand resistor R10 because if Q9A is on, ~12V are applied on Q6 gate, which will keep Q6 closed. How does this inrush control circuit work?

Question 2 Why are there several capacitors (C3,C4,C23,C24,C44) and why can't they be replaced with just one large capacitor?

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  • \$\begingroup\$ The multiple capacitors are probably decoupling multiple chips, or multiple power pins on one chip, and will be placed close to those power pins. \$\endgroup\$
    – Hearth
    Commented Jul 28, 2019 at 14:46

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1: When CHG_OK is high the R10 is open-circuit and both series pass FETs conduct with a slow turn-on time. When this signal goes low R10 shunts Vgs quickly as the 0.1uF is discharged in 1us then FETs are shut off.

2: low ESR caps of 10uF have an inherent \$\tau=\$ESR*C<=10us constant in any given family at some Vdc rating and size.

  • High ESR or General Purpose has a τ>200us thus high ESR
  • you can reduce the ESR by using parallel Caps of the same size but not by increase C value unless the sub-miniature SMD size is also substantially increased. However, experience dictates that shunting smaller low ESR Caps improves results immensely as the length of the cap also increases inductance 0.5nH/mm which then reduces the Series Resonant Frequency (SRF) which increase rise-time of current.

Expect a typical Panasonic 10uF low ESR to be < 1 Ohm and 10 in parallel = 0.1 Ohm ESR which thus reduces voltage rise from dV=I*(ESR+dt/C)

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  • \$\begingroup\$ 1: I made a mistake. Of course, when CHG_OK is high, then Q9A and Q9B are OFF. 2: understood. Is there any "best practice" on how to design these decoupling capacitors in terms of how many should be put in parallel? \$\endgroup\$
    – F. Heisenberg
    Commented Jul 28, 2019 at 15:13
  • \$\begingroup\$ Yes but you need to know what your operating frequency is and current thus Impedance or dV/dI \$\endgroup\$
    – D.A.S.
    Commented Jul 28, 2019 at 15:13

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