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In a previous question, I asked about an issue in this booster circuit.

schematic

I resolved that issue.

However, I have another problem. There is a 16V 4700uF capacitor (not shown in the schematic) at the output of the circuit and this capacitor charges up to 3.6V (input voltage of the booster) when the booster circuit is not active.

  1. Is this because of the inductor?
  2. I think I can isolate the circuit from the input voltage by using an external MOSFET, but can I solve this problem without using a MOSFET?
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    \$\begingroup\$ The 4700 uF capacitor will draw only leakage current. Look at the data sheet and see what this current is. It should be very small. \$\endgroup\$
    – Russell McMahon
    Dec 5, 2022 at 10:01
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    \$\begingroup\$ As Andy says, the capacitor will charge through D!. At very low currents - such as the leakage current of the 4700 uF capacitor, the drop across D1 will be small and Vout will approach Vbat. As current increases to the several mA range diode drop will rise. If Vout is almost the same as Vbat then the current must be small. \$\endgroup\$
    – Russell McMahon
    Dec 5, 2022 at 10:03
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    \$\begingroup\$ I'm a moderator: Each question should stand alone. It is a good idea to provide a link to any prior related question(s), as you have done. || Including the relevant schematic in a new question allows people to understand the question without having to look elsewhere. || In this case you say "There is a 16V 4700uF capacitor (not shown in the schematic) at the output of the circuit ." Presumably this is electrically in parallel with C17. Stating that explicitly would make it certain what was intended. (This is not essential - just helps people help you). \$\endgroup\$
    – Russell McMahon
    Dec 5, 2022 at 10:12
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    \$\begingroup\$ In shutdown R8 will draw I= V/R = 3.6/51000 =~ 70 uA. This is about 10% of eg this capacitor family at 4700 uF 16V - but worth being aware of At 3.6V cap leakage is probably more like 100 uA - so R8 current is significant. \$\endgroup\$
    – Russell McMahon
    Dec 5, 2022 at 10:22
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    \$\begingroup\$ (1) 3.6V - Diode D1 will drop voltage at operating currents. If it is a Schottky diode then it will drop typically 0.3v or more. If it is a silicon diode then 0.6V or more. IF the capacitor is at 3.6V then D1 is dropping very very little current as Vd1 ~= 0V. SO the capacitor is not wasting much battery energy. 2. R8: R8 also wastes battery energy. If R8 = 100k then it will sink about 36uA to ground if Vshdn is 0V. This MAY be more than the output capacitor is wasting. \$\endgroup\$
    – Russell McMahon
    Dec 7, 2022 at 11:30

1 Answer 1

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A voltage boost circuit produces a voltage that is higher than the input voltage so, if your input voltage is 3.6 volts, it cannot produce an output lower than this (minus the volt drop of the diode). The clue is in the name "booster". If you wish to be able to disengage the output from feeding the capacitor that you mention then, you will need to add another MOSFET or similar mechanism. Basic boost circuit: -

enter image description here

As you can see in the diagram from my basic website, the lowest voltage on the output is VIN - 0.7 volts because, with the MOSFET deactivated, the output connects to the input supply via the inductor and diode.

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    \$\begingroup\$ If my 16V 4700uF capacitor at the output will stay constant at 3.6V when the booster is not active, then my battery life will be affected a lot :/ I'll think about the idea of ​​adding another MOSFET. Thanks! \$\endgroup\$
    – harmonica
    Nov 28, 2022 at 10:58

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