enter image description here

I want to charge a capacitor with a MOSFET, but it is charging and discharging much slower than I want.

The circuit, which charges in an average of 5 seconds without using a MOSFET, is charged and discharged very slowly in about 3-4 minutes when using a MOSFET. Do you see a problem with the circuit?

The capacitor voltage is 2.7V.

  • \$\begingroup\$ I used a level drive directly on Q3 and Q5 +/- 5V, no problem. What are your MOSFETs? What is the value of super-capacitor? \$\endgroup\$
    – Antonio51
    Jun 3, 2022 at 6:59
  • 1
    \$\begingroup\$ Q3 might need more voltage on its gate. What does the datasheet tell you? \$\endgroup\$
    – Kartman
    Jun 3, 2022 at 7:24
  • \$\begingroup\$ MOSFET= AONR21321 CAP= MAL222051011E3 \$\endgroup\$
    – sinekel
    Jun 3, 2022 at 9:06
  • \$\begingroup\$ Ok. The capacitor is 5 F. A "little" lower than 100 F in my answer. \$\endgroup\$
    – Antonio51
    Jun 3, 2022 at 9:11
  • \$\begingroup\$ Q3 and Q5 are not the same ... Anyway, you must drive gate "directly" with +/- 10 V, min +/- 5 V) \$\endgroup\$
    – Antonio51
    Jun 3, 2022 at 9:16

1 Answer 1


I guess that capacitor used is a super-capacitor (100 F, 2.7 V rating).
Here is a theoretical circuit (example) that does it, see the values of resistors needed ...
And that, "hard" switching is also needed.
See the current needed through the capacitor.
Switches needed something as this MOSFET-P MOSFET-N enter image description here

Here are the waveforms with a direct drive of the switches (Half H-bridge).
NB: other MOSFETs, don't have original OP.
You can also drive the MOSFETs with two independent signals. Be careful in this case.

enter image description here

  • \$\begingroup\$ Why downvote? The answer is "wrong"? \$\endgroup\$
    – Antonio51
    Jun 3, 2022 at 11:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.