I am working on adding a super-capacitor to one of my 5V lines. Foolishly I tried adding the super-capacitor directly to the 5V line, but it over stresses my regulator to charge it all at once.

I am trying to design a very simple charging/ discharging circuit. Would this be the proper way to charge a large (2.2F 5.5V) super-capacitor?

enter image description here

  • 2
    \$\begingroup\$ If you think about it, you can probably omit D12, which will allow charging fully to 5V. \$\endgroup\$
    – user16324
    Nov 11, 2014 at 18:21
  • 1
    \$\begingroup\$ I think you need another Schottky in the 5V line just before the junction of D13 and the 5V line to the right; otherwise when the 5v line on the left falls, your cap is going to be driving it backwards, which may not be a good idea. \$\endgroup\$
    – tcrosley
    Nov 11, 2014 at 19:55
  • \$\begingroup\$ I didn't show the whole schematic. In my case I do have a mosfet separating my 5V regulator from the 5V line you see above. But if someone did not have that FET in place, that is a great peace of advice. Also Brian is definitely correct D12 is pointless. \$\endgroup\$
    – apaul
    Nov 13, 2014 at 2:44

2 Answers 2


What you have will work, although D12 is pointless. The problem is that when the cap is discharging onto the 5 V line, there will be a drop across D13. Using a Schottky as you show is a good idea, but the drop will still be around 1/4 volt. Another problem is that the voltage will go lower over time with the amount of charge drained from the cap.

You might consider putting the energy backup capacity before the power supply. That might allow the stored energy to be used more efficiently, and the voltage will stay regulated for a while.

  • \$\begingroup\$ Thank you for the feedback. I will eliminate D12. My problem with adding the capacity before the 5V regulator is that the power line feeding it is 12V. I would need to buy much more expensive capacitor / capacitors. My system will run down to 2.5v, so you are correct that with my current design I am going to lose quite a bit of my stored capacitance. \$\endgroup\$
    – apaul
    Nov 11, 2014 at 18:48

If you charge your capacitor through a resistor, the charging efficiency will be less than 50%. You should try charging your capacitor through an inductor if you are designing a power circuit.

The circuit below may not apply to your circuit because of its complexity, but you can use the forward converter topology as seen as a general solution to the efficiency problem. The major losses are only on the diodes if the MOSFET is driven properly and the transformer is designed correctly.


simulate this circuit – Schematic created using CircuitLab


Your Answer

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

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