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I am trying to select an output capacitor for an open loop buck converter to meet undershoot and overshoot requirements for a step change in load current.

Below is the equation to find the output capacitor as checked from multiple sources.

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For my buck converter simulation using LTspice, I have L = 100uH, Istep changing from 0 to 2A at t = 10ms, Vin = 12V, Vout = 6V (i.e. 50% duty ratio) and I am trying to meet undershoot of 0.5V.

The above equation gives Co > 133 uF.

I am using 150 uF in my simulation and yet it does not meet the undershoot criteria (Vundershoot is still around 1.5 V instead of 0.5V target.)

Can you please help to understand where I am going wrong? What else do I need to do to meet the target undershoot?

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    \$\begingroup\$ I think your equations are for a closed-loop buck. Even though they may assume that the transient is fast enough that the output cap has to supply most of the current, in an open-loop buck the control loop isn't there to keep the output filter from ringing. \$\endgroup\$
    – John D
    Commented May 18, 2023 at 14:57
  • \$\begingroup\$ Thanks for your response! I doubt that too if the above equations are applicable only for closed loop systems. However didn't find the note anywhere that "applicable for closed loop only". \$\endgroup\$
    – Dev
    Commented May 18, 2023 at 15:11
  • \$\begingroup\$ Simulation is probably your best bet for determining the amount of output capacitance you will need, but don't forget to include parasitics like ESR and ESL. \$\endgroup\$
    – John D
    Commented May 18, 2023 at 15:15
  • \$\begingroup\$ Yes, actually this way I can. But was very curious why these equations were not working as they are recommended by multiple reliable sources :) \$\endgroup\$
    – Dev
    Commented May 18, 2023 at 15:28
  • \$\begingroup\$ Well, the LC resonant frequency is usually well within the control bandwidth of the loop in a closed-loop buck, so you don't see the L-C ringing in that case. In open-loop operation the transient excites the L-C resonance and you get significant ringing. \$\endgroup\$
    – John D
    Commented May 18, 2023 at 16:33

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