Output voltage as a function of output current is plotted from oscilloscope data for a closed loop buck converter.

Switching frequency is kept the same in both the cases.

CASE 1 : Inductance = 2.2mH/Capacitance = 1uF [Closed Loop] enter image description here

CASE 2: Inductance in decreased to 1mH and the capacitance is increased to 22uF [Closed Loop] enter image description here

I have to compare these results. I know that increasing the inductance effectively moves the buck converter into DCM.

However, in closed loop control conditions, the waveforms look almost identical. is there any difference? am I missing something?

  • \$\begingroup\$ Which waveforms look the same? Did you check step response? LC products are different. \$\endgroup\$ – Tony Stewart EE75 May 4 '19 at 0:49
  • \$\begingroup\$ Please provide more information: controller used, switching frequency, and any other piece of information you may feel important. \$\endgroup\$ – joribama May 4 '19 at 2:23
  • \$\begingroup\$ Are you sure the inductance is in mH and not uH? Unless your switching frequency is very low, these inductances are huge. \$\endgroup\$ – joribama May 4 '19 at 2:24
  • \$\begingroup\$ actually frequency & LC values have nothing to do with load regulation error (steady state) but greatly affect step response. \$\endgroup\$ – Tony Stewart EE75 May 4 '19 at 2:26
  • \$\begingroup\$ The main concern when you reduce the inductance is that the ripple current will increase. You have to make sure that DC current + 1/2 of ripple current is below the current saturation level of your inductor. \$\endgroup\$ – joribama May 4 '19 at 2:26

These are just a few short comments.

This data is just load regulation error.

Your voltage error vs current is mainly due to resistance ratio of series R to load R. The series R includes inductor DCR, switch Ron and/or Diode Rs.

When you use such a large L in your test, realize the step load from 100% to 50% causes a great step load voltage error V=LdI/dt and one in the 1 ~ 100uH range. But a change from 2.2 to 1mH is not so significant. This also means the start time is long. the XL(f)/R= Tau time for start time.

Also a large C reduces ripple but a large step Response time Tau=RC just delays the time for a voltage error with such a large C.

So now you know reactance has little effect on steady-state load regulation error. It's the series R of those reactors.


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