I have one question about the bode plot of boost average current mode control.

Below is my schematic, the circuit is stable.

enter image description here

My design is for CCM boost PFC, and the current loop crossover frequency is 10 kHz, the voltage loop crossover frequency is 10 Hz. for current loop, I have already check the crossover frequency which is.

Below is my Mathcad calculation design when I am designing. As you can see the red line is the theory waveform, the total loop crossover frequency is actually in at the 10 Hz, the blue line is measuring from SIMPLIS, but the crossover frequency is only 2 Hz, I don't have any idea what’s going on.

I also measure the bode plot of voltage loop and current loop compensator, all the compensator are perfectly match with theory.

Do I measure the bode plot in the wrong place? If I did wrong, how should I improve it.

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I fine tune the di bias point to get a correct output voltage. enter image description here

I change the rail up voltage, but I don't very sure if it is ok. enter image description here

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  • \$\begingroup\$ Hi, can you please show the transfer function you have used for the analytical expression? For a PFC, you can restrict the horizontal axis from 0.1 Hz to 1 kHz for instance, no need to go to 100 kHz as crossover is only a few hertz. Your circuit will provide crude PFC only, for a true average mode PFC, you would need a multiplier as shown p571 in my book on SMPS simulations. \$\endgroup\$ May 1 at 12:03
  • \$\begingroup\$ @VerbalKint, I update my design process. about the compensator, I have already checked each one is good. I have read your book, I want to make sure the boost converter is stable that's why I want to do the boost first, after that I will add multiplier in my circuit. \$\endgroup\$
    – Power JJ
    May 1 at 12:27
  • \$\begingroup\$ I have seen inconsistent results sometimes with the boost converter when inserting the ac source as you did. Try to physically open the loop by setting the exact dc voltage at the resistive divider input. You can, for this exercise, lower the open-loop gain of the first op-amp to 60 dB and it will let you find the correct operating point more easily, e.g. an output voltage around a few volts (away from the upper and lower rails). \$\endgroup\$ May 1 at 12:42
  • \$\begingroup\$ @VerbalKint Do you mean that? If I do this, the output voltage is not correct. my output voltage is 400V, and I open the loop and give the constant voltage. \$\endgroup\$
    – Power JJ
    May 1 at 13:39
  • \$\begingroup\$ Yes, but the difficulty in doing this is to exactly tweak the 400-V source to maybe 399.899 V so that the output of the 1st op-amp is 2-3 V or so (linear mode). If any of the op-amps rail up or down, you need to better fine-tune the dc bias. Besides, in this mode, the input is well \$V_a\$ but node \$V_b\$ must be connected to \$V_{out}\$ of the boost converter. \$\endgroup\$ May 1 at 14:55


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