This is the control stage circuit of flyback, I want to use SIMPLIS to measure the bode plot to prove a theory is matched to simulation, I use Mathcad to draw the bode plot, the SIMPLIS result should be the same as Mathcad, but when I run the simulation, the result is wrong. but I don't know where I wrong, this issue stuck me for a long time.

This is my simulation result. enter image description here

I want to check the bode plot of Power Stage and Control Stage are correct

Hope someone can give me some help.

enter image description here

enter image description here

  • \$\begingroup\$ Can you show the complete circuit? \$\endgroup\$
    – HarryH
    May 22, 2021 at 16:05
  • \$\begingroup\$ This is my control stage circuit \$\endgroup\$
    – EEC
    May 22, 2021 at 16:21
  • \$\begingroup\$ If you do not disclose the complete SIMPLIS schematic, we cannot provide help. Your Bode plot is that of a type 2 compensator but the circuit you show is a proportional gain with a pole. Have a look at the automated SIMPLIS template for the flyback I posted on my page. It is part of a free 60+ template batch you can download. \$\endgroup\$ May 22, 2021 at 17:09
  • \$\begingroup\$ @VerbalKint Sir, your SIMPLIS file is based on a closed-loop system, if I just want to measure my control stage bode plot, I mean just control stage circuit, like my post. How do I test, because I want to learn the control stage bode plot. for example, if I want to know the Type II compensator, I can just use opa to do it, and then I can measure the bode plot and I don't need a power stage circuit. \$\endgroup\$
    – EEC
    May 22, 2021 at 17:23
  • \$\begingroup\$ @VerbalKint for the flyback, it needs to use opto to implement the Type II compensator, I want to just test the control circuit, but I don't know how to use it. right now my control stage using a P control and opto, the opto has a pole, so I think my bode plot should have a pole, but it doesn't. Is my control stage circuit wrong? \$\endgroup\$
    – EEC
    May 22, 2021 at 17:26

1 Answer 1


In your comment, you state that the flyback circuit I built in the templates described here is operated in closed-loop while you want the compensator response. It is true, the loop is closed in dc to let the converter regulate at the right dc value. However, because the ac source is inserted in series with the upper side of the resistive divider, it has the effect of perturbing the loop and because of its constant value, it effectively opens the loop in ac. So the Bode plot you draw while observing the voltage across the injection source is actually an open-loop gain response. This technique has been introduced by Dr. Middlebrook in the 70s and is described here. I have explained it in my book on transfer function with several SPICE examples. The SIMPLIS example looks like this:

enter image description here

By probing various points, you can extract:

  • the compensator transfer function. Here, this is a type 2 with an opto-coupler
  • the control-to-output transfer function. This is the power stage ac characteristic you need to think about a possible compensation strategy
  • the loop gain: you can finally check if the compensation strategy is adequate to meet crossover and phase margin goals. The final results are shown below:

enter image description here

Using this technique, you can run the switching converter in dc closed-loop operation while you open the loop in ac for the analysis you need. Using a frequency response analyzer (FRA) in the lab would not be different.

  • \$\begingroup\$ Sir, since we need the power stage characteristics first to design the compensator so in the technique you mentioned do we need to use arbitrary values for the elements used in the feedback network as long as we get the correct dc operating point/steady state values at output or am I missing something? \$\endgroup\$
    – tinkerer
    Oct 25, 2021 at 9:50
  • 1
    \$\begingroup\$ Before attempting to compensate a converter, you need its control-to-output transfer function. You can obtain it from SIMPLIS by running the circuit in open-loop conditions where you bias the FB pin with a dc source and a superimposed ac modulation. Once you have this transfer function and can think of a compensation strategy (what my automated sheets do for you), then you can run SIMPLIS in closed-loop conditions as I did in my example. \$\endgroup\$ Oct 25, 2021 at 10:23
  • \$\begingroup\$ Ok thx, got it now. \$\endgroup\$
    – tinkerer
    Oct 25, 2021 at 11:35

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