Calculation of control to output transfer function

Question

EXPLANATION

Picture 1 is control to output transfer function. (For those wondering how I obtained this formula, you can check Control to output transfer function. This formula was obtained with the help of VerbalKint.)

Picture 2 is my calculations about pole, dc gain, zero and phase.

Picture 3 is the LTspice simulation.

Note: Question is very easy to follow for those who knows the calculation of transfer function.

QUESTION

The LTspice calculation seems correct but I have mistakes in handmade calculation.

• As seen in my calculation, Q value is 1.033x10^-5, which is -99.71 dB. Why does Q have such a strange value?
• I found the phase asymptote values ​​for quadratic pole as 0 degrees and 10^48402, which is infinite. I cannot find the correct phase break points, where am I making a mistake?
• It seems the the phase of zero starts at 340 Hz and ends at 34 kHz. It doesn't match with the LTspice result.

So basically what is wrong with my calculation?

Picture 1. Control to Output Transfer Function
enter image description here

Answer to "https://electronics.stackexchange.com/questions/728110/control-to-output-transfer-function'' by Verbal Kint, edited by Mhan and posted on Software Engineering Stack Exchange is licensed under CC BY-SA 3.0.

Picture 2: My calculations enter image description here enter image description here

Picture 3: LTspice simulation enter image description here

• Verbal Kint: I would recommend you use SPICE notations for inductance or capacitance like 2 uH or 7 mF for instance, not 0.000002 or 0.007 which is difficult to read and prone to mistakes if you miss a 0. By the way, where is your ac source in the circuit?
• Mhan: The Ac source was inside the schematic just outside, I updated the picture. I tried your suggestion , nothing changed sir. I tried to ask this question to you many times but I couldn't get any answer from you. This transfer function is produced by you for my application so when I put my aplication values the mathematical calculation doesn't fit with Ltspice? Did you check my handmade calculation? they are just a standart calculation. Where is my mistake in the calculation ? @Verbal Kint
• Mhan: The Q value is weird, the phase of quadratic pole is weird and phase of the zero is not matching with the Ltspice @Verbal Kint
• Verbal Kint: I will try to have a look. Please narrow down the ac plot from 10 Hz to 100 kHz, no need to go down to the mHz.
• Verbal Kint: Before proceeding, did you run a cycle-by-cycle (transient) simulation to check whether all the values you have put in this ac model lead to a consistent operating point? This is the starting point before considering an ac sweep. Like \$V_{out}\$ is ok, the duty ratio seems fine etc. I see a 390-V input voltage with a 500-ohm load but the model implies a high-power CCM mode. What are the expected output voltage and power here?

Answer 1

The first thing to do, is to run a cycle-by-cycle simulation and verify the converter actually delivers the power. You can use the example from my free ready-made templates, it works on the free demo version of SIMPLIS:

enter image description here

So we have 12 V with a 240-mOhm load from a 390-V source so all is ok.

Then, run the averaged model now that you are confident with the adopted components values. As I said, use SPICE notations for clearer inspection:

enter image description here

The right-side plot compares the ac response from SIMPLIS and that of the averaged model. Both responses agree very well. The average model should include a pure delay as shown in my original document.

Finally, you have different expressions for plotting the responses and all are very close to each other:

enter image description here

Hope it's now ok with you and you can relax : )

• Mhan: Sir, my turns ratio is 21 , not 1/21 . When I take 21 Rd=4.N^2 .Lleak.Fsw = 5468 which is huge, thus I have unreasonable result for Q or wo etc. When you take n= 1/21 , result become reasonable because Rd is small. What should I do in this case ? could you please help me for n=21 ? @Verbal Kint
• Mhan: Hello sir, could you help me please ? why did you change my turns ratio of 21 to 1/21 @Verbal Kint
• Verbal Kint: Hello, sorry for the delayed reply. With a 390-V input voltage, I don't see how you can get a 12-V output with a turns ratio of 1:21, it would imply a peak voltage 21 x 390 V at the diodes anode. A turns ratio 1:N of 1:47.6m implies a peak of 18.6 V, which is normal for a 12-V output.