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LM5146DESIGN-CALC: Compensator network calculation

anushka khanna
Prodigy 30 points
Part Number: LM5146DESIGN-CALC
Other Parts Discussed in Thread: LM5146, LM5146-Q1

Tool/software:

For an input voltage of 28V to 60V, the output specs required are 12V, 8A output.

I calculated the compensation network based on the inductor (whose calculation was done based on buck converter inductor calculation) and output capacitance (kept same as that given in the datasheet) mentioned in the schematic below. The type three compensator given in LM5146 used pole placement method for controller calculation. I took L value as 33uH and output capacitance as 110.1uF.

In the datasheet they mentioned poles and zeros placement assumptions, I took those assumptions and calculated compensator network values. Could someone verify the same? I took Cout same as what was given in EVM of LM5146 Q1 for keeping the Resr of output capacitance same so that ωesr doesn't change, which in-turn makes sure ωp2 remains same. Can someone verify the compensator network calculations?

For ILIM calculation I kept both options of Rds(on) sensing and shunt resistor sensing, but in the board, I have populated the shunt resistor sensing option (R16, R17, C20) and (DNP: C3 and R6)

The output voltage however, is coming out to be 3.6V instead of 12V. Could someone let me know what changes can be made to the compensator network and if the power section calculations are correct. The switching frequency of the converter is 400KHz. 

  • 0 in reply to anushka khanna
    TI__Genius 11645 points

    Hi Anushka,

    Please fill in the design parameters in the quick start calculator and share with me: https://www.ti.com/tool/LM5145DESIGN-CALC.

    Also, please upload a higher resolution schematic. 

    Thanks,

    Best Regards,

    Taru

  • 0 in reply to Taru Kapoor
    Prodigy 30 points

    Hi, thanks for replying. I used the calculator however the inductor calculations were not matching when I used regular buck converters' inductor value calculation. Also, certain values like output capacitors ESR was not mentioned in the datasheet of the capacitor so I had to back- calculate RESR based on the capacitor used in the EVM LM5146-Q1.

    Here is another picture of the schematic. 

     

  • 0 in reply to anushka khanna
    TI__Genius 11645 points

    Hi Anushka,

    I would like to know your design parameters. Just populate the Cap and Resistor values that you have chosen in your schematic in the quick start calculator. I also need to know the derating of the capacitors. This is where I see an obvious problem - the rating of output capacitors is 16V - there will be significant DC bias derating of the capacitor. It is recommended to use capacitors of atleast 25V rating for your use case. 

    • What was the mismatch between the calculator and your calculation? 
    • Can you share the capacitor part numbers with me?
    • Compensator calculations are related to the output cap ESR Zero and Inductor value. The quick start tool should simplify things. You can plug-in the values that you have calculated and see what the expected Loop Response will be. 
    • Furthermore, the L value taken is quite high - do you really need that high of an inductance. 

    Thanks,

    Best Regards,

    Taru

  • 0 in reply to Taru Kapoor
    Prodigy 30 points

    Hi,

    I calculated the inductor value taking Delta IL as 10% of my output current (8A, in this case). The inductor value came out to be around 25uH. However, on changing delta IL calculation to 30%, I was getting 8.5uH. I changed the inductor value to 8.2uH on my board and still I am getting 3.6V as my output voltage.

    For my capacitor I took: CGA6P1X7R1C226M250AC. If the 16V rating is an issue due to derating, I can change the same.

    Would GCM32EC71E226KE36L (22uF, 25V) work? If yes, could you help me in figuring out how to find RESR of the capacitor, as I will require it while calculating my compensator poles. (ωP2). I could not use the Quick calculator here as none of the output cap datasheets mentioned the equivalent series resistance required as one one the parameters in the calculator. 

    Regarding the mismatch in the calculator and my calculations:

    The LM5146 datasheet mentions the assumptions taken for compensation component selection namely-

    ωz1 = 0.5·ωo, ωz2 = ωo, ωp1 = ωSW/2, and ωp2 = ωESR.

    The compensator zero values are not matching. 

    I was also getting stuck while calculating ωESR, ωP2. So I back calculated these values using Rc2 and Cc3 mentioned in the EVM application note. However, since the capacitor used in the EVM board might be different, I am unsure of the ESR value of the output cap that I have used.

    Lastly, even after changing the inductor value to 8.2uH the output voltage is still coming out to be 3.6V. 

    For the design parameters: LM25145_quickstart_calculator_A5.xlsm 

  • 0 in reply to anushka khanna
    TI__Genius 11645 points

    Hi Anushka,

    Apologies for the delay in response. 

    Please see answers to your questions:

    For CGA6P1X7R1C226M250AC - if you look at the characterization sheet on the vendor's website:

    At 12V - the capacitance derates by ~40% - this is important to take into consideration when calculating the value of the effective output capacitance. 

    Therefore, a 22uF cap will effectively be ~13uF. 

    For the GCM32EC71E226KE36L - derating is -23% (this data is also available on the cap vendor's website) - therefore, a 22uF cap will effectively be ~17uF. 

    Hope this gives you an idea on how to select output capacitors. 

    Similarly, there are also graphs available for the ESR of the capacitor. For CGA6P1X7R1C226M250AC it looks like:

    You can find the ESR at the switching frequency that you have chose, In your case, at 400kHz - ESR ~1mOhm.

    Coming to the compensation values - the quick start calculator simplifies your work and calculates values based on your inputs. The values calculated by the Quick start calculator are exactly the same that you will get from the datasheet formulae:

    Regarding the mismatch:
    Fz1, Fz2,Fp1 and Fp2 are values that you can choose. Based on the values you select - the calculator suggests the values for Rc1, Cc1, Cc2, Rc2 and Cc3. The Actual P/Z frequencies are calculated based on the values you have populated in the selected column. 

    As an example - I am choosing the same Fz1, Fz2,Fp1 and Fp2 as in your sheet and choosing the same Rc1, Cc1, Cc2, Rc2 and Cc3 that the tool suggests:
     

    You will see that the actual values are more or less the same but this will change when the selected values changes (as you have done in your case). 

     

    The Bode Plot that you see is plotted using the selected value column - you can play around with these values to reach a frequency response that is acceptable. The Baseline P/Z frequency section in the quick start calculator helps suggest some initial starting values for the compensation design. 

    Hope this clarifies the questions that you had. 

    Thanks,

    Best Regards,

    Taru