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LM5171: LM5171

Rob Hudson
Prodigy 30 points

Part Number: LM5171

Tool/software:

Dear Support,

The LM5171 datasheet shows the voltage loop compensation in Fig. 8-5, as shown below. This this correctly shows the voltage feedback divider Rlfbt and Rlfbb which represent the voltage feedback attenuation factor H.

However, the voltage loop block diagram, Fig. 8-4 does not show the voltage attenuation factor H, which implies that H=1, unity feedback which is not the case.

Also, equation (66), total open loop gain of the outer voltage loop, does not contain the voltage attenuation factor H.

The LM5171 converter design tool spreadsheet however does require the value of H to calculate the buck mode voltage loop compensation.

Please could you explain or justify the omission of H from the block diagram (Fig8-4) and the total open loop gain equation (66). Clearly the design equations provided will not result in the correct compensation component values if H is not taken into account.

Regards

Rob

  • 0
    TI__Mastermind 28290 points

    Hi Rob,

    The transfer function and the calculation are correct. RFBT is included in the compensation stage. I will think about the block diagram.

    I think you need to use "0.909" rather than "0,909".

    Best Regards,

    Feng

  • 0 in reply to Feng Ji
    Prodigy 30 points

    Hi Feng

    Thanks for the swift reply.

    The 0,909 vs 0.909 is not important to me since I am not using the excel design tool. I am designing my own control loops base on the approach described in "Fundamentals of Power Electronics, 3rd Edition, Erickson & Maksimovic". I get exactly the same answers as the LM5171 datasheet for the current loop design based on the recommended simplification for Fcv << Fci. 

    However, I still don't see where H is accounted for in equation (66), total open loop gain of the outer voltage loop.

    Rfbt is included in equation (61) Avm as a gain = Rlcomp/Rfbt which does not directly equate to H = Rfbb/(Rfbb+Rfbt). Avm is the gain required by the closed loop to cross 0dB at the selected cross over frequency.

    Please could you clarify this for me, perhaps I'm missing something?

    Regards

    Rob

  • 0 in reply to Rob Hudson
    TI__Mastermind 28290 points

    Hi Rob,

    RLFBT is included in equation 71 as shown below. In fact, RLFBB has no influence to the loop.

    You may also refer to Switch-mode power converter compensation made easy, section Type II error amplifier and Type II transconductance amplifier.

    Best Regards,

    Feng

  • 0 in reply to Feng Ji
    Prodigy 30 points

    Hi Feng

    Thanks again for the swift reply.

    I understand that Rlfbb has nothing to do with calculating the compensator component values i.e. it doesn't affect the poles and zeros of the compensator.

    Read the statement from my previous reply carefully, reproduced below, to understand that my concern is with the quantity H that does contain Rlfbb and does affect the open loop gain.

    Rfbt is included in equation (61) Avm as a gain = Rlcomp/Rfbt which does not directly equate to H = Rfbb/(Rfbb+Rfbt). Avm is the gain required by the closed loop to cross 0dB at the selected cross over frequency.

    Regards

    Rob

  • 0 in reply to Rob Hudson
    TI__Mastermind 28290 points

    Hi Rob,

    H = Rfbb/(Rfbb+Rfbt) will appear in the transfer function when a Type II transconductance amplifier is used.

    It will not appear if operational amplifier is used. You can consider H=1 in this case.

    Best Regards,

    Feng

  • 0 in reply to Feng Ji
    Prodigy 30 points

    Hi Feng

    I'm not discussing the opamp or OTA, I'm discussing the total loop gain which is equation (66).

    My question is: where is H in equation (66)?

    Regards

    Rob

  • 0 in reply to Rob Hudson
    TI__Mastermind 28290 points

    Hi Rob,

    There is no H if you use opamp which is the case here.

    Best Regards,

    Feng

  • 0 in reply to Feng Ji
    Prodigy 30 points

    Hi Feng

    There is always an H if you are sensing the feedback.

    Regards

    Rob

  • 0 in reply to Rob Hudson
    TI__Mastermind 28290 points

    Hi Rob,

    H will not appear in the transfer function when opamp is used.

    Best Regards,

    Feng