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PMP7760: Resistor Calculation

Maor Matityahu
Intellectual 610 points

Part Number: PMP7760
Other Parts Discussed in Thread: LM5022

Hello,

Regarding the PMP7760 design:

1) how to calculate R109? What is the formula?

2) The sense resistor formula is 0.5/Ipri-pk?

Thank you, 

  • 0
    TI__Mastermind 31305 points

    Hi Maor,

    1) R109 is used for external slope compensation, which prevents subharmonic oscillation from happening at duty cycles > 50% in peak current mode control.

    The idea behind the external slope compensation is that the additional slope is added onto the sensed inductor current during the switch on-time. This is done through an internal ramp current source which generates a current through the external resistor. In PMP7760, this resistor is R109.

    A general rule of thumb is that your external slope should be at least 0.5 of your inductor down-slope if slope compensation is required. 

    In this specific case:

    I would recommend looking at section 8.2.2.8 and 8.2.2.9 in the LM5022 datasheet and utilizing the procedure outlined in http://www.ti.com/lit/an/snva866/snva866.pdf to select your current sense resistor- and then in turn, your external slope compensation resistor. 

    Based on the two links, you could use this as a starting point: 

    (Rsns)*(IL_pk) + (Rsns*D/(Fsw*L))*(0.833)*(Np/Ns)*Vout +(45uA)*D*2000 = V_cl.

    This equation implies that the current-limit threshold is the superposition of [current-sense resistor*peak current limit], slope-compensation current through internal 2KOhm resistor, and slope compensation through current sense resistor, where the additional slope compensation is proportional to the downward-slope of the primary side inductor current, which is [(Np/Ns)*Vout]/L; and the optimal proportion is approximately 82%. This neglects the current-sense filter resistor, as the additional slope there would be negligible. 

    Afterwards, to select external slope compensation resistor, you would want (Rs2+2000+Rs1)*(45uA*D) = V_cl - (I_Lim)*R_sns, where Rs1 is selected to filter out leading spikes of the current sense signal. I would select Rs1 to be 100 Ohm. If your value of Rs2 is negative, you have enough slope-compensation from R_sns and R_s1. 

    2) Assuming no external slope compensation, yes- this is correct. I would add some margin to the I_pri_pk, such as an additional 30%. The selection of the current sense resistor can be an iterative process, so I would advise you to utilize the links above. 

    This may be a lot of information, so please let me know if you have any questions.

    Thanks,

    Richard