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Buck-Boost Regulator around LM5118 - ramp generation

Sonu Kurian
Prodigy 60 points
Other Parts Discussed in Thread: LM5118

Hi,

I trying to design a buck boost regulator around LM5118.

I am a bit confused about selecting Cramp capacitor(Capacitor used to generate the ramp ).

The datasheet gives an equation:(Equation 4, page number 15 of datasheet)

Cramp = (gm*L)/(A*Rs).

I am not able to understand how this equation is derived.Infact datasheet says "For proper current emulation, the sample and hold pedestal value and the ramp amplitude must have the same relative relationship to the actual inductor current." as an explanation to the above equation( page number 14 of datasheet).

Can anybody please give a bit more explanation to the above sentance or derivation?

3365.lm5118.pdf

Thanks a lot.

Sonu

  • 0
    TI__Guru 53171 points

    Hi Sonu

    If ignore the additional term (50uA) from the equation (2), it results. IRAMP=5u*(VIN-VOUT) [A]. And then we can say dVRAMP/dT = 5u*(VIN-VOUT)/CRAMP since I=C*dV/dT

    Assuming a resistor sensing which senses the rising slope of inductor current in BUCK mode and the gain of sense amplifier is A(=10), we can say dV/dT of the current sense amplifier output is (Vin-Vout)/L*RS*A  since V=L*dI/dT.

    Therefore,

    5u*(VIN-VOUT)/CRAMP = (Vin-Vout)/L* RS*A

    5u / CRAMP = RS*A /L

    CRAMP= 5u*L/RS/A  =gm * L /RS/A

    For proper current emulation, the sample and hold pedestal value and the ramp amplitude must have the same relative relationship to the actual inductor current

    ==> In other word, the sum of the pedestal value (DC)  and the ramp (AC) should be similar with the actual inductor current sensing during Ton.

    Regards,

    Eric

  • 0 in reply to EricLee
    Prodigy 60 points

    Hi Eric,

    Its clear.Thanks a lot!

    many thanks,

    Sonu