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TPS568230: Phase Margin Estimation Method

Aaron Hsu
Intellectual 565 points
Part Number: TPS568230


HI All

The SLVAF11 document explains very well that the LC stable relationship

https://www.ti.com/lit/an/slvaf11/slvaf11.pdf?ts=1626619199211&ref_url=https%253A%252F%252Fwww.google.com%252F

But I have a Phase Margin of TPS568230 is used as the example in the file Confuse

PM:(wcross)=180+PhaseDP(wcross)+PhaseRI(wcross) +PhaseESR(wcross)+PhaseDelay(wcross)

PhaseDP(wcross)=-tan-1(2a*W0*Wrocss/W0^2 *Wcross^2)

PhaseRI(wcross)=arctan(Wcross/WRI)

PhaseESR(wcross)=arctan(Wcross/WESR)

PhaseDelay(wcross)=-(TON*Wcross/2)*(180/3.14159)

In the example,

L=0.86uH at 8A loading

VIN=12V

VOUT=1.5V

FSW=600K HZ

Co=110uF ESR = total ESR is 0.6mohm
TPS568230 =270krad/s (43kHz)

ACP=29.3

Vref=0.6V

W0=√1/L*CO

W0=102814

Wesr=1/co*esr

Wesr=15151515

Wcross=(Acp*Vref*W0^2)/*WRI

Wcross=(29.3*0.6*102814^2)/270K=688271

PhaseRI(wcross)=arctan(688271/270K)=68.58

PhaseRI(wcross)=arctan(688271/15151515)=2.6

PhaseDelay(wcross)=-(0.2083u*688271/2)*(180/3.14159)=-4.107

TON =1/600KHZ *1.5V/12V=0.2083us

Since the range of PhaseDP (wcross) is -90 degrees to 90 degrees

PM:(wcross)=180+(-90~90)+ 68.58+2.6+(-4.107)

Since the range of PhaseDP (wcross) is -90 degrees to 90 degrees
It is impossible to approach it no matter what, the data is estimated to be 57.7 degrees

  • 0
    TI__Expert 5470 points

    Hi Aaron,

    Really honored that you could read my application report and did so detailed calculation.

    I noticed one error in your calculation is "Wcross=(Acp*Vref*W0^2)/*WRI". According to (13), it should be Wcross=(Acp*Vref*W0^2)/(Vo*WRI). The term Vo is missed here.

    Thanks,

    Andrew

  • 0 in reply to Andrew Xiong
    Intellectual 565 points

    HI Andrew

    Yes, my Vo is missed here

    I recalculate Wcross

    but still got the same conclusion

    Wcross=(Acp*Vref*W0^2)/Vo*WRI

    Wcross=(29.3*0.6*102814^2)/1.5*270K=458847

    PhaseRI(wcross)=arctan(458847/270K)=59.52

    PhaseRI(wcross)=arctan(458847/15151515)=1.73

    PhaseDelay(wcross)=-((0.2083u*458847)/2)*(180/3.14159)=-2.73

    TON =(1/600KHZ) *(1.5V/12V)=0.2083us

    Since the range of PhaseDP (wcross) is -90 degrees to 90 degrees

    PM:(wcross)=180+(-90~90)+ 59.52+1.73+(-2.73)

    It is impossible to approach it no matter what, the data is estimated to be 57.7 degrees

  • 0 in reply to Aaron Hsu
    TI__Expert 5470 points

    Hi Aaron,

    Got your point now. tan-1 here is not arctan. I'll give more explanation about it after I get office tomorrow.

    For the double poles, even if not considering the effect of Q factor, two poles could bring -180deg phase drop in the loop.

  • +1 in reply to Andrew Xiong
    TI__Expert 5470 points

    Hi Aaron,

    I rechecked the formula, it's from a control theory book:

    Here it's an arctan for (0,pi) range. Since for the tan-1(y/x), the y>0 and x could be negative. So the atan2 for programming can be used to get the result:

    Just set delta about 0.15, for the rough calculation:

    Thanks,

    Andrew

  • 0 in reply to Andrew Xiong
    Intellectual 565 points

    HI Andrew

    Thank you very much, detailed explanation