boost converter inductor value

Discussion in 'Homework Help' started by notoriusjt2, Oct 31, 2010.

  1. notoriusjt2

    Thread Starter Member

    Feb 4, 2010
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    [​IMG]
    [​IMG]
    where did I screw up? is the output voltage ripple % supposed to be in there somewhere?
     
  2. notoriusjt2

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    Feb 4, 2010
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    figured it out. i was using the formula for a buck-boost converter. not just a boost converter

    [​IMG]
     
  3. thatoneguy

    AAC Fanatic!

    Feb 19, 2009
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    The output ripple is usually a function of the output filter, typically a capacitor. Sometimes useless information is added to a question, sometimes it may appear useless, but some part of it is actually required.
     
  4. notoriusjt2

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    Feb 4, 2010
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    answer A 1.1uH was actually incorrect...

    maybe that output voltage ripple % was needed?
     
  5. notoriusjt2

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    Feb 4, 2010
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    after going over this again I found that my value for Io was incorrect

    I=P/V=25/15=1.667

    recalculating everything I got 4.4uH
     
  6. notoriusjt2

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    Feb 4, 2010
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    and thats still the wrong answer, well i only have 3 choices left
     
  7. thatoneguy

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    are you accounting for minimum current to be 1/2 of full current? Continuous switching mode, essentially.
     
  8. thatoneguy

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    Feb 19, 2009
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    My bad. D is from above, Vin/Vout

    L=(V_in -V_out ) * {D * Freq \over I_ripple}
     
  9. notoriusjt2

    Thread Starter Member

    Feb 4, 2010
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    here is another similar one
    [​IMG]
    Vs=5V
    Vo=15V
    R=75ohms
    f=100kHz
    T=10uS
    D=.667

    the book gives me this equation for max inductor current
    Imax=(Vs/(((1-D)^2)R)+((VsDT)/(2L))
    solving for Imax gives me 11.9uH

    is this the correct equation?
     
  10. thatoneguy

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    Feb 19, 2009
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    Click My Sigline. ;)

    Imax={V_s\over {((1-D)^2 \cdot R) + \frac{V_s \cdot D \cdot T} {2 \cdot L}}

    1st, Does the equation above match your book?
    2nd: How did you rearrange it to solve for L?
     
  11. notoriusjt2

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    Feb 4, 2010
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    here is the exact equation out of my book and how I solved for L

    [​IMG]
     
  12. thatoneguy

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    Feb 19, 2009
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    Ok, I'll work it out here.

    Vs=5V
    Vo=15V
    R=75ohms
    f=100kHz
    T=10uS
    D=.667

    I_{max}={\frac{V_s}{(1-D)^2 \cdot R}+{\frac{V_s \cdot D \cdot T}{2 \cdot L}}<br />
    <br />
2={\frac{5}{(1-0.667)^2 \cdot 75}+{\frac{5 \cdot 0.667 \cdot 10\times 10^{-6}}{2 \cdot L}}<br />

    <br />
2=601.2\times 10^{-3} + {\frac{33.35\times 10^{-6}}{2 \cdot L}<br />

    <br />
1.3988=\frac{33.35 \times 10^{-6}}{2 \cdot L}<br />

    <br />
2.7976 \cdot L=33.35 \times 10^{-6}<br />

    <br />
L = 11.921 \times10^{-6}<br />

    L = 11.921 μH

    So, we either both did it wrong in similar ways, or it isn't the correct formula they want you to use in this instance. :D
     
  13. notoriusjt2

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    Feb 4, 2010
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    the professor submitted the same question with new answers and 10uH was on there. So I went with that and by golly I got it right
     
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