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Boost converter diode dissipation?

Discussion in 'General Electronics Discussion' started by eem2am, Jul 13, 2010.

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  1. eem2am

    eem2am

    414
    0
    Aug 3, 2009
    Hello,

    I wish to calculate the loss in the boost diode of a boost converter.

    I wish to do this because I want to see if it’s worth using a synchronous boost converter instead.
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    Here is the spec:

    Switching frequency = 100KHz
    V(in) = 2.1V
    V(out) = 10V
    P(out) = 2.71W
    Inductor = 100uH

    Average boost diode forward current = 0.339A

    Boost diode = 1N5819

    ….datasheet of boost diode…….
    http://www.diodes.com/datasheets/ds30217.pdf

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    This diode, at my I(f) of 0.339A would have…………
    V(f) = 0.4V …and…
    Diode capacitance = 30pF

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    Please now could you check the diode dissipation calculation:-

    I am taking it from page 4 of…………
    Cree Applcation Note
    http://www.cree.com/products/pdf/CPWR-AN05.-.pdf



    So:-

    Conduction loss = 0.339 x 0.4 = 0.136W
    Switching Loss = 2 x ½.C.V^2 X f = 300uW


    …..Total Boost diode loss = 0.1363W

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    This seems very low,
    I am now wondering why anyone would ever consider using a synchronous boost converter.

    I cannot seem to find the "beta" value which is necessary to calculate the equivalent diode resistance “Rd” figure in the 1N5819 datasheet above, though the above Cree Application Note mentions that I should take it into account.

    So…..

    Do you know what figure I should take for the Diode’s Resistance “Rd” ?
    (-On page 3 of the above app note it’s given as:- (beta) x Tj + Rd0
    -but it doesn’t say what “beta” is)

    Also, would the switching losses be greater in a synchronous boost ?

    -because the hi-side FET would tend to give a higher capacitance than just a diode....not to mention that you always see a schottky in parallel with the hi-side FET to stop the FET's intrisic diode from conducting

    -And do you know if this diode dissipation calculation is correct ?
     
  2. Mitchekj

    Mitchekj

    288
    0
    Jan 24, 2010
    Honestly, at that low power, I don't think synchronous rectification will do you any good. You may gain ~3% efficiency (which comes out to a few milliwatts,) but only if you can overcome all the parasitics involved.

    Btw, where did you get your Id(avg) value of 339mA? If you're using that app note's equations, it will be wrong. You're using a 2.1V input... DC right? They're giving you formulas for a PFC boost stage in an off-line supply. Bit different. Your power and voltage output (2.71W at 10V) comes to a 271mA output... the diode avg current will be the same as your avg load current: 271mA. It's in series with the load, after all.

    The diode dissipation would just be Vd * Id = 108mW. Switching loss would be negligible in comparison.

    Edit: Your question: "This seems very low, I am now wondering why anyone would ever consider using a synchronous boost converter."

    In higher powered converters, you can save watts of loss (heat) with synchronous rectification. It's just not worth it in this case, in my opinion.
     
    Last edited: Jul 13, 2010
  3. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,482
    2,830
    Jan 21, 2010
    In addition, as the regulated voltage gets lower and lower, the power lost in the diode gets higher and higher (as a proportion of the output power). But that tends only to be an issue with buck regulators.
     
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