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LM2840 100mA step down inductor question.

Robin Davis1
Prodigy 140 points
Other Parts Discussed in Thread: LM2840

Hi to all.

Hoping someone can help out.

In the datasheet for the LM2840 there are a couple of Application ccts  , starting on page 9. I'd like to know how these inductor values are calculated , especially the one

in figure 2. With a 42V input and a 3.3V output the 15uH is no where near big enough to fulfill the 30% ripple current requirement. Well it not actually a requirement , more a rule of thumb , but it's not an unreasonable value.

Ignoring diode drop the peak to peak ripple current with 100mA load is over 400mA! , that's with 15uH. Even at 1.5Mhz this requirement won't be met , not even close.

I'd be using at least 100uH.

Am I missing something here? Is there some design requirement that I've overlooked.

Cheers

Robin

  • 0 in reply to Robin Davis1
    Prodigy 10 points
    Hi.

    I've the same trouble and I've burnout 5 samples.
    Looks like the L selection equation has mistakes.
    I need 5 Vdc 100 mA fixed sourcing with 24 Vdc.
    When the IC burns, sends 24 Vdc to the output. It's a big problem. I've burnout a microcontroller too.

    Once working I had a lot of ripple in the output pin.
    The waveforms were quite different from the datasheet.
    So I have changed the 10 uH and then the IC burnout.

    It's seems to be a high material dependence.

    Someone had the same problem?