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Switching Regulator Buck design parameter

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Member level 2
Mar 11, 2005
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I am design a switching buck regulator.Did anyone know how to choose the L and C design paramemter?? Have any formula to describe these values??
Thank you

The inductor in chosen based on the allowable ripple current as a percentage of the output DC current. Common ripple current values range from about 20% to 40% (assuming continuous inductor current). That is the peak to peak ripple current, Ipp.
Since the inductor current ramps up during the transistor on time, you can write:
Ipp=(Vin-Vo)*ton/L, where Vin is the input voltage, Vo the output voltage and L the inductance.
Now ton equal to the duty-cycle divided by the switching frequency, so ton=D/fsw.

And the duty cycle is approximately: D=(Vo+Vd)/(Vin+Vd), where Vd is the voltage drop across the freewheeling diode.

So Ipp=(Vin-Vo)*D/(L*fsw)

The output capacitor is selected merely based on its ESR and the allowable output voltage ripple, since the cap's ESR is the major contributor to the output voltage ripple. Basically the capacitance is almost unimportant, unless you are using really low capacitance, such as ceramics.

For the normal (Aluminum or Tantalum) caps, you select it such that
Vripple≤ESRmax*Ippmax, where ESR is the maximum ESR at the lowest temperature and Ippmax is the maximum inductor ripple (which occurs at maximum input voltage, minimum inductor value).

With that, ESR≤Vripple/Ippmax.
Select a cap based on the ESR. You can parallel two or more caps to obtain the required ESR.

Verify the current ripple rating Irms=Ippmax/√12.
Make sure the capacitor's RMS current rating is higher than the Irms above. If caps are paralelled, then the current divides between them.

Hope this helps,

The better you read the datasheet paper
or you can read the flowing book:
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