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LP8861-Q1: inductor calculation

Cathy Cai

Prodigy 360 points

Part Number: LP8861-Q1

hi team,

The customer uses LP8861-Q1 and their spec is as follows:

Vin-min=9V

Vboost= 37V

Iout= 100mA*3

current ripple 30%

f=330kHz

The recommend inductor is 22uH. But, the calculated value is 125.9uH, which is much higher than the recommend value.

Could you please explain why the recommend value is much smaller?

regards,

Cathy

over 2 years ago

0 Xiaoguang1 over 2 years ago

TI__Genius 14030 points

Cathy,

Let me know how you get 125.9uH value? There must be something wrong in your calculation and pls double check it. Thanks.

Regards,

Xiaoguang

0 Cathy Cai over 2 years ago in reply to Xiaoguang1

TI__Prodigy 360 points

Hi Xiaoguang,

Here is the calculation:

L=((Vout - Vin-min)*Vin)/(Iripple*f*Vout)=((37V-9V)*9V)/(2*30%*300mA*330kHz*37V)=114.66uH

Regards,

Cathy

+1 Xiaoguang1 over 2 years ago in reply to Cathy Cai

TI__Genius 14030 points

Cathy,

ripple is the inductor ripple, Thus, it's not 30%*300mA, but 30% * (300mA* Vout/(Vin*Effi))

Regards,

Xiaoguang

0 Cathy Cai over 2 years ago in reply to Xiaoguang1

TI__Prodigy 360 points

Hi Xiaoguang,

Oh, yes. Thanks for correction! : )

Regards,

Cathy