TPS61027 inductor parameter calculation

ping wang1

Other Parts Discussed in Thread: TPS61027, TPS61070

HI ,ti power experts.

I plan to use tps61027 in my design. Now I meet a problem how to calculate the inductor parameter
my requirement is :
          vin: 2~3v battery
          vout : 5v , 50ma.
I try to calculate the paramter according to the formula (4)&(5) in datasheet.

       IL=50ma*5/(2*0.8)=156ma
I chose 20% ripple , switch freq is 600K(typical)
L =2*（5-2）/(156ma*0.2*600*5)=64.1UH

I seems that the inductor value is too high. In the datasheet ,typical application inductor is about 6.8uh.

So can you please give some suggestion ?

over 10 years ago

0 Jasper Li over 10 years ago

TI__Mastermind 46905 points

in this small output current application, you don't need to follow the formula in the datasheet. please use the inductor in typical application.

i would suggest TPS61070 series for your application

thanks
Jasper Li

0 Helen chen over 10 years ago

TI__Mastermind 40080 points

20% ripple is for high output current application, you can choose a much higher current ratio like 60% or even more, so you can choose a 22uH inductor which is in the recommended range in the datasheet.

0 ping wang1 over 10 years ago in reply to Jasper Li

Prodigy 200 points

HI,thanks for your reply. I have another question：
why you recommend tps61070 ? what is the benifit compared with tps61027

0 Jasper Li over 10 years ago in reply to ping wang1

TI__Mastermind 46905 points

the tps61070 is higher frequency, smaller output capacitor, simpler which mean less cost

Jasper

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TPS61027 inductor parameter calculation