LM46001: LM43601PWPR

Evgeniy xxx

Part Number: LM46001

More question please. I calculate the inductor for the maximum current I=0,5A, but this current will appear only once at the time the device is turned on. The rest of the time the device consumes no more than I=0,05A, how then to choose an inductor ?

over 7 years ago

0 Samuel Jaffe over 7 years ago

TI__Mastermind 25590 points

Evgeniy,

How did you calculate 0.5A max current? Is this from startup when the device is charging the output capacitors? Or is this being pulled by a load?

-Sam

0 Evgeniy xxx over 7 years ago in reply to Samuel Jaffe

Prodigy 40 points

Hi, Sam!

At the moment the device is turned on, it self-tests. This process occurs once at the time of power-up and takes about 30 - 60 seconds. At this point, current consumption can be I = 0,4...0,5 A. After this moment, the current consumption is I = 0,03...0,05 A.

Evgeniy

0 Samuel Jaffe over 7 years ago in reply to Evgeniy xxx

TI__Mastermind 25590 points

Evgeniy,

In that case I recommend designing for 0.5A. 30-60 seconds is a long time. If it was a quick transient then you might be able to get away with designing for 0.05A but 30-60 seconds is a long time.

-Sam

0 Evgeniy xxx over 7 years ago in reply to Samuel Jaffe

Prodigy 40 points

Hi Sam!

I will calculate the inductor with maximum load current I = 0.5A. Thanks for the help.

Evgeniy.

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LM46001: LM43601PWPR