TPS65295:

TAN SB

Intellectual 400 points

Part Number: TPS65295

Hi,

The application circuit shown in the data sheet for VDDQ 1.2V 8A output uses 680nH inductor.

Is there a table or formula to calculator the inductor value for 1.2V 5A load?

Thanks,

over 1 year ago

0 Yuchang Zhang over 1 year ago

TI__Mastermind 18510 points

SB, actually no, the same inductor is recommended for both 5A and 8A, purpose is to make sure LC loop can be compensated well by internal circuit.

Commonly the inductor range 470nH to 1uH, 680uH is tradeoff of ripple and transient performance.

Yuchang

0 TAN SB over 1 year ago in reply to Yuchang Zhang

Intellectual 400 points

Hi Yuchang,

Thanks for your quick response.

The load current requirement for my design are:

VDDQ = 5A max, VPP = 100mA max

I want to pick a smaller package size inductor with lower Isat, if I keep the recommended inductance value from the reference circuit (4.7uH for VPP and 680uH for VDDQ), what will be the recommended min max DCR or any other parameters that I need to consider?

Thanks,

+1 Yuchang Zhang over 1 year ago in reply to TAN SB

TI__Mastermind 18510 points

SB, VPP 2.5V inductor could be used Maruta chip inductor, which has smaller size, and for both power rail, there is no DCR value limitation, just bigger DCR, big DCR loss.

Yuchang

0 TAN SB over 1 year ago in reply to Yuchang Zhang

Intellectual 400 points

Thanks Yuchang.

0 Yuchang Zhang over 1 year ago in reply to TAN SB

TI__Mastermind 18510 points

No problem

Yuchang

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TPS65295: