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# TPS61169: TPS61169

Part Number: TPS61169

Hi Team,

We are planning to use TPS61169 as a IR LED driver in our design. The requirement are as follows,

• Input voltage    : 5V
• Output voltage : 18.6V
• Output Current : 100mA

In datasheet, Its shown that the inductor value is 4.7-10 uH. But in my calculation,  for our requirement more that 22uf is need to minimize the ripple current below 40%.

So please let me know if i can follow the design with 10uH.

Please find the snapshot of my design below,

• Hi,

It seems that these two pictures are broken, pls. upload again.

BR,

Robin

• Hi,

You can utilize more than 10uH.

BR,

Robin

• Hi Yuting Shao,

Have one more doubt to clarify here, Can I design for  18.8V @ 100mA from a  5V supply with TPS61169? Also please suggest suitable inductor value for this design. Please note that the design is for IR LED driver application.

• Hi,

Yes, as for inductor design, it depends on your requirements such as current ripple, rated current, saturation current and Rdc and so on.

BR,

Robin

• Hi,

Understood.

the application is for IR LED powering. In these conditions, how much current ripple we select normally? As my calculation the rated current is 400mA So the saturation current= 1.25x400mA=500mA. Only the doubt is with howmuch ripple we can take in this condition. While i designed with 10 uf, i got peak to peak ripple as 300mA witch is about 80% of rated current. Is this is a issue on these condition?

Also the IC featurre is for led driver and they given the recommended inductor value as 4.7uH to 10uH. This makes  me confusion such that the ripple is not a concern. Please help me here.

• Hi,

The inductor value in datasheet is typical value under some certain test conditions, and you can design inductor according to your design requirement.

80% ripple is a little large, it's recommended below 40%.

BR,

Robin