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LAUNCHXL-F28379D: Maximal current of on-board step-up regulator

Killian
Prodigy 10 points

Part Number: LAUNCHXL-F28379D
Other Parts Discussed in Thread: LMR62421

Hi,

I want to supply an external device with 5V from the LAUNCHXL-28379D. According to the documentation I can disconnect jumpers J1-3 to isolate the board from USB, and supply 3.3V externally. The internal step-up regulator will supply the 5V rail if I connect JP6.

What is the maximum current which can be drawn from this 5V supply? According to the datasheet of the LMR62421 it should be up to 2.1A? This seems quite high, will this work without overheating?

Regards,

Killian

  • 0
    Guru 30330 points

    Killian,

    Your plan to supply 5 V from 3.3 V with JTAG/USB isolated is correct. The 2.1 A current limit is what is in the datasheet as a MIN limit. I will pull in the LMR62421 experts to see what they have to say

    [Edit] refer to Figure 6: Typical Maximum Output Current vs VIN in the LM62421 datasheet. I don't think that the output current will be near 2.1 A. likely ~1 A, but the experts should have a better idea.

    -Mark

  • 0
    TI__Intellectual 2655 points
    Hi Killian,
    2.1A is switch current limit not output current. Switch current limit is inductor peak current. Based on you working condition Vin=3.3V Vout=5V, Iout is about 1A.