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PMP40934: The PMP40934 is rated for 60W, yet the output resistors are rated for very low power. Am I interpreting it wrong?

Stephen O'Connor
Prodigy 70 points
Part Number: PMP40934
Other Parts Discussed in Thread: TPS25762-Q1

Hello,

After reading the schematic for the PMP40934 reference design, I noticed that R14 is in series with the output path (Pin 11), but it's only rated for 1W, which is much less power than the design is advertised to output (60W).

According to the datasheet for the TPS25762-Q1, I'm assuming pin 11 is the output pin which carries all of the current. How could this pin output 60W of current to the VBUS pin on the USB-C connector, if it passes through a 1W resistor? (PA_VBUS)

To add, Pin 29 is in series with R4, which is also a low power resistor, at 0.2W.

Am I interpreting the schematic incorrectly?

According the the BOM, R14 = 1W and R4 = 0.2

  • +1
    TI__Expert 5685 points

    Hi Stephen,

    The resistor is in series with the output to sense the current flowing to the load. 

    The output is rated for 60W (20V/3A), the maximum current in 3A, the value of the R14 is 0.01ohm. So this means with 3A output current, there is less than 100mW of loss in this part. 

    The resistor that the output power passes through doesn't have the full 60W in it, it just passes the current, the voltage at the output will be 20V that is how you get to the 60W of power. Hopefully this makes sense to you.

    Pin 29 and Pin 28 are only used to sense the voltage across the current sense resistor.

    Thanks,

    Robert