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TPSM828303: Resistor power rating

Nandini Muralidharan
Intellectual 820 points
Part Number: TPSM828303


Tool/software:

Hi,

I would like to understand the power rating required on below resistors. Please confirm if my understanding is correct.

1. Resistor for PG pin: Per datasheet, IPG is 100nA. Hence Power that the PU resistor need to withstand should be more than 100nA*100nA*100K=1nW
     Hence in reality 0402 which can support 63mW is sufficient for Pull up.

2. Resistor for FB pin: Per datasheet, IFB is 70nA. Hence Power that the PU resistor need to withstand should be more than 70nA*70nA*680K=3.332nW
     Hence in reality 0402 which can support 63mW is sufficient for Pull up.

3. Resistor for EN pin: Per datasheet, IEN(LKG) is 100nA. Hence similar to PG pin, PU 0-ohm Jumper which can withstand more than 100nA is sufficient.

Thanks

  • +1
    TI__Mastermind 24630 points

    Hello Nandini,

    thanks for reaching out in E2E.

    1. The PG pin is an open drain output, which is high impedance when high and connected to GND when low (see section 7.4.3 in the datasheet for more details). The maximum power dissipation is the low case, then approximately the full VIN is applied to the pull-up resistor. P = (VIN)² / RPU = (5V)² / 100kΩ = 250 µW. The current should be limited to a maximum of 1mA. IPG(low) = 5V / 100kΩ =  50µA, so this requirement is also fulfilled. For the high case the PG leakage current will cause some voltage drop across the pull-up resistor, which reduces the logic high level. VRPG-drop(high)max = RPU × IPG(LKG)max = 100kΩ × 100nA = 1mV, which should be no issue for the PG signal level. So a 100kΩ resistor, which can support 63mW is sufficient.

    2. The current thru the FB divider is dominated by the divider current itself. For R5590 this means PR5590 = (VFB)² / R5590 = (0.5V)² / 200kΩ = 1.25µW.The current is IR5590 = 0.5V / 200kΩ = 2.5µA. I think for R5589 it would be sufficient to neglect the FB leakage, but to be precise it would be PR5589 = (IR5590 + IFB(LKG)max)² × R5589 = (2.5uA + 70nA)² × 680kΩ = 4.49µW. So FB divider resistors which can support 63mW are sufficient.

    3. Your understanding is correct. The EN leakage is the only relevant current here.

    Please let me know if you have any open questions. If my explanations answered your questions then I would be more than happy if you click "Resolved".

    Best regards,

    Andreas.