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TPS922055: Clarification on Power Dissipation and Operation of TPS922055

Paul Brar
Prodigy 100 points
Part Number: TPS922055
Other Parts Discussed in Thread: TPS92640, TPS92641

Tool/software:

Hi,

I’m seeking a deeper understanding of the TPS922055 and its internal workings in my specific application. Here's the setup:

  • Input Voltage (Vin): 48V
  • Voltage drop across 12 LEDs: 33.5V
  • Maximum LED current: 4A
  • Power supply capacity: 48V @ 6A

I would like clarification on how power is dissipated by the IC and its external components. Specifically, I understand there will be switching and conduction losses from the internal FET, as well as heat dissipation across the sense resistor. However, I’m curious about what happens when the internal FET is off:

  • Where does the energy stored in the inductor go during this time?
  • How does current flow when the internal FET is off?

Additionally, my understanding is that the TPS922055 is designed to be efficient and doesn’t dissipate much power itself, but I’d appreciate more insight into this.

Thank you for your help, and I look forward to your response.

Regards,

  • 0
    TI__Mastermind 20515 points

    Hi Paul,

    Our expert is OoO and he will reply to you once he is back to office, thanks for your patience.

    BR, Jared

  • +1
    TI__Mastermind 23150 points

    Hi Paul,

    Sorry for the late response. Please refer to my comments below.

    Paul Brar said:
    • Input Voltage (Vin): 48V
    • Voltage drop across 12 LEDs: 33.5V
    • Maximum LED current: 4A
    • Power supply capacity: 48V @ 6A

    First, if you want to drive the LED at 100% duty cycle, then I would recommend TPS92640 / TPS92641 for better thermal performance and efficiency since it is very difficult for TPS922055 to work continuously in such conditions (due to thermal concern).

    Paul Brar said:
    • Where does the energy stored in the inductor go during this time?
    • How does current flow when the internal FET is off?

    When the internal FET is off, the current will flow through the free-wheeling diode (the Schottky diode) and through the path marked red in the below figure. At this time, the energy stored in the inductor will mainly be consumed by the LEDs.

    Best Regards,

    Steven