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TPS54302: Inductor Minimum Value Calculation Equation

Metehan Demirel
Prodigy 20 points
Part Number: TPS54302
Other Parts Discussed in Thread: DRV8874,

Tool/software:

Hello;

I am designing a PCB for a single channel motor driver. Which utilizes the TPS54302 to step down 4S li on battery voltage to 5V. I don't have anything that draws too current from the converter. The design consists of an MCU, DRV8874, a CANFD transceiver, and TPS79933. I am supplying the MCU and logic supply of DRV884 from the 3.3V regulator (TPS79933) and the input will be connected to TPS54302 buck converter's output (5V).The only load directly connected to the TPS54302 will be the CANFD driver. The DRV8874 is directly connected to input voltage of the PCB.

I have checked the MCU and CANFD driver's maximum suplly currents which are 150mA and 70mA at max. and I am thinking the DRV8874's logic inputs don't consume that much current only maybe 10mA judging by the values from the datasheet.

So when I am calculating the inductor size value using the Lmin equation on the datasheet of the TPS54302 (Eq 8.) with these parameters; Iout (max) = 300mA (I took it a bit higher just to be safe) I get ınductor values like 100uH to 200uH. It seems a bit too much and even for the reference design on datasheet for 3A output it shows 10uH inductance. Am I doing something wrong here? 

If you need other parameters I ll provide them.

Vin (max) = 16.8V (max voltage of 4S lion batter)
Vout = 5V (the output I want)
Imax = 300mA (output current I picked, I picked it higher than the calculation just to be safe)

and fsw of the TPS54302 says 400KHz

Thanks.

  • 0
    TI__Expert 8576 points

    Hi, 

    Your calculation appears correct. The reason you are seeing  a 10x difference in minimum inductor is because you are reducing the IOUT by 10x from 3A to 0.3A. 

    Best regards,

    Britton

  • 0 in reply to Britton Jones
    Prodigy 20 points

    So in Equation 8 Iout is not the current requirement of my system but the output current TPS54302 can give? (3A). And can you also clarify about the Iout(max) values on Equation 9 and 10?

  • 0 in reply to Metehan Demirel
    TI__Expert 8576 points

    The IOUT value that is being used by our tool and reference design refers to the device IOUT capability. Webench also suggests ~10uH for this application. Iout(Max) is your maximum loading current as seen in the Equation 4

    Best regards,

    Britton

  • 0 in reply to Britton Jones
    Prodigy 20 points

    And just to let you guys know, in Equation 9 and 10 it also says Iout(max) but if I put 3A there I get ridicolous results. So when I put the 0.3A (my system's estimated current) I get more logical results. I think Equation 9 and 10 has a typo as well. And also for equation 10 when I calculate using my inputs.


    Vin(max) = 16.8V
    Vout = 5V
    Iout(max) = 0.3V
    fsw = 400000kHz
    Lo = 10uH 

    I get around 0.85A which is kind of seems too high for my system which only draws 0.3 at best. I thought I could get away with putting a 750mA rated inductor. But it seems I have to increase the current rating according to Equation 10

  • +1 in reply to Metehan Demirel
    TI__Expert 8576 points

    Hi Metehan,

    That is correct. Thank you for clarifying. See this section of the datasheet which labels IOUT Max as your maximum loading current. Updating my previous response to reflect this

    Best regards,

    Britton