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TPS61178: TPS61178 Inductor current rating formula

Neil Donaldson
Prodigy 40 points
Part Number: TPS61178

Dear TI,

I am designing a boost power supply and am a little confused by the conflict in the formulas for calculating the current rating of the inductor.

From the datasheet on page 19 there are 2 formulas for Ipeak that give widely different answers. From Eq 8. I get a result of 12.64A while from Eq 10. I get 23.1A. This makes a big difference in inductor size for my design. Which one should I use and why the difference.

My input parameters are Vin 12V, Vout 15V Iout 7.5A, Fsw 2MHz, L 2.2uH

The datasheet also states that programmable switch peak current as "up to 10A" on the front page while saying 15A on page 12. Which one is correct.

Any help on this would be appreciated.

Neil Donaldson

Ellex Medical Pty Ltd

  • 0
    TI__Mastermind 46905 points
    I have noticed your question. i will provide my suggestion within this week.
  • 0
    TI__Mastermind 46905 points
    The equation (8) and equation (10) are the same. the Eq 10 is just more detail. I would suggest 1MHz switching frequency or lower in your application.
    if 1MHz, the peak current is 10.55A.
    the thermal is a concern for this device at 7.5A output condition. i would suggest you apply an EVM to evaluate.
  • 0 in reply to Jasper Li
    Prodigy 40 points

    Hello Jasper,

    Thanks for your reply.

    The two formulas give different answers. It is true that formula 10 is part a more detailed version of 8 but with one difference. The first part of the formula (10) which is not an expansion and it is this part that causes the current requirement to blow out.

    The denominator of first part of formula 10  (1-D)*n  increases the current far more than Iin formula (9). I can only assume there is some error in the formula.

    Here are the calculations

    Iout/(1-D)*n = 7.5/(1-0.63)*0.9  = 22.52A  Now using formula (9) which is meant to be the equivalent

    Vout*Iout/Vin*n  = 15*7.5/12*0.9 = 10.55

    I have got the EVM module that is switch current limited to 2.5A I believe, the resistor could be swapped to get a higher current limit

    Regards

    Neil

  • 0 in reply to Neil Donaldson
    TI__Mastermind 46905 points
    why D is 0.63 in your calculation? it is should be 1- VOUT/VIN.

    The input peak current limit is 8A in the EVM.
  • 0 in reply to Jasper Li
    Prodigy 40 points

    Hello Jasper,

    The duty cycle formula is not given in the datasheet so we have used the value taken from Figure 18 in the datasheet (Auto PFM) which gives approx 63% for the switch current limit of 9.2A.

    The formula you mentioned doesn't give an answer that would make any sense, 1- Vout/Vin  = 1-15/12 = -0.25.

    The upper limit is dependent upon the T off_min which is 140ns. If it were being run at 2MHz would give a max duty cycle of (500-140)/500

    or 72%, we have used the 63% for that switch current.

    Regards

    Neil

  • 0 in reply to Neil Donaldson
    TI__Mastermind 46905 points
    Sorry for the typo, it should be D = 1- VIN/VOUT.
    why must you set the frequency to 2MHz?
  • 0 in reply to Jasper Li
    Prodigy 40 points
    Hi Jasper,
    We are running it at a fast speed to keep the PCB area smaller, higher speed smaller inductor and output capacitors.
    The D value with the the formula D = 1-VIN/VOUT would not be correct as it would be only 0.2 (20%) which seems far too low and doesn't take into account the current draw or the frequency which would have to be part of the calculation.
    Regards

    Neil
  • 0 in reply to Neil Donaldson
    TI__Mastermind 46905 points
    a more accuracy formula is D=1- Eff*VIN/VOUT , Eff is the efficiency, So lower efficiency causes higher duty. the duty is not related to the switching frequency.