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LM25085 Excel Calculator: What am I missing?

Romulo Maggay
Prodigy 110 points
Other Parts Discussed in Thread: LM25085

Hello,

I have this application of dynamically adjustable output voltage with the following specs:

- Vout=16 to 32V, Vin_min=34.155, Vin_nom=34.5, Vin_max=34.845, Iout_max=3.3A, Iout_min=50mA

- minimum ripple voltage

I selected LM25085. I downloaded the Excel calculator and I also formulated my own Excel based on the formulas given in the Datasheet. To validate my Excel, I used the application example in the datasheet (Vin=7 to 42V, Vout=5V, time delay PFET=57ns, etc...) I plugged in the values and both Excel outputted the same Rt=90.9K as expected. 

But when I now calculate the L1, there are now disparities in computed values. The example given expects L1=13.5uH which is the same as my own excel outputted, but the Quick Excel Calculator is outputting a different value which is L1=16.2uH minimum. I then tried to just vary the Iout_max and the L1 value also varies. This is unexpected because based on L1 formula (eq.18), it is only dependent on Ton_min, Vin_max, Vout and Iout_min. Am I missing something here?

I went ahead with my computation using my Excel formula and results below:

Vout=16 to 32V, Vin_min=34.155, Vin_nom=34.5, Vin_max=34.845, Iout_max=3.3A, Iout_min=50mA, PFET delay=290ns, Fs=500KHz

Rt=342.8K

Ton_min=1.544uS @ PGATE

Ton_min=1.83uS @SW Node

Then at Vout_min=16V (same Rt), Fs=250.67KHz

Finally, L1 = 52uH @ 32V

              L1=345.7uH @ 16V

Using Quick Calculator, L1=7.9uH! @ 32V

  • 0
    TI__Genius 11490 points

    Thank you for selecting LM25085 for your application.

    Do you really have a PFET with 290ns delay?

     

  • 0 in reply to Vijay86929
    Prodigy 110 points

    hello Vijay,

    yes, please have a look at SI7461. I have used this since this is already in our material system.

  • 0 in reply to Romulo Maggay
    Prodigy 110 points

    hi vijay,

    on a seperate case, i'm planning to use external current limiter/detection (i use the SHUTDOWN signal when current limit is reached) where i intend to bypass the internal current comparator.

    any recommendation on how to properly terminate the ADJ and Isen pins to avoid unintended operation?

  • 0 in reply to Romulo Maggay
    TI__Genius 11490 points

    I would sugest setting the current limit to a value higher than what your external current limit circuit is tripping at.  This can be done in the standard LM25085 circuit by using a lower current sense resistor.

     

  • 0
    TI__Genius 11490 points

    My calculations match the quickstart calculator reasonably well. E.g. ~7.9uH @32V.

    Formulae: L1=(VIN-VOUT)*TON/(delta_I), where delta_I is the peak to peak inductor current ripple.

    I would however use the inductor value calculated for 16VOUT. This is because with 7.9uH, the current ripple with 16VOUT is really high (~4A pk-pk).

     

  • 0 in reply to Vijay86929
    Prodigy 110 points

    My worry here is the offset of +/-9mV and the current tolerance of 32uA to 48uA.

  • 0 in reply to Vijay86929
    Prodigy 110 points

    Hi Vijay,

    I'm surprised you were able to get 7.9uH using your formula. I proof-checked my formula by inputting the values from the example in the datasheet and got the correct value as detailed below;

    LM25085 Example per datasheet:

    Given:Vout=5V, Vin=7V to 42V, Iout_max=5A, Iout_min=600mA, Fs=300khz, Vos=5mV p-p, tdelay=57ns

    My generated excel sheet yielded the same values for all the external components. For example, in computing L1,

    ton_min=438nS @SW node

    Vin_max=42V

    Vout=5V

    Ior_max=2*Iout_min=1.2A

    Formulae for L1=(ton_min * (Vin_max-Vout))/Ior_max. Using my formula, the output is 13.5uH which is the same with the example. Now, using the same formula with the following inputs for my application;

    ton_min=1.83uS @SW

    Vin_max=34.845V

    Vout=32V

    Ior_max=0.05*2=0.1A

    these values gives me, L1=(34.845V-32V)*(1.83uS/0.1A) = 52.2uH!

    So how did we differ on these values?

  • 0 in reply to Romulo Maggay
    TI__Genius 11490 points

    Hi Romulo,

    Ior_max values are different in your and my calculations. I used Ior_max~20%*IOUT_max.

    It really depends on how much current ripple your are comfortable with.

    Regards,

    Vijay

     

  • 0 in reply to Vijay86929
    Prodigy 110 points

    Hi Vijay,

    Thank you for highlighting the discrepancy in our formulas used for computing the Ior_max.

    And now it is more confusing because based on the application note for LM25085, the Ior_max=0.2 * Iout_max is recommended for applications with minimum output load current of 0A, whereas for Iout_min > 0A, a guideline for Ior_max= 2 * Iout_max.

    So if I understand you right, you also recommend to use the "20%*Iout_max" guideline even for Iout_min > 0 Ampere?

  • 0 in reply to Romulo Maggay
    TI__Genius 11490 points

    Hi Romulo,

    It really depends. Some applications try to keep the converter always in CCM (inductor current is never discontinuous). In those applications Ior_max=2*Io_min makes sense.

    These are only guidelines and the user needs to ultimately decide the trade off in inductor value, size, ripple etc.

    Thanks and regards,

    Vijay

  • 0 in reply to Vijay86929
    Prodigy 110 points

    Hi Vijay,

    I think you have fully answered my queries.

    Lastly, I tried to use the 20% guideline to compute the L1 for my dynamically adjustable output from 16V to 32V.

    I used Rt computed from 32V @184.5k (hence ton_min=1.15uS).

    At 32V, L1=5uH, but the ripple at 16V would be around 4.4A pk-pk.  The ripple is too high.

    At 16V, L1=72.2uH, and the ripple at 32V is only around 0.0453A. So I'm gonna use L1=72.2uH.

    Do you agree with the values? If yes, then I have to start searching for this inductor that is rated by at least 3.5A.

  • 0 in reply to Romulo Maggay
    TI__Genius 11490 points

    Seems reasonable.

  • 0 in reply to Vijay86929
    Prodigy 110 points

    Hi Vijay,

    Seems like the 72uH inductor SMDs with Irms rating of at least 3.5A is hard to find commercially, 

    But this value is based on using the Iout_max @16V which is 1.5A estimated (Iout_max=3.3A @32V out).  I can improve the time delay of PFET from 290ns to 57ns but the L1 decrease is insignificant. I maximized the Fs @ 1,000 kHz and reduced L1 from 72uH to 58uH, but still high.

    Any other recommendation how to reduce L1 so I can use Coilcrafts XAL/XFL power inductors or other small power inductors? 

  • 0 in reply to Romulo Maggay
    TI__Genius 11490 points

    Well I think you will have to live with higher ripple current at 16VOUT. Using Iripple=40-50% of IOUT will give you something more reasonable.

    Regards,

    Vijay

  • 0 in reply to Vijay86929
    Prodigy 110 points

    Hi Vijay,

    Thanks for the support. We may now close this topic. 

    best regards

  • 0 in reply to Romulo Maggay
    Prodigy 110 points

    Hi Vijay,

    I forgot to ask you about the very low ripple current when the L1 computed at 16V is used to compute at 32V output.

    Assuming L1=57.8uH, then my ripple at 32V is around 45.3mA. Is there any lower limit when it comes to inductor ripple current?

  • 0 in reply to Romulo Maggay
    TI__Genius 11490 points

    Hi Romulo,

    You mentioned you are using minimum output ripple configuration. This uses ripple generated by resistor and capacitor. Since it does not use inductor current ripple, there is no lower limit of inductor current ripple.

    Regards,

    Vijay

  • 0 in reply to Vijay86929
    Prodigy 110 points

    Hi Vijay,

    Yes, i'm using the minimum output ripple voltage configuration.

    Anyway, this sufficiently answers my question. Thanks