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LM5067: calculation of Rin

Oren Eliezer
Prodigy 30 points
Part Number: LM5067

The formula for Rin in the spreadsheet provided by TI is:   " =IF(VINMIN>35, 5.1, (VINMIN-15)/2) "

However, this is discontinuous around VINMIN=35V.   and the current for a 5.1k resistor would be much higher than the 2mA minimum.   

I believe this is erroneous.  

Furthermore, the internal voltage is closer to 13V than to 15V (according to the datasheet), and if it is known then why assume 15V? 

  • 0
    TI__Guru** 105611 points

    Hi Oren,

    Welcome to E2E!

    Let me check and get back to you before tomorrow.

    Best Regards,

    Rakesh

  • 0 in reply to Rakesh
    TI__Guru** 105611 points

    Hi Oren,

    Minimum 2mA current should be allowed under any conditions. So Rin < (Vin_min - Vz_max)/2mA

    Ideally, Vz_max should be 13.65V but taking 15V gives more margin -> I mean higher than 2mA current.

    I agree at Vin=35V there is jump in current but it is not a concern.

    Best Regards,

    Rakesh

  • 0 in reply to Rakesh
    TI__Guru** 105611 points

    Hi Oren,

    If there are no further questions, please close this thread as resolved.

    Best Regards,

    Rakesh

  • 0 in reply to Rakesh
    Prodigy 30 points

    Hi Rakesh,

    I didn't notice the response from yesterday until the email from today.   I wish we could simply communicate directly via email. 

    Thanks for the explanation. I understand the use of Vz=15V for magin.

    But I insist that the conditional statement with Vin>35V doesn't make sense, since Rin will be stuck at 5.1k even for very high Vin.   Perhaps the original intention was to not allow Rin to go below 5.1k, but for Vin=60V it may make sense to have a higher Rin.  Otherwise the power dissipated on this resistor can be high.   And anytime a function is defined with ranges, there would be continuity around the boundary between ranges. 

    Thanks,

       Oren

  • 0 in reply to Oren Eliezer
    TI__Guru** 105611 points

    Hi Oren,

    Generally the operating voltages are wide from 36V to 60V. So at 36V, we get Iz=(36V-15V)/5.1kOhm = 4.1 mA (has 2x margin from min 2mA requirment), I think, this is the intention to choose 5.1k. I agree on power dissipation at max Vin level.

    Best Regards,

    Rakesh

  • 0 in reply to Rakesh
    Prodigy 30 points

    Hi Rakesh,

    Thanks for explaining, but this still doesn't make sense.  If the voltage at the input is much higher (e.g., 60V), then the current will further increase, which will cause unnecessary high power dissipation on this resistor. 

    And the discontinuity around 35V should also be corrected.  Anyway, I set this resistor at 10k, which satisfies the 2mA in our range of Vin, which is 36-60V. 

    Thanks and have a nice weekend,

      Oren