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LM5176: Enable/UVLO, hysteresis, VIN/VCC

d_zero
Genius 4795 points
Part Number: LM5176

Dear *,

we are designing the schematic for LM5176 and have some questions.

Our input voltage for normal operation is from 12V to 24V.

1) What is the minimal voltage that needs to be applied to Vin (pin28 QFN) to get the VCC=7.35V?

Vin = Vcc + Vs + Vinternal

Vs -> voltage drop on series blocking diode

Vinternal -> voltage drop on internal linear regulator 

We are not using BIAS pin connected to Vout, nor we are supplying BIAS voltage -> BIAS is NC(not connected)

2) Does it make sense to connect the BIAS pin to VIN pin?

3) If we don't use BIAS pin and UVLO is set to 9V do we then need the series blocking diode?

4) In the DS on page 23 under typical application 8.2.2.9 the EN/UVLO hysteresis is selected to be 0.8V , why is this so? Why didn't you select 0.4V (lower) or 1.6V(grater)?

5) Of we want that the LM5176 turns off when Vin=10V or below, and the hysteresis is 0.8V then the turn on voltage is UVLO+Vhys

=10V +0.8V is that correct?

6) if we use additional Rs(series resistor to EN/UVLO pin like this https://e2e.ti.com/support/power-management/f/196/p/680351/2506288?tisearch=e2e-sitesearch&keymatch=lm5176%2520uvlo%2520hysteresis#2506288)

how is then the UVLO and Vhys calculated?

Best Regards,

David.

  • 0
    TI__Mastermind 21720 points

    Hello d_zero,

    Thanks for reaching out with your questions and for using the LM5176 in your design.1)

    When the VIN voltage is above the VCC regulator voltage the VCC pin will regulate to 8V. Figure 4 of the datasheet shows the VCC voltage vs the VIN voltage. Since your application has an input voltage of 12V the VCC pin will regulate to 8V.

    2) I don't recommend connecting the BAIS pin to the VIN rail. The BIAS pin is intended to lower the power dissipation of the LDO. This is for application when the output voltage is lower than the typical input voltage range.

    30, the blocking diode is not needed.

    4) This fits a typical application that he evaluation board was designed for. The hysteresis can be changed based on the resistors selected. The hysteresis can also be increased by adding a resistor in series with the UVLO pin.

    5) They hysteresis will change based on the resistors that are selected. I suggest taking a look at the LM5176 quick-start calculator to calculate the correct resistors.

    6) If a series resistor is added then the hysteresis is estimated as below.

    Vhyst = Ihyst*(Ruv2+Rhyst) (Rhyst is the added series resistance. IHYST is 3.5uA)

    Please let me know if you have any questions.

    Thanks,

    Garrett

  • 0 in reply to Garrett Roecker
    Genius 4795 points

    Hello Garrett,

    thank you for your feedback.

    1) ok i want that the regulator turns on at 10V and turns off at 9V, so the hysteresis is 1V is that correct?

    2) i'm calculating Ruv2 per EQ2 from DS and the RUV2 = 1V / 3.15uA = 286kOhm 

    also i'm calculating Ruv1 per EQ1 from DS   RUV1 = 42kOhm is that correct?

    Best Regards,

    David.

  • 0 in reply to d_zero
    TI__Mastermind 21720 points

    David,

    you can use the attached document

    UVLO_Resistor_Calculations.xlsx