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LM2735: LM2735X SEPIC converter design problem

Dmil
Intellectual 390 points
Part Number: LM2735
Other Parts Discussed in Thread: TPS61170, TINA-TI

Hello!

I've calculated component values for example circuits from LM2735 documentation to check the design equations before designing my converter.

8.2.12 LM2735X WSON SEPIC Design Example 12

(1.6 MHz): VIN = 2.7 V - 5 V, VOUT = 3.3 V at 500 mA
Let's use T = 25C, Switch resistance = 0.19 Ohm, current = 3A,

"A good design practice is to design the inductor to produce 10% to 30% ripple of maximum load." - p.16, LM2735 doc.
"A rule of thumb is to use 20 to 40% of the input current," - slyt309.pdf

Let's try the range of inductor ripple: 6% to 33% of Iswitch.

vout = Vout + Vdiode = 3.3V + 0.5V = 3.8V

Vin.max = 5V - Vswitch = 5V - 3A * 0.19 Ohm = 4.43V
DCmin = vout / (vin.max + vout) = 3.8 / (4.43V + 3.8V) = 0.462
Lmin = vin.max * DCmin / (Fosc * 0.33 * Isw) = 4.43V * 0.462 / (1.6MHz * 0.33 * 3A) = 1.3uH
Lmax = vin.max * DCmin / (Fosc * 0.06 * Isw) = 4.43V * 0.462 / (1.6MHz * 0.06 * 3A) = 7.1uH

Vin.min = 2.7V - Vswitch = 2.7V - 3A * 0.19 Ohm = 2.13V
DCmax = vout / (Vin.min + vout) = 3.8 / (2.13V + 3.8V) = 0.641
Lmin = Vin.min * DCmax / (Fosc * 0.33 * Isw) = 2.13V * 0.641 / (1.6MHz * 0.33 * 3A) = 0.86uH
Lmax = Vin.min * DCmax / (Fosc * 0.06 * Isw) = 2.13V * 0.641 / (1.6MHz * 0.06 * 3A) = 4.74uH

As result: 1.3uH < L < 4.74uH

The example circuit uses 6.8uH.

If I use "10% to 30% ripple of maximum load" (from 0.1 to 0.3 of 0.5A) than the result is 7.8 uH < L < 28.44uH

So the circuit inductor value 6.8uH is also out of the range.

What's wrong with the calculation?

  • 0
    TI__Mastermind 21720 points
    Hi Dimitriy,

    Where are you getting the current of 3A if the output current is 500mA? From simplicity I suggest initially doing the calculations assuming that the switch is ideal. This assumption will make it easier to spot the issue in the equations.

    -Garrett
  • 0 in reply to Garrett Roecker
    Intellectual 390 points

    Hello Garrett,

    the IC switch current limit is 3A, so I do calculation for Lripple 6% and 33% of the max switch current.
    Is this a mistake?
    What current should I use?
    To get the L value around 6.8uH the current should be ~1.5A, but where should I get this value?

    Or another variant: Lripple should be ~1...3% of the max switch current.

  • 0 in reply to Dmil
    TI__Intellectual 2655 points
    Hi Dmitriy,
    you can take the application note "Design DC/DC converters based on SEPIC topology" as a reference.
    www.ti.com/.../slyt309.pdf
  • 0 in reply to Jing Ji
    Intellectual 390 points

    Hi Jing Ji,

    in my first post I am referring to this document:
    "A rule of thumb is to use 20 to 40% of the input current," - slyt309.pdf

    The input current is either 3A (max switch current) or 0.867A (max input current for 2.7V input voltage at 3.3V@500mA output).

    At these current values you can't get L = 6.8uH.

  • 0 in reply to Dmil
    TI__Intellectual 2655 points

    Hi Dmitriy,

      In the application note

    http://www.ti.com/lit/an/slyt309/slyt309.pdf

      You should use equation 4 to calculate minimum L. I found you didn't use 1/2 ratio.

      Based on the minimum L, you can choose a higher inductor value.

      And you should use equation (3) to calculate inductor current ripple, and you shouldn't use 3A. 3A is IC current limit, not inductor current at normal work.

      If you have other question, please tell me.

     

  • 0 in reply to Jing Ji
    Intellectual 390 points

    Yes, it is a problem!

    And I've found that different manuals use different formulas.

    For example, a manual from other manufacture says:


    Given an operating input voltage range, and having chosen ripple current in the inductor, the inductor value (L1 and L2 are independent) of the SEPIC converter can be determined using the following equation:
    L1= L2 = VIN(MIN) * DMAX / (0.5 * dISW * fOSC)

    By making L1 = L2, and winding them on the same core, the value of inductance in the preceding equation is replaced by 2L, due to mutual inductance:
    L = VIN(MIN) * DMAX / (dISW * fOSC)

    Which variant is correct?

  • 0 in reply to Dmil
    TI__Intellectual 2655 points
    I will be on weekend and I will reply later.
    Thanks
    Jing
  • 0 in reply to Jing Ji
    Intellectual 390 points

    Even in TI docs:

    slyt309.pdf
    Figure 6. TPS61170, SEPIC design with 9- to 15-V VIN and 12-V VOUT at 300 mA
    Inductor value is 22uH

    TPS61170.pdf
    Figure 19. 12-V SEPIC (Buck-Boost) Converter, 9- to 15-V VIN and 12-V VOUT at 300 mA
    Inductor value is 10 uH

    TINA-TI, TPS61170 TRANSIENT SIMULATION, Boost, 12V to 24V 300mA
    Inductor value is 22uH

    TPS61170.pdf, Figure 18. 12-V to 24-V DCDC Power Conversion
    Inductor value is 10uH

    The same factor of 2 there is in the formula for Cout.

    slyt309.pdf (7) says:
    Cout >= Iout * DC / (Fosc * dVout)

    The formula from another manufacture:
    Cout >= Iout * DC / (0.5 * Fosc * dVout)

    And there is a mistake at using this equation in slyt.309.pdf, p.22, step (7):
    you should calculate 2 values of Cout, first for Iout at Vin.min and DCmax, second for Iout at Vin.max and DCmin and than use the maximum value.
    Different conditions will give different answer for both values.

    Both documents shows oscillograms so they both use Fosc without any internal divider, and if manufactures use an internal frequency divider then you would not be able to use Fosc to choose a controller.


    I am looking forward to your answer!

    P.S.

    Today I found a variant from 3d manufacture, that has this factor of 2 for uncoupled inductors!
    So their variant leads to 4 times less coupled inductor value than from 2nd manufacture!

  • 0 in reply to Dmil
    Intellectual 390 points

    I've understood a mistake in my calculation, thank you!

    The denominator should be (1.6MHz * 0.33 * 2 * 0.867A) and (1.6MHz * 0.06 * 2 * 0.867A).

    But still, why slyt309.pdf and TPS61170.pdf have the different L value?