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TLV62084 inductor value and output cap

Antony Lin
Genius 13480 points
Other Parts Discussed in Thread: TLV62084, TPS62361B, TLV62080

Hi,

Customer would like to use TLV62084 under below condition.  From webench, we proposed 1uF and 22uF output cap.  Now customer prefer to use 0.47uH.  is it ok for them?  What is the suggested output cap value?  Is 88uF ok for them?  THanks!

  • VIN: 3~4.8V
  • VOUT: 0.55V~1V
  • IOUT: 2A.

Antony

  • 0
    TI__Guru**** 150930 points
    0.47uH is not recommended for these conditions. The higher ripple current could trip current limit at lower load currents than 2A.

    Yes, 88uF Cout is ok for a DCS-Control device.

    They might also consider the TPS62361B family. It can support a 0.47uH coil and adjust Vout in that range via I2C. Cost is much higher but size is smaller.
  • 0 in reply to Chris Glaser
    TI__Genius 13480 points

    Hi Chris,

    Thanks for your comments.  May we know how much is the current ripple would be?  Customer need to know how big the impact is.

    THanks!

    Antony

  • 0 in reply to Antony Lin
    TI__Guru**** 150930 points
    See equation 2 in the D/S. Assuming 1 MHz frequency, this is 800 mA peak to peak with 1 uH.
  • 0 in reply to Chris Glaser
    TI__Genius 13480 points
    Hi Chris,

    What will the Fsw be for 1uH inductor? It seems equation generate a pretty big current ripple which looks strange to me? Usually what is the expected current ripple you expect from this equation under the application scneario defined by my customer?

    Thanks!

    Antony
  • 0 in reply to Antony Lin
    TI__Guru**** 150930 points
    How much output current do they need for this application? Is it 2.00A or something less? Maybe 2A has some margin built in already?

    A smaller inductance, like 0.47uH, will give an even bigger ripple. However, the ripple itself is not important. What is important is making sure that the full output current needed by the application can be delivered. This requires making the ripple small enough.
  • 0 in reply to Chris Glaser
    TI__Genius 13480 points
    Hi Chris,

    Yes, they need 2A. How did we jusdge if they can deliver 2A from the output? From equation2, it calcualte the saturation current rating required for the inductor. How do we link this to the capability of delivering 2A from the output?

    Considering below case, Ipp is around 1.684A which should be too high as you mentioned. What is this Ipp value small enough here to deliver 2A output?
    Vin=4.8V
    Vout=1V
    L=0.47uH
    Fsw=1MHz

    THanks!

    Antony
  • 0 in reply to Antony Lin
    TI__Guru**** 150930 points
    Ah, now the question is clear. The output current plus half of the ripple current is the peak current in the inductor. This peak current is what is limited by the current limit circuitry in the IC. So, the peak current should remain at or below the minimum current limit level. A higher ripple current makes this impossible at heavier loads.
  • 0 in reply to Chris Glaser
    TI__Genius 13480 points
    Hi Chris,

    Thanks a lot for the reply.

    Regarding the minimum current limit level you mentioned, is it LIM that we define in the datasheet in the range of [1.6A 2.8A 4A]?
    If this is true, does that mean the output current plus half of the ripple current should be less than 1.6A or 4A?

    Considering the used case from customer I calculate from last reply, half of the ripple current is 1.684A/2=0.842A. If so, the peak current would be 2A+0.842A=2.842A. Do yiou think this number is ok for our chip? If so, can we say customer's design with 0.47uH should be workable for 2A load current?

    THanks!

    Antony
  • 0 in reply to Antony Lin
    TI__Guru**** 150930 points
    The values you gave are for the TLV62080, which is rated for 1.2A. Just below this line are the values for the TLV62084. This is the difference between the devices: the current limit which dictates the rated output current.

    Yes, the 2.842A number should be less than the 2.3A minimum value. With 1uH, it is very close to this value.