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I am trying to design a buck/boost converter with the LTC3780 chip from Linear. My specifications are:

  • Vin = 6 to 32V
  • Vout = 24V at 100mV ripple
  • Ioutmax = 2A
  • Frequency = 300kHz

The datasheet uses the terms "inductor current ripple" and "maximum inductor current at boost mode."

What are these terms and how can I calculate them?

JRE
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abdullah kahraman
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3 Answers3

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Linear has excellent datasheets with lots of application information, calculations and component selection. This datasheet is no exception. On page 16:

The inductor current ripple ΔIL is typically set to 20% to 40% of the maximum inductor current at boost mode VIN(MIN).

Yes, they're long datasheets, but worth reading from beginning to end.

stevenvh
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    [Are datasheets really that important?](http://electronics.stackexchange.com/questions/16868/whats-this-importance-of-datasheets).... :-) – Kevin Vermeer Jul 18 '11 at 13:58
  • @Kevin - In the olden days they were more important when you got them in print. A few thick data books could get you through a cold winter, heater-wise! ;-) – stevenvh Jul 18 '11 at 14:03
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The inductor current ripple is a function of the inductor you intend to use in the application and the operating conditions. If you haven't chosen an inductance value yet, you cannot calculate this parameter.

The fundamental formula \$ v_{t} = L \frac{di}{dt} \$ applies. \$ dt \$ is the on-time of the converter and \$ v_{t} \$ is the voltage applied to the inductor. Once you know \$ L \$, it's simple algebra to find \$ di \$ which will be the inductor current ripple parameter you're looking for.

The maximum inductor current at boost mode will occur at minimum input voltage, and again is a simple calculation based on the above formula.

If you don't understand how to estimate the on-times of the converter for a given input, output and operation mode, you should spend a little time and do some research on the topology.

Adam Lawrence
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  • Thanks, but the datasheet says: "The inductor current ripple ΔIL is typically set to 20% to 40% of the maximum inductor current at boost mode VIN(MIN). For a given ripple the inductance terms in continuous mode are as follows:". And then, it gives formulas for Lboostmin and Lbuckmin in Henries. – abdullah kahraman Jul 18 '11 at 14:04
  • Every datasheet that involves buck or boost design will recommend that you use some percentage of some parameter as a an inductor starting point (that's why they say 'typically'). The actual value is up to you, depending on many criteria (how much ripple voltage you can live with, the DCM-CCM threshold, etc.) - this is why it is important to understand the topology. – Adam Lawrence Jul 18 '11 at 15:55
  • Then, does this mean that inductor current ripple percentage is 20% to 40% ? Because the percentage of the current ripple in the inductor in boost mode is = ΔIL / IL_max ? – abdullah kahraman Jul 18 '11 at 16:16
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    The ripple current percentage is whatever you want it to be. An infintely large inductor makes \$ \frac {\Delta I_L}{I_L(max)} \$ approach zero. If you design with a (relatively) large inductance, the ratio will be small; if you design with a (relatively) small inductance, the ratio will be large. If the inductor is sufficiently small, the converter will be discontinuous 100% of the time and the ratio will be maximized. – Adam Lawrence Jul 18 '11 at 16:58
  • @AdamLawrence: "The maximum inductor current at boost mode will occur at minimum input voltage, and again is a simple calculation based on the above formula." ---According to the formual above, di = (Vin/L)*Ton so the maximum inductor current ripple will occur at maximum input voltage NOT minimum input voltage as you said. Coudl you explain why? – emnha Sep 06 '17 at 06:13
  • The converter is a constant-power device - if the input voltage is lower, the input current must be higher in order to meet the load current requirements. When you are operating at minimum input voltage the current through the inductor must therefore be maximal. Ton is not going to be fixed for differing Vin, it will vary. – Adam Lawrence Sep 06 '17 at 16:46
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Considered Iout min 0.5A

T=1/f

=3usec

Dmin = Vout/Vinmax

Dmin= 0.75

L=(vout+vdiode drop)(1-Dmin)T/2*Ioutmin

L=(24+1)(1-0.75)3usec/20.5

L=18.75uH

∆I=100mV

I max= 2A+(∆I=100mV/2)

Maximum inductor current= 2.05A

Ananthesh Acharya
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