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I found the minimum inductor required for a buck converter using the following formula -

$$L_1 = \frac{V_{in\,max} - V_{out}}{I_o \times K_{ind}} \times \frac{V_{out}}{V_{in\,max} \times fsw}$$

We usually take 40% buffer on the calculated minimum inductance value (so that inductance at rated current will not vary much and ripple will be under control) and choose the inductor.

I think lower inductor values will have more ripple, good transient response, as inductor value increases there will be less ripple and less pressure on output capacitors.

Is there any higher limit on the inductor value for a buck converter?

My requirements are the following:

  • \$V_{in}\$ = 10.8 to 13.2V, typical = 12V
  • \$V_{out}\$ = 0.9V
  • \$I_{out}\$ =15A
  • ripple = 30%
  • \$V_{out}\$ ripple = 2.5%
  • and possible transients can occur up to 10A/2us.

I have seen in WEBENCH from TI/National Instruments where maximum value of inductor is specified, but I'm not sure how they have calculated it.

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Increasing the value of the inductor will sooner or later turn your circuit into Continuous Conduction Mode (CCM), where there is a constant DC current flowing through the inductor, and the ripple current (the triangle waveform) being superimposed on this base DC current.

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This means that the current of the inductor in this mode never goes down to zero. The energy stored in the inductor is indeed \$\dfrac{LI^2}{2}\$, however, only a part of this energy gets transferred in a cycle to the consumer: \${\dfrac{LI_{max}^2}{2}} - {\dfrac{LI_{min}^2}{2}} = \dfrac{L(I_{max}^2-I_{min}^2)}{2}\$, the rest stays in the inductor (more or less permanently). The amount of energy used by the consumer will be the sum of the energy transferred through storage in the inductor, and the energy constantly being transferred by the DC current flowing through the inductor.

The amount of the maximum possible ripple current is determined by the inductance value, the switching frequency, the ON-OFF duty ratio (which more or less equals the output to input voltage ratio), and the voltages connected to the inductor (the input voltage minus the output voltage in the ON phase, and the output voltage in the OFF phase). Increasing the inductance value decreases the maximum possible ripple current. As soon as the output current needed is more than half of the maximum possible ripple current, operation will turn into CCM.

What disadvantages are there for having a larger inductance?

  • More turns, therefore higher DC resistance in the inductor, meaning larger copper losses, or having to increase wire thickness/number of strands to compensate for this, thereby increasing inductor size & cost.

  • Having a larger inductance, the feedback control loop will be slower, meaning that the power supply will be less flexible adapting to quickly changing loads. This will probably show itself as larger overshoots in response to a unit-step load change.

  • When using an asynchronous buck converter, in OFF-state, the free-running diode will be conducting. In DCM, the diode is conducting only as long as the energy stored in the inductor gets fully removed. In CCM, the diode is conducting during the whole OFF-phase, as the inductor current never goes to zero. This means higher losses on the free-running diode - which may be a problem especially because the losses on the FET can be decreased by using a FET with a lower Rdson, but you cannot do the same with the diode. The smaller the output to input voltage ratio is, the more grave this problem could be.

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    \$\begingroup\$ It's OK you don't have to disagree with me any more I've deleted my answer following a re-appraisal - thanks for putting me straight on this +1 \$\endgroup\$ – Andy aka Nov 14 '13 at 12:36
  • \$\begingroup\$ I understand the possible pros and cons of increasing inductance value. Practically/theoretically, is there any upper limit on inductance value for a particular design. Any equation which gives limit of max inductance. \$\endgroup\$ – user19579 Nov 15 '13 at 5:19
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    \$\begingroup\$ @user19579: No, there is no such upper limit, above which you may not go. That's the reason why every data sheet/application note calculates only the lower limit. \$\endgroup\$ – Laszlo Valko Nov 15 '13 at 6:52

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