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This question is regarding the inductor design (uncoupled) in a multiphase buck converter.

In a multiphase converter, the inductance per phase requirement is higher than the corresponding single phase solution.

The ripple current per phase is lower for the same Vin and Vout, hence the corresponding inductance requirement is higher.

enter image description here

Now there are 'n' such inductors with each inductor value higher than the corresponding single phase case. Agree that the input and output capacitance requirement has come down due to ripple cancellation.

The inductor size depends on both inductance and current through inductor.

Is that the reason why we prefer multi-phase inductors for high current application- even though we need 'n' inductors having 'n' times higher inductance?

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  • \$\begingroup\$ Due to ripple cancellation. The effect is huge if you need to halve the input voltage with a two phase buck, divide by three with three phase and so on. \$\endgroup\$
    – winny
    Aug 1 at 11:36
  • \$\begingroup\$ I have a follow-up question: Sec 3.4 of TI document ( link below) says multiphase converter improves the transient response since the 'n' inductors act in parallel and effective inductance decreases to L/n. But Iam unable to understand how the equivalent inductance is reduced (by a factor of n) since already the per phase inductance is scaled by a factor of 'n'. Theoretically the equivalent inductance gets backs to same as single phase VR inductance -during a transient event. \$\endgroup\$
    – Divya K.S
    Aug 2 at 18:28
  • \$\begingroup\$ ti.com/lit/slva882 \$\endgroup\$
    – Divya K.S
    Aug 2 at 18:29
  • \$\begingroup\$ Have you tried to simulate it? \$\endgroup\$
    – winny
    Aug 2 at 20:26
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If you are comparing (a) a single buck vs (b) a multi-phase buck where both (a) and (b) are delivering the same maximum loading output power at the most efficient duty cycle then sure, the inductors in (b) will be greater in value than the single inductor value in (a).

After all, each of the bucks in (b) receive the same input voltage, produce the same output voltage and, due to the design being synchronized, switch to to a common shared cycle AND therefore, must have the same duty cycle because each of the bucks in (b) shares the full load current current.

Is that the reason why we prefer multi-phase inductors for high current application- even though we need 'n' inductors having 'n' times higher inductance?

One of the major reasons for (b) is that the input ripple current (p-p) is reduced compared to one big brute converter (a). The output ripple current is also reduced by the same argument. These are the main reasons.

To make (a) have the same low ripple current as (b) requires (a) to be operated at a much higher switching frequency and then switching losses might become a big problem.

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The primary reason for multi-phase is higher current efficiency when load variation is also reduced from Pmax:Pmin ratio.

e.g. 10:1 for \$1\phi\$ and 2:1 for \$3\phi\$ and up might be reasonable limits for high efficiency, while losses reduce with more phases at max load sharing.

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So the added cost of increased phases is offset by reduced power loss.

Ipp = Imax / n so L is proportional to N, and C reduces with n so does LC remain constant?

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The inductor size depends on both inductance and current through inductor. Is that the reason why we prefer multi-phase inductors for high current application

Multi-phase DC-DC converters are used in high-current applications, like 10A or 20A for example. The current is shared between each phase, thus reducing the stresses on the inductors and semiconductors in each phase.

In addition to this, the effective output ripple voltage is n times the switching frequency of the FETs, so a smaller output capacitance is needed.

This also means that 'slower' semiconductors can be used in each phase, driven at a relatively low switching frequency, but because there are n phases, the effective ripple frequency is n times the base frequency.

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