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In a multi-phase buck converter, when and how does the inductor breakdown?

Lets say the multi-phase buck converter has the following parameters:

V_in = 50V, V_out = 25V, P_in = 225W (so max amps = 225V/15V = 15A), R_load = 1 Ohm, Switching_frequency_per_phase = 62500Hz and the converter is 4 phases (lets call them phase A, B, C and D respectively)

Inductor wire is rated at 15A current and the saturation current rating is also rated at 15A.

When the phase A first turns on (phase B is off) the current going through the inductor will be (50V-25)/1Ohm = 25A. Will this current breakdown the inductor? Keep in mind that when the phase B turns on, the inductors on phase B will start drawing the rest of the current and so both inductors will have 25A/2 = 12.5A which the inductors can handle.

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    \$\begingroup\$ your assertions about how a buck converter draws current are false, so your question as it stands is unanswerable. Google a bit about how they work, then ask a more focussed question. Note a buck converter will have a higher output current than input, so as you are concerned about max current, you're looking at the wrong port anyway. \$\endgroup\$
    – Neil_UK
    May 29, 2017 at 3:10
  • \$\begingroup\$ I understand how a basic (single phase) buck converter works.I have designed a basic buck converter already and that works, i've tested it by scoping and using it multiple times. My question is about advanced (multi-phase) buck converters, I have already googled multiple times about buck converters, read research papers on buck converters and they do not answer my questions. Also "you're looking at the wrong port anyway" - Which port should I be looking at? I appreciate your comment and hopefully we can find an answer. \$\endgroup\$
    – Veda Sadhak
    May 30, 2017 at 18:37
  • \$\begingroup\$ 'When the first phase turns on', the inductr current will probably be quite small, most SMPS have a soft start. When all phases are sharing current, the output current is ramped up to full load. You can consider each phase to be responsible for 1/Nth of the total load current, so you can consider just one phase in isolation, at that load, to figure out inductor currents (however the output ripple will of course be better than fr a single phase). \$\endgroup\$
    – Neil_UK
    May 30, 2017 at 19:09
  • \$\begingroup\$ Thanks for the quick response Neil! So if I understand it right when the phase A first turns on the current will slowly ramp up on inductor A so it will not get saturated instantaneously and when phase B turns on that will share the current so no inductor ends up saturating unless the duty cycle is inappropriate. That makes sense based on the inductor equation: V = L*(dI/dt). So the key knowledge I did not know is that SMPS have a soft start. \$\endgroup\$
    – Veda Sadhak
    May 30, 2017 at 19:24
  • \$\begingroup\$ Also Neil, why not put down your comment as the answer? I would be happy to give you the correct answer mark. \$\endgroup\$
    – Veda Sadhak
    May 30, 2017 at 19:24

1 Answer 1

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My premise that "When the phase A first turns on (phase B is off) the current going through the inductor will be (50V-25)/1Ohm = 25A" is incorrect. Below is how you solve how much current will be flowing through the inductor.

Initially the inductor current is 0, inductor is rated at 15uH also my switching frequency per phase is 62500 Hz.

The change in current between the time when phase A first turns ON and when phase B turns ON is calculated as follows.

V = L*(dI/dt)

dI = (V/L)*dt

dI = ((Vin-Vout)/L)*dt

dI = ((50V-25V)/15uH)*dt

Note: dt = (1/62500)/4 = 4uS -> (With 4 phases phase B turns on after 1/4 of the switching period)

dI = ((50V-25V)/15uH)*(4uS)

dI = 6.67A

So when phase B first turns on the current going through the inductor in phase A will be 6.67A. Since 6.67A < 15A (15A is the rating of the inductor) then inductor will not breakdown when the SMPS first turns on.

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  • \$\begingroup\$ What you've written there could be correct (I haven't checked the maths details), for a particular design of converter, for some particular control assumptions. The 'startup transient for a multi-phase' is never an issue, because they're not controlled in a way that would make your detail above valid. 1) Split the output current between all phases, so a 4 phase buck would be designed as far as current control is concerned as 4 off 6.25A output converters. 2) Mean of 6.25A is peak of more than that, so plenty of room below 15A if designed properly. Then handle startup, and everything else. \$\endgroup\$
    – Neil_UK
    Jun 2, 2017 at 5:06

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