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When a synchronous converter operates in high load condition,if the inductor current is always positive, the switching loss of the Synchronous mosfet is minor in comparison to the switching loss of the Control mosfet. This is because the Synchronous mosfet turns on and turns off in "soft switching" while the Controller mosfet turns on and turns off in "hard switching" (I have summarized a lot). But when a synchronous converter operates in low load condition and the inductor valley current become negative , Control mosfet and Synchronous mosfet turn on in "soft switching" and turn off in "hard switching". If I suppose that the 2 mosfets are matched, the switching loss for both devices, is the same? I'm sorry for my bad English, it's not my mother tongue, and it is not easy to express the concept that I want. Thank you all for the attention.

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  • \$\begingroup\$ High load conditions do not mean current is always positive. That's the point I gave up reading your question. \$\endgroup\$ – Andy aka Oct 24 '16 at 7:25
  • \$\begingroup\$ Yes, I made a mistake..of course if you use a current ripple >2IL (average inductor current) current became negative..Otherwise my question is about light load,where conduction losses are negligible. \$\endgroup\$ – jaster Oct 24 '16 at 8:17
  • \$\begingroup\$ OK feel free to modify your question but what do you mean by soft and hard switching? \$\endgroup\$ – Andy aka Oct 24 '16 at 8:26
  • \$\begingroup\$ if the current for the mosfet that is leading the current is negative when occurs the turn off,his Vds is the Vf of the body diode..this lead to a minor switching loss (soft switching). \$\endgroup\$ – jaster Oct 24 '16 at 8:51
  • \$\begingroup\$ It's fairly complicated but if you look into various syncronous controllers, you will see different schemes how to handle light load to save losses. You can go into burst mode, change frequency, go asyncronous and so on. There are also benefits to use forced continous conduction where you go to negative current during part of the time. \$\endgroup\$ – winny Oct 24 '16 at 8:53
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For purposes of this discussion there are three types of synchronous buck controllers/converters:

  1. Those that do not have a light load PFM, pulse skipping, burst mode or diode emulation mode. These will operate in CCM all the time, and when the load current doesn't exceed 1/2 the p-p ripple current the inductor current will go negative each cycle and you will have increased switching losses over one of the light load modes as Olin point out.

e.g. TPS54622

  1. Those that do have some type of light load operation.

e.g. TPS54623

  1. Those that have an input or programmable option for forced CCM (if for example you want to avoid the switching frequency changing, maybe to stay out of the audio band at certain times.) or give the option of a light load mode.

e.g. LM5160

When inductor current is positive, the top switch turns off first, followed by a dead time. During the dead time if there's enough current to slew the switch node capacitance to ground (or a body-diode drop below ground), the lower switch will turn on with very little voltage across it. Therefore zero voltage switching or soft switching.

If you are always in CCM and the inductor current goes negative, say at no load, then as the OP points out when the bottom switch turns off with negative current, the top switch soft switches, and if the FETs were the same part number they would both have about the same switching loss. However, it's common to choose different FETs for the control and sync FETs to optimize conduction and switching losses depending on the conversion ratio, so there could be asymmetrical switching losses due to extra gate charge and transition losses on one FET vs. the other.

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  • \$\begingroup\$ Thank you so much, I recognize that it was difficult to understand what I asked because I've explained very badly. As I said to Olin, I have to design a boost with output voltage 9 V and with very small output currents (up to a minimum of 10 mA) and I will use LM5122 controller that allows me FPWM (just as I have suggested in other question I asked. I suppose, given the low currents and consequently the low conduction losses, I will choose same devices withvery small gate charges so as to minimize the switching losses, which, you have confirmed, are practically equal. \$\endgroup\$ – jaster Oct 25 '16 at 9:06
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It's not clear what you are asking, but the inductor current never goes negative in a properly controlled switching power supply.

With synchronous rectification and in discontinuous mode, you do run the risk of leaving the synchronous switch on too long and thereby allowing reverse inductor current. So don't do that.

In discontinuous mode near the end of the synchronous on interval, the inductor current is low anyway since it goes to exactly zero at the end of the interval. The extra losses are therefore minor if you turn off the synchronous switch a little early. You get one diode drop additional voltage drop for the inductor current when it is low and about to be zero anyway, so really not much of a problem.

In general, shoot-thru and reverse current from leaving the synchronous switch on too long are worse than the extra diode drop for a short time, especially at the end of a discontinuous conduction pulse.

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    \$\begingroup\$ Why is inductor current going negative in a sync regulator a problem Olin? In the case of a very light load I can see this happening in order to keep output voltage regulation. \$\endgroup\$ – Andy aka Oct 24 '16 at 11:00
  • \$\begingroup\$ @Andy: Negative inductor current (drawing current from the output) causes loss. That negative current must ultimately be offset with more positive current to yield the same average output current. All that extra current causes more loss across inevitable resistances and voltage drops. Under light load, it's usually a better strategy to stretch out the time between pulses than to do very short pulses, or than to do longer pulses that include a negative component. \$\endgroup\$ – Olin Lathrop Oct 24 '16 at 11:12
  • \$\begingroup\$ It's not my down vote BTW! \$\endgroup\$ – Andy aka Oct 24 '16 at 11:29
  • \$\begingroup\$ @OlinLathrop thanks for the answer, I will better introduce what I have to do: I have to design a synchronous boost converter whose output specs are : Vout=9V and Io_min=10mA Io_max=500mA. \$\endgroup\$ – jaster Oct 24 '16 at 12:24
  • \$\begingroup\$ (continue) In this situation it's nearly impossible to avoid negative current in the inductor because @10mA i should have a too small ripple current.At the same time I need that the frequency remains the same and I can't use Burst mode, skip cycle or diode emulation..By the way what i'm asking is: in low load operation when the inductor current (not the averaged) is negative, synch and control mosfet have the same switching loss? \$\endgroup\$ – jaster Oct 24 '16 at 12:31

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