the nitty gritty of buck switching losses : -

Above is the classic buck switching cell

This has been analysed to death over the years, but rarely do you read a concise account of the detail of the switching and associated losses

In a more or less empirical fashion, here we go : -

Consider the mosfet is on and we are bucking from 100V to 24V and the current in the choke is 10 amp

Now we turn the mosfet off - what happens ?

For the current to commute to the diode, the diode must be forward biased - i.e. its kathode must be at lower volts than its anode

Until that condition exists, no current will flow in the diode ( excepting capacitive current for a large schottky as the kathode falls )

So as we turn the mosfet off, it will carry full current until its source voltage ( in the case above - N fet ) goes low enough for the diode to become forward biased [ some capacitive current may flow through the diode as its kathode falls relative to anode - & through any attached snubber likewise ]

The losses during this period are approx: 0.5 x ( Vin + Vf + Vf_rec ) x I x T_volt fall time x freq - for a linear fall of voltage across the diode

[ the choke is powering this drop in kathode volts as it attempts to keep its current constant ]

The fall time of this voltage ( voltage increase for the mosfet ) depends on: - the peak current being switched off, the gate drive circuit, the capacitance of the device D-S ( non linear ), diode capacitance ( also non linear ) any snubbers on the devices.

As the diode becomes forward biased it will exhibit some forward recovery voltage, ( very low to zero for schottkies ) higher for larger currents trying to ramp up quickly through the diode - this adds to the losses in the mosfet [ V_fwd_rec ]

Now when the diode is forward biased and piling up charge carriers in the junction, it can begin to pick up current ( from the mosfet )

The main limiting factor as to how fast the diode can pick up the current from the mosfet is the wiring inductance of the power loop on the RHS

During this period the current is falling in the mosfet and commuting over to the diode - let's say this takes 30nS

Once the current has fallen to zero in the mosfet it is effectively off ( excepting Cds which can carry VHF ringing as the diode voltage reduces to its normal Von )

The losses in the mosfet during this last transition period are approx: 0.5 x I x 30nS x (Vin + Vf) x freq - for a linear current fall in the mosfet, pick up in the diode

Any diode snubber present will help pass current earlier and reduce diode turn on losses - and any mosfet snubber will allow it to turn off under less stress as some of the load and other currents can flow in the snubber

[ we should note that if the mosfet switch is near perfect and can cut the current in 1nS say, then the main inductor is able to drive up its own terminal volts ( limited in dV/dt by its self capacitance ) to any level required to overcome:

• the wiring inductance of the RHS power loop
• the ESL of the electrolytic capacitor
• the turn on voltage required for the diode ( forward recovery )

causing the kathode of the diode to go some way below the zero volt rail,

[ you often see this in higher power inverter stages ( which are really buck converters ), where the mid point of a switching leg goes several volts below gnd ( for outgoing current at high side turn off ) and upsets some gate drive chips ! ]

of course a too perfect fast turning off mosfet will suffer from high over-volts across its D-S due to the energy stored in the wiring inductances on the LHS power loop and ESL of Cin,

you begin to see how important tight layout and low ESL parts are to higher switching frequencies - to the buck - and other - converters ]

Just one caveat to the above:-

If we do turn the mosfet off really quite fast ( < 30nS for 10V to zero Vgs ), then the channel capacitance can carry the load current, and the channel itself can be well and truly off, we may see volt undershoot on the diode kathode as the choke forces current around the loop through the diode - however the mosfet turn off losses will be very low indeed.

Now the diode is on, carrying our 10 amps, and a short time later we desire to turn the MOSFET back ON again - what happens ?

If we turn the mosfet on really fast, we are effectively slamming the full input volts - in reverse, across the diode, which was, an instant ago, carrying 10A forward, - pretty hard on the diode, but even if we do this - there are several steps that occur: -

First the voltage across the mosfet must say about the same until we reach the point the diode starts to turn off, losses approx: I-load x 0.5 Vin x Tvolt fall x freq for a linear fall of voltage, then:

• the mosfet continues to see the full input volts D-S until it is carrying all the inductor current - because - if the diode is conducting any current at all, its kathode must be ~ 0.5 - 1.5 volts below its anode.
• the mosfet must also supply capacitive and reverse recovery current to the diode to allow it to be swept of charge carriers, to allow it to begin its own turn off and support reverse volts, reverse recovery currents can be 10's of amps for aggressive turn on gate drive of the mosfet

[ a quirky note here - the reverse Irr current in the diode is in the opposite direction to the current we are trying to establish from the main inductor loop - it is possible for these currents to add to zero at some point(s) in the transition - & this can cause non-intuitive shapes in the voltage transition ]

The mosfet losses during this interval are then approx: Vin x 0.5 x I_load x T_current rise x freq - for a linear rise of load current PLUS: Vin x I_rev_rec x T_rev_rec x freq, where the recovery current curve is like an inverted V and the time is a function of the diode type, its starting current, temperature and how fast we turn on the mosfet ( falling rate of dif/dt ) - so just a little complex to easily compute - measuring / observing is better

So - for just an instant - at the end of the above interval the mosfet is supporting full Vin and I_load and near the peak of the now falling reverse recovery current in the diode - just for an instant - before the voltage collapses across the mosfet and increases across the diode ( reverse recovery current falls to zero ) - diode is now able to support reverse volts ( turning fully off ).

At voltage collapse across D-S the energy in the Cd-s is absorbed by the mosfet channel, so more losses to account for - esp at high freq

This voltage transition ( collapse ) can be very rapid indeed, limited by:

• the capacitance of the diode as it turns off under rising volts, and
• the capacitance of the choke through the output cap back to the diode ( often negated by wiring inductance )
• any snubber across the mosfet - note that at mosfet turn on ( volt collapse D-S ) any snubber across the mosfet D-S will supply current & losses to the mosfet ( a faster turn on reduces energy losses from this source )
• any snubber across the diode - the mosfet must supply this current too, to charge up the snubber cap - more losses ( again faster turn on reduces these losses )

The diode is now fully off and the mosfet fully on and the process is free to repeat

Note - if there is capacitance in the main choke this current has to be supplied by the mosfet too at turn on - so a lower capacitance choke build is always a thing to strive for.

For more fundamentals of power switching circuits explained - go to - pwrtrnx.com