For I know, the reason why the efficiency of synchronous buck converters is higher than that regular ones is because they use a MOSFET instead of a freewheeling diode, and MOSFETs generally have higher efficiency than diodes. However, the power dissipation on the MOSFET is R(dson)×I^2, while on the diode it is VF×I. So, when the current is high enough, will the efficiency of synchronous buck converters be lower?
The above content is my own thinking. Does such a situation exist in reality?
RDS_on and V_F describe the same thing, but different aspects of it:
V_F is the approximate voltage across the semiconductor junction when the nominal operative current has been reached, acting as if the voltage wouldn't rise further when you increase the current. That's a lie, but it's OK to make, because the voltage grows slowly with growing current in a diode at saturation.
RDS_on is the slope of the voltage-vs-current curve. As said above, that's not going to be large.
So, V_F is the vertical "offset" of the voltage-vs-current curve, RDS is the slope. Go, open any diode datasheet and look at the I/V curve:
IV curve from the SDT30B100D1 datasheet
This diode is advertised for 30 A current, so V_F is the voltage at 30 A! See how this is an exponential plot. The voltage does still have to rise when you increase the current, but it increases slowly. That's why diodes are often just characterized by their V_F. (but it is a simplification, and if you need to learn about the power dissipation, you need to look at the curve)
So, your VF×I formula simply linearizes a formula that actually has a quadratic component, but because V_F is so large for diodes, the quadratic component doesn't matter.
MOSFETs, on the other hand, have a V_F that's low enough that they only specify the RDS_ON in their "marketing claims". Again, if you care about the power dissipation at different loads, you need to look at the curve:
I_D/V_DS curve from the TPH7R006PL datasheet
Assume we're using the optimal 10 V to control the gate: you can see that the MOSFET really looks like a resistor there: the voltage is just the product of some slope times the current. Like Ohm's law! So, to characterize this MOSFET's losses quite exactly, specifying the RDS_on is totally sufficient.
However, because RDS_on is so low (4.5 mΩ or so!), that quadratic behaviour is much much better than the linear(ized) behaviour of the diode. You always lose less in a MOSFET rated for some current than in a similarly rated diode, assuming you control it with a gate-source voltage high enough:
If you look at the curve for VGS=4 V, that does indeed look a lot more like the curve from a diode! And it indeed is a diode's curve. There is some V_F that it stops growing very steeply after. But you'd avoid operating this specific MOSFET at this low V_GS!
At high output current and low output voltage the Vf of diode plays a big role in efficiency since diode drop voltage is close to output voltage.
Example, 0.7V diode drop at 5V output makes more than 14% of losses during Toff interval. And even worse the diode drop at high current goes to about 1V or so (not 0.7V).
A low Rds mosfet instead of diode eliminates this drop/losses rapidly.
However, the power dissipation on the MOSFET is R(dson)×I^2, while on the diode it is VF×I. So, when the current is high enough, will the efficiency of synchronous buck converters be lower?
Yes, of course, and that also applies to all resistive losses (inductor DCR, cap ESR, MOSFETs, etc).
Due to that squared I², above a certain level of current you need multiple phases. For example, consider a PC motherboard VRM which converts 12V into Vcore, which is in the vicinity of 1V with huge current (sometimes more than 100A). Here's a picture.
You can see 16 dual MOSFETs and 16 inductors, for a 16 phase buck converter. A bit overkill.
If you use N phases, current in each phase is divided by N, thus conduction losses in each phase (RI²) are divided by N². There are N phases, so total loss is only divided by N compared to a one phase design using the same parts. This is not usually true, because the lower current in each phase allows faster MOSFETs with higher RdsON which increases conduction losses but reduces switching losses. It also spreads the heat over a much larger area which makes cooling easier.
On the other end of voltage, high voltage MOSFETs have pretty high ESR usually, so in this case a diode may have lower losses, or perhaps some other switch type (IGBT...)
