Electrical – Synchronous buck converter in parallel with linear regulator

The average value of the output of a buck converter in discontinuous mode is still the same as continuous mode. However, the ripple voltage increases.

This happens on light loads when the switcher cannot produce pulses of low enough duty cycle and it transfers a little too much energy per switching cycle for the light load.

This means the average output voltage starts to rise too high and the control system shuts down the switcher for several cycles resulting in a slightly lower than normal output voltage. Until the load starts to take more power this situation persists.

Discontinuous mode also happens when the input voltage to the buck rises too high.

EDIT - more information:-

If the buck converter is operating at 100kHz and the inductor is 10uH. Let's also say that the supply is 12V and duty cycle is 50%. The "on" pulse of the mosfet will charge the inductor with current and this current is determined by input voltage, output voltage and inductance. Let's say the output voltage is 5V - this means the voltage across the inductor is 7V when the mosfet is "on".

V = L\$\dfrac{di}{dt} \therefore 7 = 10\times 10^{-6} \times\dfrac{di}{dt} \therefore\dfrac{di}{dt} =\space \$700,000

Because dt is 50% of 10us we can calculate I, which is \$700000\times 5\times 10^{-6}\$ = 3.5A.

The energy transferred per cycle is therefore \$\dfrac{L\times 3.5^2}{2}\$ = 57.8uJ.

This transfers 100k times per second \$\therefore\$ the power to the load is \$ 57.8\mu J \times 100000\$ = 5.78W.

If the load resistor is too high to take that power at 5V then the simple fact is that the output voltage will rise and the buck convertor will enter discontinuous mode unless the control loop reduces the duty cycle.

This duty cycle reduction will of course happen because it would be a poor buck converter that couldn't produce less than 50% duty but, it will have a minimum value and at this point, if the energy-load equation isn't in equilibrium the converter will enter discontinuous mode.

IGBTs have high voltage drop and are not suitable for low voltage switching. They only exist because reliable high current high voltage MOSFETs are hard to make.

You should use high current MOSFETs rated at 30~40V. Maximum current is often package limited, but several FETs can be wired in parallel to increase current handling. For example the IRFP7430PbF is rated at 404A, but package limited to 195A. Rdson is 1.3 milliohms, so two in parallel should drop 0.13V at 200A (= 13W per FET). In comparison an IGBT module could drop 2V or higher, resulting in 400+ Watts of heat to get rid of!

The advantage of a module is less wiring and simpler installation. One disadvantage is that it's toast if even one transistor dies. Also an IGBT module may be much more expensive than a few discrete FETs.