How it works
Buck converter: Vout = D × Vin, where D = ton/T is the duty cycle (0 < D < 1). An inductor L = 100 µH stores energy when the switch is ON (D × T seconds) and releases it when the switch is OFF through a freewheeling diode. Boost converter: Vout = Vin/(1−D); at D = 0.6, a 5 V input boosts to 12.5 V. Buck-boost: Vout = −D/(1−D) × Vin (inverts polarity). In continuous conduction mode (CCM) the inductor current never reaches zero; in DCM it does, and the conversion ratio becomes load-dependent. Critical inductance Lcrit = (1−D)R/(2f), below which the converter enters DCM at a given load R.
Key points to remember
Switching frequency f trades off inductor size (higher f → smaller L and C, but higher switching losses). Output voltage ripple ΔV = D(1−D)Vin/(8LCf²) for a buck in CCM. Transformer-isolated converters — flyback, forward, push-pull — provide galvanic isolation and are used in off-line SMPS. The flyback converter (based on buck-boost topology) stores energy in the transformer core during ON-time and releases it during OFF-time; transformer coupling ratio sets the voltage conversion. Efficiency is limited mainly by MOSFET switching loss (∝ f), diode reverse recovery, and inductor core loss.
Exam tip
The examiner always asks you to derive the voltage conversion ratio for the boost converter — apply volt-second balance on the inductor (VL during ON = Vin × DT, VL during OFF = (Vin − Vout)(1−D)T, set average to zero) to get Vout/Vin = 1/(1−D).