Power Electronics
A boost produces a HIGHER voltage than its input — it works like a slingshot: while the switch conducts, the inductor is "stretched" with energy from the input (storing current); when the switch is released, the inductor voltage stacks on top of the input and feeds the load through the diode.
The formula is striking: Vout = Vin/(1−D). Duty 50% → 2×, 75% → 4×. As D→1 the theory goes to infinity, but in the real world losses pull the limit down to 4-5× — that is how 5 V is made from a battery, or 12 V from 3.3 V.
A subtle difference: a buck draws pulsed input/continuous output current, a boost draws continuous input/pulsed output current. The output capacitor swallows these pulses; this asymmetry matters when designing EMI filters.
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