Boost Converter Duty Cycle⚠ unverified

Understanding the Parameters

• $V_{in}$, $V_{out}$ — with $V_{out} \ge V_{in}$. As $D\to1$ the ideal gain $1/(1-D)\to\infty$ (real converters are limited by losses well before that).

• $D$ — the fraction of the cycle the switch shorts the inductor to ground to store energy.

Derivation (Approaching a Proof)

Apply inductor volt‑second balance (average inductor voltage = 0 in steady state):

• Switch on ($DT$): inductor connected across the input, $V_L = V_{in}$.
• Switch off ($(1-D)T$): inductor drives the output through the diode, $V_L = V_{in} - V_{out}$.

Setting the average to zero,

$$V_{in}\,D + (V_{in} - V_{out})(1-D) = 0 \;\Rightarrow\; V_{in} = V_{out}(1-D) \;\Rightarrow\; V_{out} = \frac{V_{in}}{1-D}.$$

Solving for the duty cycle gives $D = 1 - V_{in}/V_{out}$. The inductor stores energy during the on‑time and releases it in series with the source during the off‑time — which is how the output exceeds the input.

History

Boost topologies are essential wherever a voltage must be raised from a lower source: battery‑powered electronics, LED drivers, and photovoltaic maximum‑power‑point trackers. Like the buck, the design rests on volt‑second balance and fast switching devices.

Related Concepts: Buck Converter Duty Cycle, Buck Boost Duty Cycle, Switching Loss, Conduction Loss

Notes: Registry calculator boost-converter-duty-cycle (unverified). Ideal continuous‑conduction model; real gain is limited by parasitic resistances, so very high step‑up ratios are impractical.