Boost Converter Duty Cycle⚠ unverified
Electrical / Power · Compute the duty cycle of an ideal boost converter
Parameters
| Input | Symbol | Unit | Default | Description |
|---|---|---|---|---|
| V_out | Vout | V | 1.0 | Output voltage |
| V_in | Vin | V | 1.0 | Input voltage |
| Output | Symbol | Unit | Description |
|---|---|---|---|
| result | D | — | Duty cycle as a dimensionless ratio between 0 and 1. Returns 0.0 if ``V_out`` is not greater than ``V_in`` or if ``V_in`` is not positive |
The science & history
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.