Buck Boost Duty Cycle⚠ unverified
Electrical / Power · Compute the duty cycle of an ideal buck-boost (inverting) converter
Parameters
| Input | Symbol | Unit | Default | Description |
|---|---|---|---|---|
| V_out | Vout | V | 1.0 | Output voltage, in volts (V); may be negative for the inverting topology |
| 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_in`` is not positive |
The science & history
Understanding the Parameters
-
$V_{in}$, $V_{out}$ — $|V_{out}|<V_{in}$ gives step‑down ($D<0.5$), $|V_{out}|>V_{in}$ gives step‑up ($D>0.5$); at $D=0.5$ the magnitudes are equal.
-
$D$ — the on‑time fraction; the output polarity is inverted relative to the input.
Derivation (Approaching a Proof)
Using inductor volt‑second balance:
- Switch on ($DT$): inductor across the input, $V_L = V_{in}$.
- Switch off ($(1-D)T$): inductor drives the output, $V_L = V_{out}$ (with $V_{out}$ negative).
Average to zero:
$$V_{in}\,D + V_{out}(1-D) = 0 \;\Rightarrow\; V_{out} = -V_{in}\,\frac{D}{1-D}.$$
Taking magnitudes, $|V_{out}| = V_{in}\dfrac{D}{1-D}$; solving for $D$,
$$D = \frac{|V_{out}|}{|V_{out}| + V_{in}}.$$
Because the inductor is charged from the input and then discharged into the output on its own (never directly connected input‑to‑output), the buck‑boost can both step up and step down — and inverts polarity.
History
The buck‑boost and its transformer‑isolated cousin, the flyback, are staples of low‑power supplies (wall adapters, bias rails) where a single, simple topology must handle input voltages both above and below the desired output.
Related Concepts: Buck Converter Duty Cycle, Boost Converter Duty Cycle, Switching Loss, Conduction Loss
Notes: Registry calculator buck-boost-duty-cycle (unverified). Ideal continuous‑conduction model;
the classic inverting topology reverses output polarity.