Formula and Method for the Inverting Buck-Boost Converter
An inverting buck-boost converter is a non-isolated switch-mode DC-DC converter built from one switch, one diode, one inductor, and one output capacitor. It stores energy in the inductor while the switch is closed, then releases that energy to the load while the switch is open — through a diode oriented so the output voltage appears with the opposite polarity from the input, referenced to the same ground. Because the inductor can be charged for any fraction of the switching period, the output magnitude can land above or below the input voltage, which is what separates a buck-boost converter from a plain buck (step-down only) or boost (step-up only) converter.
Deriving the duty cycle and output voltage
In continuous conduction mode (CCM), the inductor's average voltage over one full switching cycle must be zero — this is volt-second balance. While the switch is on, for a fraction D of the period, the inductor sees +Vin across it; while it is off, for the remaining (1-D) of the period, it sees the output voltage. Setting Vin·D = |Vout|·(1-D) and rearranging gives the converter's defining relationship: |Vout| / Vin = D / (1 - D), or solved for duty cycle, D = |Vout| / (Vin + |Vout|). This calculator uses that formula to find the duty cycle needed to reach your target output, then reports Vout as a negative number to reflect the inverted polarity.
Inductor current, ripple, and the CCM/DCM boundary
By charge balance, the average inductor current equals the output current divided by (1-D): IL = Iout / (1-D). While the switch is on, current ramps up by ΔIL = Vin·D / (L·f); the peak current is the average plus half that ripple. This ripple-current model, and the simple D/(1-D) voltage formula above, only hold in continuous conduction mode, where the inductor current never reaches zero. The boundary inductance is L_min = Vin·D·(1-D) / (2·f·Iout); choose an inductor above this value for your expected minimum load, or the converter drops into discontinuous conduction mode (DCM), where the output voltage depends on inductance and load as well as duty cycle.
Practical design notes
- Component ratings matter: the diode must block Vin + |Vout| in reverse, and both the switch and diode see the peak inductor current, not just its average.
- Higher switching frequency shrinks the inductor — L_min scales as 1/f — but it also increases switching losses in the diode and switch, so real designs balance the two.
- This model is ideal: it ignores switch and diode voltage drops and conduction losses, so a real converter's efficiency will be below 100%, and the true duty cycle needed will run slightly higher than the ideal value calculated here.
- Light loads push toward DCM even with a well-sized inductor, since L_min grows as load current Iout falls — check the CCM condition at your minimum expected load, not just the nominal one.