For buck, boost and buck-boost converters, calculate the duty cycle, inductor ripple and peak current, and output voltage ripple. The tool also detects discontinuous conduction mode, where the inductor current falls to zero during each cycle at light load, and plots the inductor current waveform.
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Results
Operating mode
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Duty cycle D
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Average inductor current
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Inductor ripple current ΔI
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Peak inductor current
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Output current at the CCM/DCM boundary
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Output voltage ripple (peak to peak)
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Inductance for 30% ripple
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The blue line is the inductor current and the dashed line its average (in CCM). The current rises while the switch is on and falls while it is off. Values are for an ideal steady state, ignoring switch and diode voltage drops and losses.
See it as a graph
Other inputs stay at their current values while the chosen input varies from 1/4 to 4 times its current value. The red dot is the current value.
Duty cycle and output voltage
A switching DC-DC converter turns its switch on and off every period T, and the fraction of time it is on, the duty cycle D, sets the output voltage. In steady state the average inductor voltage over a cycle is zero, so in ideal continuous conduction mode Vout = D·Vin for a buck, Vout = Vin/(1 − D) for a boost and |Vout| = D·Vin/(1 − D) for a buck-boost. For the 12 V to 3.3 V buck example, D = 3.3/12 = 27.5%. In practice D is slightly larger to make up for switch and diode voltage drops and losses.
Inductor ripple current and choosing L
While the switch is on, a constant voltage is applied across the inductor, so its current rises linearly. In a buck converter the peak-to-peak ripple is ΔI = (Vin − Vout)·D/(L·f). In the example it is (12 − 3.3) × 0.275/(4.7 µH × 500 kHz) ≈ 1.02 A, about 51% of the 2 A average. A ripple of about 20–40% of the average current is a common target, and the tool also shows the inductance for 30%. Choose an inductor whose saturation current exceeds the peak current, the average plus half the ripple.
Continuous and discontinuous conduction
As the load gets lighter, the average current falls; once the bottom of the ripple reaches zero, the inductor current stays at zero for part of each cycle. This is discontinuous conduction mode (DCM), and for a buck the boundary output current is ΔI/2. In DCM the duty cycle needed for a given output voltage depends on the load and is smaller than in CCM. Reducing the example buck's load to 0.2 A takes it below the boundary of about 0.51 A, so it enters DCM with D of about 17%. A synchronous converter can carry negative inductor current, so it can also be run in CCM even at light load.
Output voltage ripple and the capacitor
In a buck converter the inductor ripple current flows into the output capacitor, so the capacitive part of the output ripple is ΔV = ΔI/(8·f·C). In the example it is 1.02 A/(8 × 500 kHz × 22 µF) ≈ 11.6 mV. In boost and buck-boost converters the capacitor alone supplies the output current while the switch is on, so ΔV = Iout·D/(f·C) and the ripple tends to be larger. The capacitor's equivalent series resistance (ESR) times the current swing adds to this, which is why low-ESR ceramic capacitors are common. Some ceramic capacitors lose capacitance under DC bias, so check the effective capacitance.
The output voltage of an ideal boost converter is Vout = Vin/(1 − D). 5 V ÷ (1 − 0.6) = 5 ÷ 0.4 = 12.5 V. The formula follows from the balance of volt-seconds across the inductor, with Vin applied while the switch is on and (Vout − Vin) while it is off: Vin × D = (Vout − Vin) × (1 − D). The 3 V of D × Vin is the formula of the buck converter.
In an inverting buck-boost converter, energy from the input is stored in the inductor while the switch is on and released to the output through the diode while it is off. From the volt-second balance of the inductor, Vin × D = Vout × (1 − D), the magnitude of the output voltage is Vout = Vin × D/(1 − D) = 12 × 0.6/0.4 = 18 V. The output polarity is opposite to that of the input (negative with respect to ground). It works as a step-down converter for D < 0.5 and a step-up converter for D > 0.5, and at D = 0.5 the output magnitude equals the input.
The duty cycle is D = 3/12 = 0.25, so the on-time is D/f ≈ 0.833 µs. During the on-time 12 − 3 = 9 V appears across the inductor, so the current rises by ΔI = (Vin − Vout) × D/(L × f) = 9 × 0.25/(15 µH × 300 kHz) = 2.25/4.5 = 0.5 A. During the off-time it falls by Vout × (1 − D)/(L × f) = 3 × 0.75/4.5 = 0.5 A, confirming that it returns to its starting value every period. Doubling either the inductance or the frequency halves the ripple.
The inductor current is a triangular wave around its average (the output current) with a peak-to-peak ripple of ΔI = 9 × 0.25/(15 µH × 300 kHz) = 0.5 A. Its valley is "output current − ΔI/2", so when the output current falls below ΔI/2 = 0.25 A, the valley reaches zero and the converter enters discontinuous conduction mode. For the 3 V output this happens when the load resistance exceeds 12 Ω (a lighter load). In a synchronous buck converter the inductor current can also flow in the negative direction, so it can keep operating in continuous conduction mode even at light load.
The current into the capacitor is a triangular wave with an amplitude of ΔI/2. The charge that enters the capacitor during the half period T/2 when this current is positive is the area of a triangle, (1/2) × (T/2) × (ΔI/2) = ΔI × T/8. The voltage ripple is therefore ΔV = ΔI/(8fC) = 0.6/(8 × 300 kHz × 25 µF) = 10 mV. In a real capacitor the voltage ΔI × ESR from the ESR is added, which is why ceramic capacitors with low ESR are often used.
FAQ
What is the duty cycle formula for a buck converter?
In ideal continuous conduction mode, D = Vout/Vin; for 12 V to 3.3 V it is 27.5%. For a boost converter D = 1 − Vin/Vout, and for an inverting buck-boost D = |Vout|/(Vin + |Vout|). Real circuits need a slightly larger duty cycle to cover losses.
How much inductor ripple current should I design for?
About 20–40% of the average current is a common target. Lower ripple needs a larger inductor and slows the response to load changes; higher ripple increases the peak current, output voltage ripple and core loss. Choose an inductor with a saturation current comfortably above the peak current.
Is discontinuous conduction mode bad?
Not necessarily. Some controllers deliberately switch to DCM or pulse skipping at light load to keep efficiency up. However, the relation between duty cycle and output changes with load, and so do the control dynamics, so the design must be stable in both modes.