How it works
A boost converter stores energy in the inductor while the switch is on, and delivers it to the output while the switch is off. The inductor ripple grows with the input voltage, so the larger of the two input voltages sets the inductor. The input current is largest at the lowest input voltage, so the peak current and Isat are checked at Vin minimum.
D = 1 − (Vin × η) / VoutIin = (Vout × Iout) / (Vin × η)Lcalc = Vin × D / (fsw × ΔILtarget), ΔILtarget = Iin × ripple%ΔIL = Vin × D / (fsw × Lstd)Ipeak = Iin + ΔIL/2, IRMS = √(Iin² + ΔIL²/12)ΔVout = Iout × D / (fsw × Cout) + Ipeak × ESR
The DCM boundary is the point where the input current falls below ΔIL/2. Converted to output current, it is Iout,boundary = (ΔIL/2) × Vin × η / Vout.
Worked example
Input 3.0 to 4.2 V, output 5 V at 1 A, 1 MHz, 30 % ripple, 90 % efficiency. At Vin 3.0 V the duty cycle is 46.0 % and the average input current is about 1.85 A.
The ideal inductor is about 2.58 µH, so the E12 value above, 2.7 µH, is chosen. At Vin 3.0 V the ripple ΔIL is about 0.51 A and the peak current about 2.1 A. The recommended Isat is about 2.5 A and the RMS current about 1.9 A. The DCM boundary is about 0.14 A, so at a 1 A load the converter is in continuous conduction. The output ripple is about 63 mV, most of it from the peak-current and ESR terms. The loss estimate is about 0.24 W, giving an efficiency of about 95 %.
Design tips
- The input current of a boost is always larger than its output current. Size the switch, inductor and input connector by peak current, not average current.
- The output capacitor supplies the load alone while the switch is on. Ripple depends more on ESR than on capacitance in many cases, so choose a low-ESR part.
- Efficiency and ripple get much worse above a duty cycle of about 0.7 to 0.8. For a high step-up ratio, consider multi-stage designs or another topology.
- There is no direct path from input to output, so an output short leads to a runaway input current. Add an output disconnect switch or overcurrent protection.
Limitations
- This is a continuous-conduction approximation. At light load the converter enters DCM, and the results change.
- The loss estimate uses the minimum input voltage only and leaves out gate drive, dead time and core loss. The diode option treats the forward voltage as a constant.
- Efficiency η is a constant for the duty-cycle calculation. Real efficiency varies with input voltage and load.
- Inrush current and soft start are not included in this calculation.
Frequently asked questions
Why is the peak current checked at the minimum input voltage?
Because the average input current, Iin = Vout·Iout / (Vin·η), rises as the input voltage falls. The current ratings of the switch and inductor must be set for this worst case.
What happens when the duty cycle goes above 0.85?
The switch-on time gets long, which reduces conduction and switching margin, and the efficiency drops sharply for the voltage gain. In practice losses limit the boost ratio before anything else. Reduce the step-up ratio, or consider a multi-stage design or a different topology.
Why is the ESR component large in the output ripple?
Because the output capacitor supplies the load alone while the switch is on, and the rectified current pulses flow through the ESR. Using a low-ESR capacitor often improves the ripple more effectively than adding capacitance.