Boost Converter Design: Part 2 — Design Goals and Input Current / Duty Cycle Calculations

Semiconductor Integrated Circuits (ICs)
, ,
1

In a light-load design, switch current is not the same as load current; it is the inductor current viewed from the input. This article starts from output power, works backward to input power using efficiency, calculates the average input current, and finally gives the maximum duty cycle. These values become hard specifications for selecting the IC and inductor later.

Example Design Specifications

Assume that a system requires the boost converter to provide:

  • Output voltage: V_{OUT}=34\,\text{V}.
  • Output current: the original description specifies 80\,\text{mA}, but the subsequent power calculation uses 0.1\,\text{A} (100\,\text{mA}). The equations below follow the original calculation.
  • Switching frequency: f_{SW}=2\,\text{MHz}.
  • Two possible input rails: V_{IN1}=12\,\text{V} and V_{IN2}=28\,\text{V}.
  • Core requirements: clean output, very low switching noise, and a compact solution.

Source: Analog Devices technical article Boost Converter Design for Sensitive Loads at Light Load.

Step 1: Calculate the Output Power

Output power is the power actually delivered to the load:

P_{OUT}=V_{OUT}\times I_{OUT}=34\times0.1=3.4\,\text{W} \tag{1}

Step 2: Work Backward to Input Power, Including Efficiency

A boost converter is not 100% efficient. Assuming an efficiency of \eta=90\%, the input must provide additional power to cover the losses:

P_{IN}=\frac{P_{OUT}}{\eta}=\frac{3.4}{0.9}=3.77\,\text{W} \tag{2}

Engineering conclusion: P_{IN} is the total power that the source must provide, so it is greater than P_{OUT}. The difference represents the converter’s internal losses, including switching loss, conduction loss, and quiescent current.

Step 3: Calculate the Average Input Current

Because the boost converter’s inductor is connected to the input, the average inductor current equals the average input current. Calculate the two input cases separately:

I_{IN1}=\frac{P_{IN}}{V_{IN1}}=\frac{3.77}{12}=0.314\,\text{A} \tag{3}
I_{IN2}=\frac{P_{IN}}{V_{IN2}}=\frac{3.77}{28}=0.134\,\text{A} \tag{4}

The rule in one sentence: The lower the input voltage, the higher the input current. This relationship is especially important when selecting the operating mode later.

Step 4: Estimate the Peak Inductor Current

Inductor current is not a flat DC current; it is a triangular waveform with ripple. To provide margin, estimate the peak current as 1.2 times the average input current, or 20% above it. This is also the peak switch current:

I_{PEAK1}=I_{IN1}\times1.2=0.314\times1.2=0.377\,\text{A} \tag{5}
I_{PEAK2}=I_{IN2}\times1.2=0.134\times1.2=0.16\,\text{A} \tag{6}

The peak current determines two important component requirements:

  1. IC switch-current rating: It must be greater than the peak current.
  2. Inductor saturation current I_{SAT}: It should generally be greater than 1.2 times the IC overcurrent threshold, as discussed in Part 4.

In this example, neither input condition produces a peak current above 0.5\,\text{A}. Therefore, an IC with a 0.5\,\text{A} switch-current rating and a voltage rating above 40\,\text{V} can cover both cases.

Step 5: Calculate the Maximum Duty Cycle

The duty cycle D is the fraction of each switching period during which the switch is on. It is determined by the input and output voltages:

D_{MAX}=\frac{V_{OUT}-V_{IN(MIN)}}{V_{OUT}} \tag{7}

Substituting the two input voltages gives:

D_{MAX1}=\frac{34-12}{34}=0.647 \tag{8}
D_{MAX2}=\frac{34-28}{34}=0.176 \tag{9}
  • 12 V input: The duty cycle is approximately 65%, so the switch spends more than half of each cycle charging the inductor.
  • 28 V input: The duty cycle is only approximately 18%.

The rule in one sentence: The greater the duty cycle, the longer the inductor charging interval and the larger the required inductance. Part 3 will use this result.

Summary

At this point, we have:

  • Input power: 3.77\,\text{W}.
  • With a 12 V input: average input current 0.314\,\text{A}, peak current 0.377\,\text{A}, and duty cycle 0.647.
  • With a 28 V input: average input current 0.134\,\text{A}, peak current 0.16\,\text{A}, and duty cycle 0.176.

These values are the foundation for selecting the IC, inductor, and operating mode.

Key Takeaways

  • Switch current is not the same as load current. Derive the average and peak input current from P_{OUT}, \eta, and V_{IN}.
  • The lower the input voltage, the higher the input current, making CCM easier to maintain.
  • Maximum duty cycle is determined by V_{IN} and V_{OUT}. A higher duty cycle generally requires a larger inductor.
  • These values are the basis for selecting the IC, inductor, and light-load operating mode.

Related Products

Additional References

  1. LT8336 data sheet (Analog Devices)
  2. LT8330 data sheet (Analog Devices)

More Content

Boost Converter Design: Part 1 — The Light-Load Challenge: Choosing Between Efficiency and Noise