Boost Converter Design Calculations: A Step-by-Step Worked Example

Learn how to choose the main parts of a DC-DC step‑up converter. This step‑by‑step design calculations example shows how to set the duty cycle, pick an inductor, size the output capacitor, and select semiconductor ratings for continuous conduction mode.

Design Specifications and Project Prerequisites

To build a working DC-DC step‑up circuit, you first need clear performance limits. In this boost converter design calculations example, we will design a converter that raises a nominal 12V battery voltage to a steady 24V output for a 2A load. This means the converter must deliver 48W of power.
We will aim for continuous conduction mode (CCM) so that peak currents stay low and the transfer behavior is predictable.
The key operating parameters are:
– Nominal input voltage (Vin) = 12V
– Target output voltage (Vout) = 24V
– Maximum output current (Iout) = 2A
– Switching frequency (fsw) = 100 kHz
– Desired inductor ripple current (dI_L) = 30% of the average input current
– Desired output voltage ripple (dV_out) = 1% of the output voltage (0.24V)
– Estimated efficiency (n) = 90% (0.90)

For a closely related example, see Power loss model and efficiency analysis of grid-connected seven-switch boost-type photovoltaic current source inverter using two power switches configurations.

Determining Duty Cycle and Average Input Current

The first step is to find the operating duty cycle (D). The duty cycle tells how much of each switching period the power MOSFET is on. For an ideal boost converter the formula is
D = 1 – (Vin / Vout). Real components cause losses, so we add the estimated efficiency 90% to the calculation. The loss‑adjusted formula becomes
D = 1 – (Vin × n / Vout). Plugging in our design numbers gives
D = 1 – (12 × 0.90 / 24) = 1 – 0.45 = 0.55 (or 55%). Next we find the average input current (I_in_avg) to see how much load the source and inductor see. The output power is
Pout = Vout × Iout = 24V × 2A = 48W. The input power needed is
Pin = Pout / n = 48W / 0.90 = 53.33W.

Finally, the average input current is
I_in_avg = Pin / Vin = 53.33W / 12V = 4.44A.

For a closely related example, see Model Predictive Current Control and Direct Torque Control for SRM With Cost Function‐Based Switching States to Optimize System Performance.

Inductor Sizing and Peak Current Calculations

The inductor stores the main energy in a boost converter. Choosing the right size stops the core from saturating and keeps the converter running in continuous conduction mode. First we set the desired ripple current (dI_L). We aim for 30% of the average input current, so
dI_L = 0.30 * I_in_avg = 0.30 * 4.44A = 1.33A. Next we use the basic inductor voltage rule while the switch is on to get the minimum inductance:
L = (Vin * D) / (fsw * dI_L). Plugging in the numbers gives
L = (12V * 0.55) / (100,000 Hz * 1.33A) = 6.6 / 133,000 = 49.6 µH. In a real design we pick the closest standard part, which is either 47 µH or 56 µH.

If we use a 47 µH inductor, the ripple current will be a bit higher, so we need to check it. Solving the equation for L = 47 µH gives the actual ripple current:
dI_L_actual = (12V * 0.55) / (100,000 Hz * 47e-6 H) = 1.40A. This value is still well below the limit. The peak inductor current (I_L_peak) tells us what saturation current the inductor must handle:
I_L_peak = I_in_avg + (dI_L_actual / 2) = 4.44A + 0.70A = 5.14A. So the inductor should have a continuous DC current rating of at least 5A and a magnetic saturation current (Isat) of at least 6A to avoid core saturation during start‑up or sudden load changes.

For a closely related example, see Sliding-Mode Control for IT2 Fuzzy Nonlinear Singularly Perturbed Systems and Its Application to Electric Circuits: A Dynamic Event-Triggered Mechanism.

Output Capacitor Selection and ESR Constraints

The output capacitor supplies the load current when the MOSFET is turned on and the diode is reverse‑biased. It must be sized to keep the output voltage ripple at our target of 1% (0.24V). The smallest capacitance (C_out) needed is found from the output current, duty cycle, switching frequency, and desired ripple voltage:
C_out = (Iout * D) / (fsw * dV_out). Plugging in our numbers:
C_out = (2A * 0.55) / (100,000 Hz * 0.24V) = 1.1 / 24,000 = 45.8 µF. We choose a standard 47 µF part. But the capacitance value alone does not guarantee low ripple. The capacitor’s Equivalent Series Resistance (ESR) has a big effect on the actual output voltage ripple.

The total ripple is roughly the sum of the capacitive ripple and the drop caused by the ESR:
dV_out_total = (Iout * D / (fsw * C_out)) + (I_L_peak * ESR). To stay under 0.24V, the ESR must be kept very low. Because of this, ordinary aluminum electrolytic capacitors are usually not suitable unless you use several in parallel. High‑quality ceramic (X7R) or solid‑polymer capacitors with an ESR under 10 milliohms are strongly recommended.

Power Semiconductor Ratings: MOSFET and Diode Sizing

Both the active switch (MOSFET) and the passive switch (diode) must handle the peak voltages and currents present in the circuit. The peak voltage across both components during the off-state is equal to the output voltage plus any parasitic inductive ringing. As a rule of thumb, apply a 1.5x safety margin to avoid overvoltage failure. MOSFET Voltage Rating (V_DS) >= Vout * 1.5 = 24V * 1.5 = 36V. A standard 40V or 60V rated N-channel MOSFET is appropriate. The MOSFET must also handle the RMS current. A conservative estimate for the RMS current through the switch is: I_SW_RMS = I_in_avg * square_root(D) = 4.44A * square_root(0.55) = 3.29A.

Select a MOSFET with a low on-state resistance (R_DS_on) to minimize conduction losses. The diode must withstand the same reverse voltage of 36V. The average forward current through the diode is equal to the output current (2A), while the peak current is equal to the peak inductor current (5.14A). A Schottky diode with a 40V reverse voltage rating and a 3A to 5A continuous forward current rating is ideal because of its low forward voltage drop (typically 0.4V to 0.5V) and negligible reverse recovery time.

Validation of Continuous Conduction Mode Boundary

To ensure our design equations remain valid, we must verify that the converter does not enter Discontinuous Conduction Mode (DCM) under nominal load conditions. The boundary between CCM and DCM occurs when the inductor ripple current is exactly twice the average input current, meaning the inductor current touches zero at the end of each switching cycle. The critical inductance (L_crit) required to maintain CCM at a given load resistance (R) is calculated as: L_crit = (R * D * (1 – D)^2) / (2 * fsw). The equivalent load resistance at nominal load is: R = Vout / Iout = 24V / 2A = 12 ohms. Substituting this into our boundary equation: L_crit = (12 * 0.55 * (1 – 0.55)^2) / (2 * 100,000) = (6.6 * 0.2025) / 200,000 = 1.3365 / 200,000 = 6.68 microhenries.

Since our chosen inductor value of 47 microhenries is significantly larger than the critical inductance of 6.68 microhenries, the converter will operate deep within CCM, ensuring stable control dynamics and lower output ripple.

Simulation Setup and Hardware Verification Steps

Before assembling a physical prototype, validate your calculations using a SPICE circuit simulator such as LTspice. Create a schematic with a voltage source of 12V, a 47 microhenry inductor with a realistic series resistance (DCR) of 20 milliohms, a 40V N-channel MOSFET, a 40V Schottky diode, and a 47 microfarad output capacitor with 5 milliohms of ESR. Configure a pulse voltage source at the MOSFET gate with a frequency of 100 kHz, a duty cycle of 55%, and rise/fall times of 10 nanoseconds. Run a transient simulation for 10 milliseconds to allow the circuit to reach steady-state.

Verify that the output voltage settles close to 24V, the peak-to-peak inductor ripple current matches your calculated 1.4A, and the output voltage ripple stays within the 240mV target. If the output voltage is lower than expected, check the simulated voltage drops across the diode and the MOSFET’s on-resistance.