How do you calculate the duty cycle of a boost converter?

The duty cycle is D = 1 − (Vin × η / Vout). For an ideal converter (η = 1) it simplifies to D = 1 − Vin/Vout. It is the fraction of each switching period that the switch is ON and the inductor is storing energy.

How do you choose the inductor for a boost converter?

First find input current Iin = Pout / (Vin × η). Choose ripple ΔIL = 20–40% of Iin. Then L = Vin × D / (ΔIL × fsw). The inductor saturation current rating must exceed Ipeak = Iin + ΔIL/2.

How do you size the output capacitor of a boost converter?

Cout = Iout × D / (fsw × ΔVout). In a boost converter the capacitor alone supplies the load during the switch-on time, so it is typically larger than in an equivalent buck converter.

Why is the boost converter input current higher than the output current?

Because the boost converter steps voltage up and current down, conserving power: Pin = Pout / η. At higher Vout for the same power level, the output current is lower, meaning more input current is needed to supply the same power from a lower voltage.

What is CCM in a boost converter?

CCM (Continuous Conduction Mode) means the inductor current never reaches zero during a switching cycle. It occurs when ΔIL < 2 × Iin. At lighter loads the converter may enter DCM (Discontinuous Conduction Mode), where the duty cycle relationship changes.

What is the maximum duty cycle for a boost converter?

Theoretically D can approach 1 for large step-up ratios, but in practice most designs limit D to 85–90%. Very high duty cycles increase switch stress, losses, and inductor peak current significantly.

What is the efficiency of a boost converter?

Typical boost converters achieve 80–95% efficiency. Losses come from MOSFET conduction and switching losses, diode forward drop, inductor winding resistance, and core losses. Higher duty cycles and step-up ratios generally reduce efficiency.

What is the difference between a boost and a buck converter?

A boost converter steps voltage UP (Vout > Vin) with D = 1 − Vin/Vout. A buck converter steps voltage DOWN (Vout < Vin) with D = Vout/Vin. In a boost the inductor is on the input side; in a buck it is on the output side.

Why does a boost converter need a larger output capacitor than a buck?

Because during the switch-on phase the diode is reverse-biased and the output capacitor must supply 100% of the load current alone. The formula Cout = Iout × D / (fsw × ΔVout) shows the capacitor must be larger for high duty cycles or high output currents.

Boost Converter Calculator – Duty Cycle, Inductor, Capacitor & Efficiency | Step-Up DC-DC

📋 Input Parameters

Input Voltage (V_IN)

Output Voltage (V_OUT)

Output Current (I_OUT)

Switching Frequency (f_SW)

kHz

Inductor Ripple (% of I_IN)

Voltage Ripple (% of V_OUT)

Efficiency (η)

✦ Results

Duty Cycle (D)

0.500

Inductor (L)

33.0µH

Ripple Current (ΔI_L)

0.600A

Peak Inductor Current

4.74A

Output Capacitor (C_OUT)

222µF

Voltage Ripple (ΔV_OUT)

0.240V

Input Current (I_IN)

4.44A

Input Power (P_IN)

53.33W

Output Power (P_OUT)

48.00W

Power Loss (P_LOSS)

5.33W

✅All calculations are ideal. Add design margin for real-world conditions.

⚡ Boost Converter Topology

🧮 Detailed Calculation Breakdown

Step 1 — Duty Cycle

D = 1 − (V_IN × η / V_OUT)

Step 2 — Input & Ripple Current

I_IN = P_OUT/(V_IN×η) ; ΔI_L = ripple% × I_IN

Step 3 — Inductor

L = (V_IN × D) / (ΔI_L × f_SW)

Step 4 — Output Capacitor

C_OUT = (I_OUT × D) / (f_SW × ΔV_OUT)

Step 5 — Peak & Output Power

I_peak = I_IN + ΔI_L/2 ; P_OUT = V_OUT×I_OUT

Step 6 — Input Power & Loss

P_IN = P_OUT/η ; P_LOSS = P_IN−P_OUT

📐 Key Formulas

Duty Cycle (D)

D = 1 − V_IN × ηV_OUT

Inductor (L)

L = V_IN × DΔI_L × f_SW

Output Capacitor (C_OUT)

C_OUT = I_OUT × Df_SW × ΔV_OUT

🔤 Symbol Legend

V_IN = Input Voltage (V)

V_OUT = Output Voltage (V)

I_OUT = Output Current (A)

f_SW = Switching Frequency (Hz)

L = Inductor (H)

C_OUT = Output Capacitor (F)

ΔI_L = Inductor Ripple Current (A)

ΔV_OUT = Output Voltage Ripple (V)

η = Efficiency (%)

📈 Live Waveforms & Performance Curves

🔺 Inductor Current (triangular ripple)

⬛ Switch-Node Voltage (V_SW)

📈 Duty Cycle vs. Output Voltage

🔥 Power Loss vs. Efficiency

ℹ️

The inductor (input) current swings ±0.30 A around the 4.44 A average (peak 4.74 A). The switch node toggles between 0 and V_OUT at 100 kHz with duty 0.50. All results assume continuous conduction mode (CCM).

Boost Converter Calculator: Complete Design Guide

A boost converter - also called a step-up converter - is a switching DC-DC power supply that converts a lower input voltage into a higher regulated output voltage. It stores energy in an inductor while a switch is closed, then releases that energy in series with the input when the switch opens, pushing the output above the input. Boost converters are essential wherever a circuit needs more voltage than its supply can provide: battery-powered devices, LED backlights, USB power delivery, and photovoltaic systems. This calculator computes every key design parameter and shows the result on a live schematic and four dynamic waveforms.

How a Boost Converter Works

During the switch-on phase, the switch shorts the inductor to ground, so the input voltage drives a rising current through the inductor, storing energy in its magnetic field; the load is supplied entirely by the output capacitor during this time. During the switch-off phase, the inductor current is forced through the diode to the output, and the inductor voltage adds to the input voltage, charging the capacitor to a higher voltage than the input. The fraction of each cycle the switch is on - the duty cycle - sets how much the voltage is stepped up.

Duty Cycle

The boost duty cycle increases as the required step-up ratio increases. For an ideal converter it depends only on the voltage ratio; including efficiency increases it slightly to cover losses.

D = 1 − (V_IN × η / V_OUT)

• D = Duty cycle (0-1)
• V_IN = Input voltage (V)
• V_OUT = Output voltage (V)
• η = Efficiency (fraction)

Inductor Selection

In a boost converter the inductor carries the input current, which is larger than the output current. The inductor sets the input ripple current, typically chosen as 20-40% of the average input current.

I_IN = P_OUT / (V_IN × η)

ΔI_L = ripple% × I_IN

L = (V_IN × D) / (ΔI_L × f_SW)

Peak current: I_peak = I_IN + ΔI_L / 2

Because the input current is higher than the output current, boost inductors and switches must handle larger currents than an equivalent buck converter.

Output Capacitor Selection

In a boost converter the output capacitor must supply the entire load current during the switch-on time, when the diode is reverse-biased. This makes the output capacitor relatively large compared with a buck converter.

ΔV_OUT = ripple% × V_OUT

C_OUT = (I_OUT × D) / (f_SW × ΔV_OUT)

Use low-ESR capacitors, since ESR contributes additional ripple, and place them close to the diode and load.

Input Current, Power, and Efficiency

A boost converter steps voltage up and current down: the input current is higher than the output current. Conservation of power (minus losses) means the input power equals the output power divided by efficiency.

P_OUT = V_OUT × I_OUT

P_IN = P_OUT / η

I_IN = P_IN / V_IN

P_LOSS = P_IN − P_OUT

Quick Reference: All Boost Converter Formulas

Parameter	Formula	Typical Range
Duty Cycle	D = 1 − (V_IN × η / V_OUT)	0.1 – 0.9
Output Power	P_OUT = V_OUT × I_OUT	—
Input Power	P_IN = P_OUT / η	—
Input Current	I_IN = P_IN / V_IN	Always > I_OUT
Ripple Current	ΔI_L = ripple% × I_IN	20 – 40% of I_IN
Inductor	L = V_IN × D / (ΔI_L × f_SW)	µH – mH range
Peak Current	I_peak = I_IN + ΔI_L / 2	Always > I_IN
Output Capacitor	C_OUT = I_OUT × D / (f_SW × ΔV_OUT)	Larger than buck equiv.
Power Loss	P_LOSS = P_IN − P_OUT	5 – 20% of P_OUT
CCM Boundary	ΔI_L < 2 × I_IN	Trough current > 0 A

Worked Example: 12 V → 24 V at 2 A

🔆 Example — Boost Converter: 12 V → 24 V, 2 A, 100 kHz

Step 1Duty cycle: D = 1 − (12 × 0.9 / 24) = 1 − 0.45 = 0.55 (55% on-time)

Step 2P_OUT = 24 × 2 = 48 W | P_IN = 48 / 0.9 = 53.3 W | I_IN = 53.3 / 12 = 4.44 A

Step 3Ripple: ΔI_L = 0.30 × 4.44 = 1.33 A | Peak: I_peak = 4.44 + 0.67 = 5.11 A

Step 4Inductor: L = (12 × 0.55) / (1.33 × 100 000) = ≈ 50 µH (rated > 5.11 A sat.)

Step 5Output cap: C_OUT = (2 × 0.55) / (100 000 × 0.24) = ≈ 46 µF (low-ESR ceramic/polymer)

ResultD = 0.55 | L = 50 µH | C_OUT = 46 µF | I_peak = 5.11 A | P_LOSS = 5.3 W

Reading the Waveforms

Inductor Current: The triangular ripple riding on the average input current, rising while the switch is on and falling while it is off.
Switch-Node Voltage: The square wave at the switch/diode junction, near 0 V when the switch conducts and near V_OUT when the diode conducts.
Duty Cycle vs. Output Voltage: A rising curve showing how duty cycle grows as the output target increases above the input.
Power Loss vs. Efficiency: A curve showing how losses fall as efficiency improves, with a marker at your chosen efficiency.

Boost vs. Buck Converter

Aspect	Buck (Step-Down)	Boost (Step-Up)
Output voltage	Lower than input	Higher than input
Duty cycle	D = V_OUT/V_IN	D = 1 − V_IN/V_OUT
Input current	Lower than output	Higher than output
Inductor position	Output side	Input side
Output capacitor	Smaller	Larger (supplies load during t_on)

Design Tips and Best Practices

Rate components for input current: Inductor and switch carry the higher input current, not the output current.
Use a larger output capacitor: It alone supplies the load during switch-on; size it for ripple and transient response.
Watch high duty cycles: Very large step-up ratios push duty toward 1, increasing stress and reducing efficiency.
Low-ESR capacitors: Reduce output ripple and improve transient behaviour.
Stay in CCM: Keep ripple moderate so the inductor current stays continuous.
Use a synchronous rectifier: Replacing the diode with a MOSFET improves efficiency at high currents.

Common Applications

Battery-powered devices boosting a single cell up to system voltage
LED backlight and string drivers needing high forward voltages
USB-PD and power banks stepping 3.7V cells up to 5V/9V/12V
Photovoltaic and energy-harvesting maximum power point tracking
Automotive start-stop rail stabilization during voltage dips

CCM vs DCM in a Boost Converter

	CCM (Continuous Conduction Mode)	DCM (Discontinuous Conduction Mode)
Inductor current trough	> 0 A — never reaches zero	= 0 A — reaches zero each cycle
Condition	ΔI_L < 2 × I_IN	ΔI_L ≥ 2 × I_IN
Duty cycle formula	D = 1 − (V_IN × η / V_OUT)	More complex — load dependent
Capacitor size	Standard formula applies	Larger capacitor often needed
When it occurs	Normal/heavy load	Light load or oversized inductor
This calculator	✅ All formulas valid	⚠️ Results approximate only

Boost vs Buck vs Linear Regulator

Feature	Boost Converter	Buck Converter	Linear Regulator (LDO)
Output vs Input	V_OUT > V_IN	V_OUT < V_IN	V_OUT < V_IN
Efficiency	80 – 95%	85 – 98%	(V_OUT/V_IN) × 100%
Output capacitor	Larger (supplies load during t_on)	Smaller	Very small
Input current	Higher than output	Lower than output	Equal to output
Component count	L, C, switch, diode	L, C, switch, diode	IC only (no inductor)
Best for	Battery step-up, USB-PD	Point-of-load regulators	Low-noise analog rails

Design Tips and Best Practices

Goal	Action
Lower ripple current	Increase inductor value or switching frequency
Smaller components	Raise switching frequency (watch switching losses)
Lower output ripple	Larger / lower-ESR output capacitor
Higher efficiency	Use a synchronous rectifier (MOSFET) instead of a diode
Reliable inductor	Rate saturation current above I_peak with margin
Avoid very high D	Keep duty cycle below ~85% to limit stress

Frequently Asked Questions

Can a boost converter output a voltage lower than its input?

No. A boost converter can only step voltage up (V_OUT > V_IN). For step-down use a buck converter, or a buck-boost topology for both directions.

Why is the input current higher than the output current?

Power is conserved: P_IN = P_OUT / η. Since V_IN < V_OUT, the input current I_IN must be larger than I_OUT to supply the same power from a lower voltage.

Why does the output capacitor need to be larger than in a buck?

During the switch-on phase the diode is reverse-biased, so the output capacitor alone supplies 100% of the load current. The formula C_OUT = I_OUT × D / (f_SW × ΔV_OUT) shows the requirement grows with duty cycle.

What is the maximum practical step-up ratio?

Most boost converters are practical up to about 4:1 (e.g., 5 V → 20 V). At higher ratios the duty cycle approaches 1, increasing switch stress, peak current, and losses significantly. For extreme ratios, consider a SEPIC or flyback converter.

How do I reduce output ripple in a boost converter?

Use a larger, lower-ESR output capacitor, increase the switching frequency, or reduce the ripple current by using a larger inductor. In a boost the output ripple is typically larger than in an equivalent buck at the same specification.

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