Boost Converter Calculator

Design and analyze boost (step-up) DC-DC converters for power electronics.

Enter Design Parameters

Boost Converter Circuit

Input Voltage (V_in):

Output Voltage (V_out):

Output Power (P_out) [W]:

Inductor Ripple (%) :

Switching Frequency (Hz):

Enter all parameters and click Calculate.

Boost Converter Key Equations:
Duty Cycle: D = 1 - (V_in / V_out)
Output Current: I_out = P_out / V_out
Input Current: I_in = I_out / (1-D)
Inductor Ripple: ΔI_L = (Ripple %) × I_in
Inductor Value: L = (V_in × D) / (ΔI_L × f_sw)
Capacitor Value: C = ΔI_L / (f_sw × ΔV_out) (ΔV_out ≈ 10% V_out)
Load Resistor: R = V_out / I_out

Boost Converter Examples

Example 1

Given: V_in=12V, V_out=48V, P_out=120W, Ripple=10%, f_sw=50kHz

Duty Cycle: D = 1 - (12/48) = 0.75 or 75%
Output Current: I_out = 120/48 = 2.5A
Input Current: I_in = 2.5/(1-0.75) = 10A
Inductor Ripple: ΔI_L = 0.1×10 = 1A
Inductor: L = (12×0.75)/(1×50000) = 0.00018 H = 180 μH
Capacitor: C ≈ 2.08 µF (ΔV_out = 4.8V)
Load Resistor: R = 48/2.5 = 19.2 Ω

Example 2

Given: V_in=24V, V_out=48V, P_out=240W, Ripple=20%, f_sw=100kHz

Duty Cycle: D = 1 - (24/48) = 0.5 or 50%
Output Current: I_out = 240/48 = 5A
Input Current: I_in = 5/(1-0.5) = 10A
Inductor Ripple: ΔI_L = 0.2×10 = 2A
Inductor: L = (24×0.5)/(2×100000) = 0.00006 H = 60 μH
Capacitor: C ≈ 2.08 µF (ΔV_out = 4.8V)
Load Resistor: R = 48/5 = 9.6 Ω

Example 3

Given: V_in=48V, V_out=96V, P_out=240W, Ripple=15%, f_sw=75kHz

Duty Cycle: D = 1 - (48/96) = 0.5 or 50%
Output Current: I_out = 240/96 = 2.5A
Input Current: I_in = 2.5/(1-0.5) = 5A
Inductor Ripple: ΔI_L = 0.15×5 = 0.75A
Inductor: L = (48×0.5)/(0.75×75000) = 0.000427 H = 427 μH
Capacitor: C ≈ 0.694 µF (ΔV_out = 9.6V)
Load Resistor: R = 96/2.5 = 38.4 Ω