rftools.io

Calculadora de Convertidor Reductor (Buck)

Diseña convertidores reductores conmutados: calcula inductancia, condensador de salida y ondulación.

Loading calculator...

Fórmula

D = \frac{V_{out}}{V_{in}},\quad L_{min} = \frac{(V_{in}-V_{out})\cdot D}{f_{sw}\cdot \Delta I_L}

Referencia: Erickson & Maksimovic, "Fundamentals of Power Electronics" 3rd ed.

DDuty cycle
VᵢₙInput voltage (V)
VₒᵤₜOutput voltage (V)
fₛwSwitching frequency (Hz)
ΔILInductor current ripple (A)

Cómo Funciona

Buck converters are a fundamental topology in switch-mode power supply design, allowing efficient DC voltage step-down through pulse width modulation (PWM). The core principle involves switching a high-side MOSFET at high frequency, effectively 'chopping' the input voltage to create a lower average output voltage. During the switch's on-time, energy is transferred to the inductor, which acts as an energy storage element. The key design parameters include duty cycle (D), which determines the voltage reduction, switching frequency (fsw), and the inductor and capacitor values that control output voltage ripple and current fluctuation. The converter's efficiency stems from the near-lossless switching of power MOSFETs and minimal resistive losses in the magnetic and capacitive components. Key design considerations include selecting components with appropriate voltage and current ratings, managing switching losses, and ensuring stable feedback control to maintain consistent output voltage under varying load conditions.

Ejemplo Resuelto

Consider a buck converter design with Vin = 12V, desired Vout = 5V, maximum load current of 2A, and switching frequency of 100 kHz. First, calculate the duty cycle: D = 5V/12V = 0.417 or 41.7%. For the minimum inductor value, use the formula Lmin = (12V-5V)·0.417/(100 kHz·ΔIL), assuming a 30% current ripple. With ΔIL = 0.6A, Lmin calculates to 47 µH. To determine output capacitance, assume a 50 mV voltage ripple: Cout = 0.6A/(8·100 kHz·0.05V) ≈ 150 µF. The input capacitance calculation yields Cin = 2A·0.417·(1-0.417)/(100 kHz·0.05V) ≈ 22 µF.

Consejos Prácticos

  • Always use low-ESR capacitors to minimize ripple and improve transient response
  • Include a compensation network in the feedback loop for stable voltage regulation
  • Consider using Schottky diodes to reduce switching losses

Errores Comunes

  • Neglecting inductor saturation current when selecting core size
  • Overlooking parasitic resistances in component selection
  • Inadequate thermal management leading to reliability issues

Preguntas Frecuentes

What determines buck converter efficiency?

Efficiency depends on MOSFET switching losses, inductor DC resistance, diode forward voltage, and component quality. Modern converters can achieve 90-95% efficiency.

How do I choose the right switching frequency?

Higher frequencies allow smaller magnetic components but increase switching losses. Typical frequencies range from 100 kHz to 500 kHz depending on the application.

Can I use a buck converter for high-power applications?

Yes, but you'll need carefully selected components with appropriate current and thermal ratings. Paralleling MOSFETs can help distribute power handling.

What causes output voltage instability?

Poor feedback loop compensation, inadequate capacitance, and significant load current variations can cause voltage instability in buck converters.

How do I protect against overcurrent?

Implement current-mode control, use current-sense resistors, and add protective circuits like crowbar protection and current limiting.

Shop Components

Affiliate links — we may earn a commission at no cost to you.

Buck Converter ICs

Synchronous and non-synchronous step-down regulators

LDO Regulators

Low-dropout linear voltage regulators

Electrolytic Capacitors

Aluminum electrolytic capacitors for power filtering

Power Inductors

Shielded and unshielded inductors for switching supplies

Related Calculators