Switching Regulator Basics

A switching regulator converts DC to DC by rapidly switching a transistor on and off and storing energy in an inductor or capacitor between pulses. The LM2596 step-down switcher achieves 75% to 92% efficiency where a linear LM7805 under the same conditions wastes more than half the input power as heat.

Core Concept

A buck converter (step-down switcher) connects the input voltage to an inductor for a fraction of each switching cycle (the ON time), then disconnects it. During the OFF time, the inductor's stored energy freewheels through a diode, keeping current flowing to the load. A capacitor at the output smooths the current ripple into a nearly steady DC voltage.

A boost converter does the reverse: it stores energy in the inductor while the switch is ON (shorting the inductor to ground), then releases it through the output diode when the switch is OFF. This pushes the output voltage above the input. The LM2577 and MC34063 are popular boost controller ICs for battery-powered systems.

The duty cycle D = ton / T sets the output voltage. For a buck: Vout = D * Vin. For a boost: Vout = Vin / (1 - D). Efficiency in a real switcher (75% to 95%) is much higher than a linear regulator because the switching transistor dissipates power only during transitions, not continuously.

Key Equations

Buck converter output voltage: Vout = D * Vin

Boost converter output voltage: Vout = Vin / (1 - D)

Buck-boost converter output: Vout = -Vin * D / (1 - D) (inverted polarity)

Duty cycle: D = ton / T = ton * fsw where T is the switching period and fsw is the switching frequency.

Inductor current ripple: ΔIL = (Vin - Vout) * D / (L * fsw)

Efficiency: η = Pout / Pin = (Vout * Iout) / (Vin * Iin)

Given:
  Buck converter (LM2596-based)
  Vin  = 24V
  Vout = 5V
  IL   = 1A (load)
  fsw  = 150 kHz
  L    = 68 µH
  Find: Duty cycle D and inductor current ripple ΔIL

Why this formula:
  In steady state, Vout = D * Vin for a lossless buck converter.

Formula:
  D    = Vout / Vin
  ΔIL  = (Vin - Vout) * D / (L * fsw)

Substitution:
  D = 5 / 24 = 0.2083

  ΔIL = (24 - 5) * 0.2083 / (68e-6 * 150e3)
       = 19 * 0.2083 / (10.2)

Calculation:
  Numerator: 19 * 0.2083 = 3.958
  Denominator: 68e-6 * 150e3 = 10.2
  ΔIL = 3.958 / 10.2 = 0.388 A

Final Answer:
  D   = 20.8%
  ΔIL = 388 mA peak-to-peak ripple in inductor current

Exam Tip: GATE questions on switching regulators almost always ask you to identify the converter type from the output-to-input relationship, or to compute D. Remember: Buck has Vout < Vin and Vout = D*Vin. Boost has Vout > Vin and Vout = Vin/(1-D). For a boost, D approaches 1 as Vout approaches infinity, but in practice D is limited to about 0.9. Students confuse the buck-boost polarity inversion, which gives a negative output from a positive input.

Key Properties

• A buck converter always steps voltage down (Vout < Vin), with efficiency typically between 85% and 95% at full load.
• A boost converter always steps voltage up (Vout > Vin), with efficiency typically 75% to 90%, limited by diode and inductor losses.
• The LM2596 is a popular fixed-frequency (150 kHz) buck controller IC rated for 3A output, with adjustable or fixed output voltages.
• Switching frequency is a key design parameter: higher fsw reduces the required inductance and capacitance, shrinking the circuit, but increases switching losses.
• Inductor current ripple is typically designed to be 20% to 40% of the average load current to balance efficiency and component size.
• A Schottky diode (e.g., 1N5819) is used in the freewheel position because its fast recovery time and low forward voltage (0.3V) minimize losses.
• Continuous conduction mode (CCM) occurs when inductor current never reaches zero in a cycle; discontinuous conduction mode (DCM) occurs at light loads and changes the Vout vs D relationship.

Quick Revision

• Buck: Vout = D * Vin (step-down only).
• Boost: Vout = Vin / (1-D) (step-up only).
• Buck-boost: Vout = -Vin*D/(1-D) (inverted, step up or down).
• Duty cycle D = ton/T; higher D means higher Vout in a buck.
• Switching efficiency (85-95%) far exceeds linear regulator efficiency at large Vin-Vout gaps.
• Schottky diodes and low-DCR inductors are key to high efficiency.
• CCM: inductor current stays positive all cycle. DCM: it falls to zero at light loads.
• Exam trap: Students apply Vout = D*Vin to a boost converter and get a voltage lower than Vin. The boost formula is Vout = Vin/(1-D), which always gives a value greater than Vin for 0 < D < 1.