Linear Voltage Regulator

A linear voltage regulator accepts an unregulated DC input and delivers a steady, precise output voltage regardless of load current or input fluctuations. The LM7805 is inside almost every breadboard power supply, holding 5V steady even as load current swings from 0 to 1A.

Core Concept

A linear regulator works by placing a controlled transistor (the series pass element) between the input supply and the load. An error amplifier continuously compares the output voltage to an internal reference (usually a bandgap reference at 1.25V). If the output drops, the amplifier increases the transistor's base or gate drive, reducing its resistance and pulling the output back up.

The internal reference in modern regulators like the LM317 or LM7805 is a bandgap reference, which produces a stable 1.25V independent of temperature. The LM7805 uses a Zener-based reference and a built-in NPN pass transistor. It regulates to within 2% of 5V from input voltages of 7V to 25V.

The major cost of a linear regulator is heat. Every milliamp that flows through the pass transistor also flows through the input-to-output voltage drop. A 9V input delivering 1A to a 5V load dissipates (9 - 5) * 1 = 4W as heat in the IC, which is why a heatsink is required at high currents.

Key Equations

Power dissipation in pass transistor: Pd = (Vin - Vout) * IL

Efficiency: η = Vout / Vin * 100% (ignoring quiescent current Iq)

Line regulation: ΔVout / ΔVin [mV/V]

Load regulation: ΔVout / ΔIL [mV/mA or Ω]

Dropout voltage: Vdropout = Vin_min - Vout. For LM7805, Vdropout_min ≈ 2V, so Vin must be at least 7V.

Given:
  LM7805 linear regulator
  Vin  = 12V
  Vout = 5V
  IL   = 800 mA
  Iq   = 8 mA (quiescent current)

Why this formula:
  All excess voltage appears across the pass transistor.
  Power = voltage drop * current through transistor.

Formula:
  Pd = (Vin - Vout) * (IL + Iq)

Substitution:
  Pd = (12 - 5) * (0.800 + 0.008)
     = 7 * 0.808

Calculation:
  Pd = 5.656 W

  Efficiency:
  η = Vout / Vin * 100
    = 5 / 12 * 100
    = 41.7%

Final Answer:
  Power dissipated in LM7805 = 5.66 W (heatsink required)
  Efficiency = 41.7%

Exam Tip: GATE problems on linear regulators often ask for efficiency or power dissipation. Remember that efficiency is simply Vout/Vin when quiescent current is ignored. A regulator with Vin=12V and Vout=5V can never be more than 41.7% efficient, no matter how good the design. Also watch for dropout: the LM7805 needs Vin at least 7V. Below that, it stops regulating.

Key Properties

• The LM7805 regulates to 5V ± 4% with input voltages from 7V to 25V and output currents up to 1.5A with an adequate heatsink.
• Dropout voltage for the LM7805 is approximately 2V, meaning Vin must exceed Vout by at least 2V for regulation to hold.
• Low dropout (LDO) regulators such as the AMS1117 reduce the minimum dropout to 1.2V or less, enabling use with a 6V input to regulate 5V.
• Line regulation specifies how much Vout changes for a 1V change in Vin. For the LM7805 it is typically 3 mV/V.
• Load regulation specifies how much Vout changes from no load to full load, typically 15 mV for the LM7805 over 0 to 1.5A.
• Linear regulators have excellent noise performance compared to switching regulators, making them the preferred choice for analog and RF supply rails.
• Built-in protection in the LM7805 includes thermal shutdown (at 150°C junction temperature) and current limiting at approximately 2.2A.

Quick Revision

• Linear regulator operation: error amplifier drives pass transistor to hold Vout = Vref.
• Power dissipation: Pd = (Vin - Vout) * IL, always appears as heat.
• Maximum efficiency: η = Vout / Vin (ignoring Iq).
• LM7805 requires Vin ≥ 7V (2V dropout above 5V output).
• LDO regulators (AMS1117) drop as little as 1.2V between input and output.
• Good for low-noise analog rails; poor efficiency at large Vin-Vout differences.
• Exam trap: Students calculate efficiency as IL*Vout / (IL*Vin) but forget to add Iq to the input current, slightly overestimating efficiency in exam questions that explicitly provide Iq.