A linear power supply is an AC-to-DC conversion circuit that regulates output voltage by dissipating excess electrical energy as heat. Unlike switching power supplies that rapidly turn transistors on and off to transfer energy, a linear supply uses a transformer to step down the AC mains voltage, rectifies and filters it into rough DC, and then passes it through a linear regulator operating in its active (ohmic) region. The regulator acts like a smart, continuously variable resistor, dropping the exact voltage needed to maintain a rock-solid, ultra-low-noise DC output.
While they are heavy, inefficient at high power, and run hot, linear power supplies remain the undisputed choice for audio DACs, precision RF equipment, and sensitive analog sensor arrays where switching noise (EMI) is unacceptable. Below, we break down the architecture, run a complete bench design with thermal math, and compare the topology against modern switching alternatives.
The Core Architecture: How a Linear Power Supply Works
Every traditional linear power supply follows a strict four-stage pipeline. Understanding this sequence is critical for debugging ripple issues or sizing components.
- Step-Down Transformer: Isolates the circuit from mains and steps the high AC voltage (e.g., 120VAC) down to a lower AC RMS voltage (e.g., 12VAC).
- Rectifier Bridge: Converts the bipolar AC waveform into a unipolar pulsating DC waveform. A full-wave bridge utilizes four diodes, dropping the peak voltage by roughly 1.4V (two diode forward voltage drops).
- Bulk Filter Capacitor: A large aluminum electrolytic capacitor that charges to the peak DC voltage and discharges into the load during the AC zero-crossings, smoothing the pulsating DC into a relatively flat voltage with residual ripple.
- Linear Regulator: An integrated circuit (like the classic LM7805 or an LDO like the LT3042) that trims the rippling DC down to a precise, fixed output voltage, rejecting the remaining AC ripple via its Power Supply Rejection Ratio (PSRR).
Ripple and Noise Expectations
The raw DC before the regulator will always have ripple. You can calculate the peak-to-peak ripple voltage ($V_{ripple}$) using the formula: $V_{ripple} = \frac{I_{load}}{f \times C}$, where $f$ is the ripple frequency (120Hz for full-wave 60Hz mains) and $C$ is the capacitance in Farads. A well-designed linear supply will have pre-regulator ripple of 2V to 4V, which the linear regulator then attenuates by 60dB to 80dB, resulting in output noise measured in microvolts ($\mu V$).
Topology Showdown: Linear vs. Switching Power Supplies
When deciding between a linear power supply and a Switch-Mode Power Supply (SMPS), you are trading efficiency and weight for signal purity and simplicity. Here is how the two topologies stack up across four critical engineering criteria.
| Criteria | Linear Power Supply | Switching Power Supply (SMPS) |
|---|---|---|
| Efficiency | 30% – 60% (Highly dependent on input/output differential) | 80% – 95% (Consistent across wide input ranges) |
| Heat Dissipation | High; requires massive heatsinks for loads >500mA | Low; heat is mostly confined to switching MOSFETs and diodes |
| Output Noise / EMI | Ultra-low ($\mu V$ range); no high-frequency switching spikes | High (mV range); generates broadband EMI and switching ripple |
| Cost & BOM Size | Cheap for <10W, but heavy/expensive iron-core transformers at >50W | Moderate BOM cost, but requires complex PCB layout and EMI filtering |
The Verdict: Choose a linear topology when your load draws less than 1A and demands ultra-clean power (e.g., 24-bit audio ADCs, strain gauges, PLL synthesizers). Choose a switching topology for battery charging, motor drives, or any digital logic load drawing >1A where efficiency and thermal management dictate the design.
Bench Design Example: 12VAC to 5V DC at 1A
Let’s design a linear supply to power a 5V, 1A analog sensor array from a 120VAC wall outlet. We will use a standard 12VAC transformer, a KBP206 bridge rectifier, a 2200µF bulk capacitor, and an LM7805 linear regulator.
Design Specifications
| Parameter | Value | Component / Notes |
|---|---|---|
| Mains Input | 120VAC, 60Hz | Primary side: 1A slow-blow fuse |
| Transformer Secondary | 12VAC RMS @ 1.5A | Toroidal or E-I core, 18VA rating |
| Rectified Peak DC | 15.57V | $(12 \times 1.414) - 1.4V$ bridge drop |
| Bulk Capacitance | 2200µF, 25V | Low-ESR aluminum electrolytic |
| Pre-Regulator Ripple | 3.78V p-p | $1A / (120Hz \times 0.0022F)$ |
| Minimum DC to Regulator | 11.79V | $15.57V - 3.78V$ (Must exceed dropout) |
| Target Output | 5.0V DC @ 1A | LM7805 in TO-220 package |
Dropout and Headroom Math
The LM7805 requires a minimum "dropout voltage" of 2.0V to maintain regulation. This means the input voltage must never fall below 7.0V. In our design, the bulk capacitor discharges down to 11.79V at the bottom of the ripple trough. Because 11.79V is well above the 7.0V minimum, the regulator has 4.79V of headroom and will easily reject the 3.78V ripple without passing it to the output.
Thermal Derating and Heatsink Sizing
Linear regulators dissipate excess voltage as heat. Failing to calculate thermal resistance will result in silicon junction temperatures exceeding 125°C, triggering internal thermal shutdown or catastrophic failure.
To find the power dissipated ($P_D$) by the LM7805, we use the average input voltage ($V_{in(avg)}$). The average input is the peak minus half the ripple: $15.57V - (3.78V / 2) = 13.68V$.
$P_D = (V_{in(avg)} - V_{out}) \times I_{load}$
$P_D = (13.68V - 5.0V) \times 1A = \mathbf{8.68W}$
A bare TO-220 package has a junction-to-ambient thermal resistance ($\theta_{JA}$) of roughly 65°C/W. Dissipating 8.68W without a heatsink would raise the junction temperature by 564°C above ambient—instantly destroying the silicon. We must add a heatsink.
Assuming a maximum ambient temperature of 40°C and a target junction temperature of 110°C (leaving a 15°C safety margin below the 125°C limit):
$\theta_{JA(max)} = \frac{110°C - 40°C}{8.68W} = 8.06°C/W$
Subtracting the TO-220 junction-to-case resistance ($\theta_{JC} \approx 5°C/W$) and a thermal pad ($\approx 0.5°C/W$), the heatsink itself must have a thermal resistance of 2.5°C/W or lower. For reference, a standard 2-inch extruded aluminum heatsink with vertical fins typically achieves this. For deeper thermal modeling, refer to the Texas Instruments linear regulator thermal design guide.
Input Protection and Filtering Requirements
A linear power supply connected to the AC mains is vulnerable to surges, reverse polarity on the DC side, and transformer inrush currents. Implement these protection layers on your PCB or chassis:
- Primary Inrush Limiting: Toroidal transformers have massive inrush currents due to low primary resistance. Use an NTC thermistor (e.g., Ametherm SL32 2R008) in series with the primary winding to limit turn-on surge.
- Fusing: Use a slow-blow fuse on the AC primary to survive transformer inrush, and a fast-acting fuse on the secondary AC side to protect the rectifier bridge from short circuits.
- Transient Voltage Suppression (TVS): Place a bidirectional TVS diode (like the 1.5KE18CA) across the secondary AC lines before the bridge to clamp inductive kickback and mains spikes.
- Reverse Polarity Protection: If your supply accepts external DC input instead of AC, place a P-channel MOSFET or a series Schottky diode before the regulator to prevent reverse-bias destruction of the internal junctions.
Frequently Asked Questions
What is a linear power supply used for?
Linear power supplies are primarily used in applications where ultra-low output noise and high transient response are critical. Common use cases include high-fidelity audio amplifiers, precision medical instrumentation, RF communication receivers, and laboratory test equipment. They are also used as "post-regulators" in hybrid supplies, where a switching pre-regulator handles the heavy lifting, and a linear LDO cleans up the final 1V of headroom to eliminate switching ripple.
How do you calculate linear power supply efficiency and heat?
Efficiency is calculated by dividing the output power by the total input power: $\eta = \frac{V_{out} \times I_{out}}{V_{in(avg)} \times I_{in}}$. Because the input current is roughly equal to the output current (ignoring the regulator's small quiescent current), the formula simplifies to $\eta \approx \frac{V_{out}}{V_{in(avg)}}$. Heat (power dissipated) is simply the voltage drop multiplied by the load current: $P_{heat} = (V_{in(avg)} - V_{out}) \times I_{load}$. The wider the gap between input and output voltage, the lower the efficiency and the higher the heat.
What is the difference between a linear and switching power supply?
The fundamental difference lies in how they regulate voltage. A linear supply uses a transistor in its active (linear) region to act as a variable resistor, burning excess voltage as heat. A switching supply (SMPS) uses a transistor as a switch, rapidly turning fully on and off, and relies on inductors and capacitors to store and transfer energy. Switching supplies are vastly more efficient and lighter, but they generate high-frequency electromagnetic interference (EMI) that linear supplies do not.
Why does my linear regulator get so hot and shut down?
Your regulator is entering thermal shutdown because its internal silicon junction has exceeded its maximum rated temperature (typically 125°C or 150°C). This happens when the power dissipated ($V_{drop} \times I_{load}$) exceeds what the package can safely shed into the surrounding air. To fix this, you must either lower the input voltage to reduce the voltage drop, decrease the load current, or attach a larger heatsink with a lower thermal resistance ($\theta_{JA}$) to improve heat transfer. Always verify your heatsink sizing with thermal math before powering on the circuit.






