Linear Voltage Regulators
A rectifier and filter capacitor can produce a reasonably smooth DC voltage, but "reasonably smooth" is not good enough for most ham radio equipment. When a 100 W HF transceiver switches from receive to transmit, the current draw can jump from less than 1 A to 20 A or more in milliseconds. With an unregulated supply, this sudden load change causes a large voltage sag — the output voltage drops by several volts. When the operator releases the key and returns to receive, the current drops back and the voltage surges upward again. This voltage bounce causes audio distortion, frequency instability, and in extreme cases, component damage.
A voltage regulator solves all of these problems by maintaining a precisely constant output voltage regardless of changes in load current or input voltage. It does this by continuously adjusting a series transistor — the pass transistor — to absorb the difference between the raw input voltage and the desired output voltage. The price of this precision is heat: all of the excess voltage multiplied by the load current is converted to heat inside the regulator. Understanding this trade-off, calculating how much heat is generated, and designing adequate cooling are the central skills in linear regulator design.
Series-pass linear voltage regulator schematic. The error amplifier compares a sample of V_out (taken by the R1/R2 divider) against the stable voltage reference. Any difference drives the base of Q1, adjusting the transistor's conduction to correct the output voltage. The feedback loop acts continuously, keeping V_out constant despite changes in load or input voltage.
View LargerWhy a Filter Capacitor Alone Is Not Enough
Imagine you are operating an HF transceiver at a contest, calling CQ repeatedly. Each time you key the transmitter, the current draw of your 100 W rig jumps from about 0.5 A (receive) to 20 A (transmit). Your unregulated supply's output voltage is V = V_peak − I × (R_transformer + R_wiring). At 20 A, a transformer with 0.5 Ω of source resistance drops 10 V — your supply might sag from 18 V down to 8 V under full transmit load. That is not going to power a transceiver that needs 13.8 V.
Even if the sag were less severe, the changing voltage would modulate the frequency of VFO oscillators, change the gain of audio stages, and potentially reset microcontrollers. Equipment designed to run from a stable 13.8 V supply does not tolerate large, rapid voltage swings. A voltage regulator holds the output at exactly 13.8 V (or whatever voltage it is set for) from no-load to full-load, making the supply appear as an ideal voltage source to the transceiver.
What Regulation Means
Voltage regulation is quantified by three key specifications that you will find in every power supply and regulator datasheet.
Load Regulation
Load regulation describes how much the output voltage changes between the no-load condition (zero current drawn by the load) and the full-load condition (maximum rated current). It is expressed as a percentage:
A high-quality bench supply might specify ±0.1% load regulation. For a 13.8 V supply, that means the output stays within ±13.8 mV from no load to 20 A full load — a remarkable performance that no unregulated supply can approach.
Line Regulation
Line regulation describes how much the output changes for a given change in the input (line) voltage. This is relevant because the utility voltage from the wall outlet is not perfectly constant — it can fluctuate ±10% or more during the day. Line regulation is typically expressed as mV per volt of input change, or as a percentage:
A good regulator IC might specify 0.01% per volt of input change. For a 13.8 V supply with V_in varying from 16 V to 24 V (an 8 V range), the output would change by only 0.01% × 8 × 13.8 = 11 mV.
Ripple Rejection
Ripple rejection (or supply ripple rejection ratio, SRRR) quantifies how well the regulator attenuates the 120 Hz ripple on its input and prevents it from appearing on its output. This specification is given in decibels. A typical linear regulator IC achieves 40–80 dB of ripple rejection. At 60 dB, a 2 V ripple on the input produces only 2 mV of ripple at the output — a 1000-fold reduction.
The Series-Pass Transistor Concept
The core of a linear regulator is a transistor placed in series with the current path from input to output. Think of this transistor as a variable resistor: when its resistance is low, it passes more current and the output voltage rises; when its resistance is high, it passes less current and the output voltage falls. The regulator's control circuit continuously adjusts this "variable resistance" to keep the output voltage exactly where you want it.
Why is this called a "linear" regulator? Because the pass transistor operates in its linear (active) region — the region between full cutoff and full saturation where collector current is proportional to base current. In the active region, the transistor can be any resistance from nearly zero to many kilohms, giving the regulator full control over the output voltage. This is in contrast to a switching regulator, where the transistor is either fully on (saturated, near-zero resistance) or fully off (cutoff, infinite resistance).
How the Feedback Loop Works
The series-pass transistor does not operate alone — it is part of a negative feedback control loop. Here is how the loop responds to a sudden increase in load current:
- Load current increases. Output voltage tends to fall.
- The voltage divider (R1/R2) samples the output and feeds a lower voltage to the error amplifier's non-inverting input.
- The error amplifier compares this sample against the stable reference voltage. The sample is now lower than reference, so the error amplifier's output increases.
- The increased error amp output drives the base of the pass transistor harder, increasing the transistor's collector current.
- Increased collector current supplies the higher load demand. Output voltage recovers to its set point.
The entire correction happens in microseconds. The loop continuously hunts for the exact transistor bias point that keeps the output voltage at the set point. If the load current decreases, the loop works in reverse — the error amp reduces base drive, the transistor conducts less, and the output voltage remains constant.
This same feedback loop also corrects for input voltage changes. If the raw DC input voltage rises (for example, because the utility power fluctuated), the transistor automatically increases its effective resistance to absorb the extra voltage, keeping the output unchanged.
Power dissipation in a linear regulator. The pass transistor must absorb the difference between input and output voltage multiplied by the load current. At V_in = 18 V, V_out = 13.8 V, and I = 10 A, the transistor dissipates (18 − 13.8) × 10 = 42 W. This heat must be removed by a properly sized heat sink.
View LargerThe Error Amplifier and Voltage Reference
The error amplifier is the "brain" of the regulator. It compares two voltages — a sample of the output and a stable reference — and generates an error signal that drives the pass transistor. The quality of the reference voltage determines the long-term accuracy and temperature stability of the regulated output.
The Voltage Reference
Early regulators used a simple zener diode as a reference. Zener diodes have reasonable temperature coefficients, but they are not highly stable and vary from unit to unit. Modern regulators use a bandgap voltage reference — a clever circuit that exploits the predictable physics of semiconductor junctions to produce a reference voltage that is almost independent of temperature and supply voltage. The bandgap reference voltage for silicon is approximately 1.25 V, which is why you will notice that the LM317 adjustable regulator has a 1.25 V reference between its output and adjust pins.
The Sampling Divider and Output Voltage Setting
The output voltage is set by a resistor divider that feeds a fraction of V_out to the error amplifier's input. At the correct output voltage, the sampled voltage exactly equals the reference. Solving for V_out:
Where:
V_ref = reference voltage (1.25 V for LM317)
R1 = upper resistor (between V_out and adjust pin)
R2 = lower resistor (between adjust pin and ground)
Example: LM317 with R1 = 240 Ω and R2 = 1 kΩ:
V_out = 1.25 × (1 + 240/1000) = 1.25 × 1.24 = 1.55 V
Example: To set V_out = 13.8 V with V_ref = 1.25 V and R2 = 240 Ω:
13.8 = 1.25 × (1 + R1/240)
1 + R1/240 = 11.04
R1/240 = 10.04
R1 = 2410 Ω → use 2.4 kΩ standard value
This formula is used every time you set the output voltage of an LM317, LM338, or similar adjustable regulator IC. The circuit is elegant in its simplicity: change two resistors to change the output voltage.
Dropout Voltage
A linear regulator can only maintain regulation as long as the input voltage exceeds the output voltage by a minimum amount called the dropout voltage. This is the minimum voltage the pass transistor needs across it to remain in its active (linear) region. If the input drops below V_out + V_dropout, the transistor saturates and the regulator loses control — the output voltage drops below the set point and tracks the input.
Standard vs Low-Dropout Regulators
| Type | Typical Dropout Voltage | Example ICs | Pass Transistor Type | Ham Radio Application |
|---|---|---|---|---|
| Standard NPN Darlington | 2–3 V | LM78xx series, LM317, LM338 | NPN Darlington | Fixed-voltage supplies with large voltage difference between raw DC and output |
| Low-Dropout (LDO) | 0.1–1.5 V | LM1117, LT1086, MIC29302 | PNP or PMOS pass transistor | Battery-powered equipment; circuits where input/output difference is small |
| Ultra-Low-Dropout | 50–200 mV | AMS1117, LP2950 | PMOS FET | 3.3 V from a 3.7 V Li-ion battery; microcontroller supply circuits |
Why Dropout Matters for Ham Radio Supplies
Consider a 13.8 V regulated supply using a standard LM317/LM338 with a 2.5 V dropout voltage. The minimum input voltage that ensures regulation is 13.8 + 2.5 = 16.3 V. Now remember the ripple from the filter capacitor: the input to the regulator swings from V_peak down to V_peak − V_ripple twice every cycle. At the trough of the ripple, the regulator input must still be at least 16.3 V. If the ripple trough dips below 16.3 V even briefly, the regulator drops out of regulation for that instant.
For the previous example of a 15 A supply with a 10,000 µF filter capacitor, the ripple trough can be several volts below the peak. Designing the system requires ensuring the minimum input (at peak ripple) exceeds the dropout threshold, with some margin.
Power Dissipation and Heat
The pass transistor in a linear regulator operates as a controlled power resistor. The power it dissipates is the product of the voltage across it (V_in − V_out) and the current through it (I_out):
This power is converted entirely to heat inside the transistor. It cannot be avoided in a linear regulator — it is the fundamental cost of the simplicity and low noise of linear regulation.
Efficiency of a Linear Regulator
The efficiency of a linear regulator is straightforward to calculate:
The remaining (100% − efficiency%) is wasted as heat in the pass transistor.
Example: V_in = 18 V, V_out = 13.8 V
Efficiency = 13.8 / 18 × 100 = 76.7%
23.3% of input power becomes heat.
Worked Examples
V_in = 18 V, V_out = 13.8 V, I_out = 10 A
P_diss = (18 − 13.8) × 10 = 4.2 × 10 = 42 W
Efficiency = 13.8 / 18 × 100 = 76.7%
Example 2: 13.8 V supply at 5 A
V_in = 18 V, V_out = 13.8 V, I_out = 5 A
P_diss = (18 − 13.8) × 5 = 4.2 × 5 = 21 W
Example 3: 5 V supply from 12 V at 2 A
V_in = 12 V, V_out = 5 V, I_out = 2 A
P_diss = (12 − 5) × 2 = 7 × 2 = 14 W
Efficiency = 5 / 12 × 100 = 41.7% — a poor efficiency, typical of large voltage differences
These numbers make it immediately clear why high-power linear regulated supplies need substantial heat sinking. A 42 W dissipation is equivalent to a bright light bulb worth of heat, continuously generated by the pass transistor on heavy transmit loads.
Pass Transistor Power Dissipation Calculator
Linear Regulator Power Dissipation Calculator
Calculate transistor power dissipation, efficiency, and junction temperature rise for a linear voltage regulator. V_in must be greater than V_out.
Heat Sink Selection
Heat flows from the transistor junction (the hottest point) through a series of thermal resistances to the ambient air (the coolest point). Each step in this path has a thermal resistance θ (theta), measured in degrees Celsius per watt. Just as electrical resistors in series add their resistances, thermal resistances in series add their values.
T_junction = T_ambient + P_diss × (θ_JC + θ_CS + θ_SA)
θ_JC = junction to case (given in transistor datasheet, e.g. 1.67 °C/W for a 2N3055)
θ_CS = case to heat sink (depends on mounting: 0.1 °C/W with thermal grease and insulator, up to 1 °C/W without)
θ_SA = heat sink to ambient (the heat sink's thermal resistance — lower = bigger/better heat sink)
How to Choose a Heat Sink
The design process works backwards from the maximum allowable junction temperature:
- Establish the maximum junction temperature. For most power transistors (2N3055, TIP35C, etc.) T_j(max) = 150°C.
- Establish the worst-case ambient temperature inside your enclosure. Inside a metal box with no forced cooling, allow 50°C on a hot day.
- The maximum allowable junction-to-ambient temperature rise is: ΔT = 150 − 50 = 100°C.
- The total allowable thermal resistance is: θ_total = ΔT / P_diss.
- Subtract θ_JC and θ_CS to find the required θ_SA (heat sink rating).
Maximum junction temperature: 150°C
Maximum ambient inside enclosure: 50°C
Allowable temperature rise: 100°C
Total allowable thermal resistance: θ_total = 100 / 42 = 2.38 °C/W
Typical transistor θ_JC (2N3055, TO-3 case): 1.52 °C/W
Typical θ_CS (insulated mounting with thermal grease): 0.5 °C/W
Required θ_SA = 2.38 − 1.52 − 0.5 = 0.36 °C/W
A 0.36 °C/W heat sink is a large, finned aluminum extrusion, approximately 150 × 150 mm or larger. For forced-air cooling (a small fan blowing across the fins), heat sinks can achieve 0.2 °C/W or better with much smaller physical dimensions.
Thermal Compound
Thermal compound (also called heat sink paste or thermal grease) is a grease-like material filled with thermally conductive particles (usually zinc oxide or silver). Applying a thin, uniform layer between the transistor case and the heat sink surface fills microscopic air gaps and dramatically reduces θ_CS from approximately 1°C/W (without compound, metal to metal) to 0.1–0.5 °C/W. Always apply thermal compound when mounting a power transistor to a heat sink.
Electrical Isolation
The metal cases of many power transistors (TO-3, TO-220) are electrically connected to the collector terminal. If the collector is not at ground potential — and in most linear regulator circuits it is not — the transistor case must be electrically isolated from the heat sink using an insulating washer (mica, anodized aluminum, or silicone rubber). Use a thermally conductive insulating pad and ensure no metal hardware bridges the insulation. Failure to isolate properly can result in short circuits and immediate destruction of the transistor or the power supply.
Ham Radio Application: Bench Power Supply Design
The classic ham radio bench power supply produces 13.8 V at 20 A — enough to run most modern solid-state HF transceivers at full power. Designing one from scratch ties together every concept in this module.
Full Design Example: 13.8 V at 20 A
Step 1 — Select input voltage:
Need V_in ≥ V_out + V_dropout + V_ripple_peak
V_in_min = 13.8 + 2.5 (LM338 dropout) + 1.5 (ripple allowance) = 17.8 V
Choose V_in nominal = 18–20 V from the unregulated supply.
Step 2 — Transformer:
V_secondary_RMS for 18 V DC peak: V_RMS = (18 + 1.4) / 1.414 = 13.7 V RMS
Use a standard 15–18 V RMS secondary (18 V preferred for margin)
V_peak = 18 × 1.414 − 1.4 = 24.1 V DC
Current rating: 20 A DC × 1.5 (form factor correction) = 30 A secondary RMS
VA rating: 18 V × 30 A = 540 VA. Use a 600 VA toroidal transformer.
Step 3 — Bridge rectifier:
PIV = 25.5 V. Use a KBPC5010 bridge module (50 A, 1000 V).
Step 4 — Filter capacitor:
At 20 A load and 120 Hz, for 1.5 V ripple: C = 20 / (120 × 1.5) = 0.111 F = 111,000 µF
Use 2 × 56,000 µF (or 4 × 22,000 µF) 50 V electrolytic capacitors in parallel.
Step 5 — Pass transistors:
P_diss at 20 A: (24.1 − 13.8) × 20 = 206 W total
Single 2N3055 is rated 115 W maximum — insufficient. Use multiple transistors in parallel.
Use 4 × 2N3055 transistors (or equivalent), each with an emitter resistor (0.1 Ω, 5 W) for current sharing.
Total dissipation per transistor: 206 / 4 = 51.5 W each — within TO-3 ratings with adequate heat sinking.
Step 6 — Heat sink:
Total P_diss = 206 W. T_ambient = 50°C (inside chassis with fan). T_j(max) = 150°C.
θ_total = (150 − 50) / 206 = 0.49 °C/W for all four transistors combined.
With four transistors sharing one large heat sink, individual θ_SA = 0.49 × 4 = 1.96 °C/W per transistor.
Subtract θ_JC (0.7 °C/W per transistor, TO-3) and θ_CS (0.4 °C/W): required θ_SA = 1.96 − 0.7 − 0.4 = 0.86 °C/W per transistor.
Use a large extruded aluminum heat sink with a cooling fan. This is why commercial 20 A supplies have significant chassis-mounted heat sinks and often include a cooling fan.
Step 7 — Regulator IC:
The LM338 is rated for only 5 A. To boost current, add external pass transistors driven by the LM338's current sense output, or use a dedicated controller IC designed for high-current regulation. Commercial designs often use purpose-built PWM regulator ICs or high-current versions of adjustable regulators.
⚖ Experiment: Measure Linear Regulator Power Dissipation
This experiment gives you direct experience measuring the power dissipation of a linear regulator and confirming the P = (V_in − V_out) × I formula. You will also feel the heat generated and confirm the thermal behavior of the regulator body.
- An LM7812 or LM7805 three-terminal voltage regulator (or an LM317 with resistors set for 12 V or 5 V output)
- An unregulated 18–20 V DC power source (or a 12 V source for the LM7805)
- A 100 Ω and 47 Ω resistor as loads (2 W rating each)
- A digital multimeter (or two multimeters if available)
- A small piece of aluminum as an improvised heat sink (or the regulator's TO-220 body alone for low-power tests)
- Thermal paste (optional but recommended)
- Connect the LM7812 (or LM317 set to 12 V) with the 100 Ω load resistor across the output. Apply the unregulated input voltage.
- Measure V_in (across the regulator input to ground) and V_out (across the output to ground). Record both. Calculate V_drop = V_in − V_out.
- Measure I_out by measuring the voltage across the 100 Ω resistor and dividing by 100: I = V_load / 100.
- Calculate P_diss = V_drop × I_out. Compare to the formula prediction.
- Carefully touch the top (metal tab) of the TO-220 regulator after one minute of operation. It should feel warm to hot depending on the dissipation. Do not touch if the package feels excessively hot.
- Replace the 100 Ω load with the 47 Ω resistor (doubling the load current). Repeat measurements. Observe the increase in dissipation and temperature.
- Add the aluminum heat sink to the regulator (apply thermal paste if available). Repeat the 47 Ω test and observe how much cooler the package runs with heat sinking.
The measured power dissipation should agree closely with P = (V_in − V_out) × I_out. The regulator body temperature will increase noticeably when you double the load current, confirming that dissipation is proportional to current. Adding the heat sink will keep the body noticeably cooler for the same load — demonstrating the importance of adequate thermal management in power supply design.
Frequently Asked Questions
Why does a linear regulator get so hot?
A linear regulator maintains a constant output voltage by absorbing the difference between the input and output voltages across the pass transistor. Every watt of power that the transistor absorbs is converted entirely to heat — there is no way to avoid it in a linear design. The power dissipated is P = (V_in − V_out) × I_out. At large voltage differences or high currents, this can be tens or even hundreds of watts. This is the fundamental trade-off of linear regulation: low noise, simple design, and stable output, but poor efficiency and significant heat generation.
What is the difference between a linear regulator and a switching regulator?
A linear regulator uses a transistor operating in its active (linear) region as a variable resistor to drop the excess voltage. It is simple, low-noise, and produces clean output, but it is inefficient — the efficiency is approximately V_out / V_in, and the rest is wasted as heat. A switching regulator rapidly switches the transistor fully on and fully off and uses an inductor and capacitor to store and transfer energy, achieving efficiencies of 85–95% with much less heat. Switching regulators are more complex, generate switching noise, and require careful filtering. Linear regulators are preferred for noise-sensitive ham radio circuits (receiver front ends, audio stages), while switching regulators are standard in modern high-power equipment.
What does "dropout voltage" mean and why does it matter?
Dropout voltage is the minimum voltage difference required between the regulator's input and output for it to maintain regulation. Standard linear regulators (NPN Darlington type) need 2–3 V of headroom. If the input voltage falls below V_out + V_dropout — even briefly during a ripple trough — the regulator loses control and the output voltage drops. This is why the filter capacitor must keep the unregulated voltage high enough throughout the ripple cycle. Low-dropout (LDO) regulators need only 0.1–1.5 V of headroom, making them better suited for battery-powered equipment where the input voltage is close to the output voltage.
Why does my power supply output drop when I key the transmitter?
If the output of a regulated supply drops significantly when you key the transmitter, the most likely cause is that the input to the regulator — the unregulated voltage from the rectifier and filter capacitor — is dropping below the regulator's minimum input (V_out + V_dropout). This happens when the filter capacitor is too small for the load current, causing excessive ripple, or when the transformer's source impedance causes too much voltage drop under heavy load. The fix is to increase the filter capacitance, use a larger transformer, or increase the transformer secondary voltage to provide more headroom. Check that the unregulated voltage at full transmit load is at least 2–3 V above the regulated output voltage.
Test Your Knowledge
Answer the questions below to check your understanding. Every answer can be found in the lesson above.