🔋 Linear Voltage Regulator

An LDO linear regulator maintains output voltage despite load and line variations. Pass transistor dissipates (V_in−V_out)·I_load as heat. See dropout voltage, PSRR, and thermal limits.

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Efficiency vs load current · Thermal operating point · Dissipation heatmap

How it Works

A linear voltage regulator places a variable resistance (the pass transistor) in series between input and output. A feedback loop continuously adjusts this resistance so that V_out remains constant at the setpoint. The voltage difference (V_in − V_out) falls entirely across the pass transistor, which dissipates P_diss = (V_in − V_out) × I_load as heat.

The simulation shows: (left) efficiency vs load current curve — efficiency η = V_out/V_in, independent of load; (center) a thermal diagram showing junction temperature rise above ambient; (right) a heatmap showing how P_diss depends on V_in and I_load, with the 125°C limit marked.

P_diss = (V_in − V_out) × I_load
η = V_out / V_in × 100%
T_j = T_amb + P_diss × θ_JA
V_out = V_ref × (1 + R1/R2)

Frequently Asked Questions

What is a linear voltage regulator?

A linear voltage regulator maintains a constant output voltage by varying the resistance of a pass element (transistor) in series with the load. It works by dissipating excess voltage as heat: P_diss = (V_in − V_out) × I_load.

What is an LDO regulator?

LDO stands for Low Drop-Out. An LDO regulator can function with very small difference between input and output voltage (as low as 100–300 mV), unlike older regulators that needed 2–3V headroom. This allows operation when supply voltage is close to the desired output.

What is dropout voltage?

Dropout voltage V_do is the minimum voltage difference V_in − V_out needed for the regulator to maintain regulation. Below this, the pass transistor saturates and output follows input. LDOs have V_do typically 100–500 mV.

How efficient is a linear regulator?

Efficiency η = V_out/V_in × 100%. A 5V regulator from 12V input is only 42% efficient. The rest is wasted as heat. This is the main disadvantage vs switching regulators which can achieve over 90% efficiency.

What is PSRR in a voltage regulator?

Power Supply Rejection Ratio (PSRR) measures how well the regulator rejects noise and ripple on the input. At DC, PSRR is typically 60–80 dB. It degrades at higher frequencies. LDOs generally have excellent PSRR at low frequencies compared to switching regulators.

What causes thermal shutdown in a regulator?

When the pass transistor junction temperature exceeds typically 150°C, a thermal shutdown circuit turns off the regulator. T_j = T_ambient + P_diss × θ_JA where θ_JA is the thermal resistance junction to ambient in °C/W.

What is the role of the feedback network?

A resistor divider from V_out feeds back to the error amplifier, which compares it to an internal bandgap reference (~1.2V). The error amplifier drives the pass transistor to maintain V_out = V_ref × (1 + R1/R2) constant regardless of load or input variations.

When should I use an LDO vs a switching regulator?

Use an LDO when low noise is critical (RF, audio, ADC supplies), the voltage difference is small, or simplicity matters. Use switching regulators when efficiency is critical, large voltage conversion is needed, or high currents are involved.

What is the quiescent current of an LDO?

LDOs have quiescent current I_Q flowing even at zero load (through error amplifier and bias circuits). This ranges from microamps in ultra-low-power LDOs to milliamps in older designs. Total I_in = I_load + I_Q, affecting efficiency at light loads.

How do I calculate the heatsink required for an LDO?

Power dissipated is P = (V_in − V_out) × I_load. With thermal resistance θ_JA, T_j = T_amb + P × θ_JA. If T_j exceeds 125°C, a heatsink is needed. Required heatsink thermal resistance: θ_SA = (T_j_max − T_amb)/P − θ_JC.

About this simulation

This simulation models an LDO linear regulator dropping V_in down to a target V_out by burning the difference as heat in the pass transistor. As you change load current and ambient conditions, the right-hand thermal bar climbs toward the 125°C and 150°C limit lines, and a Status readout tells you whether the part is safely within spec, in a thermal warning zone, or shut down from overtemperature.

🔬 What it shows

A left efficiency-vs-load panel (η = V_out/V_in, constant regardless of I_load, with a red P_diss curve and a white operating-point dot), and a right thermal bar showing junction temperature T_j = T_amb + P_diss·θ_JA against 125°C/150°C thermal-shutdown thresholds.

🎮 How to use

Set V_in, V_out target, and I_load to change dissipation, then adjust T_ambient and θ_JA (thermal resist) to see how enclosure and heatsinking affect junction temperature. Watch V_drop, P_diss, Efficiency, T_junction, and Status update live; click Refresh to recompute.

💡 Did you know?

A linear regulator dropping 12V to 5V is stuck at a hard efficiency ceiling of just 41.7% no matter how well it's designed — the remaining 58.3% of input power is unavoidably converted to heat, which is exactly why switching regulators (often over 90% efficient) dominate in battery-powered and high-current applications.

Frequently asked questions

Why does the efficiency line stay flat no matter how I move the I_load slider?

For a linear regulator, efficiency η = V_out/V_in depends only on the input and output voltages, not on load current — so moving I_load changes power dissipated (and heat) but never changes where the orange efficiency line sits on the left panel.

Why does the Status readout flip to OVERTEMP even at moderate I_load?

Junction temperature T_j = T_amb + P_diss·θ_JA combines three things multiplicatively — a large V_in−V_out drop, high load current, and a high thermal resistance θ_JA (no heatsink) — so even moderate current can push T_j past 150°C if the other two factors are unfavorable.

What happens if I set V_out target very close to V_in?

The regulator enters the LDO's dropout region — the code clamps V_out to at most V_in−0.3V, mimicking the real minimum voltage difference a regulator needs to stay in regulation. V_drop shrinks toward its minimum, and P_diss and thermal stress drop correspondingly.

Why does raising θ_JA (thermal resist) push T_junction up so fast?

θ_JA multiplies directly against P_diss in the T_j formula, so a poorly-heatsinked part (high θ_JA, e.g. 150°C/W) can reach dangerous junction temperatures at a fraction of the power dissipation that a well-heatsinked part (low θ_JA) could handle safely.

Why is P_diss shown as a separate curve from the efficiency line?

Efficiency is a ratio (V_out/V_in) that's independent of current, but P_diss = (V_in−V_out)·I_load scales directly with load — the two curves together show that a linear regulator's percentage efficiency stays fixed while its absolute wasted power (and heat) grows with every extra milliamp drawn.