Voltage Regulator Power Dissipation Calculator

Calculate the power dissipated as heat in a linear voltage regulator, estimate junction temperature, and check efficiency. Enter input voltage, output voltage, and load current — thermal inputs are optional.

Input Voltage V

Output Voltage V

Load Current

Thermal Parameters (optional)

Thermal Resistance R_θJA °C/W

From regulator datasheet. Enables junction temp estimate.

Ambient Temperature °C

Default 25 °C if left blank.

Voltage Drop (V_DROP) —

Power Dissipation —

Efficiency —

Heat Level —

Est. Junction Temperature —

💡 Why do linear regulators get hot? Unlike switching regulators, linear types reduce voltage by burning off the excess as heat. The more headroom between V_IN and V_OUT, and the higher the current, the more heat is generated — it's essentially a resistor dropping the surplus voltage.

How to Use This Calculator

Enter the input voltage supplied to the regulator, the regulated output voltage, and the current drawn by your load. Select whether your current is in milliamps (mA) or amps (A).

Power dissipation, voltage drop, and efficiency are calculated instantly. For a junction temperature estimate, also enter the thermal resistance (R_θJA) from your regulator's datasheet — a per-device value typically in the range of 30–100 °C/W for common packages like TO-92 and TO-220. Ambient temperature defaults to 25 °C if left blank.

If the estimated junction temperature exceeds 125 °C, a warning will appear — this is the typical maximum for most linear regulators, though the exact limit varies by device. Always check your datasheet.

Formula

Voltage drop across the regulator:

V_DROP = V_IN − V_OUT

Power dissipated as heat:

P_D = (V_IN − V_OUT) × I_LOAD

Regulator efficiency:

η = (V_OUT / V_IN) × 100

Estimated junction temperature (requires R_θJA):

T_J = T_A + (P_D × R_θJA)

Where: V_IN = input voltage (V), V_OUT = output voltage (V), I_LOAD = load current (A), P_D = power dissipation (W), η = efficiency (%), T_A = ambient temperature (°C), T_J = junction temperature (°C), R_θJA = thermal resistance junction-to-ambient (°C/W).

Example

Given: V_IN = 12 V, V_OUT = 5 V, I_LOAD = 500 mA, R_θJA = 50 °C/W, T_A = 25 °C

Voltage drop: 12 − 5 = 7 V

Power dissipation: 7 × 0.5 A = 3.5 W

Efficiency: (5 / 12) × 100 = 41.7%

Junction temperature: 25 + (3.5 × 50) = 200 °C

⚠️ At 200 °C estimated junction temperature, the regulator would almost certainly overheat. A heatsink is essential at this power level, or switch to a buck (switching) regulator for much better efficiency.

Frequently Asked Questions

What is power dissipation in a linear regulator?
It's the electrical energy converted to heat inside the regulator chip. Because linear regulators reduce voltage by "absorbing" the excess, every volt of headroom multiplied by load current becomes heat — wasted energy that must be managed thermally.

What is R_θJA and where do I find it?
Thermal resistance junction-to-ambient (R_θJA) describes how efficiently a component dissipates heat to the surrounding air. It's listed in the regulator's datasheet, typically under "Thermal Characteristics." A lower value means better heat dissipation — a TO-220 package with a heatsink can be much lower than its bare R_θJA.

Why is my linear regulator's efficiency so low?
Linear regulators are inherently inefficient when there's a large gap between V_IN and V_OUT. If you need 5 V from 12 V, nearly 42% of input power is wasted as heat. For high-efficiency needs, a switching (buck) regulator can achieve 85–95% efficiency for the same conversion.

At what junction temperature should I add a heatsink?
As a rule of thumb, once estimated junction temperature exceeds around 80–100 °C under typical conditions, adding a heatsink is prudent. Most regulators have a maximum junction temperature of 125–150 °C — operating close to that limit shortens device life significantly.

Does adding a heatsink change the power dissipation calculation?
No — P_D is fixed by the circuit conditions. A heatsink lowers the effective R_θJA, which lowers junction temperature for the same power, keeping the chip cooler. The heat still has to go somewhere; a heatsink just moves it into the air more efficiently.