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Notes / Electronic Circuits

Linear regulator: simple, but it gets hot

An LDO throws away the difference between input and output as heat. Dropout voltage, heat calculation and capacitor requirements.

GROUND EMBEDDED · NOTEElectronic Circuits 09Linear regulatorsVCC 3.3V10kΩButtonMCUGPIOHIGH when released

A linear regulator places a transistor that acts like a variable resistor between input and output, and keeps adjusting that resistance so the output stays constant.

Heat calculation

Input 12 VRegulator7 V × 0.5 A = 3.5 W heatOutput 5 V · 0.5 A
A linear regulator throws away the excess voltage as heat. 12 V to 5 V at 0.5 A is 3.5 W.

P = (Vin - Vout) × Iout

12 V → 3.3 V at 200 mA: (12 - 3.3) × 0.2 = 1.74 W. In a SOT-223 package (thermal resistance about 60 °C/W) that is a rise of more than 100 °C. In this case it is usual to make 5 V first with a switching converter and then 3.3 V with an LDO.

Efficiency is Vout / Vin, so 12 V → 3.3 V is about 27%.

Dropout voltage

The smallest input-output difference at which the output can be held. To make 3.3 V from a 3.7 V battery you need an LDO with a dropout below 0.4 V. When the battery falls to 3.5 V, the output falls with it.

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Why it is still used

  • There is no switching noise on the output, which suits ADC references, analog sensors and RF supplies.
  • Few parts and an easy design.

Things to check

Meet the output capacitor's capacitance and ESR range required by the datasheet. If the conditions are not met, the LDO oscillates. Put 1 µF or more close to the input as well.

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