Lesson 5.1 · Module 5: Power Regulation & Circuit Protection

LDO vs Buck Converter: Efficiency and Heat

Two ways to turn 12 V into 3.3 V for a chip. One burns the extra as heat, the other switches cleverly and wastes very little.

The live, interactive simulation for this lesson opens here when JavaScript is on.

Key ideas

A linear voltage regulator (LDO) drops the extra voltage and turns it straight into heat.
A buck converter switches on and off fast and uses an inductor to smooth the result — about 90 % efficient.
Wasted power = (input voltage − output voltage) × current for an LDO. With a big drop, that gets hot fast.

Try this in the simulation

With the LDO, raise the input to 12 V and the load to 500 mA: watch the junction temperature climb.
Switch to Buck and compare the efficiency.
Hover any wire or part in the circuit to see its live voltage and current. Click to pin the readout.

What is the difference between an LDO and a buck converter?

Both turn a higher voltage into a lower one. A linear regulator (LDO) works like an adjustable resistor and burns the difference as heat: the wasted power is (input voltage − output voltage) × current. A buck converter switches on and off many thousands of times a second and uses an inductor to smooth the result, so it can be around 90 % efficient. For a large voltage drop, an LDO gets hot while a buck converter stays cool.

What you’ll learn

How an LDO regulates, and how to calculate the heat it wastes.
How a buck converter uses a switch, an inductor and a diode to step a voltage down.
How to compare efficiency, and when an LDO is still the better choice.
How to estimate the junction temperature of a hot regulator.

The formula, letter by letter

\begin{aligned}\htmlData{sym=P_loss}{P_{\mathrm{loss}}} &= (\htmlData{sym=U_in}{U_{\mathrm{in}}} - \htmlClass{mv mv-v_out}{\htmlStyle{font-size:1.00em}{\htmlData{sym=U_out}{U_{\mathrm{out}}}}})\,\htmlData{sym=I}{I} = (\htmlData{funit=V,v=5}{\text{5.00 V}} - \htmlData{funit=V,v=3.3}{\text{3.30 V}})\cdot \htmlData{funit=mA,v=0.3}{\text{300 mA}} = \htmlData{funit=mW,v=0.51025}{\text{510 mW}}\\ \htmlData{sym=eta}{\eta }&= \frac{\htmlData{sym=U_out}{U_{\mathrm{out}}}}{\htmlData{sym=U_in}{U_{\mathrm{in}}}} = \htmlData{funit=percent,v=65.98900183302783}{\text{66.0 \%}}\end{aligned}

Shown with this lesson’s starting values. In the simulation the numbers change as you move the sliders.

$P_{\mathrm{loss}}$: Power lost as heat (W · watt)
$U_{\mathrm{in}}$: Input voltage (V · volt)
$U_{\mathrm{out}}$: Output voltage (V · volt)
$I$: Current (A · ampere) : How much charge flows each second.
$\eta$: Efficiency η (eta) (% · percent) : How much of the input power reaches the output. The rest becomes heat.

Common questions

Why does my linear regulator get so hot?

It drops the extra voltage across itself while the full load current flows through it. Dropping 12 V to 3.3 V at 0.5 A wastes (12 − 3.3) × 0.5 ≈ 4.4 W as heat, far more than a small package can lose safely.

When should I use an LDO instead of a buck converter?

When the voltage drop or the current is small, or when you need very low noise. LDOs are simple, cheap and quiet, but waste a lot of power in large steps.

What does efficiency mean for a regulator?

It is the output power divided by the input power. A regulator with 90 % efficiency wastes the other 10 % as heat.