LDO vs buck: choosing a 3.3 V rail
Two ways to make 3.3 V for a 100 mA load from 12 V. The AMS1117 linear regulator burns the difference as heat: its efficiency can never beat V_out ÷ V_in. The buck module switches, and keeps most of the power. The LDO wins on noise, cost and parts count when the voltage gap is small.
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How it works
LDO: the same current flows in as out (plus about 5 mA for the chip), so efficiency = 3.3 × 100 mA ÷ (12 × 105 mA) = 26 %, and the AMS1117 dissipates 870 mW.
Buck: input current ≈ output power ÷ (efficiency × V_in); expect 75–90 %.
The AMS1117 needs about 1.1 V of headroom: below roughly 4.4 V in, its output sags.
Try it on the bench
- Set V_in to 5 V: the LDO is 64 % efficient, good enough for a small load.
- Set V_in to 4 V: below the LDO’s dropout, its output sinks under 3.3 V.
Parts on the bench
- Bench power supply
- AMS1117 regulator
- Ceramic capacitor
- Electrolytic capacitor
- 2 × Resistor
- Step-down converter
In the course
Electronics Technician, module “Power supplies”.