A linear regulator makes heat
The step-down makes a lower voltage by burning off the difference. The current out equals the current in, so the heat is the voltage it drops times the current it carries. At tens of milliamps that is nothing. At hundreds, at a low voltage, it is more than a part this size can shed.
Current in equals current out
A linear regulator works like a valve. It sits between the input and your rail and lets through exactly as much current as your circuit draws, dropping whatever voltage it must to hold the output where the trimmer set it. The current going in is the current coming out. The voltage it drops, times that current, is heat.
With the input at about 4.6 V:
| Set to | Current | Dropped | Heat |
|---|---|---|---|
| 3.3 V | 150 mA | about 1.3 V | about 0.2 W |
| 2.5 V | 200 mA | about 2.0 V | about 0.4 W |
| 1.8 V | 300 mA | about 2.7 V | about 0.8 W |
The first row is warm. The last is more than a small SOT-89 package on a board this size can get rid of.
What the regulator does about it
The AMS1117 has thermal protection. When it gets too hot it cuts its output, cools, and comes back — so an overloaded side does not fail outright, it flickers. The rail sags, recovers, sags again, and whatever is wired to it resets each time. If a circuit on the step-down misbehaves after a minute of running and not before, heat is the first suspect.
When to use something else
A linear regulator is the right choice when the current is small: sensors, a microcontroller that sleeps, a reference voltage for an analog circuit. Its output is clean, and it has none of the switching noise a buck converter makes.
For hundreds of milliamps at a low voltage it is the wrong one. A switching step-down (a buck converter) moves energy rather than burning it, and typically wastes a tenth or two of the power instead of more than half. Use one of those, or run the part from the standard board's 3.3 V if 3.3 V will do.
Both sides, one fuse
The two regulators share the input fuse, so whatever both sides draw together has to stay under half an amp. The back of the board says MAX 1A@3V8, which is the AMS1117's own rating and not what this board can deliver: 1 A would trip the fuse, and at 3.8 V the regulator would have no room left to regulate.
When it does not work
The regulator is overheating and cutting back, then cooling and trying again. Work out the heat: the input (about 4.6 V) minus the output, times the current. Over about half a watt on either side, lower the current, raise the voltage if the part allows it, or use a switching regulator for that load.
Because the regulator drops more. From about 4.5 V, 1.8 V out leaves 2.7 V to burn; 3.3 V out leaves 1.2 V. Same current, more than twice the heat. A lower voltage from a linear regulator is always the hotter job.
No. Both regulators take their input through the same half-amp fuse, so half an amp is the total for both sides together, and the heat at that current would stop either regulator long before the fuse did at most voltages.
That is the AMS1117's own rating, which assumes a much higher input and a much bigger heatsink than a breadboard adapter has. On this board every milliamp comes in through a fuse that holds 0.5 A, and 3.8 V is only reachable at light load. Plan on a few hundred milliamps at most, and less at low voltages.
The booster board, and a knob whose top end is different on every board.
Setting the booster →Edit this page — content/books/bbpwd/linear-means-heat.mdx
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