AMS1117/Headroom and heat/06. Where the extra volts go
Headroom and heat · 06 of 11

Where the extra volts go

Heat is the drop times the current. The chip warms by about 120 °C for every watt, so it can shed a little over 0.8 W before passing its rating. From 5 V that allows about 480 mA; from 12 V, about 90.

Volts dropped, times amps

The part gets 3.3 V. Whatever the supply gives above that is dropped across the transistor inside the chip, and a voltage dropped at a current is power:

heat in watts = (VIN − 3.3 V) × current in amps

Add a little for the 5 mA the chip uses to run. From 5 V at 200 mA that is 1.7 V × 0.2 A, plus 0.025 W: about 0.37 W. From 12 V at the same current it is 1.8 W.

How hot that makes it

UMW's datasheet gives the package 120 °C per watt: every watt raises the chip 120 °C above the air around it. The chip is rated to 125 °C, so in a 25 °C room it can shed a little over 0.8 W. Try some numbers:

Every volt above 3.3 V, times the current, is heat
5.0 V in · 200 mA
Input5.0 V
Load200 mA
Heat
0.37 W
Chip, roughly
69 °C
Most at this input
480 mA
Efficiency
64 %
Warm. 0.37 W is comfortable. Only 64% of what the supply delivers reaches your circuit, and a linear regulator cannot do better than 3.3 V divided by the input.

That 120 °C per watt is the datasheet's number. This board is 9 × 16.5 mm and has little copper to spread heat into, so treat every temperature here as the best case.

What the heat allows

InputMost current at 3.3 V out
5 Vabout 480 mA
9 Vabout 140 mA
12 Vabout 90 mA

The chip is rated for 1 A. The heat reaches its limit long before that at any input above about 4.1 V, which is to say at every input that has the headroom to regulate at 1 A in the first place.

Efficiency is fixed by the voltages

The current in equals the current out, plus the chip's own 5 mA. So the share of power that reaches your circuit is 3.3 V divided by the input: two thirds from 5 V, a little over a third from 9 V, just over a quarter from 12 V. Nothing about the board changes that; it is what a linear regulator is.

When it gets too hot

Near 150 °C the chip switches its output off. It cools, switches back on, heats up again, and the output flickers on and off. That protects the chip. It does not protect a project that needed the power.

When it does not work

It is too hot to touch at 200 mA from 9 V

5.7 V times 0.2 A is about 1.2 W, well over what the chip can shed. It is at or near its thermal shutdown. Feed it from 5 V instead, or use a buck converter for the big drop and let this board do only the last step.

The output keeps cutting out and coming back

That is the thermal shutdown. Near 150 °C the chip switches OUT off, cools, and switches it back on. Reduce the current or the input voltage until it stays on; the figure above shows how much.

Can I put two boards in parallel to share the heat?

Not reliably. Each holds its own idea of 3.3 V, anywhere from 3.234 to 3.366 V, and the higher one ends up supplying most of the current. Split the load instead: one board per group of parts, grounds joined.

Why does the board draw current with nothing connected?

The chip uses about 5 mA just to run, whatever the load, and the LED adds up to 1.5 mA. It does not matter on USB. On a battery that sleeps most of the time, it is usually the biggest drain in the project.

Where this goes next

Seven common jobs, worked out, and what to use for the ones this board cannot do.

When to use something else

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