AMS1117/Headroom and heat/05. The volt it needs to work
Headroom and heat · 05 of 11

The volt it needs to work

The AMS1117 needs its input about a volt above its output, more as the current rises. Below that it stops regulating and the output follows the input down. USB's 5 V is enough for 3.3 V out. A lithium cell is not.

About a volt, more under load

The transistor inside the chip cannot open all the way. Even flat out it drops a volt or so, and that sets the lowest input that still gives a steady output. UMW's datasheet gives these, for the 3.3 V board's chip:

CurrentTypical dropoutWorst caseSo VIN must be at least
100 mA1.00 V1.20 V4.3 V
500 mA1.05 V1.25 V4.35 V
1 A1.20 V1.30 V4.5 V

Slide the input down and watch where the output leaves 3.3 V:

The volt or so it needs above 3.3 V
5.0 V in · 100 mA
Input5.0 V
Load
OUT
3.30 V
Needs at least
4.30 V
Worst case
4.50 V
Regulating: 3.30 V. Anything above 4.30 V works, and everything above 3.3 V becomes heat — 0.20 W at 100 mA. A higher input buys nothing but that.

Right of the knee the output is flat: the chip has headroom and uses it. Left of the knee the chip is as open as it goes and the output is simply the input less the dropout. Nothing is wrong with the board; it has run out of volts.

Which supplies are enough

  • USB, 5 V. 1.7 V of headroom. Enough, with margin, at any current the heat allows.
  • 9 V or 12 V. Plenty of headroom — far more than needed. The next page is about what that costs.
  • One lithium cell. 4.2 V fully charged, and most of its charge is spent near 3.7 V. From 3.7 V at 240 mA the output is about 2.7 V: below the 3.0 V an ESP32 is rated to run from, and not far above the 2.43 V where it resets.
  • Two AA cells. About 3 V. A regulator like this only ever lowers a voltage.

USB is not always 5 V

USB promises between 4.75 V and 5.25 V at the port, and a long, thin cable loses more on the way. 4.75 V still leaves 1.45 V of headroom, which covers the worst-case dropout. A cheap cable under a heavy load can take that margin away, and it shows up as an output that is fine on the bench and low in the project.

When it does not work

My 3V3 board gives 2.7 V from a lithium battery

That is dropout, not a fault. At 3.7 V in and a couple of hundred milliamps out it needs about 4.3 V, so the output is the input less a volt. A single cell needs a regulator specified for a few hundred millivolts of dropout, or a buck-boost.

OUT is 3.3 V with nothing connected and low with the part running

Two things grow with current: the dropout, and the voltage your supply loses in its cable. Measure VIN while the part is running. If VIN is under about 4.4 V, the input is the problem — a shorter or thicker USB cable, or a better supply.

Can I get 3.3 V from two AA cells?

No. Two fresh cells give about 3 V, and a regulator like this can only lower a voltage, never raise it. The output would be around 2 V. Raising a voltage needs a boost converter.

Is 4.5 V from three AA cells enough?

Just, while they are fresh and the current is small, and not for long: three alkaline cells fall below 4.4 V well before they are flat. Four cells, or USB, leave margin.

Where this goes next

Every volt above the output, times the current, is heat. How much the chip can take.

Where the extra volts go

Edit this page — content/books/ams1117/the-volt-it-needs.mdx

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AMS1117 LDO Regulator Kit, 24 Boards (18 × 3.3 V, 6 × Adjustable)

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