Why the 5 V rail reads low
Put a meter on the standard board's 5 V rail and it reads something like 4.7 V. Nothing is wrong. The fuse and the diode between the socket and the rail take a few tenths of a volt between them, more as the current rises, and every 5 V part is built to expect it.
The 5 V is USB, less two parts
On the standard board, 5 V is not made by anything. It is the USB supply, passed through the fuse and the diode from the last article and out to the rails. So the rail is whatever the USB port gives, less whatever those two parts take.
How much they take
The diode is a Schottky, chosen because it takes less than an ordinary diode. Even so, from the maker's typical curve it takes about 0.17 V at 10 mA, 0.3 V at 100 mA and 0.38 V at half an amp. The fuse adds its resistance times the current: 0.15 to 0.7 Ω, so between 15 and 70 mV at 100 mA, and up to a third of a volt at half an amp.
From a port putting out 5.00 V, that leaves:
| The breadboard draws | The 5 V rail reads |
|---|---|
| 10 mA | about 4.8 V |
| 100 mA | 4.6 to 4.7 V |
| 300 mA | 4.4 to 4.6 V |
| 500 mA | 4.3 to 4.5 V |
The range in each row is the fuse: two boards off the same reel can differ by that much. A charger usually sits a little above 5 V, which moves every row up.
Why that is fine
A 5 V sensor, display or microcontroller board is almost never built for exactly 5.00 V. It is built for a range, and the range is nearly always about 4.5 to 5.5 V, because USB itself only promises 4.75 to 5.25 V at the port and everything designed to run from USB has to cope with that. The rail at ordinary loads sits inside that range.
So a rail that reads 4.5 to 4.9 V is working. The diode costs a few tenths of a volt, and what it buys is a charger that cannot be pushed backwards into by whatever else is on your breadboard.
Where it stops being fine
The bottom rows. Half an amp from a laptop port that already starts at 4.75 V can leave the rail near 4 V, and that is where a display flickers, a servo twitches or a dev board browns out and resets. If a circuit misbehaves only when something heavy switches on, measure the rail at that moment. Under 4.5 V, use a stronger charger or draw less from the board.
When it does not work
Either the load is heavy, the charger is weak, or both. At half an amp the diode and the fuse can take three quarters of a volt between them, and a laptop port that starts at 4.8 V leaves you near 4 V. Try a phone charger, and measure the rail with the load disconnected: if it comes back to about 4.8 V, the adapter is fine and the load is the question.
You would get perhaps a quarter of a volt back and lose the protection it gives your charger against a second supply on the rail. If a part really needs 5.0 V exactly, it needs a regulated 5 V supply, not a USB port with nothing in the way: a USB port is itself allowed to be anywhere from 4.75 to 5.25 V.
When it is plugged straight into the laptop there is no diode or fuse in the way. Through the adapter it gets a few tenths less. A module on the edge of its own supply range notices. Check what the module is rated for, and try a phone charger rather than the laptop port, which often sits lower.
No. The 3.3 V comes from a regulator fed from the 5 V rail, and a regulator only needs about a quarter of a volt more at its input than it gives out. At 4.7 V in, it holds 3.3 V out exactly. The drop is only on the 5 V side, which has no regulator.
Each side of the breadboard gets its own switch, and its own LED to tell you what it is.
Two switches, two rails →Edit this page — content/books/bbpwd/why-5v-reads-low.mdx
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