Why it needs 12 volts
A bar this long is several LEDs in a row inside one piece of plastic, and voltages in a row add up. Five volts does not make it dim — it makes it dark, which is why the board has a second power input.
Why a big LED is not just a bigger LED
A small LED lights at around 2 volts. It would be reasonable to assume that a bar a hundred millimetres long is the same thing scaled up, and it is not: making a bar that long uniformly bright means putting several LED dice inside one piece of plastic, end to end, and parts in a row each want their own forward voltage before any current flows.
Drag the supply down and watch what happens at the threshold. The bar does not fade. Below the total it does nothing at all, because there is no path through until every die in the string has its share. Above it, the bar lights and the resistor on the module decides how hard.
That is the whole argument for the 12 V input. Five volts, the most an Uno has to offer, is under the floor.
The chip cannot do it either
Even at the right voltage, the shift register is the wrong thing to ask. Texas Instruments rates the 74HC595 for a 2 to 6 volt supply and about six milliamps out of any one pin — enough for an indicator LED on a breadboard, and not in the same conversation as a segment this size.
So the board puts a driver between them. The shift register decides which segments should be on; something with more muscle actually switches the 12 V through them. You never touch that part, and it is the reason the module needs two supplies rather than one.
Three connections, and one of them is the one that bites
Twelve volts and its ground on the two-pin header. Your board's 3.3 V or 5 V and its ground on the five-pin header. And those two grounds must be the same wire.
Leave out the 12 V and everything else works perfectly — bytes arrive, the latch fires, and the display is dark. Leave out the logic rail and the board stops listening. Leave out the shared ground and you get the bad one: a circuit that half works, because a HIGH is a voltage measured against ground and the two halves no longer agree where zero is.
What to buy
Any 12 V adapter. Centre-positive barrel jacks are the common ones; screw terminals and bare leads are fine too. What matters is that its negative lead ends up on the same ground as your microcontroller, which usually means a wire from the supply's ground to a GND pin on your board as well as to the module.
When it does not work
It may light and it will not stay lit. A PP3 battery holds a couple of hundred milliamp-hours and sags under load, so a display that is bright for ten minutes goes dim and then stops. Use a mains adapter with a 12 V output; almost any of them will do for one or two modules.
Measure rather than look. Put the black probe on your board's GND — not the supply's — and the red probe on the module's 12V pin. Twelve volts there means the supply and the shared ground are both good. Zero means one of them is not.
Only if you put it on the wrong header. The 12 V pins are a separate two-pin block at the top of each edge, well away from the five-pin signal row, and nothing on the signal row goes above your own logic rail. Check which header you are in before the supply goes on.
It needs more current, not more volts. The 12 V passes across each join to the next module, so every module in the chain draws from the same supply. Two is comfortable on a small adapter; if you are building six, size the supply for six.
Both edges carry the same seven connections. Which is which, and what the ten bare pads are for.
IN, OUT, and the arrow between them →Edit this page — content/books/4-inch-7-segment/why-it-needs-12-volts.mdx
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