Band 1 · Signal Tower

Bench Series and parallel

The two ways to join anything to anything. One of them is what every build in this course already is, and nobody has said so until now.

Time: 75 min
You need: 3 LEDs, 2–3 resistors, 6 jumper wires. No sketch: this one runs from the 5V pin.
At once: Everybody at once
Before this: Do bench-ohms-law first. You need I = V ÷ R to explain what you are about to see.

Sketches for this project

Every file opens with a plain-English header saying what it does and how to wire it, hole by hole.

Why

You have been building one of these all along and nobody said which

Your signal tower has three lights. Each one has its own resistor, its own pin, and its own wire back to the same − rail.

That arrangement has a name, and this course has never once used it. Which is a real gap, because it is the arrangement of almost everything you will ever build, and the other arrangement behaves so differently that mixing them up wastes an afternoon.

There are exactly two ways to put two components in a circuit. Two. That is the whole subject.

Predict first, and write it down

Two red lights, one battery, and no maths yet. Just say what you think.

You wire them one after the other in a line, so all the current goes through the first and then through the second. Are they brighter, dimmer, or the same as one light on its own?
You wire them side by side, each with its own path back. Brighter, dimmer, or the same?
In which of the two does taking one light out leave the other still lit?

Answer all three on paper now. You are about to build both, and the value of this lesson is almost entirely in whether your guess was right.

Build

One after the other: a series circuit

30 min

One loop. The current has one path, so it is the same in every part of it, and one resistor is enough for both lights.

Before you startUSB unplugged. This lesson has no sketch in it at all. Every circuit here runs from the 5V pin, because electricity does not need a program and this is about the electricity.

Which rails and which half. Everything sits in rows a to e, so use the pair of rails along the bottom edge, nearest row a. That leaves Band 1's tower undisturbed in the top half if it is still there. Wire the + rail to the 5V pin and the − rail to a GND pin. Then check by eye before you plug in: no wire runs from the + rail to a GND pin, and no wire runs from the − rail to 5V. Joining 5V straight to GND is the one mistake that damages the board. Getting the two rails the other way round, on the other hand, does no harm at all: nothing will simply light.

Part	Holes	Note
Power in	wire `a5` to the + rail
First LED	long leg `b5`, short leg `b7`
Second LED	long leg `a7`, short leg `b9`	This is the whole idea. Its long leg is in column 7, the same column as the first LED's short leg, so the two are joined underneath. Different hole, same column.
Resistor	`a9` to `a11`	One resistor for the pair, and one is right here
Back to ground	wire `b11` to the − rail	`b11`, not `a11`: the resistor leg is there

Check the direction of both LEDs before you plug in. Long leg towards the + rail, short leg towards the − rail, all the way down the line. In a series circuit they all point the same way, like people queueing.

One of them backwards and nothing lights at allNot “one lights”. Nothing, because the current has to get through both and a backwards LED stops it dead. That is your first clue about how a series circuit behaves.

Plug in the USB.

You should seeBoth lights on, and both noticeably dimmer than the single light you built in Band 1. Compare them with the one you remember, not with one on the board: the 5V pin is already feeding your rail, and with no sketch running there is nothing to light a pin.

Now work out why, with the sum you already have.

Two red LEDs take about 2 V each = 4 V gone
5 − 4 = 1 V left for the resistor

I = 1 ÷ 220 = 0.0045 A = 4.5 mA

Against 13.6 mA for one light. A third of the current.

You should seeThe arithmetic matching your eyes. The lights are dimmer because there is genuinely less current, and there is less current because the two LEDs between them ate four of the five volts and left almost nothing to push with.

That 4.5 is a rougher answer than the 13.6 wasAnd for an interesting reason. With only one volt left over, a small error in the two-volts-per-LED guess moves the answer a long way: guess 1.85 volts each instead and you get six milliamps, which is a third more. When the leftover is large, the guess hardly matters. When it is small, it matters enormously. Treat this one as “a few milliamps, far less than before” rather than a number to defend.

Unplug the USB, then take one LED out. Pull the second one straight up, both legs. Plug back in.

You should seeThe other one goes out too. There is one path and you have broken it.

Put it back, and now add a third. Unplug the USB first, and do these four moves in this order, because the holes the third LED needs are currently full:

Take the resistor out of a9 and a11.
Take the − rail wire out of b11.
Now put the third LED in: long leg a9, short leg b11.
Put the resistor back, a11 to a13, and the − rail wire in b13. Plug in.

You should seeAlmost nothing. A dim glow at best, and on some LEDs nothing you can see at all. Three red LEDs want about six volts between them and you have five, so there is barely any push left over.

Whatever glow you do see is worth understandingTwo volts per LED was never a fixed fact. It is the figure at normal brightness. With almost nothing flowing, each LED drops nearer one and a half volts, three of those just about fit inside five, and a trickle gets through. The numbers in this course describe parts doing their normal job, and they drift once a part is working far outside that.

This is the most useful thing in the lessonA series circuit has a ceiling. Add one part too many and you do not get “a bit dimmer”, you fall off a cliff: a useless trickle, with no warning and nothing broken. Remember it when somebody tells you to wire lights in a line to save resistors.

The three rules of a series circuit

The current is the same everywhere. One path, one flow. You saw this in Band 1 with the rope: squeeze anywhere and the marker slows everywhere at once.
The voltages add up to the total. 2 + 2 + 1 = 5. Each part takes a share and the shares must come to what you started with.
The resistances add up. Two 220s in a line behave as one 440.

Build

Side by side: a parallel circuit

25 min

Two complete loops from the same two rails. Same parts as the series build, different shape, and each branch needs its own resistor.

Start cleanUnplug the USB and take the whole series circuit off the board, both ends of every wire, except the two that power the rails: leave the + rail to 5V wire and the − rail to GND wire exactly where they are. Everything else comes off. You are building something with the same parts and a completely different shape.

Part	Holes	Note
First LED	long `b5`, short `b7`. Resistor `a7` to `a9`. Wire `a5` to the + rail, wire `b9` to the − rail.	A complete loop on its own
Second LED	long `b15`, short `b17`. Resistor `a17` to `a19`. Wire `a15` to the + rail, wire `b19` to the − rail.	Another complete loop, sharing only the rails

Its own resistor each. That is not tidiness and it is not optional; the last part of this lesson is about what goes wrong when people share one.

Plug in.

You should seeBoth lights at full brightness, each looking like the single light from Band 1. Not dimmer than each other, and not dimmer than one on its own. If you want a third to compare against, wire it on its own branch from the same two rails: b25 long, b27 short, resistor a27 to a29, wire a25 to the + rail and b29 to the − rail. All three stay bright.

Take one out.

You should seeThe other one carrying on, completely unbothered. Two separate loops that happen to start and end in the same place.

Work out the current. Each loop has the full 5 volts across it, so each one is exactly the circuit you did on paper in the last lesson: 13.6 mA. Two of them.

Each branch: 13.6 mA
Coming out of the 5V pin: 13.6 + 13.6 = 27.2 mA

You should seeWhy they stayed bright: nothing was shared out. The voltage is shared in series and the current is shared in parallel, and that one sentence is most of what this lesson is for.

The number in that sum is a warning, and it is about your Arduino

27 milliamps. An Arduino pin is rated for 20.

You are safe here because you took power from the 5V pin, which supplies the whole board and can give hundreds of milliamps. But hang two lights off one digital pin in parallel and you have asked that pin for 27, which is over its limit, quietly, with everything looking perfectly normal.

This is why Band 1's tower gives each light its own pin. Not for the code's convenience. Because three lights on one pin would be 41 milliamps and would slowly damage the chip.

The three rules of a parallel circuit

The voltage is the same across every branch. Each one gets the full push.
The currents add up. The source has to supply the lot.
Taking one branch out leaves the others working. Which is why almost everything real is wired this way.

Look closer

Now go and look at what you have already built

20 min

With those two shapes in your head, three things you have already done stop being arbitrary.

Your signal tower is parallel

Three lights, three resistors, three pins, one shared − rail. Each light is its own loop from its own pin back to the same ground. That is why you can light any one without the others, why they are all equally bright, and why a dead LED in the middle does not take the other two with it.

Every build in this course from Band 1 onward puts its branches in parallel. The four buttons, the two lights of the night guard, the lights and the buzzer of the lock. All of them: separate branches sharing the rails, and each branch with its own parts in series inside it. The two shapes are not rivals. Real circuits are made of both, nested.

Why the RGB LED needs three resistors and not one

The RGB fault, built on purpose. One resistor for two lights fixes the total current, and the two colours divide it between them.

Band 2 warns you not to put a single resistor on the shared leg. It gives the symptom, which is that the brightness drifts as the colour changes. Now you can say why.

The three colours are three parallel branches. Put one resistor on the leg they all share and that resistor fixes the total: about 13.6 milliamps, however many colours are lit. So the colours have to share one allowance between them.

One colour on and it gets the lot. Two on and they split it. Three on and each gets roughly a third. Every colour dims the moment you add another one, which is the drift the Band 2 page warns you about.

And they do not split it evenly. Red needs the least push of the three, so red takes more than its share and blue takes less. Your white ends up warm and yellowish rather than white, and nothing you can change in the code will fix it, because the fault is a missing resistor rather than a wrong number.

Try it, since it costs nothingUnplug the USB. Rebuild the two parallel lights with the resistors taken out: LED1 long b5 short b7, LED2 long b15 short b17, wire a5 and a15 both to the + rail. Now join the two short legs with a wire from a7 to a17, and give the pair one resistor from c17 to c19, with a single wire a19 to the − rail. Plug in. Both light, dimmer than before. Now pull one out and watch the other get brighter. That is the fault, in your hands, in thirty seconds.

Where series is the right answer

Series is not the wrong shape, it is the shape for a different job. The resistor and the LED in every build you have made are in series with each other: that is the whole point, because you want the same current through both and you want the resistor to take a share of the voltage.

Series is also how a torch stacks its batteries: three 1.5 V cells in a line make 4.5 V, because in series the voltages add. In Band 6 you will be handed a battery pack with four cells in a line, 1.5 volts each, six volts out. It is a series circuit with a lid on it, and when that six volts starts mattering very much, you will already know where it came from.

Done

What to keep

Your three predictions from the start of the lesson, and which ones were wrong
What happened with three LEDs in a line, in your own words
The two sentences: voltage is shared in series, current is shared in parallel, and 27 milliamps is more than a pin should give
Your answer to this: your Band 5 game has four lights and a buzzer. If they were all on at once, on four pins and one buzzer pin, is that a problem? Work it out.

The answer to that last one, once you have tried itFour lights at 13.6 mA each is 54 mA, but on four separate pins, so 13.6 each, which is fine. The limit is per pin, not per light. There is a second limit for the whole chip added together, around 200 mA, and four lights and a buzzer is nowhere near it. Knowing which limit applies is the skill, and it is the sort of question you can now answer instead of guess.

Ship it

This one is a measurement, so post the measurement. A photograph of your meter showing what your 220 ohm resistor actually reads, or your table with the predictions you got wrong still visible, is a better post than a finished build. Almost nobody publishes the part where they checked.

The five lines still work. What you measured, what you expected, the gap between them, the number itself, and what you will do differently now you know.

Show your work has the template. Tag it #BozomaBuilds.