Level 1 · Light · project 15 of 24 · 15 min · ●●○○○
Rainbow Loop
Red, yellow and green LEDs on three parallel 220 Ω lanes under one slide. One flip lights all three, each lane running its own Ohm's law.
new here? Start with First Light — one battery, one resistor, one LED. It teaches the loop, LED direction, and why the 220 Ω resistor matters.
What you need
- 1× 3×AA battery pack (4.5 V)
- 1× Solderless breadboard
- 7× Jumper wires
- 1× Slide switch the master gate — one flip arms every lane at once
- 3× 220 Ω resistor red-red-brown × 3 — one 220 Ω speed bump per lane, never a shared single resistor
- 1× Red LED (5 mm) the warm end of the rainbow — Vf ≈ 2.0 V
- 1× Yellow LED (5 mm) Vf ≈ 2.0 V — electrically a near-twin of the red
- 1× Green LED (5 mm) Vf ≈ 2.1 V — eats a slightly bigger voltage bite than the other two
you'll learn: Circuits are loops · Ohm's law · Series & parallel · Switches · LED polarity
Circuit map
The circuit you'll build, running on the Workbench right now.
simulating…
If it doesn't work
The boring ways this circuit actually breaks — check these first before you unwire everything.
- Only the red and yellow light; the green stays dark. Green LED is almost certainly flipped. Pull it and reinsert with the long leg toward the + rail, flat-side toward the − rail. If still dark, swap it with the red LED — if the red lane lights the green LED in the green position, your green resistor is fine but the LED itself is dead. Rare, but possible.
- All three LEDs light but one is clearly much dimmer than the other two. That lane has the wrong resistor. 220 Ω is red-red-brown. If one band reads red-red-red (= 2.2 kΩ), the lane is 10× undercurrent — a dim glow instead of a lamp. Swap for a real 220 Ω and the lane joins its siblings at 11 mA.
- All three LEDs are dark. The master slide is open, or the master or battery wiring is broken. Flip the slide, then trace from the battery + terminal through the master to the top rail — any loose jumper there kills every lane at once. The other direction (bottom rail → battery −) is just as critical; a loose cathode-side jumper also kills every lane.
- One LED glows but the other two are completely dark, even though all three resistors look right. The two dark lanes share a broken piece of + or − rail between them. Trace the top rail (row 0 jumpers from col 1 to col 6) and the bottom rail (row 3 jumpers) — if the one lit LED is between two breaks, the two dark LEDs are stranded. Reseat every rail jumper.
- LEDs all light but very faintly — like Night Pin, not First Light. Battery pack voltage is sagging below 4 V (dead or dying AAs), or the master slide has a bad contact (grimy, bent). Replace the batteries. If the master works intermittently when wiggled, the switch itself is dirty — swap it for a fresh one.
What you're building
A three-colour signal bar — one red, one yellow, one green LED standing side by side, each with its own 220 Ω resistor, each on its own lane straight to the battery, and one master slide switch that lights the whole row. Flip the slide: three colours fire together. Flip it back: three colours vanish together.
This is Three Sisters with the twins replaced by siblings. The topology is identical — three parallel lanes feeding one shared pair of rails — but swapping two red LEDs for a yellow and a green introduces something new to the circuit's physics: forward voltage is a property of the LED, not the lane. Red and yellow drink ~2.0 V each; green drinks ~2.1 V. That difference is small enough that the eye can't see it, but it's the first hint of why Twin Reds glows and Two Greens almost dies: in series, that extra 0.1 V eats a whole visible chunk of brightness; in parallel, it barely moves the needle.
The big idea: parallel lanes do their own math
Three parallel LEDs is one of the first circuits where you cannot think of the loop as a loop. There are now three loops stitched together at the rails:
- Loop 1: battery + → master → red lane 220 Ω → red LED → battery −
- Loop 2: battery + → master → yellow lane 220 Ω → yellow LED → battery −
- Loop 3: battery + → master → green lane 220 Ω → green LED → battery −
Each loop is a mini-circuit that follows Ohm's law on its own. The rails are shared, but the parts in each lane only know about their lane's current. The battery sees the sum.
I_total = I_red + I_yellow + I_green
That rule has a name — Kirchhoff's Current Law, or KCL — and it's the parallel sibling of the voltage-balancing rule you've been using since First Light. KCL says: whatever current flows into a shared point must flow out again, so the lanes just add up at the + rail. Think of a highway on-ramp — three one-car-wide lanes merging into one two-lane freeway. The freeway carries the sum.
Do the math — one lane at a time
The master slide carries no resistance when closed, so each lane sees the full 4.5 V of the battery. Each lane's current is its own little Ohm's-law problem:
Red lane (long leg into 220 Ω, cathode to −):
- Battery push: 4.5 V
- Red LED eats: ~2.0 V (datasheet Vf at ~10 mA)
- Leftover for the 220 Ω: 4.5 − 2.0 = 2.5 V
- Current: I = V ÷ R = 2.5 ÷ 220 ≈ 11.4 mA
Yellow lane — same math as red. Yellow 5 mm LEDs have essentially the same forward drop as red (both are AlGaInP semiconductors with similar bandgaps, around 2.0 V at 10 mA). So:
- Current: ≈ 11.4 mA
Green lane — green LEDs are a different semiconductor (gallium nitride, GaN). The bigger bandgap means a bigger Vf:
- Green LED eats: ~2.1 V
- Leftover for the 220 Ω: 4.5 − 2.1 = 2.4 V
- Current: I = 2.4 ÷ 220 ≈ 10.9 mA
Total battery current: 11.4 + 11.4 + 10.9 = 33.7 mA. The 3×AA pack supplies the sum effortlessly — AAs are rated for over 1 000 mA.
The subtle lesson: colour barely shows in parallel
Red: 11.4 mA. Green: 10.9 mA. That's a 5 % difference — the eye can't tell. Hold your thumb half over each LED and all three look equally bright.
Now compare Twin Reds and Two Greens — the series builds:
| Build | Topology | Red current | Green current | Difference |
|---|---|---|---|---|
| First Light / Rainbow lane | 1 LED + 220 Ω | 11.4 mA | 10.9 mA | 5 % — invisible |
| Twin Reds / Two Greens | 2 LEDs in series + 220 Ω | ~2.3 mA | <1 mA | 60 %+ — stark |
In parallel, each LED gets its OWN voltage headroom (4.5 V minus its own 2.0-ish V). Spare 2.4 to 2.5 V vs. 220 Ω → similar currents. In series, the LEDs stack their Vf and share the leftover (0.5 V for reds, 0.3 V for greens) → tiny currents, and the 0.1 V difference between red and green cuts the leftover in half. That's why Two Greens looks like a candle while Twin Reds looks dim-but-clearly-on.
The rule: colour matters a lot when the voltage headroom is tight (series stacks of LEDs, battery-sagging circuits). Colour barely matters when each LED has its own lane and plenty of headroom (parallel at 4.5 V with a 220 Ω each).
Where you'll see this pattern
- Status bars on real appliances. Printers, routers, dashboards — any time three LEDs light together under one power rail, they're on three parallel lanes with three resistors (one per LED, never shared, usually sized per colour to balance brightness).
- Traffic lights. Three colours, three independent lanes, three independent controllers. Rainbow Loop is a traffic light with the master "night" switch flipped on, firing all three at once (which never happens on the road but is useful during the yearly test).
- Theatre and photography lighting. RGB floodlights fire three colour arrays simultaneously, often from parallel buck-converter channels. The arrangement scales — change 220 Ω to 2.2 Ω and the LED to a 1 W Cree, and the diagram is identical.
- Christmas lights with mixed colours. Modern LED strings use parallel colour groups driven from one controller. Pulling one bulb kills just its lane. If pulling one kills the whole string, it's a cheap series design — and now you know why.
Build it
- Power column. Battery + at the top of column 1, − at the bottom. A jumper from the + terminal to the + rail.
- Master slide (col 1, top). Slide switch sitting vertically from the + rail to the next row — one side on the + rail, other side on the top horizontal rail that feeds all three lanes.
- Top rail (row 0 continuation). A jumper from the master's output across cols 2, 3, 4, 5 — this is the shared "+ after the switch" rail.
- Lane A — red (col 2). A 220 Ω (red-red-brown) from the top rail down to the LED row. Red 5 mm LED with the long leg (anode) in the resistor's output row, short leg (cathode) one row below. Jumper from the cathode row down to the bottom rail.
- Lane B — yellow (col 4). Repeat with the yellow LED. Same 220 Ω, same orientation, long-leg-up.
- Lane C — green (col 6). Repeat with the green LED. The green LED's cathode goes to the bottom rail just like the others.
- Bottom rail. A jumper runs under the LEDs along row 3, from col 1 to col 6, connecting every lane's cathode row and the battery − terminal.
- Flip the master on. All three LEDs light together — a red, a yellow and a green stripe. If one is dark, that one lane is broken (bad LED orientation, missing resistor, loose jumper). The other two will still work — that's the point of parallel lanes.
Troubleshoot by watching the dark lane
If one or two LEDs don't light but the others do, that lane is broken somewhere between the + rail and the − rail. The other lanes keep working, because they're independent. Trace the dark lane with your finger: + rail → 220 Ω → long leg → LED → short leg → − rail. Any break anywhere in that path kills just that lane.
| What you see | What it is | Fix |
|---|---|---|
| All three dark | Master slide is open, or the + / − rail jumper to the battery is loose | Flip the master; wiggle the battery jumpers |
| Red + yellow lit, green dark | Green LED flipped, or the green lane's 220 Ω is missing | Pull and reinsert the green LED long-leg-up; check the 220 Ω's bands |
| One LED is way brighter than the others | That lane's resistor is a smaller value by accident (100 Ω, 47 Ω) | Read the colour bands: red-red-brown = 220 Ω |
| All three glow but very faintly | The battery pack is dying (voltage sagging below 4 V); the master slide may be a push button being half-pressed | Fresh batteries; make sure the master is a slide, not a wobbly push |
🧠 Your challenge
No single right answer. That's the point.
- Add a fourth LED — any colour you have. Give it its own 220 Ω lane in parallel with the other three. What do you expect to happen to the other three when you add it? (Nothing: KCL guarantees each lane runs independently. The battery just supplies more total current — 44.6 mA instead of 33.7 mA. The battery pack can easily do this; AAs are rated for over 1 000 mA.)
- Pull the red LED's resistor and jumper the gap. What happens to the red LED? (It pops. 4.5 V − 2.0 V = 2.5 V across what is now essentially 0 Ω, drawing hundreds of mA. The LED's semiconductor junction overheats and shorts in a few seconds — don't actually do this; just notice that you're one jumper away from killing the LED. The resistor is a 25-mA-wide moat around a 20-mA LED.)
- Starve the green lane to match the red. Swap the green LED's 220 Ω for one that makes the green lane carry exactly the same current as red. (You want 10.9 mA matched to the 11.4 mA — a tiny difference. Required R = (4.5 − 2.1) ÷ 0.0114 = 210 Ω. Closest standard value: still 220 Ω. The lanes are already matched within a resistor's tolerance; this is why mass-produced status bars rarely bother sizing per-colour. For a bigger brightness difference — say, dim down a bright blue — you'd use 470 Ω or 1 kΩ on just that lane.)
- Make a half-dimmed lane. Add a second 220 Ω in series with the yellow lane's existing 220 Ω. Lane total is now 440 Ω. What's the current? (I = 2.5 ÷ 440 ≈ 5.7 mA — about half. The red and green lanes don't notice; their math never involved the yellow lane. Each lane is its own little universe.)
- Design the Christmas-tree-topper. Add a push button in series with just the red lane, so the whole row shows yellow + green by default, and the red "star" only fires when the button is held. What's the schematic? (Red lane becomes: + rail → push button → 220 Ω → red LED → − rail. The push button is a user-switch in series with just that lane; the master slide is still in series with the whole top rail. Two switches, two scopes.)
- What if the green LED is actually blue? Blue 5 mm LEDs have a Vf around 3.1 V. Does the blue lane still work on 4.5 V? (Just barely — I = (4.5 − 3.1) ÷ 220 ≈ 6.4 mA, visible but noticeably dimmer than red/yellow/green at 10+ mA. Blue is right at the edge of what 4.5 V can push, and it's why "RGB" LED strings often need 5 V or 6 V to balance brightness across colours. The physics is the same math.)
For grown-ups: safety notes
- Battery-safe at every lane count. Three parallel 220 Ω lanes draw ~33 mA total — about the current of a single bright LED, well under what a 3×AA pack can supply. Nothing in this build gets warm, and the battery lasts for the usual LED-circuit duration (hundreds of hours).
- Each LED needs its OWN 220 Ω. Never share a single 220 Ω across all three — that's the One for All failure where the LEDs dim-split the current and the green one goes dark because it eats more voltage than the others. The point of three parallel lanes is three parallel Ohm's-law problems, which only works if each lane has its own speed bump.
- LED polarity, three times. All three LEDs must sit long-leg-up (anode toward the 220 Ω, flat-side-down toward the − rail). A flipped LED stays dark forever — a current limiter with no LED current in it just has 2.5 V sitting across it, 0 mA. If one lane doesn't light, that LED is almost always flipped. The other two keep working, so the "something's wrong" signal is just the one dark LED.
- Resistor colour code reminder. 220 Ω is red-red-brown. It's easy to confuse with 2.2 kΩ (red-red-red — the third band is red instead of brown). A 2.2 kΩ on one lane drops current to ~1 mA — the lane will glow faintly like Night Pin, not die, but it'll look visibly dimmer than its siblings. If one LED is 10x dimmer than the others, suspect the resistor before suspecting the LED.
- Master slide is in series with ALL lanes. This is deliberate — one flip disarms everything. If you want each lane to have its own switch, add three slide switches (one per lane) and remove the master, like Step Up. If you want the master AND per-lane switches, you need both — the master as the overall gate, the per-lane switches as the lane selectors. Four switches, three lanes.
- Small parts stay away from pets and toddlers. LED legs, resistor leads and jumper tips are all the same swallow-hazard shape as every other breadboard part. Never wire this to mains.
Checked against
Keep building
You might also like
shares ideas with this one