Level 1 · Light · project 11 of 24 · 15 min · ●●○○○
One for All
Three LEDs on ONE shared 220 Ω — the shortcut that fails. Red and yellow dim-split the current; green sits dark. Lesson: each LED needs its own resistor.
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 arm switch — one for the whole bank
- 1× 220 Ω resistor the shared throttle — the whole point of this build is that ONE resistor is NOT enough for three LEDs
- 1× Red LED (5 mm)
- 1× Yellow LED (5 mm)
- 1× Green LED (5 mm) the "losing" lane — its slightly higher forward voltage keeps it starved
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.
- The green LED is dark. That's the whole project. Green's forward voltage is 0.1 V higher than red/yellow, so when they share one resistor red and yellow hog the current and green gets starved. Not a fault — it's the lesson. Build *Three Sisters* with the same three LEDs to confirm the green works fine in its own lane.
- Red and yellow are dark too. Then it IS a wiring fault. Check the master slide is ON, the battery pack switch is ON, the 220 Ω connects master output to the shared LED row, and the LED cathodes reach the − rail.
- All three LEDs are very bright (close to First Light brightness). You accidentally wired per-lane resistors instead of ONE shared one. That's *Three Sisters* — great, build this one again with just one 220 Ω feeding all three LEDs.
- One or more LEDs are completely dark but the others look First-Light bright. The dark one is flipped (long leg must be at the shared node, flat side at −). But also: if the OTHERS are at First-Light brightness ~11 mA, you probably have per-lane 220 Ωs — see above.
- Red and yellow glow at obviously different brightnesses. They have slightly different Vf — exactly the hogging effect, scaled down. Try with two reds from the same bag: the brighter one has the lowest Vf. Swap positions to confirm the hog follows the bulb, not the wire.
- The 220 Ω feels warm. Normal if ambient is high — it's dissipating V² ÷ R = (4.5 − 2.0)² ÷ 220 ≈ 28 mW, which is tiny but on a thin lead can be slightly warm. If it feels HOT (uncomfortable to touch), you accidentally bypassed one of the LEDs and the resistor is now across 4.5 V directly — pull power and trace the LEDs.
What you're building
Three LEDs — red, yellow, green — hanging in parallel from a single 220 Ω resistor. One slide switch arms the whole thing. Flip it on and… you see red and yellow glowing visibly dimmer than First Light, and the green LED completely dark. All three LEDs are electrically "on". Only two of them light up.
This is a failure demo on purpose. The project exists so you can watch, with your own eyes, why the "I'll just share one resistor between the LEDs" instinct is wrong — and understand the physics well enough that you never do it again. Its sibling Three Sisters is the right-way-around version; build both and compare.
The big idea: parallel LEDs through a shared resistor fight each other
In Three Sisters, each LED had its own 220 Ω. Each lane ran the Ohm's-law equation independently and each LED got 11 mA, no questions asked. The lanes were electrically separate branches with nothing to argue about.
Here, there is no "lane C's resistor" and "lane A's resistor" because there's only one resistor to argue over. The three LEDs sit on the same node — their anodes are tied together, their cathodes are tied together — and that single resistor hands them whatever current the Ohm's-law bottleneck allows. The current they receive is divided by their own physics, not by any wire pattern you chose.
The division is not fair.
Do the math — the hogging effect
Call the voltage at the LED node V. The three LEDs are all in parallel across the battery − rail, so each one has V volts across it. The current the resistor supplies is (4.5 − V) ÷ 220. Each LED draws its own current based on how far V is above its own forward voltage.
Red and yellow LEDs both have a forward voltage of about 2.0 V at normal currents. Green has about 2.1 V. That tiny 0.1 V difference is the whole story.
- If V is around 2.0 V: red and yellow are happily conducting ~5 mA each. Green needs ~2.1 V to conduct any current worth seeing — at 2.0 V it's basically off (its current is 15× smaller because of its lower reverse-saturation current).
- The system settles at V ≈ 1.97 V: resistor supplies (4.5 − 1.97) ÷ 220 = 11.5 mA. Red takes about 5.3 mA, yellow another 5.3 mA, green a mere 0.8 mA.
Here are the brightnesses in a single picture:
- Red: ~5 mA — about half of First Light. A visible, soft glow.
- Yellow: ~5 mA — same brightness as red, same reason.
- Green: ~0.8 mA — well below the 1 mA threshold where a 5 mm LED starts looking "on" in daylight. In a dark room, maybe a hint of green. In a lit room, you'd swear it was dead.
Two conclusions, both useful:
- The resistor is the bottleneck. The LED node settles wherever makes the arithmetic work: resistor current in equals total LED current out. 11.5 mA splits unevenly across three LEDs with slightly different "resistance to turning on". The fairness depends on how similar the LEDs are — identical ones split roughly equally, mismatched ones don't.
- A tiny Vf difference becomes a huge brightness difference. The exponential diode equation is vicious. A 100 mV shift in forward voltage can be the difference between 5 mA and 0.5 mA — a 10× brightness ratio. You cannot win this fight by picking "similar enough" LEDs.
Even three identical LEDs misbehave in real life
The sim above shows the mismatched-color case because the mismatch is big and the lesson is loud. But even if you build this with three LEDs of the same colour from the same bag, you'd see the same hogging pattern — just subtler, and sadly less predictable.
Why: real LEDs have a manufacturing spread in their forward voltage of about ±50 mV. The one with the lowest Vf in your batch becomes the hog; its brighter current heats it slightly, which lowers its Vf further, which widens the gap. The feedback is positive, not negative. Give it enough current and the hogging LED actually burns out first.
Datasheets for LED strings always say: "Do not parallel LEDs without individual current-limiting resistors." You just saw why.
The fix: one resistor per LED
Build Three Sisters next, with the same three LEDs — three individual 220 Ωs. Watch all three light up at 11 mA each, red and yellow and green side by side, no fighting. The two projects form a before/after pair. Keep both wired and alternate between them until the lesson is wordless.
Build it
- Power column. Battery + at the top of column 1, − at the bottom. A slide switch at the very top of column 1 — this is your master.
- The one resistor. From the master's output, drop a 220 Ω straight down to a free row — this row becomes the shared LED node. Mark it mentally; everything LED touches this row.
- Three LEDs across the shared node. Insert the red LED long-leg into the shared node, short-leg into a lower row. Insert the yellow LED the same way, two or three columns over. Insert the green LED the same way, in a third column. All three anodes land on the same shared row; all three cathodes land on the breadboard's − rail.
- Bottom rail. A jumper from each LED's short-leg hole back to the breadboard's − power rail; another from that rail to the battery − terminal.
- Flip the master slide ON. Watch what happens. Red glows softly. Yellow glows softly. Green stays dark (or just barely visibly on if you cup your hand around it in a dim room).
- Pull the green LED out. Red and yellow brighten slightly — now there are only two mouths to feed from the same 11.5 mA. (New split: ~5.7 mA each, still noticeably dim.)
- Pull the yellow LED out too. Red brightens again — it's now alone, running the full 11.5 mA, the same current it would carry in First Light. The resistor was always going to hand out 11.5 mA; the only question was how many LEDs had to share it.
Where this bites in the real world
- "Why is one LED on my string light dimmer than the others?" You bought a cheap LED decoration where the manufacturer parallels LEDs without per-device resistors. The dim one has the highest forward voltage in that batch; it's starving. In expensive strings, every LED (or small series group) has its own resistor and they all match.
- "My DIY LED torch pops after a while." You tied two or three 5 mm LEDs in parallel through a single resistor to make it brighter. The lowest-Vf LED hogged current, heated up, lowered its Vf further, and ran away. Three LEDs with individual resistors gives the same brightness and lives forever.
- Interview question: "Can you parallel LEDs?" The textbook answer is "only with per-LED resistors." This project is the demo you can describe when you give that answer.
One resistor can only control one LED's current. Memorize that; it's a Level 1 rule that reappears every level after.
🧠 Your challenge
No single right answer. That's the point.
- Replace all three LEDs with reds from the same bag. Measure the dim-and-brighter spread — some reds will clearly outshine others when wired this way. Can you rank them by Vf without a meter? (Yes: the hog is the lowest-Vf LED. Mark it with tape; try again with two different reds — the "stronger" one wins.)
- Short the green LED's cathode to its anode (bypass it). Now the shared node only has red and yellow fighting. Each should jump from ~5 mA to ~5.75 mA. Can you see the difference? (Barely — the increase is about 15%, right at the edge of what the eye picks up. But the battery current rises from 11.5 to 11.5 mA basically unchanged — the resistor was the bottleneck, not the LEDs.)
- Fix one lane. Add a second 220 Ω in series with the green LED only — so now green has 220 Ω → green LED, while red and yellow still share the original 220 Ω. Does green light up now? (No — green still sees a 2.0 V node from red+yellow via the original path, which isn't enough to turn green on. The real fix is to give green its own 220 Ω and its OWN top-rail connection, not share with the others. That's Three Sisters again.)
- Try the right-way-around. Build Three Sisters: same three LEDs, three separate 220 Ωs, each in its own lane straight from the + rail. Confirm all three light at ~11 mA. The side-by-side comparison is the whole lesson.
- Design a 10-LED bar graph. Using what you now know, sketch (don't build) a circuit that lights 10 LEDs from a single battery. How many resistors do you need? (Ten — one per LED. The battery supplies 10 × 11 mA = 110 mA, which the 3×AA pack handles cool.)
For grown-ups: safety notes
- Battery-safe. The 220 Ω still throttles the total current to ~11.5 mA no matter how many LEDs you dangle off its output. The build never gets near the LED's 20 mA pop threshold because the one resistor hard-limits the whole bank. (This is the one thing the shared-resistor trick does right: it prevents LED damage. It just doesn't achieve equal brightness.)
- The green LED looking dead is NOT a wiring fault. New builders will flip the green LED, swap it, touch it, pull it, swear it's broken. It isn't — it's starving on 0.8 mA. Confirm by moving the green to a working First Light with its own 220 Ω: it lights perfectly there.
- LED polarity still matters per lane. Even in the dim/dark soup, the LEDs are directional: long leg up (toward the shared node), flat side down (toward −). A flipped LED makes that lane go completely dark — different from the "starved by hogging" dark. Pull the suspect LED and reinsert.
- Resistor colour code is red-red-brown (220 Ω). If you grabbed a 220 kΩ (red-red-yellow) by mistake, the entire bank would be too dim to see — about 11 µA total. Easy to debug: the whole bank is dark, not just the green.
- Do not add more LEDs to prove the lesson gets worse. The dim-and-dimmer effect compounds — four LEDs through one resistor each get ~3 mA on average, five get ~2 mA. It stays safe (because the resistor still clamps the total), but the lesson is already landed with three.
- Small parts stay away from pets and toddlers. Never wire this to mains.
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