Level 1 · Light · project 4 of 24 · 15 min · ●○○○○
Two Greens
Twin Reds with green LEDs — bigger voltage bites, less leftover for the resistor, and a great lesson in how far a 4.5 V battery can actually push.
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
- 4× Jumper wires
- 2× Green LED (5 mm) long leg toward + for BOTH — same direction, stacked end to end
- 1× 220 Ω resistor the same 220 Ω safety bump as First Light and Twin Reds
- 1× Slide switch
you'll learn: Circuits are loops · LED polarity · Ohm's law · Series & parallel
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.
- Both LEDs dark, batteries fresh. One green LED flipped. Series means one broken step kills the whole loop — pull each LED and try it the other way. The bump in the long leg is the anode; it goes toward +.
- Both LEDs dark, batteries older. A drooping 4.2 V pack leaves *nothing* for the resistor after two green bites. Twin Reds still works at 4.2 V; Two Greens does not. Swap in fresh AAs.
- Only one LED glows. Can't happen in a real series loop — if one lights, the other must too. Look again: is one leg touching a rail it shouldn't, forming a side lane? Confirm both green LEDs sit on the SAME column with no stray jumpers.
- LEDs snap on for a heartbeat then die. You dropped the 220 Ω or grabbed a jumper by mistake. Even with two greens stacked, they need a limiter — the LEDs' curve blocks most of the current, but the safety resistor is still the responsible move. Put it back before power on.
- Slide switch does nothing. Middle leg plus ONE outer leg — the other outer is dead in every position. Same as First Light and Twin Reds.
What you're building
The same series loop as Twin Reds — battery, switch, resistor, two LEDs in a line — but with green LEDs instead of red. Slide the switch and both glow together. Then compare them side by side with your Twin Reds: even dimmer, even faster to drop off if a battery droops. That gap is Kirchhoff's voltage law doing its thing.
The one-line change from Twin Reds
Red LEDs "eat" about 2.0 V each. Green LEDs eat about 2.1 V. Only a tenth of a volt more per LED — but this is a 4.5 V battery, and tenths of a volt matter when you're stacking two loads onto one loop.
- Battery push: 4.5 V
- Green LED #1 eats: ~2.1 V
- Green LED #2 eats: ~2.1 V
- Leftover for the resistor: 4.5 − 2.1 − 2.1 = 0.3 V
That 0.3 V is what Ohm's law has to divide by 220:
I = V ÷ R = 0.3 V ÷ 220 Ω ≈ 0.0014 A = 1.4 mA
Predicted current: 1.4 mA. Compared to Twin Reds' 2.3 mA, that's another 40 % of the light gone — because you spent another 0.2 V total on the LEDs and had less left over to push electrons through the resistor.
What actually happens (the LED cheats a little)
Put this on the Workbench and you'll see closer to 2 mA, not the 1.4 the math predicted. Why? A green LED's 2.1 V "bite" isn't fixed — it's the drop at the LED's happy 10 mA current. When we push less current, the LED helpfully lets a little less voltage bite too.
At 2 mA, each green LED gulps roughly 2.0 V instead of 2.1 V. Two greens = 4.0 V together, leaving 0.5 V for the 220 Ω, which pushes 2.3 mA — and the loop settles right there.
Real LEDs don't need the math to be perfect. The exponential curve inside them (its fancy name is the Shockley diode equation) makes them self-adjust. That's why you always see LEDs quoted as "Vf at some current" — the number depends on how hard you push.
Build it
If you have Twin Reds already breadboarded, this is a five-second swap: pull the two red LEDs, put two green ones in their place, done.
Fresh build:
- Battery + switch: battery + → top row, then the slide switch. Same start as First Light and Twin Reds.
- The 220 Ω: after the switch, drop the resistor along the top row.
- LED #1: long leg on the top row, short leg one row down. Green LED — you'll see it's noticeably taller-glassed than the red one.
- LED #2: long leg on that middle row, short leg on the bottom row. Both LEDs point the same way.
- Return: bottom row → battery −.
- Slide the switch. Both greens glow gently — dimmer than Twin Reds, which was already dimmer than First Light. That's three levels of "the math is running in real time on your retina."
Debug like a series builder
Same rules as Twin Reds: series loops are one road with no branches, so one broken step kills the whole chain. If either green LED is flipped, both go dark. Swap the suspect LED and, if you get light, you found the flipped one.
The one new failure mode: weak batteries kill this project faster than Twin Reds. Once your AAs droop from 4.5 V to about 4.2 V, the leftover for the resistor slips near zero and the LEDs struggle to reach visible brightness. Twin Reds coasts along on tired batteries; Two Greens is finicky. That's a real lesson about "headroom" — how much slack the math has before it gives up.
The takeaway
Every extra LED in series takes another ~2 V bite out of your supply. On a 4.5 V pack, two greens is right at the edge; a third green would want 6.3 V total and, on a fresh 4.5 V pack, nothing would light at all — the loop simply can't build up enough push to open even one diode's door. When designers want long chains of LEDs (a bike-light row, a strip of holiday lights), they either run them from a bigger supply or wire them in parallel lanes — each lane a short series stack, each with its own resistor. Same tools you already know.
🧠 Your challenge
No single right answer. That's the point.
- Three greens in a row. Add a third green LED in series before power-on. Predict first: 3 × 2.1 V = 6.3 V of LED bite, and your battery is only 4.5 V. The math says negative leftover. What should you see on the bench? (Answer: nothing. The pack can't push hard enough to open three diodes at once. The Workbench will show microamps of current — invisibly small.)
- One green + one red. Swap LED #1 back to red, leave LED #2 green. Redo the math: 4.5 − 2.0 − 2.1 = 0.4 V for the resistor → 1.8 mA. Both LEDs light, at almost identical brightness, because the same current flows through both — the color of the light differs, but the current is the single number for the whole loop.
- Weaker resistor? Try the 1 kΩ in place of the 220 Ω. Now the math says 0.3 V ÷ 1000 Ω = 0.3 mA — five times fainter than Two Greens' 2 mA. Do you notice a difference? At what current do green LEDs stop looking "on" at all? (Around 0.05 mA for the human eye, roughly — but this depends heavily on your room's brightness.)
- Add a knob. Slot a 10 kΩ potentiometer in series with the 220 Ω. Now you can dial the leftover voltage down toward zero and find the exact point where each LED gives up. That's the forward-voltage knee — the moment the diode's exponential curve bottoms out. You just built an LED-threshold experiment.
For grown-ups: safety notes
- Battery-safe, exactly like Twin Reds: 4.5 V and roughly 2 mA of loop current mean a fresh AA pack runs this for weeks.
- Same-direction rule for both greens. Long legs both point toward +, short legs both point toward −. A reversed LED anywhere breaks the whole loop; it won't hurt the LED, but nothing will glow until you spot the flip.
- Do NOT stack a third green. Nothing will burn, but you'll waste an evening wondering why nothing lights. The math above says why: two greens is right at the edge of what 4.5 V can drive.
- Weak batteries look identical to a broken circuit here. If Twin Reds still works but Two Greens doesn't, your AAs are tired, not your build. Swap the batteries first before you take the circuit apart.
- Small parts stay away from pets and toddlers — LEDs and slide-switch legs are the same swallow-hazard shape as any other breadboard piece.
Checked against
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