Level 2 · Sound · project 19 of 19 · 15 min · ●●○○○
Shared Knob
One pot sits above the merge of two parallel LED lanes (red + yellow). Twist it and BOTH LEDs fade together — the knob throttles the shared rail.
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) L1
- 1× Solderless breadboard L1
- 7× Jumper wires L1
- 1× Slide switch the master gate — on means the knob is live L1
- 1× 10 kΩ potentiometer the shared knob — sits on the top rail, upstream of the merge where both lanes tap in
- 2× 220 Ω resistor one per LED lane — same per-lane safety speed bump as Three Sisters L1
- 1× Red LED (5 mm) L1
- 1× Yellow LED (5 mm) same Vf as the red (2.0 V) — the two lanes track each other exactly L1
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 one LED lights at any knob position. The other LED is reversed — long leg must face + (top rail). Flip it. Both lanes must be identical in orientation: 220 Ω toward +, LED long leg toward 220 Ω, short leg toward − rail.
- Both LEDs stay dark at every knob position with the slide closed. Battery pack switch off, battery dead, or the slide switch itself is open. Trace from + → slide → pot → merge; the slide must be the only gate.
- Twisting the knob changes brightness, but only the red LED is affected. The pot is wired INSIDE the red lane instead of on the top rail. Move it to the shared + rail above the lane split so both lanes drink from its output.
- LEDs are already dim at knob = 0 and get darker from there. A 220 Ω is in the wrong place — probably you accidentally put a 1 kΩ or 10 kΩ instead of 220 Ω in one or both lanes. Check the colour bands: brown-black-red for 220 Ω.
- Battery pack feels warm after a few minutes. Likely a short — probably the slide switch straddles + and − instead of sitting between + and the pot. Open the slide and check the trace: + → slide → pot → merge → LED lanes → −, no shortcut.
- Both LEDs barely glow even at knob = 0. Pot is wired as a 3-leg voltage divider instead of a 2-leg rheostat. Only the wiper (middle) and ONE end leg should be connected into the circuit; the other end leg hangs free.
What you're building
Two LEDs — one red, one yellow — sitting side by side in parallel, with their own 220 Ω each. That part is already familiar from Three Sisters: two independent lanes, same voltage, each running at about 11 mA.
The new thing is where the pot sits. The 10 kΩ knob isn't inside the red lane or the yellow lane. It sits on the top rail before either lane branches off — above the point where the + rail splits in two.
Twist the knob and both LEDs fade together. Twist it back and both come up together. One hand, two LEDs, same story. The knob is a shared throttle.
The big idea: a part on the rail above the split affects everything downstream
Up until now, every pot you've met has lived inside one lane:
- Dial-a-Brightness — pot in a single LED lane. Dims one LED.
- Dial-a-Buzz — pot in a single buzzer lane. Dims one buzzer.
- Whisper Alarm — pot inside the buzzer lane of a doorbell, so the LED stays constant while only the buzzer gets quieter.
The lesson there is lane independence: a part inside a lane only touches that lane.
Flip that around. Put the pot above the point where two lanes split, and the opposite is true: every lane downstream gets the throttle, equally, because every lane's current has to pass through the pot first on its way from + to −.
This is the pattern you see in a real dimmer switch for a ceiling fixture that drives several bulbs: one knob at the wall, every bulb dimming together. Same idea here, in miniature.
Why red and yellow track each other exactly
Both LEDs in the kit's red and yellow 5 mm parts share the same forward voltage of ~2.0 V — that's the small bite each one takes out of the loop before it lets current through. (Green is slightly higher at 2.1 V, which would throw off the symmetry — that's why this project pairs red and yellow.) Same forward voltage, same 220 Ω per lane, same rail above them: whatever current the circuit picks for the red lane, it picks the same for the yellow. They're twins drinking from one tap.
Do the math — knob at the low end (pot ≈ 0 Ω)
A pot "at the low end" is just a wire. Each lane sees the full battery.
- Red lane. 4.5 V − 2.0 V red Vf = 2.5 V across 220 Ω → I_red = 2.5 ÷ 220 ≈ 11.4 mA.
- Yellow lane. Identical arithmetic. I_yellow ≈ 11.4 mA.
- Battery total: 11.4 + 11.4 ≈ 23 mA. Both LEDs at full First-Light brightness.
The pot being at 0 Ω means it's doing nothing. The two LEDs behave exactly like a two-lane Three Sisters.
Do the math — knob a tenth of the way (pot = 1 kΩ)
Now the pot is a real speed bump sitting above the split. Here's the setup: call the shared rail voltage downstream of the pot V_rail. Each lane carries (V_rail − 2.0) ÷ 220 of current, so the total current through the pot is twice that. Set up Kirchhoff's voltage law — battery = pot drop + rail — and solve:
- V_rail × (1 + 2·pot ÷ 220) = 4.5 + 4.0 × pot ÷ 220
- At pot = 1 kΩ: V_rail = (4.5 + 18.2) ÷ (1 + 9.09) = 2.25 V.
- I per lane ≈ (2.25 − 2.0) ÷ 220 ≈ 1.1 mA. Dim but clearly visible.
- Battery total ≈ 2.2 mA.
Both LEDs lose brightness together. The red doesn't stay bright while the yellow dims, or vice versa — because they both drink from the same rail downstream of the pot.
Do the math — knob at the high end (pot = 10 kΩ)
Same analysis, bigger pot.
- V_rail = (4.5 + 182) ÷ (1 + 90.9) = 2.03 V — the LEDs are just barely over their 2.0 V knee.
- I per lane ≈ (2.03 − 2.0) ÷ 220 ≈ 0.14 mA. A pinprick glow — Night Pin territory from Level 1.
- Battery total ≈ 0.28 mA. A fresh AA pack (~2500 mAh) could sustain this for over a year without noticeably draining.
Here's the kicker: the LEDs never go fully off, even with the pot cranked all the way. There's always some current, because 10 kΩ in front of two 220 Ω × LED lanes still lets a sliver of a milliamp through. If you want "off", open the slide.
Why the two LEDs always track each other
Two LEDs in two separate lanes, both tapping the same rail voltage, both with the same 220 Ω and the same forward-voltage knee — the arithmetic for each lane is identical. Whatever V_rail is at this moment, both lanes convert it the same way into current.
That's lane independence running underneath lane dependence: each lane makes its own tiny math problem out of the rail voltage, but the rail voltage itself is shared, so turning the knob moves both lanes in lock-step.
If you wanted the two LEDs to fade independently, you'd put two separate pots inside the two lanes — one knob per LED.
Where "shared knob" shows up
- Household dimmer switches that drive multiple bulbs on one line all do this — one wall knob, every bulb tracking.
- Stage theatre fader rows have the same architecture: one fader throttles the whole bank of lights hanging off its channel, every bulb on that bank changing together.
- Thermostat dimmers in buildings turn a whole row of LED down-lights up and down with one trim pot upstream.
Learning to read "where does the knob sit?" is a key electronics skill, and this project is the smallest-possible place to practise it.
🧠 Your challenge
No single right answer. That's the point.
- Swap the yellow LED for a green one. Predict what you'll see as you twist. (The pair stops tracking exactly, because green's Vf is 2.1 V — a hair higher knee. At the low end of the knob, green runs a bit dimmer than red; at the high end, V_rail drops close to 2.0 V and the green starves out while the red keeps a faint glow. Same wiring, different physics — the forward-voltage mismatch breaks the symmetry.)
- Add a third parallel lane (any colour LED, its own 220 Ω). Predict how all three behave when you twist. (All three track together, slightly dimmer than before at any knob position because the shared pot now has three lanes drinking from it instead of two — V_rail at any pot position drops a little. The idea generalises to any number of lanes.)
- Swap one 220 Ω for a 1 kΩ in just one of the two lanes. Twist the knob. Does the pair still track? (Yes — both LEDs still dim together when the knob moves, because V_rail is still shared. But the lane with the 1 kΩ is now permanently dimmer than the other at every knob position. The ratio between the two LEDs stays the same as you twist — only the overall brightness changes.)
- Move the pot INSIDE the red lane only — put it between the merge and the red LED's 220 Ω. Now twist. (The red fades; the yellow stays constant. You just rebuilt Whisper Alarm's "pot-in-one-lane" wiring with two LEDs instead of LED-and-buzzer. "A part inside a lane only touches that lane.")
- Can the knob kill both LEDs completely? Try it. (No — the pot's max is 10 kΩ, which still lets ~0.14 mA through each lane. For hard-off you need the slide. This is why real appliances with dimmer knobs usually have a separate on/off switch: a dimmer alone can never be fully off.)
- Guess what happens with a 100 Ω fixed resistor instead of the pot. (Each lane at ~10 mA — nearly as bright as a plain two-lane Three Sisters. The upstream resistor would steal only about 10 mA × 100 Ω = 1 V, dropping V_rail from 4.5 to ~3.5 V — barely enough to notice.)
For grown-ups: safety notes
- Battery-safe end to end. Max current with the knob at the low end is about 23 mA at 4.5 V — a 100 mW load, trivially inside what three AAs deliver all day without warming up.
- The 10 kΩ pot never runs hot in this circuit. Even at its hottest (knob at 0 Ω, carrying 23 mA), the pot has 0 V across it so dissipates nothing. At 10 kΩ carrying 0.28 mA, it drops ~2.5 V × 0.00028 A = 0.7 mW. Standard pots handle 125 mW+ — this is 200× under spec.
- The two 220 Ω each carry at most 11 mA = 27 mW. A standard quarter-watt resistor is rated 250 mW. If either resistor feels warm to the touch, something is miswired — probably one of the LEDs is bypassed somewhere and the resistor is sinking the full battery current.
- LED polarity doubles up here, one per lane. A reversed red stays dark at every knob position (the yellow still fades normally). A reversed yellow does the same from the other side. If only one LED tracks the knob and the other is always dark, polarity is the first suspect.
- Three-leg pot caution: a potentiometer has three legs. In this project
you're using two of them — the wiper (middle) and one of the end
terminals, wired as a variable resistor. The third leg hangs free. If
you accidentally include all three (wiper to one rail, both end legs to
- and −), you've built a voltage divider — a different wiring that drains the pack at ~0.5 mA no matter the knob position. Not dangerous, just not what this project does.
- Never wire this to mains. Battery-only, every project on this site, no exceptions.
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