Level 1 · Light · project 12 of 24 · 15 min · ●●○○○
Step Up
Three 1 kΩ lanes in parallel into one red LED, each with its own slide switch. One slide = candle, two = glow, three = full lamp. The lanes stack.
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
- 8× Jumper wires
- 3× Slide switch one per lane — each closes an independent parallel path to the LED
- 3× 1 kΩ resistor brown-black-red — three identical speed bumps, one per lane
- 1× Red LED (5 mm) the shared lamp — long leg toward the merge
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 — shown with the switches closed.
simulating…
If it doesn't work
The boring ways this circuit actually breaks — check these first before you unwire everything.
- All three slides closed and the LED is barely visible. A lane is probably not reaching the merge row. Open two of the three slides and leave only one closed; if that lane works solo (candle-dim glow), the other two have broken connections. Re-seat each resistor so its bottom lead lands squarely on the merge row.
- LED is bright with ONE slide closed and dark otherwise. The merge row is only connected to that one lane. You missed a merge jumper. Add row-2 jumpers from column 2 to column 3, column 3 to column 4, and column 4 to column 5 so every 1 kΩ ends on the same horizontal node.
- LED is very bright at one slide — brighter than it should be. That lane has a 220 Ω or no resistor at all. Pull the lane and check: brown-black-red is 1 kΩ, red-red-brown is 220 Ω, a bare wire is a short. Swap in the right 1 kΩ before anything pops.
- Zero slides closed and the LED is dim-but-visible. A slide is physically stuck closed, or a jumper is accidentally shorting around one slide. Pull all three slides out and see if the LED goes dark; if it stays lit, follow + rail to merge with your finger and find the sneak path.
- LED stays dark no matter which slide you close. Classic flipped LED. Pull it, rotate 180°, re-insert so the long leg sits on the merge row (row 2) and the short leg sits on the bottom rail (row 3). Confirm with the plastic rim: flat side faces −.
- LED gets dimmer as you add slides (opposite of expected). You wired the merge row to a resistor-side instead of to the LED anode. The lanes should ALL end on the LED anode row; the LED cathode goes to −. Trace from the LED up: if your finger passes through a resistor, you have the merge backwards. Rebuild with the long leg of the LED touching the row that holds the resistor bottoms.
What you're building
A single red LED with a three-position brightness ladder that you climb by flipping slide switches. No knob, no code — just three identical lanes wired side by side, each one adding its own trickle of current to the LED.
- One slide closed: candle-dim glow (~2.5 mA). Readable in a dark room.
- Two slides closed: reading-glow (~5 mA). Twice the current, visibly brighter.
- Three slides closed: full lamp (~7.5 mA). Three times the first step, still safely under the LED's 20 mA pop limit.
- Zero slides closed: off.
This is the opposite move from Twin Reds, where you stacked loads (two LEDs) in series on one lane and the brightness collapsed. Here you stack lanes (three identical resistor paths) in parallel, and the brightness climbs in evenly-spaced steps. Same parts bag, opposite physics.
The big idea: parallel resistors share the current — and lower the combined speed bump
Each slide-plus-resistor pair is a complete road from the battery + rail down to the LED's long leg. Three roads, meeting at one lamp. The current the LED sees is the sum of all currents the roads deliver.
Because the roads are identical — same 1 kΩ, same leftover voltage after the LED eats its share — each lane carries the same amount: 2.5 mA. With one road open, 2.5 mA reaches the LED. With two, 5 mA. With three, 7.5 mA. Linear, predictable, no surprises.
Flip that fact sideways and you've just discovered the parallel resistor rule: three 1 kΩ speed bumps side by side act like one much smaller speed bump. The formula for identical resistors in parallel is dead simple:
R_parallel = R ÷ n (where n is the number of equal resistors in parallel)
- One 1 kΩ: 1 000 Ω
- Two 1 kΩ in parallel: 500 Ω (half)
- Three 1 kΩ in parallel: 333 Ω (a third)
The current through the LED comes straight out of Ohm's law, just with the combined parallel R in place of a single resistor:
I = (V_battery − V_LED) ÷ R_parallel = (4.5 − 2.0) ÷ R_parallel = 2.5 V ÷ R_parallel
- One lane open: I = 2.5 ÷ 1 000 = 2.5 mA (candle)
- Two lanes open: I = 2.5 ÷ 500 = 5 mA (glow)
- Three lanes open: I = 2.5 ÷ 333 = 7.5 mA (lamp)
Each lane carries exactly 2.5 mA regardless of what the other lanes are doing — because the voltage across every lane is the same (4.5 V battery push minus the LED's 2.0 V bite = 2.5 V), and the resistor on each lane is 1 kΩ. The lanes stay in their own lanes. The current they all drop into the LED's anode is where the sum happens.
That's the whole project in one sentence: parallel resistors give you a smaller combined speed bump, and the LED sees the sum of the lane currents.
Why this is the opposite of Twin Reds
Twin Reds stacked two LEDs in series on one resistor lane. Each LED added its own 2 V bite to the voltage budget, and the leftover for the resistor shrank. The current dropped for each extra LED. Series stacks make the whole lane fight for less voltage, and brightness collapses.
Step Up stacks three resistors in parallel feeding one LED. Each lane adds its own 2.5 mA to the current the LED receives. Parallel stacks let the lanes work independently, and brightness climbs linearly with lane count.
The rule you can keep forever:
- Stacking in SERIES → voltages add → the component sees LESS of each piece of the budget → effect multiplies down
- Stacking in PARALLEL → currents add → the component sees MORE as you add paths → effect multiplies up
Every time you look at a circuit, ask "series or parallel" at every junction. The right answer unlocks the math.
Where you'll see this pattern
- A dimmer that works by counting switches, not by turning a knob. Elevator "number of floors" indicators sometimes work this way — each extra floor passed adds a lamp lane into the display's brightness, so the display is dim near the bottom and bright near the top. The lanes stack.
- Charger status LEDs that get brighter as more cells reach full. Each cell that finishes closes its own parallel path to the status lamp. You read the number of active lanes by looking at how bright it is.
- Multiple-source lighting on a shared bulb. A low-voltage halogen desk lamp with two independent AC adapters wired in parallel runs cool from one and brightens up when the second is plugged in. Same stacking math, scaled to mains-adjacent voltages.
- Resistor substitution. You need an "about-333 Ω" part for a one-off repair and your bag only has 1 kΩs. Three 1 kΩs in parallel = 333 Ω. The bag just grew a new value. Series does the same trick upward: two 1 kΩs in series = 2 kΩ.
Build it
- Power column. Battery + at the top of column 1, − at the bottom. A jumper from column 1 bottom to the breadboard's − rail.
- Top rail. Run a jumper along the top row (row 0) from column 1 across to column 4. This row is now at +.
- Lane A (column 2). From the top rail, drop a slide switch vertically into column 2 (its two leads one above the other). Below the slide, insert a 1 kΩ resistor (brown-black-red), the top lead on the row just below the slide, the bottom lead two rows further down. That bottom lead lands on your merge row — row 2.
- Lane B (column 3). Repeat exactly: top rail → slide → 1 kΩ → merge row.
- Lane C (column 4). Repeat again: top rail → slide → 1 kΩ → merge row.
- The merge. Run jumpers along row 2 from column 2 all the way across to column 5. Every 1 kΩ's bottom lead is now tied to the same horizontal node, which also carries over to column 5.
- The LED. At column 5, insert the red LED vertically with the long leg (anode) in row 2 (on the merge) and the short leg (cathode) two rows down in row 3.
- Bottom rail. Jumpers along row 3 from column 5 back to column 1, where the battery − waits.
- Flip the slides. Start with all three off — LED dark. Flip slide A — candle. Flip slide B — glow. Flip slide C — lamp.
Hold a hand around the LED in a dim room and the three brightness steps are obvious. In full room light the two upper steps look very similar; a dark room makes the ladder visible.
Troubleshoot by counting lanes
If the LED is dimmer than you expect, count how many slides are closed and compare to the table:
| Lanes closed | Expected current | How it looks |
|---|---|---|
| 0 | 0 mA | fully dark |
| 1 | 2.5 mA | candle-dim glow |
| 2 | 5 mA | reading-glow |
| 3 | 7.5 mA | full lamp (not First-Light bright — that'd be ~11 mA with a single 220 Ω) |
If one lane is closed but the LED is dark, that lane's slide is broken, that lane's resistor is in the wrong hole (not reaching the merge row), or that lane's merge jumper is missing. Open the other slides and test each lane solo — easy divide-and-conquer because the lanes are independent.
🧠 Your challenge
No single right answer. That's the point.
- Predict a fourth lane. If you added one more 1 kΩ + slide-switch lane to the row, what would the brightest-mode current become? (Four lanes × 2.5 mA each = 10 mA, with R_parallel = 1 000 ÷ 4 = 250 Ω and I = 2.5 ÷ 250 = 10 mA. Still safely under the 20 mA LED pop limit — you could keep adding lanes up to seven before you'd need to worry.)
- Mix the resistors. Swap one of the three 1 kΩ for a 220 Ω. Which lane does what now? (That lane alone carries I = 2.5 ÷ 220 ≈ 11.4 mA — First Light brightness. The other two still carry 2.5 mA each. With all three closed, the LED sees 11.4 + 2.5 + 2.5 ≈ 16 mA — nearly max. The lane with the smallest resistor dominates — this is the parallel-resistor flip side: the SMALLEST resistance wins, the opposite of series where the biggest resistance wins.)
- Compute the parallel of two different values. Mix one 1 kΩ and one 220 Ω in parallel (only two lanes). The combined resistance is given by the general formula: R_total = (R1 × R2) ÷ (R1 + R2) = (1 000 × 220) ÷ 1 220 ≈ 180 Ω. Current through the LED: I = 2.5 ÷ 180 = 13.9 mA. Check it against the sum of the two lanes: 2.5 mA + 11.4 mA = 13.9 mA. They match — because KCL says the LED's current is the sum of lane currents, no matter what. Two ways to think, same answer.
- Rebuild in series. Take the same three 1 kΩs and stack them end-to-end on one lane — one slide switch, three resistors, one LED, done. What's the current now? (Series: resistances add → 3 000 Ω total. I = 2.5 ÷ 3 000 ≈ 0.83 mA — well below the 1 mA "visible glow" threshold. The LED is almost dark. Parallel multiplied up to 7.5 mA; series dropped to 0.83 mA. Nine times brighter in parallel. Flip the arrangement, flip the physics.)
- Replace the LED with a second 1 kΩ and measure the merge voltage. If you had a multimeter, what voltage would you read on the merge row with all three lanes closed? (Three 1 kΩs in parallel = 333 Ω from +rail to merge. Then 1 kΩ from merge to −. The 333 Ω and 1 kΩ form a voltage divider: V_merge = 4.5 × 1 000 ÷ (333 + 1 000) = 4.5 × 0.75 ≈ 3.38 V. Voltage dividers aren't a Level 3 thing — they're everywhere, including here.)
- Design a 5-step dimmer. Using what you now know, sketch a circuit (don't build it) that uses five parallel lanes and gives five visibly-distinct brightness steps. What total parallel-R and current do you need at full brightness to still be LED-safe? (5 lanes × 1 kΩ = 200 Ω parallel, 2.5 ÷ 200 = 12.5 mA — still safe. Or 5 lanes × 2.2 kΩ ≈ 440 Ω parallel, giving ~5.7 mA at max for a very soft dimmer. The design choice is yours.)
For grown-ups: safety notes
- Battery-safe at every lane count. All three slides closed = 7.5 mA — well under the LED's 20 mA pop threshold. You'd need to add seven more 1 kΩ lanes before the LED would see its max current. Nothing in this build gets warm; if a resistor feels warm you've probably built a short that bypasses one of them.
- Each lane gets its OWN 1 kΩ. The temptation when you have three slides is "I'll just share one resistor on the output side." That's the One for All mistake, scaled to switches instead of LEDs. One shared resistor means every lane's current fights for the same bottleneck — you'd lose the clean "+ 2.5 mA per lane" lesson, and the brightness jumps would get mushy. Keep one resistor per lane.
- Red LED polarity. Long leg (anode) at the top on the merge row; flat side (cathode) at the bottom on the ground rail. A flipped LED means no current no matter how many slides you close — a classic L1 miss. The LED can't tell "three lanes" from "zero lanes" if it's facing the wrong way.
- 1 kΩ colour code is brown-black-red. It's easy to confuse with 10 kΩ (brown-black-orange) or 100 Ω (brown-black-brown). A 10 kΩ ends up in a lane means ten times less current on that lane (0.25 mA — invisible). A 100 Ω ends up in a lane means ten times more current on that lane (25 mA — pops the LED immediately when that slide closes). Read the colours twice; if you're not sure, keep the resistor out and measure with a continuity tester or multimeter.
- Slide switches don't care which way they go. Unlike LEDs, slide switches work the same whichever lead you call "input" vs "output." All three slides are interchangeable; swap any pair and nothing changes.
- Small parts stay away from pets and toddlers. Switches, resistor leads and LED legs are the same swallow-hazard shape as every other breadboard piece. Never wire this to mains — a 3×AA pack is the whole supply and keeps the parallel-R math safely bounded.
Checked against
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