Level 2 · Sound · project 8 of 12 · 20 min · ●●○○○
Arm & Ring
A doorbell that ignores you unless a hidden slide switch is armed first — series-AND with sound, plus an 'armed' LED that shows the state.
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
- 6× Jumper wires L1
- 1× Slide switch the "arm" switch — hidden inside a box or under the desk L1
- 1× Push button (tactile) the "ring" button — the visible one anyone can press L1
- 1× Active buzzer (3–5 V)
- 1× Red LED (5 mm) the "armed" indicator — glows whenever the alarm can fire L1
- 1× 220 Ω resistor LED safety resistor L1
you'll learn: Circuits are loops · Switches · Series & parallel · Ohm's law · 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.
- Buzzer silent even when the button is held. Slide switch isn't armed. The buzzer only rings when BOTH the slide is closed AND the button is pressed — that's the whole point.
- 'Armed' LED never lights. Flip the LED (long leg toward +) and check the 220 Ω is in the LED lane. Then confirm the LED taps power AFTER the slide, not before.
- Buzzer rings with just the button, no slide needed. You wired the button in parallel with the arm switch. They should be in series: + → slide → button → buzzer → −.
- Buzzer stays silent even when armed and pressed. Buzzer polarity. The '+' mark on top faces the + rail; wired backwards it stays silent.
- Nothing at all. Battery pack switch first, then both battery leads reaching the rails.
What you're building
A doorbell with a safety catch. There's a slide switch (the arm) and a push button (the ring). The ring button alone does nothing — press it and you get silence. Only after you flip the arm switch does the ring button start working. Then, and only then, pressing the ring button sounds the buzzer.
An "armed" LED tells you which state you're in without having to guess. Slide off: LED dark, ring button dead. Slide on: LED glowing, ring button armed. Press ring: buzzer beeps.
This is The Vault rebuilt in the Sound level: two switches that both have to be closed before the loop can carry current. The extra trick this time is that one of the two switches also branches off to an LED that reports the state of the arm, without needing to press anything.
The big idea: series-AND with a status branch
Back in The Vault both switches sat in one straight line before the LED — the loop was only complete when both were closed. Here you're building the same series-AND, but with a twist: the arm switch sits at the top of two parallel lanes. Both lanes hang off the same "armed" node, and only the ring switch controls one of the two lanes.
- Lane 1 — the armed indicator. The LED (with its 220 Ω safety resistor) drops straight from the armed node to ground. Nothing else is in its lane, so it lights the instant the arm switch is on.
- Lane 2 — the ring lane. The push button lives inside this lane, in series with the buzzer. Even with the arm switch on, this lane stays open until you press the button.
The result: the LED reports whether the alarm is ready to fire; the buzzer only fires when someone actually presses ring. Two useful observations from one clean circuit.
Do the math (armed, no ring — the "ready" state)
The arm switch is closed, the ring button is up. Only the LED lane carries current:
- Battery push: 4.5 V
- Red LED forward drop: 2.0 V
- Voltage left for the 220 Ω: 4.5 − 2.0 = 2.5 V
- Current through the LED lane: 2.5 ÷ 220 ≈ 11 mA
Same 11 mA as First Light. The ring lane is broken open at the push button, so it draws zero. Total from the battery: 11 mA. A red LED at 11 mA is comfortably bright and the whole rig can sit armed for days on fresh AAs without a wobble.
Do the math (armed AND ring — the "beep" state)
Both switches are now closed. Both lanes carry current in parallel, each with its own voltage-drop story:
- LED lane: unchanged. Still 11 mA, still 2.5 V across 220 Ω.
- Buzzer lane: the active buzzer looks like a ~150 Ω load at 4.5 V (from its datasheet spec of about 30 mA at 5 V). Ohm's law: I = 4.5 ÷ 150 = 30 mA.
- Battery total: 11 + 30 = 41 mA.
Notice what did not change: adding the buzzer lane did not dim the LED. The two lanes are independent — a lesson you first met in Do Not Disturb and Whisper Alarm. Each lane sees the full 4.5 V across itself; each does its own math. The battery just has to supply the sum.
Do the math (disarmed)
The arm switch is open, so the "armed" node has no way to reach the battery's plus side. Both lanes are dead: LED dark, buzzer silent, no matter what happens to the ring button. Battery current is 0 mA. The alarm is fully off-line.
Where "arm-then-ring" shows up in the world
- Fire extinguishers have a pin you pull (arm) before you squeeze the handle (fire). Same series-AND — if you skip the pin, the handle does nothing.
- Fireworks and stage pyrotechnics use a key switch (arm) plus a fire button. The key stays in the operator's pocket while the wire runs to the fire button on a table.
- Missile-launch panels in movies really do have two-key arm-then-fire. It's a real pattern, not just cinema.
- Test buttons on smoke alarms are always after a battery-connect step — you can't test-fire a smoke alarm with the battery out.
Anywhere a designer wants "no accidental triggers", the arm-then-fire pattern is what they reach for.
🧠 Your challenge
No single right answer. That's the point.
- Two rings, one arm. Add a second push button in parallel with the ring button, so either button rings the buzzer once armed. You already know this move — it's the Panic Doorbell trick. Draw the change before you wire it. What does the armed-LED do while neither button is pressed? (Still lit — it doesn't care which button rings.)
- Silent-armed mode. Rewire so the armed LED is also on the ring lane (after the push button, before the buzzer). Now the LED only lights while someone is ringing, not while merely armed. Which of the two behaviours would a stealthy detective want?
- Two arms in series. Add your second push button as a second "arm" that must also be held down. That's three switches in series to sound the alarm — genuinely burglar-proof. What's the practical downside? (Someone has to keep a finger on both arm switches to hear the ring — a real system solves this with a latching relay, which you can build once you own transistors.)
- Trade the LED for a second buzzer. Move the LED and its 220 Ω aside and drop the second active buzzer into that lane. Now the moment you arm, you get a continuous "armed" tone as well as a ring-triggered one. What's the total battery current in that configuration? (30 + 30 = 60 mA — twice as loud in draw, not much louder in perceived volume.)
For grown-ups: safety notes
- Battery-safe end to end. 4.5 V won't shock anyone, and the worst case (armed + ringing) draws about 41 mA — well inside what fresh AAs deliver for hours on end.
- If the arm LED never lights, the slide switch is probably in the wrong slot. Slide switches close between adjacent pins on the breadboard — the third pin is a decoy. Trace the loop with a finger from + through the slide, across to the LED, down through the resistor, and back to −.
- If the ring button rings the buzzer even with the slide off, the ring lane is probably wired straight to the + rail instead of to the "armed" node between the slide switch and the two lanes. That means the arm switch isn't actually in the loop at all — go back to step 3 of the build.
- The active buzzer is loud at close range. Point it away from ears when you first arm and ring, and warn anyone nearby before you push the button.
- Never wire this to mains power. The whole site is battery-only for good reason; wall power is hundreds of times what these parts can survive.
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