🌡️ Warm Hand Alarm

What you're building

A red LED that stays dark in a normal room, but glows within a few seconds of you cupping your hand over the sensor. Take your hand away and it fades back off in about a minute. Nothing to press, no code. The sensor is a thermistor — a resistor that changes with temperature — and your body heat is enough to trigger it.

Once this works, you know the same sensing pattern behind every home thermostat, every 3D-printer hot-end controller, and every car engine temperature gauge.

Meet the thermistor

A thermistor ("THUR-mistor" — thermal resistor) is a small bead of a special metal-oxide material that changes its resistance with temperature. Yours is a 10 kΩ NTC thermistor:

• NTC = Negative Temperature Coefficient. Hotter → less resistance. (There are PTC thermistors that go the other way, but the ones you get in a beginner kit are NTC.)
• 10 kΩ at 25 °C — that's its "nameplate" value at room temperature.

Rough numbers from the typical NTC datasheet curve:

Temperature	Approx resistance
5 °C (fridge)	~25 kΩ
25 °C (room)	10 kΩ
40 °C (warm hand)	~3.6 kΩ
60 °C (hot mug)	~1.5 kΩ

Your body sits at 37 °C on the inside and your fingertips are usually around 30–35 °C. So palming the sensor pushes it maybe 5–15 degrees above the room — enough to drop its resistance by roughly half. That's what we're going to detect.

The voltage divider (from Sunrise Alarm — same trick)

You already know this pattern from Automatic Night Light and Sunrise Alarm: two resistors in a series stack, and the point between them sits at a voltage decided by the ratio of the two:

V_middle = 4.5 V × R_bottom ÷ (R_top + R_bottom)

Same as before. What's new is the sensor: this time the top resistor is the thermistor, and the bottom is a small 1 kΩ (not 10 kΩ like the night light). Here's why the size matters:

• Room, thermistor ≈ 10 kΩ: V_middle = 4.5 × (1 ÷ (10 + 1)) = 4.5 × 0.091 = 0.41 V. Below the transistor's 0.7 V wake-up line. LED off. ✓
• Warm hand, thermistor ≈ 3.6 kΩ: V_middle = 4.5 × (1 ÷ (3.6 + 1)) = 4.5 × 0.217 = 0.98 V. Above the wake-up line. LED comes on. ✓

If you'd used a 10 kΩ bottom (like the night light), the room-temp middle voltage would be 4.5 × (1/2) = 2.25 V — already way past the wake-up line, so the LED would be on all the time. The whole project would fail. The bottom resistor is a threshold knob: smaller bottom = harder to trigger, needs more heat. Pick it to match the sensor's swing.

Do the transistor math

Once V_middle rises above 0.7 V, the transistor's base starts to conduct. The 1 kΩ base resistor on the way in limits the base current politely — and you can solve exactly where things settle.

With warm-hand thermistor ≈ 3.6 kΩ, the base clamps at 0.7 V and KCL at the middle node gives V_middle ≈ 0.86 V (worked out from the divider plus the base's tiny drain), so:

• I_base = (0.86 − 0.7) ÷ 1 kΩ ≈ 0.16 mA
• With β ≈ 200 (from the 2N2222 datasheet), the transistor could push up to 31 mA through the collector — plenty for our LED.
• The LED lane wants: (4.5 − 2.0 − 0.2) ÷ 220 Ω = 10.5 mA (that's battery minus LED forward drop minus the transistor's saturation drop V_CE(sat), divided by the 220 Ω safety resistor).
• 31 mA available, 10.5 mA needed → transistor saturates and the LED glows at a comfortable ~11 mA. Same brightness as First Light.

At room temperature, base current is essentially zero and the LED draws only microamps of leakage. Battery drain at rest is tiny — you can leave this armed all afternoon and barely dent an AA.

Build it

1. Power rails: battery + to a top row on the breadboard, battery − to a bottom row. Same as always.
2. The divider (sensing half): battery + → thermistor → middle row → 1 kΩ → battery −. The thermistor has no polarity (it's just a resistor), so either leg goes to which side. Mount it so the bead is exposed to the air, not tucked under other parts.
3. The base resistor: middle row → 1 kΩ → transistor Base leg. This is the second 1 kΩ in your BOM — don't confuse it with the divider one.
4. The transistor: flat face toward you, legs left to right are Emitter, Base, Collector. Emitter → − rail.
5. The output lane: battery + → 220 Ω → LED long leg → LED short leg → Collector.
6. Power it up in a normal room. LED off. Now cup your palm over the thermistor for 5–10 seconds. Watch — the LED fades on. Take your hand away and it fades back off over 30–60 seconds as the bead cools.

The fade is the interesting part. Because it's a slow analog signal — no snap, no click — you can see the transistor working through its in-between region as your temperature swings past the 0.7 V line. Analog sensors show the change while it happens. Digital switches only flip states.

Debug like an engineer

Same idea as Automatic Night Light: test the halves separately. Here the divider's middle row only ever reaches about 1 V, which is too little to light a test LED, so use the other half as the probe instead. Pull the thermistor out and bridge its two holes with a plain wire: the middle row jumps to nearly 4.5 V, the base gets all the push it could want, and the LED lane should fire at once. If it does, the transistor lane is fine and your bug is on the sensing side (thermistor legs, the 1 kΩ at the bottom, the 1 kΩ into the base). If it doesn't, check E-B-C order first. With a multimeter you can go further: measure the middle row — about 0.4 V at room temperature, rising toward 1 V under a warm hand. Splitting the loop in half turns "nothing works" into "this part works, that part doesn't" — which is 90% of debugging.