555 Astable Oscillator
Difficulty: Beginner | Time: 30 minutes | Cost: < $2
Background
The 555 timer is the most-produced integrated circuit in history. Billions are manufactured every year, over fifty years after it was first designed, and the chip in its original 8-pin package still costs less than a cup of coffee.
Its origin is a story of elegant constraint. In 1970, Signetics Corporation asked freelance engineer Hans Camenzind to design a new timing IC. Camenzind was given an unusual brief: he could use no more than twenty transistors. The number-of-components limit was a manufacturing cost target, not a design suggestion, but it shaped everything. Working alone over the better part of a year, Camenzind developed a circuit that fit the constraint while covering an astonishing range of timing tasks. He submitted his design in 1971 and it was released as the NE555 — the name coming from the three 5 kΩ resistors forming its internal voltage divider.
The chip was an immediate hit. Applications appeared faster than Signetics could catalog them: precision timers, pulse generators, missing-pulse detectors, switch bouncers, motor speed controllers, tone generators, and simple analog-to-digital converters. Within a few years it was standard inventory in every electronics lab. Camenzind designed an improved CMOS version in 1994 (the ICM7555), which draws a thousand times less current and is still in production. He later wrote what remains one of the best accounts of practical analog IC design, partly as a record of what went into that original chip.
The circuit in this project — the astable multivibrator — is the 555’s simplest and most satisfying configuration. It runs without any external trigger, producing a square wave whose frequency you control by choosing two resistors and a capacitor. With a 10 µF capacitor it blinks an LED at a leisurely human-visible rate; swap in a 100 nF capacitor and the same circuit produces an audible tone. The only change is one component.
Components
| Qty | Part | Value / Notes |
|---|---|---|
| 1 | NE555 timer IC | or LM555, TLC555, ICM7555 — all pin-compatible |
| 1 | Resistor R1 | 1 kΩ (brown–black–red) |
| 1 | Resistor R2 | 47 kΩ (yellow–violet–orange) |
| 1 | Capacitor C1 | 10 µF electrolytic — observe polarity |
| 1 | Capacitor C2 | 10 nF (0.01 µF) ceramic — any type |
| 1 | Resistor R3 | 470 Ω (yellow–violet–brown) — LED current limiter |
| 1 | LED | any colour |
| 1 | 9 V battery + clip connector | |
| 1 | Breadboard | half-size (400 tie points) or larger |
| — | Jumper wires |
For the audio variant (Part 2): swap C1 for a 100 nF ceramic capacitor and connect a small 8 Ω speaker or piezo buzzer in place of (or in parallel with) the LED + R3.
Circuit
+9V
│
[R1 1k]
│
├──── pin 7 (DISCHARGE)
│
[R2 47k]
│
├──── pin 2 (TRIGGER)
│ pin 6 (THRESHOLD)
[C1]
│
GND
pin 8 (VCC) → +9V
pin 1 (GND) → GND
pin 4 (RESET) → +9V (disables reset — tie high)
pin 5 (CONTROL) → C2 (10nF) → GND (noise bypass)
pin 3 (OUTPUT) → [R3 470Ω] → [LED] → GNDThe 555 is a DIP-8 package. Pin 1 is at the notch end, left side, and numbers run anti-clockwise when the chip is viewed from above.
Assembly
Place the 555 straddling the breadboard centre gap so each pin lands in its own row. The notch (or dot) marks pin 1.
Power rails: connect the top rail to +9 V and the bottom rail to GND. Use a jumper from pin 8 to +9 V, and from pin 1 to GND.
Reset and control: jumper pin 4 to +9 V. Connect C2 (10 nF ceramic) between pin 5 and GND — orientation does not matter for ceramics.
Timing network:
- R1 (1 kΩ) between +9 V and pin 7.
- R2 (47 kΩ) between pin 7 and the junction of pins 2 and 6.
- C1 (10 µF electrolytic) between that junction and GND. The longer leg of the capacitor (positive) goes to the junction; the shorter leg (negative, often marked with a stripe) goes to GND.
- Jumper pins 2 and 6 together if they are not already in the same row.
Output LED: from pin 3, place R3 (470 Ω) in series, then the LED (longer leg = anode toward R3), then to GND.
Connect the battery. The LED should begin blinking immediately at roughly 1–2 Hz. If it does not, check the C1 polarity and confirm pins 2 and 6 are connected.
How It Works
Internal structure
The 555 contains three 5 kΩ resistors in series between VCC and GND, forming a voltage divider that sets two reference levels: ⅓ VCC and ⅔ VCC. These feed two comparators:
- The lower comparator monitors pin 2 (TRIGGER): if the voltage there falls below ⅓ VCC, its output sets an internal SR flip-flop, driving pin 3 (OUTPUT) high and turning off an internal discharge transistor connected to pin 7.
- The upper comparator monitors pin 6 (THRESHOLD): if the voltage there rises above ⅔ VCC, its output resets the flip-flop, driving OUTPUT low and turning the discharge transistor on.
Pin 4 (RESET) overrides everything and forces OUTPUT low when pulled to GND — we tie it high to disable this.
The oscillation cycle
With the circuit assembled and power applied, trace one complete cycle starting from a discharged capacitor:
Phase 1 — charging (OUTPUT high): C1 is discharged; voltage at pins 2/6 is near zero, below ⅓ VCC. The lower comparator fires, setting the flip-flop: OUTPUT goes high, the discharge transistor turns off. Current now flows from +9 V through R1 and R2 into C1, charging it toward VCC. The LED lights.
Phase 2 — threshold reached: When the C1 voltage reaches ⅔ VCC, the upper comparator fires, resetting the flip-flop: OUTPUT goes low, the discharge transistor turns on. The LED goes off.
Phase 3 — discharging (OUTPUT low): With the discharge transistor on, C1 now drains through R2 into pin 7 (to GND internally). R1 is bypassed — current cannot flow backward through it into VCC. The voltage falls.
Phase 4 — trigger level reached: When the C1 voltage falls to ⅓ VCC, the lower comparator fires again, setting the flip-flop — and we are back to Phase 1. The cycle repeats indefinitely.
Timing equations
The charge and discharge times follow from the RC time constant, and because the capacitor charges through R1 + R2 but discharges through R2 only:
\[t_\text{high} = 0.693 \times (R_1 + R_2) \times C_1\]
\[t_\text{low} = 0.693 \times R_2 \times C_1\]
\[T = t_\text{high} + t_\text{low} = 0.693 \times (R_1 + 2R_2) \times C_1\]
\[f = \frac{1}{T} = \frac{1.44}{(R_1 + 2R_2) \times C_1}\]
With the values in this project (R1 = 1 kΩ, R2 = 47 kΩ, C1 = 10 µF):
\[f = \frac{1.44}{(1{,}000 + 94{,}000) \times 10 \times 10^{-6}} = \frac{1.44}{0.95} \approx 1.5 \text{ Hz}\]
The LED blinks about one and a half times per second.
The duty cycle — the fraction of each period that OUTPUT is high — is:
\[D = \frac{t_\text{high}}{T} = \frac{R_1 + R_2}{R_1 + 2R_2}\]
With R1 ≪ R2, the duty cycle approaches 50%. With larger R1 relative to R2 it approaches 100%. Notice that with a standard 555 the output is always more than 50% high, because C1 charges through both resistors but discharges through R2 only. Achieving exactly 50% (or below) requires either a diode across R2 or a CMOS 555 with a modified topology.
Why C2 matters
The 100 nF capacitor on pin 5 bypasses high-frequency noise on the internal voltage divider. Without it, brief voltage spikes on the supply rail can trip the comparators prematurely, causing jitter or missed cycles. It is not part of the timing calculation — just a practical necessity on any real power supply.
Exploratory Questions
1. Predicting the audio range. Swap C1 for a 100 nF ceramic capacitor and keep R1 and R2 the same. Calculate the expected frequency using the formula, then connect a small speaker or buzzer to pin 3 (with R3 still in series if using a low-impedance speaker) and verify by ear. The note A4 is 440 Hz — what value of C1 would you need to hit that exactly with R2 = 47 kΩ?
2. Measuring the duty cycle. With the LED blink version, use a stopwatch to time 10 complete on–off cycles. From the measured period and the on-time, calculate the actual duty cycle. How closely does it match the formula? Now estimate what duty cycle you would get if you swapped R1 and R2 (1 kΩ for R1, 47 kΩ for R2 — wait, they are already that way. Try R1 = 47 kΩ and R2 = 1 kΩ instead). What does the formula predict? Does the circuit still oscillate?
3. Tunable frequency. Replace R2 with a 100 kΩ potentiometer in series with a 1 kΩ fixed resistor (to prevent R2 reaching zero). Sweep the pot from end to end. Over what frequency range does the output vary? Compare the measured range with the formula’s prediction at each extreme.
4. The discharge bypass diode. To achieve exactly 50% duty cycle, place a 1N4148 signal diode across R2 with the anode toward pin 7 and the cathode toward the C1 junction. This allows C1 to charge through R1 only and discharge through R2 only. Recalculate the expected duty cycle and timing for R1 = R2 = 10 kΩ with this modification, then build and verify.
5. Current consumption. Measure the average current drawn from the 9 V battery with a multimeter in series with the positive supply lead. Calculate the power dissipated. Now replace the NE555 with a CMOS ICM7555 and repeat. The quiescent current of the CMOS variant is specified at 60 µA versus ~3 mA for the bipolar 555 — how does your measurement compare, and what does that mean for battery life?
6. VCC sensitivity. The oscillation frequency should be independent of supply voltage, because both the charging voltage and the comparator thresholds scale with VCC. Test this: run the circuit from a 5 V supply instead of 9 V. Does the frequency change? Does the LED brightness change, and why?
7. Cascaded timers. Connect the output of this oscillator (pin 3) to the RESET pin (pin 4) of a second 555 configured as another astable at a much higher frequency (e.g., R1 = 1 kΩ, R2 = 4.7 kΩ, C1 = 100 nF). The first timer will periodically enable and disable the second. What do you observe on the output LED of the second timer? This is a simple two-tone alarm pattern — sketch the output waveform before building.