555

If you meant the 555 timer IC (a common exam topic), here’s a compact refresher you can use.

What is it

• A versatile 8-pin timer IC used to generate precise delays and oscillations (astable, monostable, bistable).

Pinout (quick reference)

• Pin 1: GND
• Pin 2: TRIG (trigger)
• Pin 3: OUT
• Pin 4: RESET (active low; tie to Vcc if unused)
• Pin 5: CONTROL (usually 0.01 µF to ground)
• Pin 6: THRS (threshold)
• Pin 7: DISCH (discharge)
• Pin 8: Vcc

Modes of operation

• Astable: free-running oscillator (no external trigger)
  • No stable state; output toggles high/low continuously.
  • Components: R1 between Vcc and DISCH (pin 7), R2 between DISCH (pin 7) and THRS/TRIG (pins 6/2) tied to C to ground.
  • Formulas:
    • t_high = 0.693 (R1 + R2) C
    • t_low = 0.693 R2 C
    • f = 1 / [0.693 (R1 + 2R2) C]
    • Duty cycle D = (R1 + R2) / (R1 + 2R2)
• Monostable: one-shot pulse on trigger
  • Trigger input (pin 2) goes low to start a timing interval.
  • The capacitor charges through R (and possibly a diode network) until 2/3 Vcc is reached.
  • Output goes high for t ≈ 1.1 R C, then returns low.
  • Basic hookup: R between Vcc and THRS/TRIG network, C from THRS/TRIG to ground; DISCH is used to discharge C after the pulse; RESET tied high.
  • Time constant: t = 1.1 R C
• Bistable (toggle flip-flop)
  • Output latches high or low until a trigger or reset toggles it.
  • No timing capacitor is required.

Key design notes

• Reset (pin 4) should be tied to Vcc if unused.
• Control voltage (pin 5) can be left open or bypassed with 0.01 µF to ground for noise immunity.
• Supply voltage range: typically 4.5–15 V (depends on device variant).
• Use decoupling capacitor near Vcc (e.g., 0.1 µF to 1 µF) for stability.
• Tolerances of R and C directly affect frequency/time; expect ±1% to ±5% depending on parts and temperature.

Example values you can memorize

• Astable example: R1 = 1 kΩ, R2 = 10 kΩ, C = 100 nF
  • f ≈ 1 / [0.693 (1k + 2×10k) × 100 nF] ≈ 686 Hz
  • D ≈ (1k + 10k) / (1k + 2×10k) ≈ 11k / 21k ≈ 52%
• Monostable example: R = 10 kΩ, C = 100 nF
  • t ≈ 1.1 × 10k × 100 nF ≈ 1.1 ms

Quick practice problem

• For an astable configuration with R1 = 2 kΩ, R2 = 47 kΩ, C = 2.2 µF, estimate:
  • a) f
  • b) Duty cycle Solution hints:
  • f = 1 / [0.693 × (R1 + 2R2) × C]
  • D = (R1 + R2) / (R1 + 2R2)

If you meant something else by “555” (e.g., a different context or a specific exam topic), tell me and I’ll tailor the notes or give you practice problems for that. Would you like a few practice problems with solutions or a quick diagram-oriented guide?