Задача Задача №3621615
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Nice—555 timer is a staple for exams. Here’s a compact refresher you can use.
What a 555 does at a glance
- Internal: two comparators, a flip-flop, a discharge transistor; used as an astable or monostable multivibrator.
- Key pins: 1 GND, 2 Trigger, 3 Output, 4 Reset, 5 Control, 6 Threshold, 7 Discharge, 8 Vcc.
- Threshold and trigger levels are 1/3 Vcc and 2/3 Vcc, respectively. Control voltage on pin 5 can shift these if you add a capacitor.
Common configurations
- Astable (free-running oscillator)
- No external trigger input; output toggles high/low periodically.
- timing: t_high = 0.693 × (R1 + R2) × C t_low = 0.693 × R2 × C Period T = t_high + t_low = 0.693 × (R1 + 2R2) × C Frequency f = 1 / T = 1 / [0.693 × (R1 + 2R2) × C] Duty cycle D = t_high / T = (R1 + R2) / (R1 + 2R2)
- Practical notes: Use diodes if you need separate charge/discharge paths for duty cycles far from 50%.
- Monostable (one-shot)
- Triggered by a negative pulse on pin 2; output goes high for a time and then returns low.
- timing: t ≈ 1.1 × R × C Here R is the external resistor (often connected to pins 7/6) and C is the timing capacitor.
- Frequency depends on trigger events (not a fixed oscillator).
Helpful quick references
- Reset pin: Tie to Vcc if not used (active-low reset).
- Control voltage (pin 5): Usually left open or decoupled with a 0.01 µF cap to ground to stabilize thresholds.
- Common pitfalls: incorrect wiring of R1/R2 in astable, forgetting the 1/3 and 2/3 Vcc thresholds, not tying Reset high.
Quick worked example (typical values)
- Example: R1 = 10 kΩ, R2 = 100 kΩ, C = 100 nF in astable. t_high = 0.693 × (R1 + R2) × C = 0.693 × 110k × 100nF ≈ 7.62 ms t_low = 0.693 × R2 × C = 0.693 × 100k × 100nF ≈ 6.93 ms Period T ≈ 14.55 ms, f ≈ 68.7 Hz Duty ≈ 7.62 ms / 14.55 ms ≈ 52.4%
Want a quick practice problem or a printable one-page cheat sheet for your course? Tell me your exam focus (analog circuits, PWM, microcontroller interfacing, etc.) and I’ll tailor it.
