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Задача Задача №3619677
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Аноним
17 июня 2026 21:58 · публичный вопрос
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Аноним

555

Нейросеть17 июня 2026 21:58

Do you mean the 555 timer IC, or something else with the number 555 (math, code, etc.)? If you’re prepping for exams, here’s a quick 555 timer refresher you can use right away.

If you mean the 555 timer IC:

  • Pinout (DIP): 1 GND, 2 Trigger, 3 Output, 4 Reset, 5 Control, 6 Threshold, 7 Discharge, 8 Vcc
  • Modes:
    • Astable (oscillator, free-running): no external trigger needed; makes a square wave.
    • Monostable (one-shot): produces a single pulse when triggered.
  • Key formulas (standard astable with R1 between Vcc and Discharge (pin 7), R2 between Discharge and Capacitor (pin 2/6), C from Capacitor to GND):
    • t_high = 0.693 (R1 + R2) C
    • t_low = 0.693 R2 C
    • f = 1 / [0.693 (R1 + 2 R2) C]
    • Duty cycle D = (R1 + R2) / (R1 + 2 R2)
  • Quick example (typical values):
    • Choose C = 10 nF, R1 = 50 kΩ, R2 = 50 kΩ
    • Then f ≈ 1 / [0.693 (50k + 2×50k) × 10nF] ≈ 1 kHz
    • D ≈ (50k + 50k) / (50k + 2×50k) ≈ 2/3 ≈ 66.7%
    • Note: to get 50% duty in a standard astable you’d use diodes or a different configuration; with R1 = 0 you’d get close to 50% but practical designs use a diode network or a 556/other arrangement.

If you meant something else (e.g., math 555, or a different context), tell me and I’ll tailor it.

Would you like a worked problem or a concrete design for a specific frequency or duty cycle?

555
Do you mean the 555 timer IC, or something else with the number 555 (math, code, etc.)? If you’re prepping for exams, here’s a quick 555 timer refresher you can use right away. If you mean the 555 timer IC: - Pinout (DIP): 1 GND, 2 Trigger, 3 Output, 4 Reset, 5 Control, 6 Threshold, 7 Discharge, 8 Vcc - Modes: - Astable (oscillator, free-running): no external trigger needed; makes a square wave. - Monostable (one-shot): produces a single pulse when triggered. - Key formulas (standard astable with R1 between Vcc and Discharge (pin 7), R2 between Discharge and Capacitor (pin 2/6), C from Capacitor to GND): - t_high = 0.693 (R1 + R2) C - t_low = 0.693 R2 C - f = 1 / [0.693 (R1 + 2 R2) C] - Duty cycle D = (R1 + R2) / (R1 + 2 R2) - Quick example (typical values): - Choose C = 10 nF, R1 = 50 kΩ, R2 = 50 kΩ - Then f ≈ 1 / [0.693 (50k + 2×50k) × 10nF] ≈ 1 kHz - D ≈ (50k + 50k) / (50k + 2×50k) ≈ 2/3 ≈ 66.7% - Note: to get 50% duty in a standard astable you’d use diodes or a different configuration; with R1 = 0 you’d get close to 50% but practical designs use a diode network or a 556/other arrangement. If you meant something else (e.g., math 555, or a different context), tell me and I’ll tailor it. Would you like a worked problem or a concrete design for a specific frequency or duty cycle?