---
title: "Frequency of 555 timer astable multivibrator"
subject: "Electronics Devices and Circuit"
chapter: 5
key: "semester_3/electronics_devices_and_circuit:555-astable-frequency"
source: https://www.bctengineer.com/q/semester_3/electronics_devices_and_circuit/555-astable-frequency
site: BCT Engineering
---

# Frequency of 555 timer astable multivibrator

Electronics Devices and Circuit · Chapter 5: Signal Generator and Waveform-Shaping Circuits

- Asked in: 2081 Bhadra, 5 marks; 2075 Ashwin, 6 marks; 2073 Shrawan, 6 marks; 2072 Chaitra, 6 marks; 2068 Chaitra, 6+2 marks; 2072 Kartik, 6 marks (asked 6 times)
- Video: [555 Astable Calculations](https://www.youtube.com/watch?v=DEZpZvxgpKU) — ElectronicsTeaching
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/electronics_devices_and_circuit/555-astable-frequency
- Whole chapter: https://www.bctengineer.com/md/semester_3/electronics_devices_and_circuit/chapter-5

## Question

Derive frequency of oscillation of 555 timer Astable Multivibrator.

## Short answer

Charging time tH = 0.693(R₁ + R₂)C, discharging time tL = 0.693 R₂ C. Frequency f = 1.44 / [(R₁ + 2R₂) C]. The duty cycle is always above 50%.

## Answer

In astable operation the 555 timer works as a free-running oscillator giving a rectangular wave. There is no stable state.

#### Circuit
```text
          +VCC
     +-----+-----+
     |     |     |
   [RA]   (4)(8)  |
     |            |
     +--- (7) discharge
     |
   [RB]
     |      (6) threshold
     +------(2) trigger
     |
   [C]       (3) output
     |
    GND      (5) via 0.01 µF to GND
```

Pins 2 and 6 are tied together at the capacitor, and the capacitor charges through RA + RB and discharges through RB into pin 7.

#### Operation
The comparators have thresholds of 2VCC/3 (upper) and VCC/3 (lower).

1. Capacitor charges from VCC/3 to 2VCC/3 through (RA + RB). Output is HIGH, and the discharge transistor is OFF.
2. At 2VCC/3, the upper comparator resets the flip-flop; output goes LOW and the discharge transistor turns ON.
3. The capacitor discharges through RB to VCC/3. At VCC/3, the lower comparator sets the flip-flop and the cycle repeats.

#### Derivation
**Charging (HIGH) time:** the capacitor voltage is vC(t) = VCC − (VCC − VCC/3) e^(−t/(RA+RB)C). It reaches 2VCC/3 at time

2VCC/3 = VCC − (2VCC/3) e^(−tH/τ1) → e^(−tH/τ1) = 1/2

**tH = 0.693 (RA + RB) C**

**Discharging (LOW) time:** vC(t) = (2VCC/3) e^(−t/RB C), reaching VCC/3:

1/2 = e^(−tL/RB C)

**tL = 0.693 RB C**

**Period and frequency:**

T = tH + tL = 0.693 (RA + 2RB) C

**f = 1/T = 1.44 / [(RA + 2RB) C]**

**Duty cycle:** D = tH/T = (RA + RB)/(RA + 2RB), which is always above 50 %. D approaches 50 % when RA << RB.

**Example:** RA = 1 kΩ, RB = 10 kΩ, C = 0.1 µF gives f = 1.44/(21 k × 0.1 µ) = 686 Hz.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Between which voltages does the capacitor swing in astable operation?
   A. 0 and VCC
   B. VCC/3 and 2VCC/3
   C. VCC/2 and VCC
   D. 0 and 2VCC/3
   Answer: B. The lower comparator triggers at VCC/3 and the upper threshold is 2VCC/3. (Key point: Thresholds)

2. Through which resistance does the capacitor charge, and what is tH?
   A. RB, 0.693 RB C
   B. RA only, 0.693 RA C
   C. RA + RB, 0.693 (RA + RB) C
   D. RA + 2RB, 0.693 (RA + 2RB) C
   Answer: C. Charging is via RA and RB from VCC/3 to 2VCC/3, taking one half-life: tH = 0.693 (RA + RB) C. (Key point: Charging time)

3. What is the discharge (LOW) time?
   A. 0.693 RB C, through the discharge pin 7
   B. 0.693 (RA + RB) C
   C. 1.44 RB C
   D. 0.693 RA C
   Answer: A. The capacitor discharges only through RB into pin 7, from 2VCC/3 to VCC/3: tL = 0.693 RB C. (Key point: Discharging time)

4. Which is correct for the frequency and duty cycle?
   A. f = 0.693/[(RA + 2RB)C], duty cycle below 50%
   B. f = 1.44/[(RA + 2RB)C], duty cycle (RA + RB)/(RA + 2RB), always above 50%
   C. f = 1.44/[(RA + RB)C], duty cycle 50%
   D. f = 1.44/(RB C), duty cycle RB/RA
   Answer: B. T = 0.693 (RA + 2RB) C gives f = 1.44/[(RA + 2RB)C]; tH always exceeds tL so D is above 50%. (Key point: Frequency and duty cycle)

## Related in Electronics Devices and Circuit

- [555 astable multivibrator and duty cycle](https://www.bctengineer.com/question.md?key=semester_3%2Felectronics_devices_and_circuit%3A555-astable-duty-cycle) — Electronics Devices and Circuit, chapter 5
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- [Barkhausen criteria and 50% duty cycle from 555](https://www.bctengineer.com/question.md?key=semester_3%2Felectronics_devices_and_circuit%3Abarkhausen-555-duty-cycle) — Electronics Devices and Circuit, chapter 5
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