In the astable multivibrator circuit shown in the figure, the frequency of oscillation (in kHz) at the output pin 3 is _____________
The frequency of oscillation for a 555 timer configured as an astable multivibrator is given by:
f = 1.44 / ((RA + 2RB)C)
Where:
- RA = 2.2 kΩ
- RB = 4.7 kΩ
- C = 0.022 μF
First, calculate the value of (RA + 2RB):
RA + 2RB = 2.2 kΩ + 2(4.7 kΩ) = 2.2 kΩ + 9.4 kΩ = 11.6 kΩ
Convert RA, RB, and C to consistent units (ohms and farads):
- 11.6 kΩ = 11600 Ω
- 0.022 μF = 0.022 × 10-6 F
Plug these values into the frequency formula:
f = 1.44 / (11600 × 0.022 × 10-6)
f = 1.44 / (0.254 × 10-3)
f = 1.44 / 0.000254
f ≈ 5669.29 Hz
Convert the frequency to kHz:
f ≈ 5.67 kHz
Finally, verify that this value is within the given range of 5.55 to 5.75 kHz. Since 5.67 kHz is indeed within this range, the computation is correct and validated.
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