Design a monostable multivibrator using a 555 timer, a capacitor with value C = 1 nF and appropriate resistor values to produce an output pulse of 20 µs duration.
The Correct Answer and Explanation is:
Monostable Multivibrator Design Using 555 Timer
Given:
- Pulse duration, T=20 μsT = 20 \, \mu s
- Capacitance, C=1 nF=1×10−9 FC = 1 \, nF = 1 \times 10^{-9} \, F
1. Formula for Monostable Multivibrator Using 555 Timer:
The output pulse duration of a monostable 555 timer is given by: T=1.1×R×CT = 1.1 \times R \times C
Solving for RR: R=T1.1×CR = \frac{T}{1.1 \times C}
Substitute the given values: R=20×10−61.1×1×10−9=20×10−61.1×10−9=18,181.8 ΩR = \frac{20 \times 10^{-6}}{1.1 \times 1 \times 10^{-9}} = \frac{20 \times 10^{-6}}{1.1 \times 10^{-9}} = 18,181.8 \, \Omega
So, use a resistor R≈18.2 kΩR \approx 18.2 \, k\Omega (standard E24 value).
2. Components Needed:
- 555 Timer IC
- Capacitor: 1 nF
- Resistor: 18.2 kΩ
- Trigger input (momentary switch or digital pulse)
- Power supply (typically 5V to 15V depending on the IC version)
3. Operation Explanation
A monostable multivibrator using a 555 timer produces a single output pulse of fixed duration in response to an external trigger. This configuration is commonly used in timing applications like pulse stretching, debouncing, or generating time delays.
In monostable mode, the 555 timer has one stable state—LOW output. When a negative trigger pulse (voltage drops below 1/3 Vcc) is applied to the TRIG pin (pin 2), the internal flip-flop of the 555 timer is set. This causes the output (pin 3) to go HIGH and initiates the charging of the external capacitor CC through the resistor RR.
The capacitor begins to charge exponentially through the resistor, and the timer monitors its voltage. When the voltage across the capacitor reaches 2/3 Vcc, the internal comparator resets the flip-flop, causing the output to return LOW. The timing duration for which the output stays HIGH is determined by: T=1.1×R×CT = 1.1 \times R \times C
In this design, with C=1 nFC = 1 \, nF and R=18.2 kΩR = 18.2 \, k\Omega, the output remains HIGH for: T=1.1×18.2×103×1×10−9≈20 μsT = 1.1 \times 18.2 \times 10^3 \times 1 \times 10^{-9} \approx 20 \, \mu s
This provides a precise 20 µs pulse each time the circuit is triggered. After the pulse, the system resets and waits for the next trigger, making it ideal for generating consistent short-duration pulses in control circuits or signal processing.
