Linear Integrated Circuit Multiple Choice Questions on “555 timer as a Monostable Multivibrator”.
1. Determine the time period of a monostable 555 multivibrator.
A. T = 0.33RC
B. T = 1.1RC
C. T = 3RC
D. T = RC
Answer: B
Clarification: The time period of a monostable 555 timer is T = RC×ln(1/3) = 1.1.RC.
2. Find monostable vibrator circuit using 555 timer.
Answer: A
Clarification: When 555 timer is configured in monostable operation, the trigger input is applied through pin2 whereas, upper comparator threshold (pin6) & discharge (pin7) are shorted and connected at the output.
3. How to overcome mistriggering on the positive pulse edges in the monostable circuit?
A. Connect a RC network at the input
B. Connect an integrator at the input
C. Connect a differentiator at the input
D. Connect a diode at the input
Clarification: To prevent the mistrigger on positive pulse edges, a resister & capacitor combined of 10kΩ and 0.001µF at the input to form a differentiator
The circuit shows the differentiator to be connected between trigger input and the +VCC.
4. A monostable multivibrator has R = 120kΩ and the time delay T = 1000ms, calculate the value of C?
A. 0.9µF
B. 1.32µF
C. 7.5µF
D. 2.49µF
Answer: C
Clarification: Time delay for a monostable multivibrator, T = 1.1RC
=> C = T/(1.1R) = 1000ms/(1.1×120kΩ) = 7.57µF.
5. Which among the following can be used to detect the missing heart beat?
A. Monostable multivibrator
B. Astable multivibrator
C. Schmitt trigger
D. None of the mentioned
Answer: A
Clarification: A monostable multivibrator can be used as a missing pulse detector by connecting a transistor between trigger inputs. If a pulse misses, the discharge trigger input goes high & transistor become cut-off and the output goes low. So, this type of circuit can be used to detect missing heart beat.
6. A 555 timer in monostable application mode can be used for
A. Pulse position modulation
B. Frequency shift keying
C. Speed control and measurement
D. Digital phase detector
Answer: C
Clarification: In monostable operation mode, if input trigger pulses are generated from a rotating wheel, the circuit will determine the wheel speed whenever it drops below a predetermined value. Therefore, it can be used for speed control and measurement.
7. How can a monostable multivibrator be modified into a linear ramp generator?
A. Connect a constant current source to trigger input
B. Connect a constant current source to trigger output
C. Replace resistor by constant current source
D. Replace capacitor by constant current source
Answer: C
Clarification: The resistor R of the monostable circuit is replaced by a constant current source. So, that the capacitor is charged linearly and generates ramp signal.
8. Determine time period of linear ramp generator using the specifications
RE = 2.7kΩ, R1 =47kΩ , R2 100kΩ , C= 0.1µF, VCC =5v.
A. 8ms
B. 4ms
C. 2ms
D. 1ms
Answer: D
Clarification: The time period of the linear ramp generator, T= [(2/3)×(VCC×RE)×(R1+ R2)×C]/{(R1×VCC)-[VBE×(R1+R2)]}
= {(2/3)×5v×[2.7kΩ×(4.7kΩ+ 100kΩ)]×(0.1µF)}/{[(47kΩ)×5v]-[(0.7)×(47kΩ+100kΩ)]}
=>T= 132.3/132.100 =1.0015×10-3 = 1ms.
9. What will be the output, if a modulating input signal and continuous triggering signal are applied to pin5 and pin22 respectively in the following circuit?
A. Frequency modulated wave form
B. Pulse width modulated wave form
C. Both pulse and frequency modulated wave form
D. None of the mentioned
Answer: B
Clarification: On application of continuous trigger at pin22 and a modulated input signal at pin5, a series of output pulses are obtained. The duration of which depends on the modulating signal. Also in the pulse duration, only the duty cycle varies, keeping the frequency same as that of the continuous input pulse train trigger.