555 Timer MCQ [Free PDF] – Objective Question Answer for 555 Timer Quiz

11. 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 with constant current source
D. Replace the capacitor with a constant current source

Answer: C

The resistor R of the monostable circuit is replaced by a constant current source. So, the capacitor is charged linearly and generates a ramp signal.

 

12. Determine the time period of the linear ramp generator using the specifications
RE = 2.7kΩ, R1 =47kΩ , R2 100kΩ , C= 0.1µF, VCC =5v.

Determine time period of linear ramp generator using the specifications

A. 8ms
B. 4ms
C. 2ms
D. 1ms

Answer: D

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.

 

13. 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?

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 waveform
B. Pulse width modulated waveform
C. Both pulse and frequency-modulated waveform
D. None of the mentioned

Answer: B

On application of a 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 the same as that of the continuous input pulse train trigger.

 

14. Free running frequency of Astable multivibrator?

A. f=1.45/(RA+2RB)C
B. f=1.45(RA+2RB)C
C. f=1.45C/(RA+2RB)
D. f=1.45 RA/( RA+RB)

Answer: A

The frequency of the Astable multivibrator is

T=0.69(RA+2RB)C.

Therefore, f = 1/T

=1.45/(RA+2RB)C.

 

15. Find the charging and discharging time of the 0.5 µF capacitors

Find the charging and discharging time of the 0.5 µF capacitors

.

A. Charging time=2ms; Discharging time=5ms
B. Charging time=5ms; Discharging time=2ms
C. Charging time=3ms; Discharging time=5ms
D. Charging time=5ms; Discharging time=3ms

Answer: B

The time required to charge the capacitor is

tHigh=0.69(RA+RB)C =0.69(10kΩ+5kΩ)x0.5µF =5ms.

The time required to discharge the capacitor is

tLow=0.69xRC

=0.69 × 5kΩ  × 0.5µF = 2ms.

 

16. Astable multivibrator operating at 150Hz has a discharge time of 2.5m. Find the duty cycle of the circuit.

A. 50%
B. 75%
C. 95.99%
D. 37.5%

Answer: D

Given f=150Hz.

Therefore,T=1/f =1/150 =6.67ms.

∴ Duty cycle, D%=(tLow/T) x 100%

= (2.5ms/6.67ms)×100% = 37.5%.

 

17. Determine the frequency and duty cycle of a rectangular wave generator.

Determine the frequency and duty cycle of a rectangular wave generator.

A. Frequency=63.7kHz; Duty cycle=50%
B. Frequency=53.7kHz; Duty cycle=55%
C. Frequency=43.7kHz; Duty cycle=50%
D. Frequency=60kHz; Duty cycle=55%

Answer: B

Frequency=1.45/(RA+RB)C .

Where RA=100Ω+50Ω=150Ω,

RB=100Ω+20Ω=120Ω.

=>∴f=1.45/((150+120)×0.1µF)

= 53703Hz = 53.7kHz.

Duty cycle, D% = [RB/(RA+RB)] × 100%

= 120Ω/(150Ω +120Ω) × 100%

= 0.55 × 100% = 55%.

 

18. How to achieve a 50% duty cycle on is adjustable rectangular wave generator? (Assume R1 –> Resistor connected between supply and discharge and R2 –> Resistor connected between discharge and trigger input.)

A. R12
B. R1 > R2
C. R1 = R2
D. R1 ≥ R2

Answer: C

The equation of duty cycle, D = R2/(R1 + R2). If R1 is made equal to R2 then a 50% duty cycle is achieved.

 

19. How to obtain symmetrical waveform in Astable multivibrator?

A. Use clocked RS flip-flop
B. Use clocked JK flip-flop
C. Use clocked D-flip-flop
D. Use clocked T-flip-flop

Answer: B

A symmetrical square wave can be obtained by adding a clocked JK flip-flop to the output of the Astable multivibrator. The clocked flip-flop acts as a binary divider to the time’s output and produces a 50% duty cycle without any restriction on the choice of resistors.

 

7. Determine the output frequency of the circuit.

Determine the output frequency of the circuit.

A. 1450Hz
B. 1333Hz
C. 1871Hz
D. 1700Hz

Answer: C

The output frequency of the frequency shift keying generator is

f=1.45/[(RA||RC)+2(RB)]xC

= 1.45/[(2.3kΩ||2.3kΩ) + (2×3.3kΩ)] × 0.1µF

= 1.45/{[(2.3×2.3)/(2.3+2.3)] + 6.6kΩ}× 0.1µF

= 1.45/(7.75×10-4)

= 1870.9 ≅ 1871Hz.

 

20. How does a monostable multivibrator used as a frequency divider?

A. Using a square wave generator
B. Using triangular wave generator
C. Using a sawtooth wave generator
D. Using sine wave generator

Answer: A

A monostable multivibrator can be used as a frequency divider when a continuously triggered monostable circuit is triggered using a square wave generator. Provided the timing interval is adjusted to be longer than the period of triggering the square wave input signal.

 

21. An astable multivibrator has:

  1. one quasi-stable state
  2. two quasi-stable states
  3. one stable states
  4. two stable states

Answer.2. two quasi-stable states

  • Astable Multivibrators are free-running oscillators that oscillate between two states continuously producing two square wave output waveforms.
  • Both states are not stable as it changes from one state to the other all the time.
  • As the Multivibrator keeps on switching, these states are known as quasi-stable or half-stable states.

 

22. Which is a voltage to frequency converter multivibrator?

  1. Bistable
  2. Astable
  3. Monostable
  4. Schmitt trigger

Answer.2. Astable

Astable multivibrators are also known as Free-running multivibrators.

It has NO stable output states as it changes from one state to the other all the time.

The astable circuit consists of two switching transistors and a cross-coupled feedback network.

The two-time delay capacitors allow oscillation between the two states with no external triggering to produce the change in state.

  • The amplitude of the output waveform is approximately the same as the supply voltage.
  • VCC with the time period of each switching state determined by the time constant of the RC.
  • If the two-time constants produced by C2 x R2 and C1 x R3 in the base circuits are the same.
  • Then, the mark-to-space ratio ( t1 / t2 ) will be equal to one-to-one making the output waveform symmetrical in shape.
  • By altering the time constant of just one RC network the mark-to-space ratio.
  • The output frequency will be altered keeping the mark-to-space ratios the same at one-to-one.

​Hence the output frequency of the Astable Multivibrator can be controlled by the input voltage.

So an astable multivibrator is used as Voltage to a frequency converter.

 

23. The frequency of oscillations of a symmetric astable multivibrator, given that R = 10 kΩ and C = 10 nF is _____

  1. 14 kHz
  2. 3.5 kHz
  3. 28 kHz
  4. 7 kHz

Answer.4. 7 kHz

The frequency of oscillations of symmetric astable multivibrator is f = 1/T

where

T = t+ t2

T = time period , t1 = charging period, t2 = discharging period

For symmetrical Astable multivibrator

t= 0.69 RC

t= 0.69 RC

T = 1.38 RC

f = 1/1.38RC

Calculation:

Given that R = 10 x 103 ohm C= 10 nF

1/(1.38 x 10 x 10x10 x 10-9)

=7246.37 = 7.2 Khz

Scroll to Top