Scribd
Upload
18 views5 pages

Quiz11 (Uf)

The document contains an assignment on electrical measurements and electronic instruments, focusing on the analysis of circuits using a 555 timer and a function generator. It includes questions about peak-to-peak values, ON/OFF times, and waveforms, with provided solutions and correct answers. The assignment is structured with specific answer ranges and calculations for various parameters in the circuits.

Uploaded by

nss rao
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as PDF, TXT or read online on Scribd
18 views5 pages

Quiz11 (Uf)

The document contains an assignment on electrical measurements and electronic instruments, focusing on the analysis of circuits using a 555 timer and a function generator. It includes questions about peak-to-peak values, ON/OFF times, and waveforms, with provided solutions and correct answers. The assignment is structured with specific answer ranges and calculations for various parameters in the circuits.

Uploaded by

nss rao
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as PDF, TXT or read online on Scribd
You are on page 1/ 5

Information

Electrical Measurements and Electronic Instruments


Assignment 11
Question 1

Not answered

Marked out of 5.00

+15V +15V +15V

2.5kΩ 5kΩ
-15V
4kΩ 5kΩ
+15V

0.25mF 5kΩ

-15V

Consider the above circuit of an astable multivibrator using a 555 timer.

Peak-to-peak value of VC =? (in V)

Answer Range: 4.75 to 5.25


Solution:
VCC 2VCC
VC varies between 3
and 3

Correct answer is: 5

When VC is increasing then Vout is

a. High

b. Low

Answer: Low

Correct answer is: Low

Estimate the OFF (low) time of Vout (in seconds)

Answer Range: 1.07 to 1.183


Solution:
OFF time = (R1 + R2 )Cln(2) = 1.1263641684099 s

Correct answer is: 1.1263641684099

Estimate the ON (high) time of Vout (in seconds)

Answer Range: 0.6585 to 0.7278


Solution:
ON time = R2 Cln(2) = 0.69314718055995 s

Correct answer is: 0.69314718055995

Estimate the frequency of Vout (in Hz)


Answer Range: 0.5221 to 0.5771
Solution:
1
Frequency = ON time+OFF time
= 0.54959811081484 Hz

Correct answer is: 0.54959811081484

[0.5+0.5+1.5+1.5+1]
Question 2

Not answered

Marked out of 5.00

500μF
6kΩ

2kΩ +12V
+12V
4kΩ

-12V
-12V
Consider the above simple function generator circuit.
The waveform of V1 is

a. sinusoidal

b. square

c. triangular

Answer: triangular

Correct answer is: triangular

The waveform of V2 is

a. sinusoidal

b. square

c. triangular

Answer: triangular

Correct answer is: square

Estimate the peak-to-peak amplitude of V1 (in V)

Answer Range: 15.2 to 16.8


Solution:
4kΩ
The peak-to-peak amplitude of V1 = Hysteresis band of the non-inverting Schmitt trigger part = 212V × 6kΩ

Correct answer is: 16

Estimate the peak-to-peak amplitude of V2 (in V)

Answer Range: 22.8 to 25.2


Solution:
The peak-to-peak amplitude of V2 = 2 × 12V

Correct answer is: 24

Estimate the time-period of V1 (in milli-second)

Answer Range: 2533 to 2800


Solution:
12V
The slope or the rate of increase of V1 = R C
1
212VR2
R3 2R 1 R 2 C
The time required for V1 to go from the LTP to the UTP of the Schmitt trigger = 12V
= R3
R1 C

2R1 R2 C
Similarly, the time required for V1 to go from the UTP to the LTP of the Schmitt trigger is also = R3
4R 1 R 2 C
∴ The time period of V1 = R3

Correct answer is: 2666.6666666667

Estimate the time-period of V2 (in milli-second)

Answer Range: 2533 to 2800


Solution:
Time-period amplitude of V2 = Time-period amplitude of V1

Correct answer is: 2666.6666666667

[0.5+0.5+1+1+1+1 = 5 marks]

You might also like