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Analog Electronic Circuits Test Solutions

This document contains the solutions to questions from Test 2 of the course EKT 214 Analog Electronic Circuit II. Question 1 involves designing a Wien bridge oscillator circuit to produce a 3 kHz sinusoidal output. Question 2 defines active filters and lists two of their advantages over passive filters. It also involves designing a multiple feedback bandpass active filter. Question 3 defines and explains the terms line regulation and load regulation, and includes calculating the load regulation of a given voltage regulator.

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204 views4 pages

Analog Electronic Circuits Test Solutions

This document contains the solutions to questions from Test 2 of the course EKT 214 Analog Electronic Circuit II. Question 1 involves designing a Wien bridge oscillator circuit to produce a 3 kHz sinusoidal output. Question 2 defines active filters and lists two of their advantages over passive filters. It also involves designing a multiple feedback bandpass active filter. Question 3 defines and explains the terms line regulation and load regulation, and includes calculating the load regulation of a given voltage regulator.

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نورول نضيره رشدي
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© © All Rights Reserved
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EKT 214 SEM 2 2011

EKT 214 ANALOG ELECTRONIC CIRCUIT II


TEST 2 (1 Hour) SOLUTION
Question 1
(a) Draw the structure of Wien-Bridge oscillator and identify two circuits in your
diagram.
[5 marks]
Ans:
Wien-Bridge Oscillator
It is a low frequency oscillator which ranges from a few kHz to 1 MHz.

Wien Bridge circuit combines a voltage devider and the lead-lag circuit.

(b) Based on this Figure 1, design the circuit to produce a sinusoidal output voltage at
frequency of 3 kHz. Use C = 22 nF and R1 = 50. Given that the frequency of
6
oscillation, f o . [3 marks]
2RC
Ans:

B = 1/29 = R 1/R2
R2 = 1.45 k

1
EKT 214 SEM 2 2011

(c) If R1 in (b) is changed to 35, determine the value of R2 necessary for the circuit in
Figure 1 to operate as oscillator. [2 marks]

Rule for oscillator to oscilate is


B = 1/29 = R 1/R2
R2 = 1.015 k

Figure 1

Question 2
(a) Explain the Active Filter and discuss TWO (2) advantages of Active Filter.
[3 marks]
Ans:
Active filters : The circuits that employ one or more op-amps in the design an addition to
resistors and capacitors [1m]

Advantages of Active Filters over Passive Filters: (choose only 2)


Active filters can be designed to provide required gain, and hence no attenuation as in
the case of passive filters [1m]
No loading problem, because of high input resistance and low output resistance of op-
amp. [1m]
Active Filters are cost effective as a wide variety of economical op-amps are available.
[1m]

2
EKT 214 SEM 2 2011

(b) Design a multiple-feedback band-pass active filter with the center frequency, fo = 2
kHz, center gain, Ao = 5 and quality factor, Q = 10. Let capacitance C1 = C2 = 2 F.
Illustrate the circuit you have designed.
[9 marks]

Ans:
The resistance R1 is determined as follows:
Q 10 [2 marks]
R1 79.62
2Ao f o C 2 3.14 5 2 x10 3 Hz 2 10 6 F

The resistance R3 is determined as follows:

Q 10 [2 marks]
R3 796.2
f o C 3.14 2 10 Hz 2 10 6 F
3

The resistance R2 is determined as follows:

R1 R3 79.62 796.2 [2 marks]


R2 2.04
4Q R1 R3 4 10 2 79.62 796.2
2

2 mF

2.04
V in
79.62 2 mF V out

796.2 [3 marks]

3
EKT 214 SEM 2 2011

Question 3
(a) Explain the following terms for voltage regulator:
(i) Line regulation
(ii) Load regulation
[6 marks]
Ans:
Line regulation as applied to a voltage regulator is a measure of its ability to maintain a
constant output voltage despite variation in the input voltage. It is the change of output
voltage per unit change in the input voltage and is normally expressed in percentage form,
thus; [2 marks]
V
Line regulation out 100% [1 mark]
Vin

Ans:
Load regulation as applied to a voltage regulator is a measure its ability to maintain a
constant output voltage despite variation in the load connected its terminal, and is normally
express in percentage form, thus; [2 marks]

V Vfullload
Load regulation noload 100% [1 mark]
Vfullload

(b) The full-load current of a voltage regulator is 1.5A and the output voltage at this
current is 14.86 V. When the load is disconnected from the regulator, its output
voltage is 15.2 V. Calculate the load regulation of the regulator.
[2 marks]

Ans:
V Vfullload
Load regulation noload 100%
Vfullload

[1 mark]
15.2 14.86
100%
14.86

2.29% [1 mark]

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