Download EXP 6 Active Filters

UNIVERSITI MALAYSIA PERLIS
ANALOG ELECTRONICS II
EMT 212
EXPERIMENT # 6
ACTIVE FILTERS
MARKS
T1
T2
T3
T4
G1
G2
Q
C
Total
2
21
2
21
3
3
10
10
72
NAME
PROGRAMME
signature
MATRIK #
EXPERIMENT 6
GROUP
DATE
100%
Analog Electronics II (EMT 212) 2007/2008
Laboratory Module
EXPERIMENT 6
Active Filters
1. OBJECTIVE:
1.1. To demonstrate the use of operational amplifiers in active filters
1.2. To specify the components required for a Butterworth low or high-pass filter
1.3. To build and test a Butterworth low or high-pass active filter for a specific
frequency
2. INTRODUCTION:
A filter is a circuit that produces a prescribed frequency response as described in the
experiment on Passive Filters. Passive filters are combination circuits containing only
resistors, inductors, and capacitors (RLC). Active filters contain resistance and
capacitance plus circuit elements that provide gain, such as transistors or operational
amplifiers. The major advantage of active filters is that they can achieve frequency
response characteristics that are nearly ideal and for reasonable cost for frequency
up to about 100 kHz. Above this, active filters are limited by bandwidth.
Active filters can be designed to optimize any of several characteristics. These
include flatness of the response in the passband, steepness of the transition region,
or minimum phase shift. The Butterworth form of filter has the flattest passband
characteristic, but is not as steep as other filters and has poor phase characteristics.
Since a flat passband is generally the most important characteristic, it will be used in
this experiment.
The order of a filter, also called the number of poles, governs the steepness of
the transition outside the frequencies of interest. In general, the higher the order, the
steeper the response. The roll-off rate for active filters depends on the type of filter
but is approximately – 20dB/decade for each pole. (A decade is a factor of 10 in
frequency). A four pole filter, for example, has a roll-off of approximately –
80dB/decade. A quick way to determine the number of poles is to count the number
of capacitors that are used in the frequency-determining part of the filter.
Figure 2.1 illustrates a two-pole active low-pass and a two-pole active high-pass
filter. Each of these circuits is a section. To make a filter with more poles, simply
cascade these sections, but change the gains of each section according to the
values listed in TABLE 1.
1
Analog Electronics II (EMT 212) 2007/2008
Laboratory Module
(a) Second Order Low-Pass Filter
(b) Second Order High-Pass Filter
Figure 2.1 Second Order Filter
TABLE 1
Poles
2
4
6
Section 1
1.586
1.152
1.068
Gain Required
Section 2
Section 3
2.235
1.586
2.483
Useful Relationships:
fc 
1
(Single-pole filter)
2 RC
1
(Two-pole filter)
2 R1 R2C1C2
V
ACL  out
Vin
ACL ( dB )  20 log ACL
fc 
2
(2.1)
(2.2)
(2.3)
(2.4)
Analog Electronics II (EMT 212) 2007/2008
Laboratory Module
3. EQUIPMENT:
3.1. 15 k (2)
3.2. 10 k (1)
3.3. 22 k (1)
3.4. 0.47 µF (2)
3.5. 0.022 µF (1)
3.6. 0.01 µF (2)
3.7. LM 741 OP-AMP
3.8. DC Power Supply
3.9. Dual-trace oscilloscope
3.10. Function generator
3.11. Voltmeter
3.12. Breadboard
4. PROCEDURE:
4.1. Second Order Low-Pass Active Filter
4.1.1. Calculate theoretical cutoff frequency for Second Order Low Pass Filter
shown in Figure 4.1 and record in TABLE 2.
4.1.2. Construct circuit shown in Figure 4.1. Connect the function generator at
input. Adjust the function generator to produce 1 Vp sine wave at 500 Hz.
4.1.3. Increase the frequency from function generator until the output voltage V0
is equal to 0.707 times the input voltage. The frequency where this occurs
is the cutoff frequency of the filter. Measure and record this frequency,
input voltage, output voltage in TABLE 2 and TABLE 3.
4.1.4. Set the frequency to the value in TABLE 2, measure and record input and
output voltage. Complete TABLE 3.
4.1.5. Use the values to calculate the voltage gain, A and AdB . Complete
TABLE 2 and plot the ideal and experiment voltage gain versus frequency
on GRAPH 1.
3
Analog Electronics II (EMT 212) 2007/2008
Laboratory Module
C2
0.022uF
7
+15V
R1
R2
2
-
3
+
6
IN PUT,Vin
OUT PUT,Vout
LM741
15k
4
15k
C1
0.01uF
-15V
Figure 4.1 Second Order Low-pass filter
4.2. Second Order High-Pass Active Filter
4.2.1. Calculate theoretical cutoff frequency for Second Order High Pass Filter
shown in Figure 4.2 and record in TABLE 4.
4.2.2. Construct circuit shown in Figure 4.2. Connect the function generator at
input. Adjust the function generator to produce 1 Vp sine wave at 1kHz.
4.2.3. Increase the frequency from function generator until the output voltage V0
is equal to 0.707 times the input voltage. The frequency where this occurs
is the cutoff frequency of the filter. Measure and record this frequency,
input voltage, output voltage in TABLE 4 and TABLE 5.
4.2.4. Set the frequency to the value in TABLE 4, measure and record input and
output voltage. Complete TABLE 5.
4.2.5. Use the values to calculate the voltage gain, A and AdB . Complete
TABLE 4 and plot the ideal and experiment voltage gain versus frequency
on GRAPH 2.
4
Analog Electronics II (EMT 212) 2007/2008
Laboratory Module
R1
10k
7
+15V
C1
C2
2
-
3
+
6
IN PUT,Vin
LM741
0.01uF
4
0.01uF
R2
22k
-15V
Figure 4.2 Second Order High-pass filter
5
OUT PUT,Vout
Analog Electronics II (EMT 212) 2007/2008
Laboratory Module
Name
:_____________________________
Matric No.
: _______________________________
Date : _______________
5. RESULT:
i. Second Order Low-Pass Filter
TABLE 2
Frequency
(Calculated)
fC LPF 
Frequency
(Experiment)
1
2 R1R2C1C2
Vout  0.707Vin
(2 marks)
TABLE 3
frequency, f (kHz)
Vin
Vout
A
A(dB)
1
fC 
12 LPF
1
fC 
8 LPF
1
fC 
4 LPF
f C LPF 
2 f C LPF 
4 f C LPF 
6 f CLPF 
(21 marks)
6
Analog Electronics II (EMT 212) 2007/2008
Laboratory Module
Name
:_____________________________
Matric No.
: _______________________________
Date : _______________
ii. Second Order High-Pass Filter
TABLE 4
Frequency
(Calculated)
fC LPF 
Frequency
(Experiment)
1
2 R1R2C1C2
Vout  0.707Vin
(2 marks)
TABLE 5
frequency, f (kHz)
Vin
Vout
A
A(dB)
1
fC 
8 HPF
1
fC

4 HPF
1
fC 
2 HPF
f C HPF 
4 f C HPF 
10 fC HPF 
50 f C HPF 
(21 marks)
7
Analog Electronics II (EMT 212) 2007/2008
Laboratory Module
Name
:_____________________________
Matric No.
: _______________________________
Date : _______________
SECOND ORDER LOW PASS ACTIVE FILTER
DIFFERENTIATOR FREQUENCY RESPONSE
30
4
28
2
26
0
24
-2
22
-4
20
-6
VOLTAGE GAIN (dbB)
d B VOLTAGE GAIN
18
-8
16
-10
14
-12
12
-14
10
-168
-186
-204
2
-22
0
-24
2
-26
4
-28
6
-30
8
10
10
100
3
1 10
4
1 10
5
1 10
FREQUENCY
GRAPH 1
8
(3 marks)
Analog Electronics II (EMT 212) 2007/2008
Laboratory Module
Name
:_____________________________
Matric No.
: _______________________________
Date : _______________
SECOND ORDER HIGH PASS ACTIVE FILTER
DIFFERENTIATOR FREQUENCY RESPONSE
30
4
28
2
26
0
24
-2
22
-4
20
-6
VOLTAGE GAIN (dbB)
d B VOLTAGE GAIN
18
-8
16
-10
14
-12
12
-14
10
-168
-186
-204
2
-22
0
-24
2
-26
4
-28
6
-30
8
10
10
100
3
1 10
4
1 10
5
1 10
FREQUENCY
(3 marks)
GRAPH 2
9
Analog Electronics II (EMT 212) 2007/2008
Laboratory Module
Name
:_____________________________
Matric No.
: _______________________________
Date : _______________
6. QUESTION:
Q1
mark
(2)
What is an Active Filter?
A1
Q2
State the cut-off frequency for Low-Pass Filter in Graph 1.
(2)
State the cut-off frequency for High-Pass Filter in Graph 2.
(2)
Determine the bandwidth of Low-Pass Filter by referring Graph 1.
(2)
How to obtain the appropriate frequency response characteristic?
(2)
A2
Q3
A3
Q4
A4
Q5
A5
10
Analog Electronics II (EMT 212) 2007/2008
Laboratory Module
Name
:_____________________________
Matric No.
: _______________________________
Date : _______________
7. CONCLUSION: (10 MARKS)
Make your overall conclusion by referring to the Low-Pass Filter and High-Pass Filter
experiment result (your answer should be in simple notes).
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