AIM: EXPERIMENT TO DEMONSTRATE ACTIVE FILTERS
OBJECTIVES
 To determine the input/output characteristics of a low pass filter
 To Explain the polarity of the capacitors
 To calculate cut off frequency of a lowpass filter.
APPARATUS
 DC power supply
 Signal generator
 Resistors R1 = 1k, R2 = 10k and 100k, R3 = R4 = 330 Ω
 Capacitors C1 = 2.2 µF ,C2=47 µF
 Op- amp UA 741
THEORY
A filter is a device that passes electric signals at certain frequencies or frequency ranges while
preventing the passage of others.
A filter is a circuit that is designed to pass a specified band of frequencies while attenuating all
signals outside this band. Filter networks may be either active or passive. Passive filter
networks contain only resistors, inductors, and capacitors. Active filters, employ operational
amplifiers (op-amps) as well as resistors and capacitors
The passband of a filter is the range of frequencies that are allowed to pass through the filter
with minimum attenuation. (Usually defined as less than – 3 dB of attenuation).
The most common kinds of frequency selective filters are:
 Low pass filter – passes signals with a frequency lower than certain cut off frequency
 High pass filters – passes signals with a frequency higher than certain cut off
frequency
 Band pass filters – passes signals with a frequency within certain band or vice versa.
The critical frequency, fc ,(also called the cutoff frequency) is defined as the end of the
passband and is normally specified at the point where the response drops – 3 dB (70.7%) from
the passband response.
Following the passband is a region called the transition region that leads into a region called
the stopband.
There is no precise point between the transition region and the stopband.
Fig 1. Frequency response of a band pass filter
Fig 2. Frequency response of a high pass filter
Fig 3. Frequency response of a low pass filter.
The frequency response of a filter is the graph of its voltage gain versus frequency. Frequency
response of an operational amplifier refers to how the amplifier responds to different input
frequencies.
In order to measure the frequency response we use input signals with a range of frequencies .
The input signal can be a sinusoidal waveform, a square wave, a triangular wave or another type
of a periodic wave. The output signal will also be a periodic waveform but the amplitude and the
phase of the output may be different from the input depending on the character of the amplifier.
Gain is given by;
𝑜𝑢𝑡𝑝𝑢𝑡 𝑣𝑜𝑙𝑡𝑎𝑔𝑒
Gain = 𝑖𝑛𝑝𝑢𝑡 𝑣𝑜𝑙𝑡𝑎𝑔𝑒
to convert gain to dB we use the following formula
gain in dB=20log|gain|
A graph of the magnitude of gain versus frequency in dB on a logarithmic scale can be plotted.
The frequency of an op amp is important because it is used to determine the amplifier’s
performance at different frequencies.
The frequency response of an op amp can affected by different factors such as the gain-band
width product, the input and output impendences and the type of feedback used in the amplifier
circuit.
THE LOW PASS FILTER
A low-pass filter is one that passes frequencies from dc to fc , and significantly attenuates all
other frequencies. This RC circuit will provide a low frequency path to the input of the
amplifier. The amplifier acts as a buffer circuit providing unity gain output. This circuit has
more input impedance value. Even though the input impedance of the op-amps high below the
cut-off frequency, this input impedance is limited by the series impedance which is equal to
R+ 1⁄jωC.
The output impedance of the op-amp which is connected in the circuit is always low. This
circuit will provide high stability to filter. The main drawback of this configuration is voltage
gain is unity. Even for this circuit also the output power is high since the input impedance is
low.
Fig 5. An active RC circuit
PROCEDURE
The circuit is arranged as shown below.
 The output of the op amp is biased under no signal conditions at half supply volts. This is
achieved by using the potential divider R3 and R4 and helps to have a maximum output
undistorted signal swing.
 The frequencies are varied from the signal generator from 100Hz to 1MHz as we obtain
the output and the results are tabulated.
TABLE OF RESULTS
WHEN R2=10KΩ
FREQUENCY
100
200
300
500
1000
2000
5000
8000
10000
20000
30000
50000
70000
100000
200000
300000
600000
900000
1000000
INPUT
1.24
1.24
1.24
1.24
1.24
1.24
1.24
1.24
1.24
1.24
1.24
1.24
1.24
1.24
1.24
1.24
1.24
1.24
1.24
OUTPUT
22.4
22.4
22.4
22.4
22.4
22.4
22.4
22.4
22.4
18.0
12.0
5.20
3.60
1.60
1.20
800m
800m
800m
800m
GAIN
18.06
18.06
18.06
18.06
18.06
18.06
18.06
18.06
18.06
14.52
9.68
4.19
2.90
1.29
0.97
0.65
0.65
0.65
0.65
Gain(dB)
25.13
25.13
25.13
25.13
25.13
25.13
25.13
25.13
25.13
23.24
19.18
12.44
9.25
2.21
-0.26
-3.74
-3.74
-3.74
-3.74
Note and explain the polarity of the capacitor C1 and C2
C1
Terminal connected to the 10k resister is negative and the other to the Vin positive.
C2
The negative terminal is connected to the op amp and the positive is connected to the output.
Explanation;
For C1 the terminal that is connected to the input is the positive terminal and the terminal
connected to the other connected to the 10k resister is the negative.
For C2 the output voltage is measured with respect to the ground therefore it is the negative and
the other is the negative.
TABLE OF RESULTS WHEN R2=1MΩ
Frequency
100
200
300
500
1000
2000
5000
8000
10000
20000
30000
50000
70000
100000
200000
300000
600000
900000
1000000
Input
1.21
1.21
1.21
1.21
1.21
1.21
1.21
1.21
1.21
1.21
1.21
1.21
1.21
1.21
1.21
1.21
1.21
1.21
1.21
Output
24.0
23.6
23.6
23.6
23.2
23.2
23.2
23.2
23.2
18.4
13.2
5.20
3.60
1.60
1.20
800m
800m
800m
800m
Gain
19.83
19.50
19.50
19.50
19.17
19.17
19.17
19.17
19.17
15.21
10.91
4.30
3.00
1.32
0.99
0.66
0.66
0.66
0.66
Gain(dB)
25.95
25.80
25.80
25.80
25.65
25.65
25.65
25.65
25.65
23.64
20.76
12.67
9.54
2.41
-0.09
-3.61
-3.61
-3.61
-3.61
OBSERVATION
At the input, one terminal is connected to the Vin therefore that side of the capacitor is positive
and the side connected to the 10KΩresistor is negative. Also at the output, voltage is measured
with respect to the ground implying that the right side of the capacitor,C2 is positive and the
one connected to the ground is negative.
CONCLUSION
The polarity of the capacitors was determined.
REFERENCES
ACTIVE_FILTERS_Theory_and_Design.pdf
https://www.electronicshub.org/active-low-pass-filter/