Ee321 lab expt 7_negative_feedback_in_ amplifiers

EE321: Analog Circuits Laboratory
Experiment 7: Negative Feedback Amplifiers using BJT
1
Objectives:
1.To study the influence of the negative feedback in BJT amplifier circuits.
2.To examine via experimentation the properties of the Current-Series, Current-Shunt, Voltage-Series and
Voltage-Shunt feedback BJT amplifiers.
3.To determine the input impedance, output impedance, gain, bandwidth of BJT amplifiers with and
without feedback.
Pre Lab Work:
Read about feedback Amplifiers from Microelectronic Circuits by Sedra, Smith & Chandorkar.
Introduction and Theory
Transistors amplifiers are commonly used in applications like RF (radio frequency), audio, OFC (optic fiber
communication) etc. Anyway the most common application we see in our day to day life is the usage of
transistor as an audio amplifier. As you know there are three transistor configurations that are used commonly
i.e. common base (CB), common collector (CC) and common emitter (CE). In common base configuration has a
gain less than unity and common collector configuration (emitter follower) has a gain almost equal to unity).
Common emitter follower has a gain that is positive and greater than unity. So, common emitter configuration is
most commonly used in audio amplifier applications.
A good transistor amplifier must have the following parameters; high input impedance, high band width, high
gain, high slew rate, high linearity, high efficiency, high stability etc.
Feedback plays a very important role in electronic circuits and the basic parameters, such as input impedance,
output impedance, current and voltage gain and bandwidth, may be altered considerably by the use of feedback
for a given amplifier.
A portion of the output signal is taken from the output of the amplifier and is combined with the normal input
signal and thereby the feedback is accomplished.
There are two types of feedback. They are i) Positive feedback and ii) Negative feedback.
Negative feedback helps to increase the bandwidth, decrease gain, distortion, and noise, modify input and
output resistances as desired. An amplifier circuit equipped with some amount of negative feedback is not only
more stable, but it distorts the input waveform less and is generally capable of amplifying a wider range of
frequencies. The tradeoff for these advantages (there just has to be a disadvantage to negative feedback, right?)
is decreased gain. If a portion of an amplifier's output signal is “fed back” to the input to oppose any changes in
the output, it will require a greater input signal amplitude to drive the amplifier's output to the same amplitude
as before. This constitutes a decreased gain. However, the advantages of stability, lower distortion, and greater
bandwidth are worth the tradeoff in reduced gain for many applications.
A current shunt feedback amplifier circuit is illustrated in the figure 1. It is called a series-derived, shunt-fed
feedback. The shunt connection at the input reduces the input resistance and the series connection at the output
increases the output resistance. This is a true current amplifier.

Voltage shunt feedback is also called shunt derived shunt feedback connection. Here a fraction of the output is
supplied in parallel with input voltage through the feedback network. This type of amplifier is also called as
trans-resistance amplifier.
In Current-Series Feedback, the input impedance and the output impedance are increased. Noise and distortions
are reduced considerably.
Figure 1 shows the basic feedback topologies. The effect of the feedback topology on the amplifier input-output
resistance levels can be summarized as follows:
Fig 1. Basic feedback topologies
Table (i): The different topologies of the feedback with their analysis a) Current-shunt b) Voltage-shunt c)
Current-series d) voltage series
Characteristics Voltage-series Current –series Current-shunt Voltage-shunt
G I  V 0
2
f
o
X
V
V
  X f
o
f
o
V
I
f
o
I
I
I
f
V
o
A  X 0
X 0
i
A  V 0
V
V i
m
i
A  I 0
i
I i
m
i
R V  I
D  1  A 1 V D    A 1 m D  G 1 i D    A 1 m D    R
if R i R D i R D / i R D / i R D
0 f R 0 / (1 ) V R   A 0 (1 ) m R   G 0 (1 ) i R    A 0 / (1 ) m R   R

Table (ii): The effect of Negative feedback on Amplifier characteristics
Characteristics Voltage-series Current –series Current-shunt Voltage-shunt
of R Decreases Increases Increases Decreases
if R Increases Increases Decreases Decreases
Gain Decreases Decreases Decreases Decreases
Bandwidth Increases Increases Increases Increases
Non-linear Decreases Decreases Decreases Decreases
3
PROCEDURE
A. INPUT AND OUTPUT IMPEDANCEMEASUREMENT FOR ANY AMPLIFIER CIRCUIT
Fig. (a) Measuring input impedance
1. Connections are made as per circuit diagram shown in Fig. (a).
2. Signal generator is set to provide a sine wave output at 1kHz. The amplitude of the input signal should
be adjusted so that the display on the oscilloscope is noise free (large enough) and distortion free (not
too large) say 200mV. Because of the attenuator the net input applied to amplifier will be 0.1 times
signal value i.e., 20 mV. The display on the oscilloscope screen should be as large as is practical and set
so that its amplitude and half its amplitude can be easily estimated.
3. The resistance at the amplifier input should then be increased until the output waveform is exactly half its
previously set value. At this setting the signal is shared equally between the test resistance and the input
impedance of the amplifier, meaning that the resistance and impedance are equal. After switching off and
removing the test resistance, measuring the variable resistor with an Ohm meter gives the value equivalent
to the input impedance of the amplifier.
Fig. (b) Measuring Output impedance

4. Connections are made as per circuit diagram shown in Fig. (b).
5. The measurement of output impedance uses the same method as for input impedance but with different
connections. In this case the amplifier load is replaced with variable resistor. Care must be taken
however, to ensure that the resistance connected in place of the load is able to dissipate sufficient power
without damage
6. Initially the output from the amplifier should be adjusted for a display similar to that used for the input
impedance test, but this time with no load connected to the output terminals. The test resistance is then
connected across the output terminals and adjusted for maximum resistance before switching on the
amplifier. The test resistance is reduced in value until the display indicates half the amplitude of that
noted with no load. This test resistance is now the same value as the output impedance.
7. Now again measure the input and output impedance with feedback in the amplifier circuit.
4
Voltage- series Feedback Amplifier
(a)
200mVrms
1kHz
0°
vi
Rc1
4.7kΩ
Q1
BC 547
68kΩ
R1
2.2kΩ
22kΩ Re
R2
100Ω
R5
R5
2.2kΩ
Q2
BC 547
47kΩ
R1
33kΩ
R2
2.2kΩ
Re
VCC
12V
10μF
C2
C2
10μF
100μF
CE CE1
100μF
10μF
CB
20kΩ
R2
180kΩ
R2
1kΩ
RL
DSO

(b)
200mVrms
1kHz
0°
vi
Rc1
4.7kΩ
R5
2.2kΩ
Fig.-2 Circuit diagram of Voltage-series feedback Amplifier (a) without feedback (b) with feedback
5
Case a: Without feed back
1. Connections are made as per circuit diagram Fig. 2(a)
2. Measure input and output impedance of the amplifier as described in section III (a)
3. Connect function generator to give sine wave input signal (Vin) and set the value of Vin at 200 mV peak
to peak and frequency 20 Hz.
4. To generate very small signal of the order of 20 mV pp, we used voltage divider (of voltage gain 1:10)
between signal generator and amplifier input.
5. To view the output and the input signal simultaneously, connect one probe of CRO/DSO to Vin and
another probe across RL. Connect ground of DSO, DC power supply and Function Generator very close
to each other at a single point on breadboard and to amplifier ground.
6. Keep the input voltage constant at 20mV peak-peak and 1 KHz frequency. Note down the output voltage
and calculate the gain by using the expression 0
10 20*log ( ) dB v
i
A V
V

(This is the gain at mid-frequency which corresponds to maximum gain)
7. Keeping the input voltage at constant at 20mV peak-peak, the frequency is slowly increased until output
voltage becomes 0.707 0 V . Stop and note down the frequency which corresponds to higher cut-off
frequency.
8. Repeat the same procedure by decreasing the frequency and note down the frequency at which output
voltage becomes 0.707 0 V , which corresponds to lower cut-off frequency.
9. The Bandwidth of the amplifier is calculated from the graph using the expression Bandwidth
. H L BW  f  f .
Q1
BC 547
68kΩ
R1
2.2kΩ
22kΩ Re
R2
100Ω
R5
Q2
BC 547
47kΩ
R1
33kΩ
R2
2.2kΩ
Re
VCC
12V
10μF
C2
C2
10μF
100μF
CE CE1
100μF
10μF
CB
20kΩ
R2
180kΩ
R2
1kΩ
RL
DSO
4.7kΩ
Rf
10μF
C2

10. The gain-bandwidth product of the amplifier is calculated by using the expression
Gain-Bandwidth Product = (3dB mid-band gain) X (Bandwidth).
6
Case (b):With feed back
11. Connections are made as per circuit diagram Fig. 2(b)
12. Measure input and output impedance as described in section III (a)
10 20*log ( ) dB v
i
A V
V

frequency.
. H L BW  f  f .
Note down the following:
Practical
Without feedback With feedback
Input Impedance
Output Impedance
Gain (Mid Band) in dB
Lower cut-off frequency ( fL )
Higher cut-off frequency ( fH )
Band width ( fH—fL )
Gain-Bandwidth Product
Fig. (c) Model Frequency response of a feedback amplifier with and without feedback

VCC
12V
68kΩ R4
R1
4.7kΩ
7
B. VOLTAGE- SHUNT FEEDBACK AMPLIFIER
(a)
68kΩ R4
R1
(b)
4.7kΩ
Q1
BC 547
1.2kΩ
R5
15kΩ
R2
22μF
C2
22μF
CB
200mVrms
1kHz
0°
vi
20kΩ
R2
180kΩ
R1
Fig.-3 Circuit diagram Voltage-shunt feedback amplifier (a) without feedback (b) with feedback
1kΩ
RL
DSO
47μF
CE
Q1
BC 547
1.2kΩ
R5
15kΩ
R2
VCC
12V
22μF
C2
22μF
CB
200mVrms
1kHz
0°
vi
20kΩ
R2
180kΩ
R1
1kΩ
RL
DSO
47μF
CE
22μF
Cf
47kΩ
Rf

8
10 20*log ( ) dB v
i
A V
V

frequency.
. H L BW  f  f .
10 20*log ( ) dB v
i
A V
V

frequency.
. H L BW  f  f .
Note the following:
Practical
Input Impedance
Output Impedance

47kΩ
Rf
9
C. CURRENT- SHUNT FEEDBACK AMPLIFIER
(a)
(b)
Rc1
10kΩ
Q1
BC 547
47kΩ
R1
5kΩ
R2
2kΩ
Re1
47kΩ
Rc2
Rc1
10kΩ
47kΩ
Rc2
Fig.-4 Circuit diagram Current-shunt feedback amplifier (a) without feedback (b) with feedback
Q2
BC 547
47kΩ
R3
5kΩ
R4
2kΩ
Re2
VCC
12V
1kΩ
RL
22μF
CB
200mVrms
1kHz
0°
vi
20kΩ
R2
180kΩ
R1
DSO
22μF
CB
22μF
CB
100μF
CB
Q1
BC 547
47kΩ
R1
5kΩ
R2
2kΩ
Re1
Q2
BC 547
47kΩ
R3
5kΩ
R4
2kΩ
Re2
VCC
12V
1kΩ
RL
22μF
CB
200mVrms
1kHz
0°
vi
20kΩ
R2
180kΩ
R1
DSO
22μF
CB
22μF
CB
100μF
CB
S1
22μF Key = Space
Cf

10 20*log ( ) dB v
10
i
A V
V

frequency.
. H L BW  f  f .
10 20*log ( ) dB v
i
A V
V

frequency.
. H L BW  f  f .
Note down the following:
Practical
Input Impedance
Output Impedance
---------------------------------------------------------------------------

Ee321 lab expt 7_negative_feedback_in_ amplifiers

More Related Content

What's hot

Viewers also liked

Similar to Ee321 lab expt 7_negative_feedback_in_ amplifiers

Recently uploaded

Ee321 lab expt 7_negative_feedback_in_ amplifiers