ECE DEPARTMENT
KITSS, HUZURABAD
1. P-N JUNCTION DIODE CHARACTERISTICS
Aim:
1. To plot the Volt-Ampere Characteristics of P-N Junction Diode under forward and
reverse bias conditions.
2. To find static and dynamic resistance in forward bias condition.
Equipments required:
1. DC Regulated Power Supply(0-30V)
2. Digital Ammeter((0-200)μA/(0-200)mA) )
3. Digital Voltmeter(0-20v)
Components required:1. Bread board
2. Resistor 1KΩ
3. Diode
Circuit Diagram:
i)FORWARD BIAS:
ii) REVERSE BIAS:
1
ECE DEPARTMENT
KITSS, HUZURABAD
THEORY:A p-n junction diode conducts only in one direction. The V-I characteristics of the diode are
curve between voltage across the diode and current through the diode. When external voltage is
zero, circuit is open and the potential barrier does not allow the current to flow. Therefore, the
circuit current is zero. When P-type (Anode is connected to +ve terminal and n- type (cathode) is
connected to –ve terminal of the supply voltage, is known as forward bias. The potential barrier
is reduced when diode is in the forward biased condition. At some forward voltage, the potential
barrier altogether eliminated and current starts flowing through the diode and also in the circuit.
The diode is said to be in ON state. The current increases with increasing forward voltage.
When N-type (cathode) is connected to +ve terminal and P-type (Anode) is connected to –ve
terminal of the supply voltage is known as reverse bias and the potential barrier across the
junction increases. Therefore, the junction resistance becomes very high and a very small current
(reverse saturation current) flows in the circuit. The diode is said to be in OFF state. The reverse
bias current due to minority charge carriers.
PROCEDURE:
(i) FORWARD BIAS :
1. Connections are made as per the circuit diagram.
2. For forward bias, the RPS +ve is connected to the anode of the diode and RPS –ve is
connected to the cathode of the diode,
3. Switch ON the power supply and increases the input voltage (supply voltage) in Steps.
4. Note down the corresponding current flowing through the diode and voltage across the diode
for each and every step of the input voltage.
5. The readings of voltage and current are tabulated.
6. Graph is plotted between voltage on x-axis and current on y-axis.
2
ECE DEPARTMENT
KITSS, HUZURABAD
(ii) REVERSE BIAS :
1. Connections are made as per the circuit diagram.
2. For reverse bias, the RPS +ve is connected to the cathode of the diode and RPS –ve is
connected to the anode of the diode.
3. Switch ON the power supply and increase the input voltage (supply voltage) in Steps.
4. Note down the corresponding current flowing through the diode and voltage across the diode
for each and every step of the input voltage.
5. The readings of voltage and current are tabulated.
6. The Graph is plotted between voltage on x-axis and current on y-axis.
Observation table:
Forward bias:
S.No
Voltage(Volts)
Current(Milli Amp)
Reverse bias:
S.No
Voltage(Volts)
Current(Micro Amp)
3
ECE DEPARTMENT
KITSS, HUZURABAD
V-I CHARACTERISTICS:
Calculations:
Static resistance (Rf)= Vf / If
Dynamic resistance (rf)= Vf / If =(V2-V1) / (I2-I1)
RESULT: The V-I Characteristics of the PN junction diode are plotted for the both Forward and
Reverse Bias conditions and calculated static and dynamic resistance in forward bias condition.
The Static forward resistance of Diode is_____________
The Dynamic forward resistance of Diode is__________
4
ECE DEPARTMENT
KITSS, HUZURABAD
2. Full Wave Rectifier with & without filters
Aim: To Rectify the AC signal and then to find out Ripple factor of Full-wave bridge
rectifier circuit with and without filter.
Equipments Required:
CRO
Transformer
BNC Probes
Bread board and connecting wires
Components Required: Resistors (1KΩ, 10KΩ)
Diodes IN4007
Circuit Diagrams:
5
ECE DEPARTMENT
KITSS, HUZURABAD
Theory:
A device is capable of converting a sinusoidal input waveform into a unidirectional
waveform with non zero average component is called a rectifier.
The Bridge rectifier is a circuit, which converts an ac voltage to dc voltage using both half
cycles of the input ac voltage. The Bridge rectifier has four diodes connected to form a Bridge.
The load resistance is connected between the other two ends of the bridge.
For the positive half cycle of the input ac voltage, diode D1 and D3 conducts whereas
diodes D2 and D4 remain in the OFF state. The conducting diodes will be in series with the
load resistance RL and hence the load current flows through RL .
For the negative half cycle of the input ac voltage, diode D2 and D4 conducts whereas
diodes D1 and D3 remain in the OFF state. The conducting diodes will be in series with the
load resistance RL and hence the load current flows through RL in the same direction as in the
previous half cycle. Thus a bidirectional wave is converted into a unidirectional wave.
Procedure:
With out filter:
1.
2.
3.
4.
5.
Connecting the circuit on bread board as per the circuit diagram.
Observe the waveforms on CRO.
Find maximum voltage(Vm) from the full wave.
Calculate Vrms and Vdc using formulas.
Calculate ripple factor.
With filter:
1.
2.
3.
4.
5.
Connecting the circuit on bread board as per the circuit diagram.
Observe the waveforms on CRO.
Measure Vmax and Vmin.
Calculate Vrms, Vdc and ripple factor.
Repeat this different values of C.
6
ECE DEPARTMENT
KITSS, HUZURABAD
Expected waveforms:
Tabular forms:
With out filter:
S.No
Vm
Vdc=2Vm/π
Vrms= Vm / 2
Ripple factor
= (Vrms/Vdc)2-1
With filter:
S.No
C
Vmax Vmin
Vdc=(Vmax+Vmin)/2
Vrms=(Vmax-Vmin)/2 2 Ripple factor=
Vrms/Vdc
Result: Observed the input and output waveforms and calculated the ripple factor of full wave
bridge rectifier with and with out filter.
Without filter: Ripple factor:
With filter :
Ripple factor:
7
ECE DEPARTMENT
KITSS, HUZURABAD
3. COMMON EMITTER AMPLIFIER CHARACTERISTICS
Aim: To study CE amplifier and to find the frequency response, band width.
Equipments required:
1. Single channel power supply(0-30V)
2. Function Generator(0-3MHz, 20m-30V)
3. CRO(0-30MHz, 0-20Vp-p)
4. BNC Probes(1:10)
Components required:
1. Bread board
2. Resistors:
3. Capacitors:
4. Transistor:
5. Connecting wires.
22KΩ
2.2KΩ
-
1 no.
1 no.
10KΩ
-
1 no.
1KΩ
-
1 no.
10μF
100μF
BC107
-
2 no.s
1 no.
1 no.
Circuit diagram:
8
ECE DEPARTMENT
KITSS, HUZURABAD
Theory:
A common emitter amplifier is one of three basic single-stage bipolar-junctiontransistor (BJT) amplifier topologies, typically used as a voltage amplifier. In The input signal is
applied across the ground and the base circuit of the transistor. The output signal appears across
ground and the collector of the transistor. Since the emitter is connected to the ground, it is
common to signals, input and output.
Common emitter amplifiers give the amplifier an inverted output and can have a very
high gain that may vary widely from one transistor to the next. The gain is a strong function of
both temperature and bias current, and so the actual gain is somewhat unpredictable. Stability is
another problem associated with such high gain circuits due to any unintentional positive
feedback that may be present.
The voltage gain depends almost exclusively on the ratio of the resistors rather than the
transistor's intrinsic and unpredictable characteristics. The distortion and stability characteristics
of the circuit are thus improved at the expense of a reduction in gain.
The common- emitter circuit is the most widely used of junction, transistor amplifiers. As
compared with the common- base connection, it has higher input impedance and lower output
impedance. A single power supply is easily used for biasing.
Current gain in the common emitter circuit is obtained from the base and the collector
circuit currents. Because a very small change in base current produces a large change in collector
current, the current gain (β) is always greater than unity for the common-emitter circuit, a typical
value is about 50.
Procedure:
1.
2.
3.
4.
5.
6.
Make the connections as per the circuit diagram.
Keep the input voltage constant at 24mV.
By changing the frequency at input, find the corresponding output.
Calculate the gain.
Plot the frequency response.
Calculate the band width from the response curve.
9
ECE DEPARTMENT
KITSS, HUZURABAD
Observation table: Keep Vin = 24mV
S. No
F(KHZ)
Vout(V)
Gain=Vo/Vin
Gain(db)=20log(gain)
Expected graph:
Calculations:
Band width = f2-f1
Result: The CE amplifier is studied and frequency response is plotted. Band width is found to be
___________ .
10
ECE DEPARTMENT
KITSS, HUZURABAD
4. COMMON BASE AMPLIFIER CHARACTERISTICS
Aim: To study CB amplifier and to find the frequency response, band width.
Equipments required:
1. Single channel power supply(0-30V)
2. Function Generator(0-3MHz, 20m-30V)
3. CRO(0-30MHz, 0-20Vp-p)
4. BNC Probes(1:10)
Components required:
1. Bread board
2. Resistors:
10KΩ
3.3KΩ
-
1 no.
1 no.
1KΩ
-
1 no.
470Ω
-
1 no.
-
3 no.s
1 no.
3. Capacitors:
1μF
4. Transistor:
BC107
5. Connecting wires.
Circuit Diagram:
11
ECE DEPARTMENT
KITSS, HUZURABAD
Theory:
In Common Base Amplifier Circuit Base terminal is common to both the input and output
terminals. In this Circuit input is applied between emitter and base and the output is taken from
collector and the base.
As we know, the emitter current is greater than any other current in the transistor, being
the sum of base and collector currents i.e. IE= IB+ IC In the CE and CC amplifier configurations,
the signal source was connected to the base lead of the transistor, thus handling the least current
possible.
Because the input current exceeds all other currents in the circuit, including the output
current, the current gain of this amplifier is actually less than 1 (notice how Rload is connected to
the collector, thus carrying slightly less current than the signal source). In other words, it
attenuates current rather than amplifying it.
With common-emitter and common-collector amplifier configurations, the transistor
parameter most closely associated with gain was β. In the common-base circuit, we follow
another basic transistor parameter: the ratio between collector current and emitter current, which
is a fraction always less than 1. This fractional value for any transistor is called the alpha ratio, or
α ratio (α= IC/IE). Since it obviously can't boost signal current, it only seems reasonable to
expect it to boost signal voltage.
Procedure:
1.
2.
3.
4.
5.
6.
Make the connections as per the circuit diagram.
Keep the input voltage constant at 100mV.
By changing the frequency at input, find the corresponding output.
Calculate the gain.
Plot the frequency response.
Calculate the band width from the response curve.
12
ECE DEPARTMENT
KITSS, HUZURABAD
Observation table: Keep Vin = 100mV
S. No
F(KHZ)
Vout(V)
Gain= Vo/Vin
Gain(db)=
20log(gain)
Expected graph:
Calculations: Band width = f2-f1
Result: The CB amplifier is studied and frequency response is plotted. Band width is found to be
___________ .
13
ECE DEPARTMENT
KITSS, HUZURABAD
5. COMMON SOURCE AMPLIFIER CHARACTERISTICS
Aim: To study CS amplifier and to find the frequency response, band width.
Equipments required:
1. Single channel power supply(0-30V)
2. Function Generator(0-3MHz, 20m-30V)
3. CRO(0-30MHz, 0-20Vp-p)
4. BNC Probes(1:10)
Components required:
1. Bread board
2. Resistors:
3. Capacitors:
4. Transistor:
5. Connecting wires.
100KΩ
1.5KΩ
-
1 no.
1 no.
1KΩ
-
2 no.
10μF
100μF
BF245
-
2 no.s
1 no.
1 no.
Circuit diagram:
14
ECE DEPARTMENT
KITSS, HUZURABAD
Theory:
A common-source amplifier is one of three basic single-stage field-effect transistor (FET)
amplifier topologies, typically used as a voltage or trans conductance amplifier.
The common-source (CS) amplifier may be viewed as a trasconductance amplifier or as a
voltage amplifier. As a trans conductance amplifier, the input voltage is seen as modulating the
current going to the load. As a voltage amplifier, input voltage modulates the amount of current
flowing through the FET, changing the voltage across the output resistance according to Ohm's
law.
The amplifier has limited high-frequency response. Therefore, in practice the output often
is routed through either a voltage follower (common-drain or CD stage), or a current follower
(common-gate or CG stage), to obtain more favourable output and frequency characteristics. The
CS–CG combination is called a cascode amplifier. The bandwidth of the common-source
amplifier tends to be low, due to high capacitance.
For the proper operation of FET, the gate must be negative with respect to source i.e.
input circuit should always be reserve biased. This is achieved by inserting a battery VCC in the
gate circuit
A small change in reserve bias on the gate produces a large change in drain current. This
fact makes FET capable of raising the strength of a weak signal. During the positive half cycle
the reverse bias on the gate decreases. This increases channel width and hence the drain current
increased.
During the negative half cycle the reverse bias on the gate increases, consequently
decreases the drain current. In this way amplified output is obtained across FET.
Procedure:
1.
2.
3.
4.
5.
6.
Make the connections as per the circuit diagram.
Keep the input voltage constant at 20mV.
By changing the frequency at input, find the corresponding output.
Calculate the gain.
Plot the frequency response.
Calculate the band width from the response curve.
15
ECE DEPARTMENT
KITSS, HUZURABAD
Observation table: Keep Vin = 20mV
S. No
F(KHZ)
Vout(V)
Gain= Vo/Vin
Gain(db)=
20log(gain)
Expected graph:
Calculations: Band width = f2-f1
Result: The CS amplifier is studied and frequency response is plotted. Band width is found to be
___________ .
16
ECE DEPARTMENT
KITSS, HUZURABAD
6. MEASUREMENT OF H-PARAMETERS OF TRANSISTOR
IN CB, CE, CC CONFIGURATIONS
Aim: To plot the Input and Output characteristics of a transistor connected in Common
Base Configuration and to find the h – parameters from the characteristics.
Apparatus:
S.No
Name
Range / Value
Quantity
1
Dual Regulated D.C Power
supply
0–30 Volts
1
2
Transistor
BC107
1
3
Resistors
1K
1
4
DC Ammeters
(0-200mA)
2
5
DC Voltmeters
(0-2V), (0-20V)
Each 1 No
6
Bread Board and connecting wires -
Circuit Diagram:
17
1 Set
ECE DEPARTMENT
KITSS, HUZURABAD
Procedure:
To Find The Input Characteristics:
1. Connect the circuit as in the circuit diagram.
2. Keep VEE and VCC in zero volts before giving the supply
3. Set VCB = 1 volt by varying VCC. and vary the VEE smoothly with fine control such
that emitter current IE varies in steps of 0.2mA from zero upto 20mA, and note down
the corresponding voltage VEB for each step in the tabular form.
4. Repeat the experiment for VCB =2 volts and 3 volts.
5. Draw a graph between VEB Vs IE against VCB = Constant.
To Find The Output Characteristics:
1. Start VEE and VCC from zero Volts.
2. Set the IE = 1mA by using VEE such that, VCB changes in steps of 1.0 volts from
zero upto 20 volts, note down the corresponding collector current IC for each step
in the tabular form.
3. Repeat the experiment for IE = 3mA and IE = 5mA, tabulate the readings.
4. Draw a graph between VCB Vs IC against IE = Constant.
5. Plot the Input characteristics by taking IE on y–axis and VEB on x–axis.
6. Plot the Output characteristics by taking IC on y–axis and VCB on x–axis.
Model graph:
18
ECE DEPARTMENT
KITSS, HUZURABAD
Tabular Forms:
Input Characteristics:
VCB = 0V
VCB =
2V
VCB = 1V
S.No
VEB (V)
IE (mA)
VEB
(V)
IE (mA)
VEB
(V)
IE (mA)
1
0.0
0.0
0.0
2
0.2
0.2
0.2
3
0.4
0.4
0.4
4
0.6
0.6
0.6
5
0.8
0.8
0.8
6
1.0
1.0
1.0
7
4.0
4.0
4.0
8
8.0
8.0
8.0
9
10.0
10.0
10.0
10
14.0
14.0
14.0
11
18.0
18.0
18.0
12
20.0
20.0
20.0
19
ECE DEPARTMENT
KITSS, HUZURABAD
Output Characteristics:
IE = 1 mA
IE = 3 mA
IE = 5 mA
S.No
VCB (V)
IC (mA)
VCB
(V)
IC (mA)
VCB
(V)
IC (mA)
1
0.0
0.0
0.0
2
0.2
0.2
0.2
3
0.4
0.4
0.4
4
0.6
0.6
0.6
5
0.8
0.8
0.8
6
1.0
1.0
1.0
7
3.0
3.0
3.0
8
5.0
5.0
5.0
9
7.0
7.0
7.0
10
10.0
10.0
10.0
11
12.0
12.0
12.0
12
15.0
15.0
15.0
To find the h – parameters:
Calculation of hib:
Mark two points on the Input characteristics for constant VCB. Let the coordinates of
these two points be(VEB1, I E1) and (VEB2, I E2)
VEB2 -VEB1
h ib
= -------------------------
IE2-I E1
20
ECE DEPARTMENT
KITSS, HUZURABAD
Calculation of hrb:
Draw a horizontal line at some constant IE value on the input characteristics. Find
VCB2, VCB1, VEB2, VEB1
VEB2 - VEB1
hrb = -----------------;
VCB2 - VCB1
Calculation of hfb:
Draw a vertical line on the Output characteristics at some constant VCB value. Find
Ic2, Ic1 and IE2, IE1 .
IC2 - IC1
hfb = ------------ ;
IE2 - IE1
Calculation of hob:
On the Output characteristics for a constant value of IE mark two points with
coordinates (VCB2 , IC2) and (VCB1 , IC1) .
IC2 - IC1
hob = --------------- ;
VCB2 -V CB1
21
ECE DEPARTMENT
KITSS, HUZURABAD
B) COMMON EMITTER TRANSISTOR CHARACTERISTICS
Aim: To plot the Input and Output characteristics of a transistor connected in Common
Emitter Configuration and to find the h – parameters from the characteristics.
Apparatus:
S.No
Name
Range / Value
Quantity
1
Dual Regulated D.C Power supply
0–30 Volts
1
2
Transistor
BC107
1
3
Resistors
120K
1
4
DC Ammeters
(0-500 A), (0-200mA)
Each 1 No
5
DC Voltmeters
(0-2V), (0-20V)
Each 1 No
6
Bread Board and connecting wires
-
1 Set
Circuit Diagram:
22
ECE DEPARTMENT
KITSS, HUZURABAD
Procedure:
To Find The Input Characteristics:
1. Connect the circuit as in the circuit diagram.
2. Keep VBB and VCC in zero volts before giving the supply
3. Set VCE = 1 volt by varying VCC and vary the VBB smoothly with fine control such that
base current IB varies in steps of 5μA from zero upto 200μA, and note down the
corresponding voltage VBE for each step in the tabular form.
4. Repeat the experiment for VCE =2 volts and 3 volts.
5. Draw a graph between VBE Vs IB against VCE = Constant.
To Find The Output Characteristics:
1. Start VEE and VCC from zero Volts.
2. Set the IB = 20μA by using VBB such that, VCE changes in steps of 0.2 volts from
zero upto 10 volts, note down the corresponding collector current IC for each step in
the tabular form.
3. Repeat the experiment for IE = 40μA and IE = 60μA, tabulate the readings.
4. Draw a graph between VCE Vs IC against IB = Constant.
Model Graphs:
1. Plot the Input characteristics by taking IB on y-axis and VBE on x-axis.
2. Plot the Output characteristics by taking IC on the y-axis and VCE on x-axis.
23
ECE DEPARTMENT
KITSS, HUZURABAD
Tabular Forms:
Input Characteristics:
VCE = 0V
VCE = 1V
VCE = 2V
S.No
VBE (V)
IB (μA)
VBE (V)
IB (μA)
VBE
(V)
IB (μA)
1
0
0
0
2
5
5
5
3
10
10
10
4
20
20
20
5
40
40
40
6
60
60
60
7
80
80
80
8
100
100
100
9
120
120
120
10
140
140
140
11
180
180
180
12
200
200
200
24
ECE DEPARTMENT
KITSS, HUZURABAD
Output Characteristics:
IB = 20 μA
IB = 40 μA
IB = 60 μA
S.No
VCE (V)
IC (mA)
VCE (V)
IC (mA)
VCE (V)
1
0.0
0.0
0.0
2
0.2
0.2
0.2
3
0.4
0.4
0.4
4
0.6
0.6
0.6
5
0.8
0.8
0.8
6
1.0
1.0
1.0
7
3.0
3.0
3.0
8
5.0
5.0
5.0
9
7.0
7.0
7.0
10
10.0
10.0
10.0
IC (mA)
Calculation of h-parameters:
Calculation of hie:
Mark two points on the Input characteristics for constant VCE. Let the coordinates of
these two points be (VBE1, IB1) and (VBE2, IB2).
VBE2 - VBE1
hie = -------------
;
IB2 - IB1
25
ECE DEPARTMENT
KITSS, HUZURABAD
Calculation of hre:
Draw a horizontal line at some constant IB value on the Input characteristics. Find
VCE2, VCE1, VBE2, VBE1
VBE2 - VBE1
hrb = ---------------;
VCB2 - VCB1
Calculation of hfe:
Draw a vertical line on the out put characteristics at some constant VCE value. Find
Ic2, Ic1 and IB2, IB1 .
IC2- IC1
= ----------
;
IB2 - IB1
Calculation of hoe:
On the Output characteristics for a constant value of IB mark two points with
coordinates (VCE2 , IC2) and (VCE1 , IC1) .
IC2 - IC1
hob = --------------- ;
VCE2 - CE1
Results:
The input and out put characteristics are drawn on the graphs and the h parameters are
calculated .hie= --------- ohms. hre= -----------hoe= -------- mhos.
hfe = -----------
26
ECE DEPARTMENT
KITSS, HUZURABAD
7. Inverting and Non-inverting Amplifiers using Op Amp
Aim: To design and study Inverting and Non-inverting amplifiers using OP-AMP IC741.
Equipment Required: Dual power supply (0-30v)
Function Generator (1 MHz)
CRO (20MHz)
Components Required: Bread board
IC 741-------------1 no.
Resistors 1KΩ-----4 no.s.
CRO probes
Connecting wires.
Circuit Diagram:
1) Inverting Amplifier:
27
ECE DEPARTMENT
KITSS, HUZURABAD
2) Non-Inverting Amplifier:
Theory:
An inverting amplifier using op amp is a type of amplifier using op amp where the output
waveform will be phase opposite to the input waveform. The input waveform will be amplifier
by the factor Av (voltage gain of the amplifier) in magnitude and its phase will be inverted. In
the inverting amplifier circuit the signal to be amplified is applied to the inverting input of the op
amp through the input resistance R1. Rf is the feedback resistor. Rf and Rin together determine
the gain of the amplifier. Inverting operational amplifier gain can be expressed using the
equation Av = – Rf/R1. Negative sign implies that the output signal is negated.
The input and output waveforms of an inverting amplifier using opamp is shown below. The
graph is drawn assuming that the gain (Av) of the amplifier is 2 and the input signal is a sine
wave. It is clear from the graph that the output is twice in magnitude when compared to the input
(Vout = Av x Vin) and phase opposite to the input.
Practical inverting amplifier using 741.
A simple practical inverting amplifier using 741 IC is shown below. uA 741 is a high
performance and of course the most popular operational amplifier. It can be used in a verity of
applications like integrator,differentiator, voltage follower, amplifier etc. uA 741 has a wide
supply voltage range (+/-22V DC) and has a high open loop gain. The IC has an integrated
compensation network for improving stability and has short circuit protection. Signal to be
amplified is applied to the inverting pi (pin2) of the IC. Non inverting pin (pin3) is connected to
ground. R1 is the input resistor and Rf is the feedback resistor. Rf and R1 together sets the gain
of the amplifier. With the used values of R1 and Rf the gain will be 10 (Av = -Rf/R1 = 10K/1K =
10). RL is the load resistor and the amplified signal will be available across it. POT R2 can be
used for nullifying the output offset voltage. If you are planning to assemble the circuit, the
power supply must be well regulated and filtered. Noise from the power supply can adversely
affect the performance of the circuit. When assembling on PCB it is recommended to mount the
IC on the board using an IC base.
28
ECE DEPARTMENT
KITSS, HUZURABAD
In the inverting amplifier only one input is applied and that is to the inverting input (V2)
terminal. The Non inverting input terminal (V1) is grounded. Since, V1=0 V& V2=Vin , Vo=
-A Vin
The negative sign indicates the output voltage is 1800 out of phase with respect to the input and
amplified by gain A.
Practical Non-inverting amplifier using 741:
The input is applied to the non-inverting input terminal and the Inverting terminal is connected
to the ground.
V1= Vin & V2=0 Volts
Vo= A Vin
The output voltage is larger than the input voltage by gain A & is in phase with the input
signal.
Procedure:
1) Check the components.
2) Setup the circuit on the bread board and check the connections.
3) Switch on the power supply
4) Give 2 Vpp /1 KHz sine wave as input.
5) Observe input and output on the two channels of the oscilloscope simultaneouly.
6) Note down and draw the input and output waveforms on the graph.
7) Verify the input and output waveforms are out of phase for Inverting amplifier and in phase
for Non inverting amplifier.
8) Verify the obtained gain is same as designed value of gain.
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ECE DEPARTMENT
KITSS, HUZURABAD
Expected Waveforms:
1) Inverting Amplifier:
2) Non Inverting Amplifier:
RESULT: Hence verified and drawn the operation and respective waveforms of inverting and
non-inverting amplifier.
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ECE DEPARTMENT
KITSS, HUZURABAD
8. Adder and Subtractor using Op Amp
Aim: To study Adder, Subtractor circuits using OP-AMP IC741 and verify their theoretical and
practical output.
Equipment Required:
Dual power Supply (0-30v)
Function Generator (1 MHz)
Multi meter
CRO (20 MHz)
Components Required: Bread Board
IC741----------------1no.
Resistors 1KΩ------4 nos.
CRO probes
Connecting Wires.
Circuit Diagram:
1) Adder:
2) Subtractor:
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ECE DEPARTMENT
KITSS, HUZURABAD
Theory:
1) Adder:
Op-Amp may be used to design a circuit whose output is the sum of several input signals
such as circuit is called a summing amplifier or summer. We can obtain either inverting or noninverting summer. The circuit diagrams shows a two input inverting summing amplifier. It has
two input voltages V1and V2, two input resistors R1, R2 and a feedback resistor Rf. Assuming
that op-amp is in ideal conditions and input bias current is assumed to be zero, there is no voltage
drop across the resistor Rcomp and hence the non-inverting input terminal is at ground potential.
By taking nodal equations,
𝑉1
𝑉2
𝑉0
+
+
= 0 − − − − − − − −(1)
𝑅1
𝑅2 𝑅𝑓
𝑅𝑓
𝑅𝑓
𝑉0 =– [( ) 𝑉1 + ( ) 𝑉2 ] − − − − − −(2)
𝑅1
𝑅2
𝐴𝑛𝑑 ℎ𝑒𝑟𝑒 𝑅1 = 𝑅2 = 𝑅𝑓 = 1 𝐾Ω
𝑉0 = − (𝑉1 + 𝑉2 )
Hence the output is inverted sum of inputs.
2) Subtractor:
A basic differential amplifier can be used as a sub tractor. It has two input signals V1 and
V2 and two input resistances R1 and R2 and a feedback resistor Rf. The input signals scaled to
the desired values by selecting appropriate values for the external resistors. From the figure, the
output voltage of the differential amplifier with a gain of ‘1’ is
𝑅
(𝑉 − 𝑉1 ) − − − − − −(3)
𝑅𝑓 2
𝑉0 = 𝑉1 − 𝑉2 − − − − − − − − − (4)
𝑉0 = −
Procedure:
1) Adder:
1. Connect the components/equipment as shown in the circuit diagram.
2. Switch ON the power supply.
3. Apply dc voltages at each input terminal for V1 and V2 from the dc supply and check the
output voltage Vo at the output terminal.
4. Tabulate 3 different sets of readings by repeating the above step.
5. Compare practical Vo with the theoretical output voltage Vo =V1+V2.
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ECE DEPARTMENT
KITSS, HUZURABAD
2) Subtractor:
1. Connect the components/equipment as shown in the circuit diagram.
2. Switch ON the power supply.
3. Apply dc voltages at each input terminal for V1 and V2 from the dc supply and check the
output voltage Vo at the output terminal.
4. Tabulate 3 different sets of readings by repeating the above step.
5. Compare practical Vo with the theoretical output voltage Vo =V2-V1.
Observation Tables:
1) Adder:
S.No.
V1
Volts
V2
Volts
Theoretical
Vo=V1+V2
Practical Vo
Volts
S.No.
V1
Volts
V2
Volts
Theoretical
Vo=V1-V2
Practical Vo
Volts
2) Subtractor:
Result:
Thus Adder, Subtractor using op-amp is studied and tested.
33
ECE DEPARTMENT
KITSS, HUZURABAD
9. Integrator circuit using IC741
Aim: To study the operation of the Integrator using op-amp and trace the output wave forms for
sine and square wave inputs.
Equipment Required: Function Generator (1MHz)
Dual power supply (0-30v)
CRO (20MHz)
Components Required: Bread Board
IC741-----------------------1no.
Resistor 100KΩ----------1no.
Resistors 10KΩ----------2nos.
Capacitor 0.01µf---------1no.
CRO Probes
Connecting wires.
Circuit Diagram:
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ECE DEPARTMENT
KITSS, HUZURABAD
Theory:
A circuit in which the output voltage is the integration of the input voltage is called an
integrator.
In the practical integrator to reduce the error voltage at the output, a resistor RF is
connected across the feedback capacitor CF. Thus, RF limits the low-frequency gain and hence
minimizes the variations in the output voltage.
The frequency response of the integrator is shown in the fig. fb is the frequency at which
the gain is 0 dB and is given by fb = 1/2 R1Cf.
In this fig. there is some relative operating frequency, and for frequencies from f to fa the
gain RF/R1 is constant. However, after fa the gain decreases at a rate of 20 dB/decade. In other
words, between fa and fb the circuit of fig. acts as an integrator. The gain limiting frequency fa is
given by
fa=1/2𝜋RfCf
Normally fa<fb. From the above equation, we can calculate Rf by assuming fa& Cf.
This is very important frequency. It tells us where the useful integration range starts.
If fin <fa - circuit acts like a simple inverting amplifier and no integration results,
If fin = fa - integration takes place with only 50% accuracy results,
If fin = 10fa - integration takes place with 99% accuracy results.
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ECE DEPARTMENT
KITSS, HUZURABAD
In the circuit diagram of Integrator, the values are calculated by assuming fa as 50 Hz.
Hence the input frequency is to be taken as 500Hz to get 99% accuracy results. Integrator has
wide applications in
1. Analog computers used for solving differential equations in simulation arrangements.
2. A/D Converters.
3. Signal wave shaping.
4. Function Generators.
Procedure:
1. Connect the components/equipment as shown in the circuit diagram.
2. Switch ON the power supply.
3. Apply sine wave at the input terminals of the circuit using function Generator.
4. Connect channel-1 of CRO at the input terminals and channel-2 at the output terminals.
5. Observe the output of the circuit on the CRO which is a cosine wave (90o phase shifted from
the sine wave input) and note down the position, the amplitude and the time period of Vin & Vo.
6. Now apply the square wave as input signal.
7. Observe the output of the circuit on the CRO which is a triangular wave and note down the
position, the amplitude and the time period of Vin & Vo.
8. Plot the output voltages corresponding to sine and square wave inputs.
Expected Waveforms:
36
ECE DEPARTMENT
KITSS, HUZURABAD
Observation Table:
Input –square wave
Amplitude (VP-P) Time period
(V)
(ms)
Output – Triangular
Amplitude (VP-P)
Time period
(V)
(ms)
Input –sine wave
Amplitude (VP-P) Time period
(V)
(ms)
Output –cosine wave
Amplitude (VP-P)
Time period
(V)
(ms)
Result:
For a given square wave and sine wave, output waveforms for integrator is observed the
output phase is leading by--------with respect to the input and the output phase is lagging by-------with respect to the input.
37
ECE DEPARTMENT
KITSS, HUZURABAD
10. Differentiator circuit using Op Amp
Aim: To study the operation of the differentiator using op-amp and trace the output wave forms
for sine and square wave inputs.
Equipment Required: Function Generator (1MHz)
Dual power supply (0-30v)
CRO (20MHz)
Components Required: Bread Board
IC741-----------------------1no.
Resistor 100KΩ----------1no.
Resistors 10KΩ----------2nos.
Capacitor 0.01µf---------1no.
CRO Probes
Connecting wires.
Circuit Diagram:
38
ECE DEPARTMENT
KITSS, HUZURABAD
Theory:
As the name suggests, the circuit performs the mathematical operation of differentiation,
i.e. the output voltage is the derivative of the input voltage. Vo = - RfC1
Both the stability and the high-frequency noise problems can be corrected by the addition
of two components: R1 and Cf, as shown in the circuit diagram. This circuit is a practical
differentiator. The input signal will be differentiated properly if the time period T of the input
signal is larger than or equal to RfC1. That is, T>= RfC1Differentiator can be designed by
implementing the following steps.
1. Select fa equal to the highest frequency of the input signal to be differentiated.
Then, assuming a value of C1<1 F, calculate the value of Rf
2. Calculate the values of R1and Cf so that R1C1=RfCf.
Differentiator has wide applications in
1. Monostable Multivibrator
2. Signal wave shaping
3. Function Generators.
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ECE DEPARTMENT
KITSS, HUZURABAD
Procedure:
1. Connect the components/equipment as shown in the circuit diagram.
2. Switch ON the power supply.
3. Apply sine wave at the input terminals of the circuit using function Generator.
4. Connect channel-1 of CRO at the input terminals and channel-2 at the output terminals.
5. Observe the output of the circuit on the CRO which is a cosine wave (90o phase shifted from
the sine wave input) and note down the position, the amplitude and the time period of Vi&Vo.
6. Now apply the square wave as input signal.
7. Observe the output of the circuit on the CRO which is a spike wave and note down the
position, the amplitude and the time period of Vin & Vo.
8. Plot the output voltages corresponding to sine and square wave inputs.
Expected Waveforms:
40
ECE DEPARTMENT
KITSS, HUZURABAD
Observation Table:
Input –square wave
Amplitude (VP-P) Time period
(V)
(ms)
Output – Spikes
Amplitude (VP-P)
Time period
(V)
(ms)
Input –sine wave
Amplitude (VP-P) Time period
(V)
(ms)
Output –Cosine wave
Amplitude (VP-P)
Time period
(V)
(ms)
Result:
For a given square wave and sine wave, output waveforms for integrator and
differentiator are observed the output phase is leading by--------with respect to the input and the
output phase is lagging by--------with respect to the input.
41
ECE DEPARTMENT
KITSS, HUZURABAD
11. CURRENT SHUNT FEEDBACK AMPLIFIER
Aim: To study and perform current shunt feedback amplifier and to measure currents and
voltage gain.
Equipments required:
1. Single channel power supply(0-30V)
2. Function Generator(0-3MHz, 20m-30V)
3. CRO(0-30MHz, 0-20Vp-p)
4. BNC Probes(1:10)
Components required:
1. Bread board
2. Resistors:
3.
Transistor:
4. Connecting wires.
Circuit diagram:
1KΩ
320Ω
-
2 no.s
1 no.
4.7KΩ
-
1 no.
100Ω
-
1 no.
BC107
-
42
2 no.s
ECE DEPARTMENT
KITSS, HUZURABAD
Theory:
In Current-Shunt feedback amplifier Feedback signal is proportional to the output current
and feedback to input in shunt.
Current-shunt feedback is 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. As this type of feedback have the least desirable effects. Thus current shunt feedback
circuit behave like a current controlled current source.
Procedure:
1.
2.
3.
4.
5.
6.
Make the connections as per the circuit diagram.
Apply input voltage of 50mV.
Measure output current (I0) and feedback current (If).
With feedback find the output voltage.
Calculate the voltage gain.
Compare voltage gain with the theoretical gain.
Expected graph:
Calculations: Practical voltage gain (AV) with feed back
AV=
Vo/p
Vi/p
R R
Theoretical voltage gain Avf = R2 R3
1 5
Result: The current shunt feedback amplifier is studied and currents and voltage gains are
calculated.
43
ECE DEPARTMENT
KITSS, HUZURABAD
12. RC PHASE SHIFT OSCILLATOR
Aim: To study the RC phase shift oscillator and to find the frequency of oscillations and
compare the theoretical and practical frequencies.
Equipments required:
1. Single channel power supply(0-30V)
2. CRO(0-30MHz, 0-20Vp-p)
3. BNC Probes(1:10)
Components required:
1. Bread board
2. Resistors:
22KΩ
3. Capacitors:
2.2KΩ
10KΩ
1KΩ
4.7KΩ/10 KΩ 100µf
-
1 no.
1 no.
1 no.
3 no.s
1 no.
10 µf
-
2 no.s
0.01µf/0.1 µf -
3 no.s
BC107
1 no.
4. Transistor:
-
1 no.
-
5. Connecting wires.
Circuit diagram:
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ECE DEPARTMENT
KITSS, HUZURABAD
Theory:
An oscillator is a circuit which takes DC voltage or current and converts into useful Ac
voltage or current. It is different from the oscillator in the sense that amplifier takes AC and
amplifiers the given Ac signal where as the oscillator takes the DC input and converts it into AC,
there are the sinusoidal and non sinusoidal oscillators
RC phase shift oscillator is a sinusoidal oscillator. The oscillator is an amplifier with a
feedback. In feedback resistor and capacitor are connected. This is called RC phase shift
oscillator because RC networks are used in feedback and each RC network gives a 60 degrees
phase shift and the total phase shift produced is 360 degrees, the frequency of oscillator depends
on the RC values.
Procedure:
1. Make the connections as per the circuit diagram.
2. Note down the time period of the oscillations for a given value of R, C in the tank
circuit.
3. Find frequency of oscillation which is the reciprocal of time period for different
values of R, C in the tank circuit.
4. Verify the observed frequency with the theoretical values.
Observations:
S. No
R(KΩ)
C(µF)
Time Period(s)
F(HZ)
F(HZ)
Practical
Practical
Theoretical
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ECE DEPARTMENT
KITSS, HUZURABAD
Expected graph:
1
Calculations: Frequency: F=2πRC√6
Result: RC Phase shift oscillator is studied. Theoretical and practical frequencies of RC phase
shift oscillator have been compared.
46
ECE DEPARTMENT
KITSS, HUZURABAD
13. HARTLEY AND COLPITT’S OSCILLATOR
Aim: To study the working of Hartley and colpitts oscillator and to calculate the frequency of
oscillations practically and compare with theoretical frequency.
Equipment required:
1.
2.
3.
4.
5.
Single channel power supply(0-30V)
CRO(0-30MHz, 0-20Vp-p)
BNC Probes(1:10)
Decade capacitance Box
Decade inductance Box
-
2 no.s
2 no.s
Components required:
1. Bread board
2. Resistors:
3. Capacitors:
4. Transistor:
5. Connecting wires.
22KΩ
2.2KΩ
10KΩ
1 KΩ
10µF
BC107
-
1 no.
1 no.
1 no.
1 no.
2 no.s
1 no.
Circuit diagram:
Hartley Oscillator:
47
ECE DEPARTMENT
KITSS, HUZURABAD
Colpitts Oscillator:
Theory:
Oscillator is a circuit which takes DC voltage or current and converts it into useful AC
voltage. Amplifier is different from the oscillator. In the oscillator DC is converted to AC. There
are two types of oscillators. They are sinusoidal oscillator and non sinusoidal oscillators.
Hartley is a sinusoidal oscillator. The construction of oscillator is an amplifier with
feedback. For Hartley oscillator in feedback there are two inductors and one capacitor. The
frequency of the oscillator depends on the values of the C, L of the feedback circuit.
Colpitts is a sinusoidal oscillator. The construction of oscillator is an amplifier with
feedback. For Colpitts oscillator in feedback there are two capacitors and one inductor. The
frequency of the oscillator depends on the values of the C, L of the feedback circuit.
Procedure:
Hartley Oscillator:
1.
2.
3.
4.
5.
Make the connections as per the circuit diagram.
Vary one of the inductance (L1 or L2), at constant capacitance C.
Note down the time period of the oscillations.
Find the frequency of oscillations.
Compare the practical frequency with the theoretical frequency.
48
ECE DEPARTMENT
KITSS, HUZURABAD
Colpitts Oscillator:
1. Make the connections as per the circuit diagram.
2. Vary one of the capacitance (C4 or C5), at constant inductance L.
3. Note down the time period of the oscillations.
4. Find the frequency of oscillations
5. Compare the practical frequency with the theoretical frequency.
Observation table:
Hartley Oscillator:
S. No
C(nF)
L1(mH)
L2(µH)
Time Period(s)
Practical
F(KHz)
Practical
F(KHz)
Theoretical
F(KHz)
Practical
F(KHz)
Theoretical
Colpitts Oscillator:
S. No L(mH)
C4(µF)
C5(µF)
Time Period(s)
Practical
49
ECE DEPARTMENT
KITSS, HUZURABAD
Expected graph:
Calculations:
Hartley Oscillator:
F=
1
2π√LeqC
, where Leq = L1 + L2
Colpitts Oscillator:
F=2π
1
√LCeq
C .C
, where Ceq = (C 1+C2 )
1
2
Result: The Hartley and colpitt’s oscillator is studied and practical frequencies are compared
with theoretical frequencies.
50
ECE DEPARTMENT
KITSS, HUZURABAD
14. CLASS A POWER AMPLIFIER
Aim: To study the working of class A power amplifier in order to achieve maximum output ac
power and to calculate efficiency.
Equipments required:
1.
2.
3.
4.
5.
Single channel power supply(0-30V)
Function Generator(0-3MHz, 20m-30V)
CRO(0-30MHz, 0-20Vp-p)
BNC Probes(1:10)
DMM(200mA-10A)
Components required:
1. Bread board
2. Resistors:
3.
4.
5.
6.
147KΩ
510Ω
-
1 no.
1 no.
Capacitors:
10μF
Inductor:
50mH
Transistor:
BC107
Connecting wires.
-
2 no.s
1 no.
1 no.
Circuit diagram:
51
ECE DEPARTMENT
KITSS, HUZURABAD
Theory:
This amplifier is called as “series fed” because the load RL is connected in series with
transistor output. It is also called as direct coupled amplifier.
ICQ = Zero signal collector current
VCEQ = Zero signal collector to emitter voltage
Power amplifiers are mainly used to deliver more power to the load. To deliver more power
it requires large input signals, so generally power amplifiers are preceded by a series of voltage
amplifiers.
In class-A power amplifiers, Q-point is located in the middle of DC-load line. So output
current flows for complete cycle of input signal. Under zero signal condition, maximum power
dissipation occurs across the transistor. As the input signal amplitude increases power dissipation
reduces. The maximum theoretical efficiency is 25%.
Procedure:
1.
2.
3.
4.
5.
Make the connections as per the circuit diagram.
Apply input voltage 50mV.
At this point measure the DC current (IDC) from multi meter.
Find the output voltage.
Calculate the efficiency.
Expected graph:
52
ECE DEPARTMENT
KITSS, HUZURABAD
Calculations: Efficiency is defined as the ratio of AC output power to DC input power
DC input power = Vcc x IDC
AC output power =
2
𝑉𝑃−𝑃
8𝑅𝐿
P
Efficiency η%= (PAC ) x100
DC
Result: Class A power amplifier is studied and the efficiency is found to be__________.
53