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5/20/2018 EC203 Lab Manual
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CSPIT EC 208 Analog Electronics Circuits
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SOLID STATE ELECTRONICS
(EC203)
LAB MANUAL
V.T.PTAEL DEPARTMENT OF ELECTRONICS AND COMMUNICATION
ENGINEERING
CHANDUBHAI S. PATEL INSTITUTE OF TECHNOLOGY, CHANGA,
CHARUSAT
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Dos and Donts in Laboratory :
1.Do not handle any equipment before reading the instructions /Instruction manuals.
2.Read carefully the power ratings of the equipment before it is switched ON, whether ratings
230 V/50 Hz or 115V/60 Hz. For Indian equipment, the power ratings are normally 230V/50Hz.
If you have equipment with 115/60 Hz ratings, do not insert power plug, as our normal supply is230V/50Hz., which will damage the equipment.
3.Observe type of sockets of equipment power to avoid mechanical damage.
4.Do not forcefully place connectors to avoid the damage.5.Strictly observe the instructions given by the Teacher/ Lab Instructor.
Instruction for Laboratory Teachers:-
1.Submission related to whatever lab work has been completed should be done during the nextlab session.
2.Students should be instructed to switch on the power supply after getting the
checked by the lab assistant / teacher. After the experiment is over, the students must hand overthe circuit board, wires, CRO probe to the lab assistant/teacher.
3.The promptness of submission should be encouraged by way of marking and
evaluation patterns that will benefit the sincere students.
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Practical-1
Aim :-To plot forward characteristics of pn junction diode (If vs. Vf)
Apparatus :-D.C.regulated power supply (0-30V) , Diode(1N4007), Resistance (820 ),voltmeter, ammeter (0-25mA) Connecting wires
Circuit Diagram :
Procedure:-1. 1. Connect the circuit as shown in fig.
2. 2. By varying applied voltage measure corresponding reading for voltage & current.3. 3. Plot the graph of voltage & forward current.
Observations:-Input voltage (Vdc) = -------V
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Applied voltage
Vdc ( V)
O/P Voltage Vf ( V) Current If ( mA)
Result:-Current increases exponentially with respect to voltage after cut in voltage as seenfrom the graph
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Practical-2, 3 & 4
Aim :-To study and perform Half-wave and full wave rectifiers with
filters
Theory: -
The diode rectifier converts the input sinusoidal voltage Vs to a unipolar out Vo. There are two
types of rectifier circuits:
1.
Half-wave Rectifier and2.Full-wave Rectifier.
Apertures: -
1.Trainer Board
2. Multimeter
3. Resistor 10K
4. Capacitor 1F, 47F
5. Diode Four Pieces
6. Oscilloscope
7. Signal Generator
8. Wire.
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1. Circuit Diagram For Half-wave
Rectifier:
The input and output of the rectifier are drawn in Figere-1. Diode conducts only when it is forward biased. For Vs =
Vm sint, DC voltage and current of a half-wave rectifier are as follows
VDC= Vm/ (1/2)VDO
IDC= {Vm/ (1/2)VDO}/R
Where VDO0.7 V
PIV (Peak Inverse Voltage): -
PIV is the Peak Reverse Voltage that appears across the diode when it is reverse-biased,
PIV = Vm
Same procedure for full wave rectifier using two diode
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Circuit Diagram For Full-wave Rectifier:-
The bridge rectifier circuit and their input and output voltage as a function of time is shown below. Peak voltage
across each diode when it is reverse-biased
PIV = Vm VDO
DC Voltage, VDC= 2Vm/ 2 VDO
Ripple Factor: -
A rectifier converting alternating currents into a unidirectional current, periodically fluctuating components still
remaining in the output wave. A measure of the fluctuating component is given by the ripple factor r, which is
defined as
R = rms value of alternating components of wave/Average value of wave
= I rms/Idc = V rms/Vdc
Where, I rms and V rms denote the rms value of the ac components of the current and voltage, respectively.
For a half-wave rectifier, r = 1.21 and for a full wave rectifier, r = 0.482
Calculating Ripple Factor for Half-wave Rectifier: -
For C = 1F,
The DC value is 0.5V
The rms value is 0.5V
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So the Ripple Factor is 0.5/0.5 = 1
For C = 47F,
The DC value is 5V
The rms value is 0.9V
So the Ripple Factor is 0.9/5 = 0.18
Calculating Ripple Factor for Full-wave Rectifier: -
For C = 1F,
The DC value is 0.9V
The rms value is 0.22V
So the Ripple Factor is 0.22/0.9 = 0.24
For C = 47F,
The DC value is 8.78V
The rms value is 6.18V
So the Ripple Factor is 6.18 /8.78 = 0.7
Procedure: -1. Construct circuit of Figure-1 without the capacitor. Observe Vi and Vo simultaneously on the oscilloscope.
Sketch input and output waveforms. Measures Vo with multimeter in dc and ac mode.
2. Connect 1F capacitor across the load resistor. BE CAREFUL about the polarity of the capacitor. Sketch inputand output waveforms. Measure Vo with multimeter.
3. Replace 1F capacitor with 47F and repeat step-2.
4. Construct the circuit of Figure-2 without the capacitor. Observe and sketch Vi, Vo. DO NOT TRY to observe Vi,
Vo simultaneously. Measure AC and DC components of Vo with multimeter.
5. Connect 1F capacitor as shown in Figure-2 and repeat step-4.
6. Replace 1F capacitor by 47F for Figure-2 and repeat step4
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Practical-5
TRANSISTOR COMMON -BASE CONFIGURATION
AIM: 1.To observe and draw the input and output characteristics of a transistor connectedin common base configuration.
2. To find of the given transistor.
APPARATUS: Transistor, BC 107
Regulated power supply (0-30V, 1A)
Voltmeter (0-20V)
Ammeters (0-100mA)
Resistor, 1000
Bread board
Connecting wires
THEORY:
A transistor is a three terminal active device. The terminals are emitter, base,
collector. In CB configuration, the base is common to both input (emitter) and output
(collector). For normal operation, the E-B junction is forward biased and C-B junction is
reverse biased.
In CB configuration, IEis +ve, IC is ve and IBis ve. So,
VEB=f1 (VCB,IE) and
IC=f2 (VCB,IB)
With an increasing the reverse collector voltage, the space-charge width at the
output junction increases and the effective base width W decreases. This phenomenon is
known as Early effect. Then, there will be less chance for recombination within the base
region. With increase of charge gradient with in the base region, the current of minority
carriers injected across the emitter junction increases.The current amplification factor of
CB configuration is given by,
= IC/ IE
CIRCUIT DIAGRAM
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PROCEDURE:
INPUT CHARACTERISTICS:
1. Connections are made as per the circuit diagram.
2. For plotting the input characteristics, the output voltage VCEis kept constant at 0V and
for different values of VEBnote down the values of IE.
3. Repeat the above step keeping VCBat 2V, 4V, and 6V.All the readings are tabulated.
4. A graph is drawn between VEBand IE for constant VCB.
OUTPUT CHARACTERISTICS:
1. Connections are made as per the circuit diagram.
2. For plotting the output characteristics, the input IE is kept constant at 10m A and for
different values of VCB, note down the values of IC.
3. Repeat the above step for the values of IE at 20 mA, 40 mA, and 60 mA, all the readings
are tabulated.
4. A graph is drawn between VCB and Ic for constant IE
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OBSERVATIONS:
INPUT CHARACTERISTICS:
S.No VCB=0V VCB=1V VCB=2V
VEB(V) IE(mA) VEB(V) IE(mA) VEB(V) IE(mA)
OUTPUT CHARACTERISTICS:
S.No
IE=10mA IE=20mA IE=30mA
VCB(V) IC(mA) VCB(V) IC(mA) VCB(V) IC(mA)
MODEL GRAPHS:
INPUT CHARACTERISTICS
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OUTPUT CHARACTERISTICS
PRECAUTIONS:
1. The supply voltages should not exceed the rating of the transistor.
2. Meters should be connected properly according to their polarities.
RESULT:
1. The input and output characteristics of the transistor are drawn.
2. The of the given transistor is calculated.
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Practical-6
TRANSISTOR CE CHARACTERSTICS
AIM: 1. To draw the input and output characteristics of transistor connected inCE configuration
2. To find of the given transistor.
APPARATUS:
Transistor (BC 107)
R.P.S (O-30V) 2Nos
Voltmeters (0-20V) 2Nos
Ammeters (0-200A)
(0-500mA)
Resistors 1Kohm
Bread board
THEORY:
A transistor is a three terminal device. The terminals are emitter, base, collector.
In common emitter configuration, input voltage is applied between base and emitter
terminals and out put is taken across the collector and emitter terminals.
Therefore the emitter terminal is common to both input and output.
The input characteristics resemble that of a forward biased diode curve. This is
expected since the Base-Emitter junction of the transistor is forward biased. As compared
to CB arrangement IB increases less rapidly with VBE . Therefore input resistance of CE
circuit is higher than that of CB circuit.
The output characteristics are drawn between Ic and VCE at constant IB. the
collector current varies with VCE unto few volts only. After this the collector current
becomes almost constant, and independent of VCE. The value of VCEup to which the
collector current changes with V CEis known as Knee voltage. The transistor always
operated in the region above Knee voltage, IC is always constant and is approximately
equal to IB.
The current amplification factor of CE configuration is given by
= IC/IB
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CIRCUIT DIAGRAM:
PROCEDURE:
INPUT CHARECTERSTICS:
1. Connect the circuit as per the circuit diagram.
2. For plotting the input characteristics the output voltage VCEis kept constant at 1V and for
different values of VBE . Note down the values of IC
3. Repeat the above step by keeping VCE at 2V and 4V.
4. Tabulate all the readings.
5. plot the graph between VBE and IB for constant VCE
OUTPUT CHARACTERSTICS:
1. Connect the circuit as per the circuit diagram
2. for plotting the output characteristics the input current IB is kept constant at 10A and
for different values of VCEnote down the values of IC
3. repeat the above step by keeping IB at 75 A 100 A
4. tabulate the all the readings
5. plot the graph between VCEand ICfor constant IB
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OBSERVATIONS:
INPUT CHARACTERISTICS:
S.NO
VCE= 1V VCE= 2V VCE= 4V
VBE(V) IB(A) VBE(V) IB(A) VBE(V) IB(A)
OUT PUT CHAREACTARISTICS:
S.NOIB= 50A IB= 75A IB= 100A
VCE(V) IC(mA) VCE(V) ICmA) VCE(V) IC(mA)
MODEL GRAPHS:
INPUT CHARACTERSTICS:
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OUTPUT CHARECTERSTICS:
PRECAUTIONS:
1. The supply voltage should not exceed the rating of the transistor
2. Meters should be connected properly according to their polarities
RESULT:
the input and out put characteristics of a transistor in CE configuration are Drawn
the plot of a given transistor is calculated
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Practical-7
Aim: To study and perform Single stage transistor amplifier.
Brief Theory:
RC-coupled CE amplifier is widely used in audio frequency applications in radio and TV receivers. It
provides current, voltage and power gains. Base current controls the collector current of a common
emitter amplifier. A small increase in base current results in a relatively large increase in collector current.
Similarly, a small decrease in base current causes large decrease in collector current. The emitter-base
junction must be forward biased and the collector base junction must be reverse biased for the proper
functioning of an amplifier. In the circuit diagram, an NPN transistor is connected as a common emitter
ac amplifier. R1and R2are employed for the voltage divider bias of the transistor. Voltage divider bias
provides good stabilization independent of the variations of . The input signal Vinis coupled through Cci
to the base and output voltage is coupled from collector through the capacitor Cc2.
The input impedance of the amplifier is expressed as Zin= R1||R2||(1 + hFEre) and output impedance as
Zout = Rc ||RLwhere re is the internal emitter resistance of the transistor given by the expression = 25
mV/IE, where 25 mV is temperature equivalent voltage at room temperature.
Circuit Diagram:
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Procedure
1. Test all the components using a multimeter. Set up the circuit and verify dc bias conditions. To
check dc bias conditions, remove input signal and capacitors in the circuit.
2.
Connect the capacitors in the circuit. Apply a 100 mV peak to peak sinusoidal signal from the
function generator to the circuit input. Observe the input and output waveforms on the CRO screensimultaneously.
3.
Keep the input voltage constant at 100 mV, vary the frequency of the input signal from 0 to 1 MHz
or highest frequency available in the generator. Measure the output amplitude corresponding to
different frequencies and enter it in tabular column.
4.
Plot the frequency response characteristics on a graph sheet with gain in dB on y-axis and logf on x-
axis. Mark log fLand log fHcorresponding to 3 dB points. (If a semi-log graph sheet is used instead
of ordinary graph sheet, mark f along x-axis instead of logf).
5. Calculate the bandwidth of the amplifier using the expression BW= fH fL.
6. Remove the emitter bypass capacitor CE from the circuit and repeat the steps 3 to 5 and observe
that the bandwidth increases and gain decreases in the absence of CE .
Observation Table:
f (Hz) Vi Vo Gain f (Hz) Vi Vo Gain
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Expected Results:
Conclusion:
With CE:
Mid-band gain of the amplifier = ..............
Bandwidth of the amplifier = ............. Hz
Without CE:
Mid-band gain of the amplifier = ...............
Bandwidth of the amplifier = ............... Hz
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Practical-9
FET CHARACTERISTICS
AIM: a). To draw the drain and transfer characteristics of a given
FET.
b). To find the drain resistance (rd) amplification factor () and
Tran conductance (gm) of the given FET.
APPARATUS: FET (BFW-11)
Regulated power supply
Voltmeter (0-20V)
Ammeter (0-100mA)
Bread board
Connecting wires
THEORY:
A FET is a three terminal device, having the characteristics of high input impedance and
less noise, the Gate to Source junction of the FET s always reverse biased. In response to
small applied voltage from drain to source, the n-type bar acts as sample resistor, and the
drain current increases linearly with VDS. With increase in ID the ohmic voltage drop
between the source and the channel region reverse biases the junction and the conductingposition of the channel begins to remain constant. The VDSat this instant is called pinch of
voltage.
If the gate to source voltage (VGS) is applied in the direction to provide
additional reverse bias, the pinch off voltage ill is decreased.
In amplifier application, the FET is always used in the region beyond the
pinch-off.
FDS=IDSS(1-VGS/VP)^2
CIRCUIT DIAGRAM
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PROCEDURE:
1. All the connections are made as per the circuit diagram.
2. To plot the drain characteristics, keep VGSconstant at 0V.
3. Vary the VDD and observe the values of VDS and ID.
4. Repeat the above steps 2, 3 for different values of VGS at 0.1V and 0.2V.
5. All the readings are tabulated.
6. To plot the transfer characteristics, keep VDSconstant at 1V.
7. Vary VGGand observe the values of VGSand ID.
8. Repeat steps 6 and 7 for different values of VDSat 1.5 V and 2V.
9. The readings are tabulated.
10.From drain characteristics, calculate the values of dynamic resistance (rd) by using the
formula
rd =VDS/ID11.From transfer characteristics, calculate the value of transconductace (gm) By using the
formula
Gm=ID/VDS
12.Amplification factor () = dynamic resistance. Tran conductance
= VDS/VGS
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OBSERVATIONS:
DRAIN CHARACTERISTICS:
S.NO VGS=0V VGS=0.1V VGS=0.2V
VDS(V) ID(mA) VDS(V) ID(mA) VDS(V) ID(mA)
TRANSFER CHARACTERISTICS:
S.NO VDS=0.5V VDS=1V VDS=1.5V
VGS(V) ID(mA) VGS(V) ID(mA) VGS(V) ID(mA)
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MODEL GRAPH:
TRANSFER CHARACTERISTICS
DRAIN CHARACTERISTICS
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PRECAUTIONS:
1. The three terminals of the FET must be care fully identified
2. Practically FET contains four terminals, which are called source, drain, Gate, substrate.
3. Source and case should be short circuited.4. Voltages exceeding the ratings of the FET should not be applied.
RESULT :
1. The drain and transfer characteristics of a given FET are drawn
2. The dynamic resistance (rd), amplification factor () and Tran conductance (gm) of the
given FET are calculated.
Recommended