Gopi 1 lab manual

1 BTL Institute of Technology EEE Dept.
15EEL 34 Electronics Laboratory Manual
B T L INSTITUTE OF TECHNOLOGY & MANAGEMENT
No.259/B, Bommasandhra Industrial Area , Hosur Road, Bangalore- 560 099
Ph: 080- 27832379, (EEE Dept)
A LAB MANUAL ON
ELECTRONICS LABORATORY
(Analog & Digital)
Subject Code: 15EEL37
(As per VTU Syllabus CBCS )
PREPERAED BY GOPINATH.B.L
APPROVED BY HOD EEE Dept.

INDEX
Sl No Contents Page No
1. INTRODUCTION 4
2. INSTRUCTION TO STUDENTS 5
3. LAB CYCLES 6
4.
DESIGN AND TESTING OF FULL WAVE – CENTER TAPPED
TRANSFORMER TYPE. DETERMINATION OF RIPPLE FACTOR,
REGULATION AND EFFICIENCY.
7
5.
BRIDGE TYPE RECTIFIER CIRCUITS WITH AND WITHOUT
CAPACITOR FILTER DETERMINATION OF RIPPLE FACTOR,
REGULATION AND EFFICIENCY.
13
6.
CHARACTERISTICS OF CE CONFIGURATION
DETERMINATION OF H PARAMETERS.
19
7.
CHARACTERISTICS OF CB CONFIGURATION
DETERMINATION OF H PARAMETERS.
25
8.
FREQUENCY RESPONSE OF SINGLE STAGE BJT RC COUPLED
AMPLIFIER AND DETERMINATION OF HALF POWER POINTS,
BANDWIDTH, INPUT AND OUTPUT IMPEDANCES
31
9.
FREQUENCY RESPONSE OF SINGLE STAGE FET RC COUPLED
AMPLIFIER AND DETERMINATION OF HALF POWER POINTS,
BANDWIDTH, INPUT AND OUTPUT IMPEDANCES
37
10.
DETERMINATION OF GAIN, INPUT AND OUTPUT IMPEDANCE
OF BJT DARLINGTON EMITTER FOLLOWER WITH AND
WITHOUT BOOTSTRAPPING.
47
11.
SIMPLIFICATION, REALIZATION OF BOOLEAN EXPRESSIONS
USING LOGIC GATES/UNIVERSAL GATES.
53
12.
REALIZATION OF HALF/FULL ADDER AND HALF/FULL SUB
TRACTORS USING LOGIC GATES.
59
13.
REALIZATION OF PARALLEL ADDER/SUB TRACTORS USING
7483 CHIP- BCD TO EXCESS-3 CODE CONVERSION & VICE -
VERSA.
63
14.
REALIZATION OF BINARY TO GRAY CODE CONVERSION
AND VICE VERSA.
67
15. DESIGN AND TESTING RING COUNTER/JOHNSON COUNTER. 72
16. DESIGN AND TESTING OF SEQUENCE GENERATOR. 74
17.
REALIZATION OF 3 BIT COUNTERS AS A SEQUENTIAL
CIRCUIT AND MOD – N COUNTER DESIGN USING 7476, 7490,
74192, 74193.
75
18. VIVA QUESTIONS 92
19. APPENDIX I (SYMBOLS, VARIABLES) 97
20. APPENDIX II(RATING OF TRANSISTOR) 99
21. APPENDIX III(COMPONENT VALUE IDENTIFICATION) 102

INTRODUCTION
“A practical approach is probably the best approach to mastering a subject and gaining
a clear insight.”
Analog Circuits and Digital Circuits Practical session covers those practical oriented electronic circuits
that are very essential for the students to solidify their theoretical concepts. This workbook provides a
communication bridge between the theory and practical world of the Electronic Laboratory. The
knowledge of these practical are very essential for the engineering students. All of these practical are
arranged on the modern electronic trainer boards.
This practical session comprises of two sections. The first section consists of Analog circuits. Some of the
very useful electronic circuits are discussed in this section. The second section consists of Digital circuits.
Each and every practical provides a great in depth practical concepts.

INSTRUCTIONS TO THE STUDENTS
1. Write Experiment number , Experiment name, Experiment date in the rough record
2. Keep your rough record/ Fare record neat and clean.
3. Draw circuit diagram neatly with HB pencil in the rough record as per circuit given in the lab
manual/available in your laboratory.
4. Wire the circuit using electronic components available in the lab.
5. Write the readings in the observation table in the given format in the lab manual.
6. Draw waveforms in the given space.
7. Prepare for vive based on the experiment that you are going to do in a day.
8. The viva questions are included in the semester hand book as well as lab manual
9. Use text book, reference book or internet to find out answers of the questions.
10. Every student must complete your rough record get sign from the faculty on the same day of your
lab or the very next day itself.
11. Fare record must be completed and submitted in the next lab.
12. Keep the work place neat and clean.

15EEL34 ELECTRONICS LABARATORY
LAB CYCLES
Cycle 1
1. Design and Testing of Full wave – center tapped transformer type and Bridge type rectifier circuits with and
without Capacitor filter. Determination of ripple factor, regulation and efficiency
2. Static Transistor characteristics for CE, CB and CC modes and determination of h parameters...
3. Frequency response of single stage BJT and FET RC coupled amplifier and determination of half power
points, bandwidth, input and output impedances
4. Design and testing of BJT - RC phase shift oscillator for given frequency of oscillation.
5. Determination of gain, input and output impedance of BJT Darlington emitter follower with and without
bootstrapping
Cycle 2
1. Simplification, realization of Boolean expressions using logic gates/Universal gates
2. Realization of half/Full adder and Half/Full Sub tractors using logic gates
3. Realization of parallel adder/Sub tractors using 7483 chip- BCD to Excess-3 code Conversion & Vice -
Versa
4. Realization of Binary to Gray code conversion and vice versa
5. Design and testing Ring counter/Johnson counter
6. Design and testing of Sequence generator
7. Realization of 3 bit counters as a sequential circuit and MOD – N counter design using 7476, 7490, 74192,
74193.
.

EXPERIMENT NO. 1
DESIGN AND TESTING OF FULL WAVE – CENTER TAPPED TRANSFORMER TYPE AND BRIDGE TYPE
RECTIFIER CIRCUITS WITH AND WITHOUT CAPACITOR FILTER. DETERMINATION OF RIPPLE
FACTOR, REGULATION AND EFFICIENCY
Full Wave – Center Tapped Transformer Type
AIM: To observe waveform at the output of full wave rectifier with and without filter capacitor. To
measure DC voltage, DC current, ripple factor with and without filter capacitor.
Introduction:
Full wave rectifier utilizes both the cycle of input AC voltage. Two or four diodes are used in full wave
rectifier. If full wave rectifier is designed using four diodes it is known as full wave bridge rectifier. Full
wave rectifier using two diodes without capacitor is shown in the following figure. Center tapped
transformer is used in this full wave rectifier. During the positive cycle diode D1 conducts and it is
available at the output. During negative cycle diode D1 remains OFF but diode D2 is in forward bias
hence it conducts and negative cycle is available as a positive cycle at the output as shown in the following
figure. Note that direction of current in the load resistance is same during both the cycles hence output is
only positive cycles.

Advantages of full wave rectifier over half wave rectifier:
1. The rectification efficiency is double than half wave rectifier
2. Ripple factor is less and ripple frequency is double hence easy to filter out.
3. DC output voltage and current is higher hence output power is higher.
4. Better transformer utilization factor
5. There is no DC saturation of core in transformer because the DC currents in two halves of
secondary flow in opposite directions.
Disadvantages:
1. Requires center tap transformer
2. Requires two diodes compared to one diode in half wave rectifier.
Practical Circuit diagram:
List of components:
1. Transformer (center tapped) 12-0-12 V AC, 500 mA
2. Diode 1N4007 ---- 2 No.
3. Resistor 10K
4. Capacitor 1000μF
5. Toggle Switch
Experiment Procedure:
1. Construct circuit on the general board.
2. Keep toggle switch OFF to perform practical of full wave rectifier without filter capacitor and ON
to connect filter capacitor.
WORKSHEET
Waveforms:
Without filter capacitor:
Input Waveform at secondary of transformer:

Output waveform:
With filter capacitor:
Output waveform:

Observations:
Peak Voltage, Vm = (From CRO for FWR with and without filter)
DC Voltage, VDC(full load) =
(From Voltmeter/ Multimeter for FWR with and without
filter)
No Load DC Voltage, VDC(No load) =
filter)
Ripple Voltage, Vr = (From CRO for FWR with filter)
Calculations:
Without filter:
Ripple factor (Theoretical) =
Ripple Factor (Practical)
With filter:

Ripple factor (Theoretical)
Where f = 50Hz, R =1K , C = 1000 F.
Ripple Factor
Percentage Regulation = %
VNL = DC voltage at the load without connecting the load (Minimum current).
VFL = DC voltage at the load with load connected.
Efficiency %u200B
PAC = V2rms / RL
PDC = Vdc / RL
Conclusion:
Important Viva Questions
1. What is the frequency of AC component at the output of full wave rectifier? Give reason.
2. What is the difference in DC output voltage in half wave and full wave rectifier for the same AC
input?
3. What is the PIV necessary for the diode if transformer of 24-0-24 V is used ?
4. What is the mathematical relationship between rms input AC voltage and DC output voltage in half
wave rectifier with and without filter capacitor?
5. What is filter?
Ans: Electronic filters are electronic circuits which perform signal processing functions, specifically to
remove unwanted frequency components from the signal.
6. Give some rectifications technologies?

Ans: Synchronous rectifier, Vibrator, Motor-generator set , Electrolytic ,Mercury arc, and Argon gas
electron tube.
7. What is the efficiency of bridge rectifier?
Ans: %
8. What is the value of PIV of a center tapped FWR?
Ans: 2Vm.
9. In filters capacitor is always connected in parallel, why?
Ans: Capacitor allows AC and blocks DC signal.in rectifier for converting AC to DC, capacitor placed in
parallel with output, where output is capacitor blocked voltage.If capacitance value increases its capacity
also increases which increases efficiency of rectifier.
10. What is the purpose of Center Tapped transformer?
11. What is Regulation?
12. What is the location of poles of filter in S-plane?
13. What is the output of FWR with filter? Is it unidirectional?
14. What are the advantages and disadvantages of center tapped full-wave rectifiers compared with
Bridge rectifiers?
15. Define Ripple factor „γ‟ and its values for the three types of rectifiers.
16. What is the value of No load voltage for all the three types of the rectifiers?
17. What are the different types of filters used for the rectifiers?
Conclusion:
Result
The Full wave rectifier circuit was done and the wave forms with and without filter capacitor are noted.
The DC voltage, DC current, ripple factor with and without filter capacitor is noted.

EXPERIMENT NO. 2
BRIDGE RECTIFIER
AIM: To observe waveform at the output of bridge rectifier with and without filter capacitor. To measure
DC voltage, DC current, ripple factor with and without filter capacitor.
Introduction:
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 circuit is shown in the following figure.
The circuit has four diodes connected to form a bridge. The ac input voltage is applied to the diagonally
opposite ends of the bridge. The load resistance is connected between the other two ends of the bridge. For
the positive half cycle of the input ac voltage, diodes D1 and D2 conduct, whereas diodes D3 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, diodes D3 and D4
conduct whereas, D1 and D2 remain OFF. The conducting diodes D3 and D4 will be in series with the

load resistance RL and hence the current flows through RL in the same direction as in the previous half
cycle. Thus a bi-directional wave is converted into a unidirectional wave.
The circuit diagram of the bridge rectifier with filter capacitor is shown in the following figure. When
capacitor charges during the first cycle, surge current flows because initially capacitor acts like a short
circuit. Thus, surge current is very large. If surge current exceeds rated current capacity of the diode it can
damage the diode. To limit surge current surge resistance is used in series as shown in the figure. Similar
surge resistance can be used in half wave as well as center-tapped full wave rectifier also.
Bridge rectifier package (combination of four diodes in form of bridge) is easily available in the market
for various current capacities ranging from 500 mA to 30A. For laboratory purpose you can use 1A
package.
Advantages of bridge rectifier:
1. No center tap is required in the transformer secondary hence transformer design is simple.
2. If stepping up and stepping down not required than transformer can be eliminated. (In SMPS used
in TV and computer, 230V is directly applied to the input of bridge rectifier).
3. The PIV of the diode is half than in center tap full wave rectifier
4. Transformer utilization factor is higher than in center tapped full wave rectifier
5. Smaller size transformer required for given capacity because transformer is utilized effectively
during both AC cycles.
Disadvantages of bridge rectifier:
1. Requires Four diodes (But package is low cost)
2. Forward voltage drop across two diodes. This will reduce efficiency particularly when low voltage
(less than 5V) is required.
3. Load resistance and supply source have no common point which may be earthed.

Practical circuit diagram:
List of components:
1. Transformer 12 V AC, 500 mA
2. Diode 1N4007 ---- 4 No. or 1 A bridge rectifier package
3. Resistor 10K [4] Capacitor 1000μF [5] Toggle Switch
1. Construct circuit on the bread board.
2. Keep toggle switch OFF to perform practical of full wave rectifier without filter capacitor
and ON to connect filter capacitor.
WORKSHEET
Waveforms:
Without filter capacitor:

Output waveform:
With filter capacitor:
Output waveform:

Observations:
Peak Voltage, Vm = (From CRO for FWR with and without filter)
DC Voltage, VDC(full load) =
filter)
No Load DC Voltage, VDC(No load) =
filter)
Ripple Voltage, Vr = (From CRO for FWR with filter)
Calculations:
Without filter:
Ripple factor (Theoretical) =
Ripple Factor (Practical)
With filter:
Ripple factor (Theoretical)
Where f = 50Hz, R =1K , C = 1000 F.
Ripple Factor
Percentage Regulation = %
VNL = DC voltage at the load without connecting the load (Minimum current).

VFL = DC voltage at the load with load connected.
Efficiency %u200B
PAC = V2rms / RL
PDC = Vdc / RL
Conclusion:
Result
The Bridge rectifier circuit was done and the wave forms with and without filter capacitor are noted. The
DC voltage, DC current, ripple factor with and without filter capacitor is noted.

EXPERIMENT NO. 3
CHARACTERISTICS OF CE CONFIGURATION
AIM: To obtain common emitter characteristics of NPN transistor
Introduction:
Transistor is three terminal active device having terminals collector, base and emitter. Transistor is widely
used in amplifier, oscillator, electronic switch and so many other electronics circuits for variety of
applications. To understand operation of the transistor, we use three configurations common emitter,
common base and common collector. In this practical, we will understand common emitter configuration.
As the name suggest, emitter is common between input and output. Input is applied to base and output is
taken from collector.
We will obtain input characteristics and output characteristics of common emitter (CE) configuration. We
will connect variable DC power supply at VBB and VCC to obtain characteristics. Input voltage in CE
configuration is base-emitter voltage Vbe and input current is base current Ib. Output voltage in CE
configuration is collector to emitter voltage VCE and output current is collector current Ic. We will use
multi-meter to measure these voltages and currents for different characteristics. Collector to emitter
junction is reverse biased and base to emitter junction is forward biased. The CE configuration is widely
used in amplifier circuits because it provides voltage gain as well as current gain. In CB configuration
current gain is less than unity. In CC configuration voltage gain is less than unity. Input resistance of CE
configuration is less than CC configuration and more than CB configuration. Output resistance of CE
configuration is more than CC configuration and less than CB configuration.

Circuit setup for input characteristics
1. Connect circuit as shown in the circuit diagram for input characteristics
2. Connect variable power supply 0-30V at base circuit and collector circuit.
3. Keep Vcc fix at 0V (Or do not connect Vcc)
4. Increase VBB from 0V to 20V, note down readings of base current Ib and base to emitter voltage
Vbe in the observation table.
5. Repeat above procedure for Vcc = +5V and Vcc = +10V
6. Draw input characteristics curve. Plot Vbe on X axis and Ib on Y axis.
Observation table
Transistor: __________

Input Characteristics
Circuit setup for output characteristics
1. Connect circuit as shown in the circuit diagram for output characteristics
2. Connect variable power supply 0-30V at base circuit and collector circuit.
3. Keep base current fix (Initially 0)
4. Increase VCC from 0V to 30V, note down readings of collector current Ic and collector to emitter
voltage Vce in the observation table.
5. Repeat above procedure for base currents Ib = 5μA, 50 μA, 100 μA. Increase base current by
increasing VBB.
6. Draw output characteristics curve. Plot Vce on X axis and Ic on Y axis.

Observation table:
Output Characteristics
1. Graph:
Input Characteristics Output Characteristics

Calculations from Graph:
1. Input Characteristics: To obtain input resistance find VBE and IB for a constant VCE on one
of the input characteristics.
Input impedance = hie = Ri = VBE / IB (VCE is constant)
Reverse voltage gain = hre = VEB / VCE (IB = constant)
2. Output Characteristics: To obtain output resistance find IC and VCB at a constant IB.
Output admittance 1/hoe = Ro = IC / VCE (IB is constant)
Forward current gain = hfe = IC / IB (VCE = constant)
Inference:
1. Medium input and output resistances.
2. Smaller values if VCE, lower the cut-in-voltage.
3. Increase in the value of IE causes saturation of the transistor of an earlier voltage.
Precautions:
1. While performing the experiment do not exceed the ratings of the transistor. This may lead to
damage the transistor.
2. Connect voltmeter and ammeter in correct polarities as shown in the circuit diagram.
3. Do not switch ON the power supply unless you have checked the circuit connections as per the
circuit diagram.
4. Make sure while selecting the emitter, base and collector terminals of the transistor.
Result:
Input and Output characteristics of a Transistor in Common Emitter Configuration are studied.
The h-parameters for a transistor in CE configuration are:
a. The Input Resistance (hie) _______________Ohms.
b. The Reverse Voltage Gain (hre) _______________.
c. The Output Conductance (hoe) _______________ Mhos.
d. The Forward Current Gain (hfe) _______________.
Outcomes: Students are able to
1. Analyze the characteristics of BJT in Common Emitter and configuration.
2. Calculate h-parameters from the characteristics obtained.
Conclusion

Result
CE Transistor configuration was set up, I/P and O/P characteristics were plotted.
1. How to check transistor with help of multimeter?
2. How to check type of transistor (NPN or PNP) with help of multimeter?
3. Define current gain of the transistor in CE configuration. What is the DC current gain you obtain in
this practical?
4. Can transistor be replaced by two back to back connected diodes?
Ans: No, because the doping levels of emitter(heavily doped), base(lightly doped) and
collector(doping level greater than base and less than emitter) terminals are different from p and n
terminals in diode.
5. For amplification CE is preferred, why?
Ans: Because amplification factor beta is usually ranges from 20-500 hence this configuration
gives appreciable current gain as well as voltage gain at its output on the other hand in the
Common Collector configuration has very high input resistance(~750K ) & very low output
resistance(~25 ) so the voltage gain is always less than one & its most important application is for
impedance matching for driving from low impedance load to high impedance source.
6. To operate a transistor as amplifier, emitter junction is forward biased and collector junction
is reverse biased. Why?
Ans: Voltage is directly proportional to Resistance. Forward bias resistance is very less compared
to reverse bias. In amplifier input forward biased and output reverse biased so voltage at output
increases with reverse bias resistance.
7. Which transistor configuration provides a phase reversal between the input and output
signals?
Ans: Common emitter configuration.
8. 5. What is the range β of a BJT?
Ans: Beta is usually ranges from 20 - 500.
9. List the current components of BJT in CE configuration
10. What is Early Effect?
11. Why the doping of collector is less compared to emitter?
12. What do you mean by “reverse active”?
13. What is the difference between CE and Emitter follower circuit?
14. What are the input and output impedances of CE configuration?
15. Identify various regions in the output characteristics?
16. What is the relation between α, β and γ?
17. Define current gain in CE configuration?
18. Why CE configuration is preferred for amplification?
19. What is the phase relation between input and output?
20. 17. Draw diagram of CE configuration for PNP transistor?
21. 18. What is the power gain of CE configuration?
22. 19. What are the applications of CE configuration?

EXPERIMENT NO. 4
CHARACTERISTICS OF CB CONFIGURATION
AIM: To obtain common base characteristics of NPN transistor
Introduction:
In a common base configuration, base terminal is common between input and output. The output is taken
from collector and the input voltage is applied between emitter and base. The base is grounded because it
is common. To obtain output characteristics, we wil l measure collector current for different value of
collector to base voltage (VCB). Input current is emitter current Ie and input voltage is Veb. To plot input
characteristics we wi ll plot Veb versus Ie . Current gain for CB configuration is less than unity. CB
configuration is used in common base amplifier to obtain voltage gain. Output impedance of common base
configuration is very high. CB amplifier is used in multi-stage amplifier where impedance matching is
required between different stages.
Circuit diagram to obtain input characteristics:
Circuit diagram to obtain output characteristics

Experiment Procedure to obtain input characteristics:
1. Connect circuit as shown in the circuit diagram for input characteristics
2. Connect variable power supply 0-30V (VEE) at emitter base circuit and another power supply 0-
30V at collector base circuit (Vcc).
3. Keep Vcc fix at 0V (Or do not connect Vcc)
4. Increase VEE from 0V to 20V, note down readings of emitter current Ie and emitter to base voltage
Veb in the observation table.
5. Repeat above procedure for Vcc = +5V and Vcc = +10V
6. Draw input characteristics curve. Plot Veb on X axis and Ie on Y axis.
Experiment Procedure to obtain output characteristics:
1. Connect circuit as shown in the circuit diagram for output characteristics
2. Connect variable power supply 0-30V at emitter circuit and collector circuit.
3. Keep emitter current fix (Initially 0)
4. Increase VCC from 0V to 30V, note down readings of collector current Ic and collector to base
voltage Vcb in the observation table.
5. Repeat above procedure for base currents Ie = 1mA, 5 mA and 10mA. Increase emitter current by
increasing VEE.
6. Draw output characteristics curve. Plot Vcb on X axis and Ic on Y axis.
Observation table for input characteristics:
Input Characteristics

Observation table for output characteristics:
Output Characteristics

Model Graph:
Input characteristics: Output characteristics:
1. Plot the input characteristics for different values of VCB by taking VEE on X-axis and IE on Y-axis
taking VCB as constant parameter.
2. Plot the output characteristics by taking VCB on X-axis and taking IC on Y-axis taking IE as a
constant parameter.
Calculations from Graph:
The h-parameters are to be calculated from the following formulae:
1. Input Characteristics: To obtain input resistance, find VEE and IE for a constant VCB on one
of the input characteristics.
Input impedance = hib = Ri = VEE / IE (VCB = constant)
Reverse voltage gain = hrb = VEB / VCB (IE = constant)
2. Output Characteristics: To obtain output resistance, find IC and VCB at a constant IE.
Output admitance = hob = 1/Ro = IC / VCB (IE = constant)
Forward current gain = hfb = IC / IE (VCB = constant)
Inference:
1. Input resistance is in the order of tens of ohms since Emitter-Base Junction is forward biased.
2. Output resistance is in order of hundreds of kilo-ohms since Collector-Base Junction is reverse
biased.
3. Higher is the value of VCB, smaller is the cut in voltage.
4. Increase in the value of IB causes saturation of transistor at small voltages.
Precautions:
1. While performing the experiment do not exceed the ratings of the transistor. This may lead to
damage the transistor.
2. Connect voltmeter and ammeter in correct polarities as shown in the circuit diagram.

3. Do not switch ON the power supply unless you have checked the circuit connections as per the
circuit diagram.
4. Make sure while selecting the emitter, base and collector terminals of the transistor.
Result:
Input and Output characteristics of a Transistor in Common Base Configuration are studied.
The h-parameters for a transistor in CB configuration are:
a. The Input resistance (hib) __________________ Ohms.
b. The Reverse Voltage Transfer Ratio (hrb) __________________.
c. The Output Admittance (hob) __________________ Mhos.
d. The Forward Current gain (hfb) __________________.
Outcomes: Students are able to
1. analyze the characteristics of BJT in Common Base Configuration.
2. calculate h-parameters from the characteristics obtained.
Discussion/Viva Questions:
1. What is transistor?
Ans: A transistor is a semiconductor device used to amplify and switch electronic signals and electrical
power. It is composed of semiconductor material with at least three terminals for connection to an external
circuit. The term transistor was coined by John R. Pierce as a portmanteau of the term "transfer resistor".
2. Write the relation between and ?
Ans:
3. Define (alpha)? What is the range of ?
Ans: The important parameter is the common-base current gain, . The common-base current gain is
approximately the gain of current from emitter to collector in the forward-active region. This ratio usually
has a value close to unity; between 0.98 and 0.998.
4. Why is less than unity?
Ans: It is less than unity due to recombination of charge carriers as they cross the base region.
5. Input and output impedance equations for CB configuration?
Ans: hib = VBE / IE, 1 / hoe = VCE / IC
6. What is carrier lifetime?
7. What is the importance of Fermi level?
8. Can the junction less transistors be realized?
9. What is the doping level of E, B and C layers?
10. List the various current components in BJT.
11. Draw the input and output characteristics of the transistor in CB configuration?
12. Identify various regions in output characteristics?
13. What are the applications of CB configuration?
14. What are the input and output impedances of CB configuration?
15. What is EARLY effect?
16. Draw diagram of CB configuration for PNP transistor?
17. What is the power gain of CB configuration?
Conclusion

Result
CB Transistor configuration was set up, I/P and O/P characteristics were plotted.
1. What is early effect? Have you observed early effect in your experiment?
2. Compare common base and common emitter configuration
3. Justify the statement: Common base amplifier is used as buffer
4. What is the value of phase shift between input and output signal in common base and common
emitter amplifier?

EXPERIMENT NO. 5
RC COUPLED CE AMPLIFIER
AIM: To observe input-output waveforms of common emitter (CE) amplifier. To measure gain of
amplifier at different frequencies and plot frequency response
Introduction:
Common emitter amplifier is used to amplify weak signal. It utilizes energy from DC power supply to
amplify input AC signal. Biasing of transistor is done to tie Q point at the middle of the load line. In the
circuit shown, voltage divider bias is formed using resistors 10K and 2.2K. During positive cycle, forward
bias of base-emitter junction increases and base current increases. Q point moves in upward direction on
load line and collector current increases β times than base current. (β is current gain). Collector resistor
drop Ic*Rc increases due to increase in collector current Ic. This will reduce collector voltage. Thus during
positive input cycle, we get negative output cycle. When input is negative cycle, forward bias of base-
emitter junction and base current will reduce. Collector current reduces (Q point moves downside). Due to
decrease in collector current, collector resistance voltage drop IcRc reduces and collector voltage
increases. Change in collector voltage is much higher than applied base voltage because less base current
variation causes large collector current variation due to current gain B. This large collector current further
multiplied by collector resistance Rc which provides large voltage output. Thus CE amplifier provides
voltage gain and amplifies the input signal. Without emitter resistance gain of amplifier is highest but it is
not stable. Emitter resistance is used to provide stability. To compensate effect of emitter resistance
emitter bypass capacitor is used which provides AC ground to the emitter. This will increase gain of
amplifier.
CE amplifier does not provide constant voltage gain at all frequencies. Due to emitter bypass and coupling
capacitors reduces gain of amplifier at low frequency. Reactance of capacitor is high at low frequency,
hence emitter bypass capacitor does not provide perfect AC ground (Emitter impedance is high). There is
voltage drop across coupling capacitor at low frequency because of high reactance at low frequencies.
Gain of CE amplifier also reduces at very high frequency because of stray capacitances. Audio frequency
transistors like AC127, AC128 works for audio frequency range. It does not provide large voltage gain for
frequency greater than 20 KHz. Medium frequency transistors are BC147/BC148/BC547/BC548 provides
voltage gain up to 500 KHz. High frequency transistors like BF194/BF594/BF200 provides gain at radio
frequencies in the MHz range.
If we apply large signal at the input of CE amplifier, transistor driven into saturation region during positive
peak and cut-off region during negative peak (Q point reaches to saturation and cut-off points). Due to this
clipping occurs in amplified signal. So we have to apply small signal at the input and ensure that transistor
operates in active region.

Circuit diagram

Experimental procedure:
1. Connect function generator at the input of the amplifier circuit.
2. Set input voltage 10 mV and frequency 100 Hz.
3. Connect CRO at the output of the amplifier circuit.
4. Observe amplified signal and measure output voltage
5. Increase frequency from the function generator and repeat above step
6. Note down readings of output voltage in the observation table for frequency range from 100 Hz to
10 MHz
7. Calculate voltage gain for different frequencies and gain in dB. Plot frequency response.
Observation table
Input voltage: Vi = 10 Mv

Conclusion
1. Design and set up an ampli_er for the speci_cations: gain = -50, output voltage = 10 VPP ; fL = 50
Hz and calculate Zi.
2. Set up an RC coupled ampli_er and measure its input and output impedances
Measurement of input resistance
Method 1:
Connect a known resistor (say 1 k) in series between the signal generator and the input of the circuit.
Calculate the current though the resistor from the potential di_erence across it. Since this current also ows
into the circuit, input resistance can be measured taking the ratio of the voltage at the right side of the
resistor to the current.
Method 2: Connect a pot in series between the signal source and the input of the circuit. Adjust the pot
until the input voltage to the circuit is 50% of the signal generator voltage. Remove the pot from the circuit
and measure its resistance using a multimeter.
Measurement of output resistance
Method 1: Measure the open circuit output voltage. This is the Thevenin voltage. Output resistance of the
circuit is actually the Thevenin resistance in series with the Thevenin voltage. Connect a known value
resistor, say 1 k and measure the voltage across it. A reduction in the output voltage can be observed.
Calculate the current through the resistor. Since this current also ows trough the Thevenin resistance,
output resistance is the ratio of the di_erence in the output voltage to the current.
Method 2: Connect a pot at the output of the circuit. Adjust the pot until the voltage across it is 50% of
the open circuit voltage. Remove the pot from the circuit and measure
its resistance using a multimeter.
3. Differentiate between ac and dc load lines?
4. Explain their importance in ampli_er analysis.
5. Why is the center point of the active region chosen for dc biasing?

6. What happens if extreme portions of the active region are chosen for dc biasing?
7. Draw the output characteristics of the ampli_er and mark the load-line on it. Also mark
8. the three regions of operation on the output characteristics.
9. Which are the di_erent forms of coupling used in multi-stage ampli_ers?
1. What will be emitter current in the given circuit diagram in absence of input AC signal?
2. Draw DC load line of CE amplifier circuit. Show Q point on it.
3. Draw output waveform when inverted sine wave is applied at the CE amplifier circuit
4. What is bandwidth? What is the approximate bandwidth of CE amplifier that you have used during
your practical.
5. What is the effect on gain of amplifier if value of Rc increases?
6. What are the different biasing methods?
7. What happens if emitter bypass capacitor is removed from the circuit?
8. MODEL GRAPH
9.

Result
With CE:
1. Mid-band gain of the amplifier =: : : : : :
2. Bandwidth of the amplifier =: : : : : : Hz
Without CE:
1. Mid-band gain of the amplifier = : : : : : :
2. Bandwidth of the amplifier = : : : : : :Hz

EXPERIMENT NO. 6
RC COUPLED SINGLE STAGE FET AMPLIFIER
AIM
To design RC coupled single stage FET amplifier and determine the gain, frequency response,
input and output impedance.
COMPONENTS REQUIRED
Sl. No Apparatus and components Range Quantity
1 Spring board 1
2 FET BFW10/11 1
3 Resistors 2.7KΩ,1KΩ,2.2KΩ 1
4 Capacitors 0.1µF,.47 µF 2+2
5 VRPS 0-30Vdc 3A 1
6 Signal generator 10Hz to 1MHz 1
7 CRO 1
8 Probes, wires 2+15
9 DRB 0 to1Mohm 1
10 Digital Multimeter 2
THEORY
The field effect transistor (FET) has a capability to amplify a.c signals like a BJT. Depending
upon the type of configuration, the FET amplifiers may be classified as:
*Common source amplifier.
*Common drain amplifier.
*Common gate amplifier.
The circuit diagram 2.5 illustrates a common source junction FET amplifier. It is quite similar
to a common emitter amplifier .Here, the resistors R1 & R2 are used to bias the FET.The coupling
capacitors (Cc1,Cc2) are used to couple the a.c. input voltage source and the output voltage
respectively, these are known as coupling capacitors. The capacitor Cs keeps the source of the FET
efficiently at a.c. ground and is known as bypass capacitor.
The operation of the circuit may be understood from the assumption that when a small a.c.
signal is made to apply to the gate, it produces variations in the gate to source voltage which in

turn,producs variations in the drain current.As the gate to source voltage increases, the drain current
also increases because of this the voltage drop across the resistor Rd also increases. This causes the
drain voltage to decrease. It means the positive half cycle of the output voltage produces the negative
half cycle of the output voltage. In other words , there is a 180 degree phase shift between input and
output amplifier. This phenomenon of phase inversion is similar to that exhibited by a common emitter
bipolar transistor amplifier.
CIRCUIT DIAGRAM
Fig 1(B).a Circuit to find the frequency response curve of FET amplifier
Fig 1(B).b Circuit to find the input impedance
Fig 1(B).c Circuit to find the output impedance

DESIGN
Given: Idss = 8mA; Vp= -4V; gm=4m mhos; VDD=+15V;
Let VGS =-2V;
Id=Idss (1-VGS/VP)2
=2mA
Vs =Id*RS( Assume Id=Is)
Vs =Is*Rs
Rs= Vs/ Is=-Vgs/ Is=2/2=1KΩ.
Let gm=4m mhos
AV=µ= gm* Rd
Rd =10/4m=2.5KΩ≈2.7 KΩ
Let Cc1= Cc2 =0.1µF
Cs=47 µF
Input impedance is high hence select RG=2.2MΩ
PROCEDURE
1. Rig up the circuit as shown in the circuit diagram and give VDD = +15V and without connecting
signal generator check the biasing conditions i.e. VDS, VS and VGS.
2. Connect the signal generator and set the input voltage constant (say 200mV) at 10 KHz.
3. For different input frequencies note the corresponding output voltage.
4. Plot the frequency v/s decibel.
5. Find the figure of merit i.e. product of maximum gain and bandwidth.
6. Find the input and output impedance of the FET amplifier.
7. Connect the circuit as shown in Fig 2.b.
8. Set the DRB value to minimum initially and start increasing the resistance in the DRB from the
minimum value until output voltage becomes half. When the output voltage becomes half of the
initial value, the corresponding resistance in the DRB is the input impedance (Zi).
9. Connect the circuit as shown in Fig 2.c.
10. Set the DRB value to maximum initially and start decreasing the resistance in the DRB from the
maximum value until output voltage becomes half. When the output voltage becomes half of the
initial value, the corresponding resistance in the DRB is the output impedance (Zo).

EXPECTED GRAPH
Fig 1(B).d Frequency response curve of FET amplifier
TABULAR COLUMN
Frequency
(Hz)
Output
Voltage
(VoP-P)
Volts
Voltage
Gain
Vo / Vi
Gain (db)=20 log
(Vo / Vi)

Table 1(B).a To record the observed values
RESULT
The RC Coupled FET Amplifier was designed and the Bandwidth (BW), Input Resistance (Zi),
Output Resistance (Zo) is
Bandwidth (BW) =
Input Resistance (Zi) =
Output Resistance (Zo) =
EXPECTED VIVA QUESTIONS
1. What happens to the gain when the amplifiers are connected in cascade?
2. What is field effect transistor?
3. Why FET is called unipolar device?
4. Differentiate between FET and BJT.
5. Mention the parameters of FET.
6. Define drain resistance (rd).
7. Define Trans-conductance.
8. Define amplification factor.
OBSERVATION AND WORK SHEET
SEMILOG GRAPH SHEET HAS TO BE INSERTED

EXPERIMENT NO.7
RC PHASE SHIFT OSCILLATOR
AIM
To design and verify the performance of RC phase shift Oscillator for the given frequency.
EQUIPMENTS REQUIRED
Sl. No Apparatus and components Range Quantity
01 Bread Board 1
02 NPN transistor SL 100 1
03 Resistors
220Ω,5.6KΩ,22KΩ,820
Ω,
6.8KΩ
1+1+1+1+
3
04 Capacitors 4.7µF,0.001 µF 2+3
05 VRPS 0-30Vdc 3A 1
06 Potentiometer 47kΩ 1
07 CRO for testing 1
09 Digital multimeter 1
THEORY:
RC phase shift Oscillator basically consists of an amplifier and feed back network consisting of
resistors and capacitors in ladder fashion. The basic RC circuit is as shown below
Fig 3.a Phase lead network and Phasor diagram
The current I is in phase with Vo, whereas the capacitor voltage Vc lags the current I by φ
(90®→Ideal value).
OR the output voltage Vo leads the I/P voltage Vi by angle φ is adjusted in practice, equal to 60®.RC
network is used in feedback path. In Oscillator, feedback network must introduce a phase shift of

180® to obtain total phase shift around a loop as 360®.Thus three Rc network each
provide 60® phase shift is cascaded, so that it produces total 180® phase shift. The
Oscillator circuit consisting amplifier and Rc feedback network is as shown below.
CIRCUIT DIAGRAM:
Fig 3.b Circuit diagram of Phase
Shift Oscillator

PROCEDURE:
1) Make the circuit connections as shown in Fig 3.b
2) The output Vo is obtained on CRO. The 10 KΩ pot is adjusted to get a stable output on the
CRO.
3) The frequency of Oscillations is measured using CRO and then compared with the
theoretical values.
4) With respect to output at point P, the waveforms at point Q, R and S are observed on the
CRO.
5) We can see the phase shift at each point being 600
, 1200
and 1800
respectively.
NOTE:
The value of all three capacitors C is changed and the frequency of Oscillation
can be changed to new value and is measured again.

Designed frequency 2 KHZ
Actual frequency got 2KHZ
Phase shift between P & Q 60
P & R 120
P & S 180
RESULTS :
Theoretical frequency of oscillations = KHz
Practical frequency of oscillations = KHz

EXPERIMENT NO.8
BJT DARLINGTON EMITTER FOLLOWER
AIM
To design BJT Darlington emitter follower with and without bootstrapping and
determine the
Voltage gain, input impedance and output impedance.
EQUIPMENTS REQUIRED:
Sl. No Apparatus and
components
Range Quantity
1 Bread board 1
2 NPN transistor SL100 2
3 Resistors 560Ω, 480KΩ,800KΩ 1 each
4 Capacitors 0.47uF , 47uF 2+2
5 VRPS 0-30V DC, 3A 1
6 Signal generator 10Hz to 3MHz 1
7 CRO for testing 1
9 DRB 0 to 1MΩ 1
10 Digital multimeter 1
THEORY
A Darlington connection is a very popular connection of two transistors for
operation as one super beta transistor. The composite transistor acts as a single unit with a
current gain equal to the product of the current gains of individual transistors.
Sometimes ,the current gain and input impedance of an emitter followed are
insufficient to meet the requirement. In order to increase the overall values of circuit
current gain and input impedance, two transistors are connected together. The result
is that emitter current of the first transistor is the base current of the second transistor.
Therefore the current gain of the pair is equal to product of individual current gain that is β
= β1*β2.
CIRCUIT DIAGRAM

Fig 2.a Darlington emitter follower
Fig 2.b Darlington emitter follower with bootstrapping
DESIGN

Let VCE = 6V, ICQ≈IEQ=10mA (Q point of transistor Q2),
ẞ=100 (SL 100) Then Vcc = 2VCE=2 x 6 =12 V
IE = Ic = 10 mA
VR3 = Vcc – VCE= 12 – 6 = 6V
RE= VR3 / IE = 6V / (10 mA) = 0.6K
=560Ω (Choose)
VR2 –VBE1 - VBE2 – VR3 = 0
i.e , VR2 = VBE1 + VBE2 + VR3
= 0.6 + 0.6 + (IERE) = 1.2 +
(10x0.6) = 7.2V Vcc = VR1 +VR2
VR1= Vcc – VR2 = 12 –
7.2 = 4.8 V IE1=IB2 ≈IE2
/ ẞ=10mA/ 100= 0.1mA
IB1=IE1 / hfe = 0.1mA /
100 = 1 μA
R1 = VR1/ (10 (IB1)) = 4.8 / (10 x 1 μA) = 480 KΩ
R2= VR2 / (9 IB1) = 7.2 / (9 x 1 μA) = 800kΩ
To find Cc1
XcC1 Ri / 10 (Ri = R1|| R2 || hie = hie) Let fL=100Hz (Lower Cut-off Frequency)
fL= 1 /
(2π*(Ri /
10)*Cc1) Ri=
R1 || R2 || hie
For the above darlington pair hie≈βD*RE
For SL100 β=150 and βD= β* β=22500
Ri≈290 KΩ
So Cc1= 1 / 2π*(Ri/10)*fL= 1 / (2π*29KΩ*100) =0.05 uF
So, Use Cc1 = 0.1 F or 0.47 F .
To find Cc2
Let fL=100Hz ( Lower Cut-off frequency ) ,RL=1KΩ
fL= 1 / 2π*(Ro+RL)*CE
But Ro=Re ≈ re Here, re=VT / Ic
=26mV/10mA=2.6Ω Re ≈ re ≈3Ω
1/(2π*fL* CE )= Re =>
Therefore, CE=1/(2*π*100*(3+1KΩ))
CE =1.59 µF
Use Cc2 ≈0.47 or 2 µF

PROCEDURE
Rig up the circuit as in the case of biasing circuit for RC coupled BJT Amplifier
(Exp 1(A)) without connecting signal generator and capacitors. Check the biasing
conditions i.e. VCC =12V and check corresponding values of VCE,VBE,VE.
1. Connect the as in Fig 2.a (without bootstrapping)with signal generator and
designed capacitor values and set the input voltage constant at 50mV(p-p) ,
1KHz.
2. Now vary the input frequency starting from 100Hz to MHz range and note the
corresponding output voltage(peak to peak).
3. Plot the graph of frequency v/s output voltage gain in decibel with frequency on X-
axis and dB
gain on Y-axis and determine the bandwidth.
4. Repeat the procedure for circuit diagram in Fig 2.b (with bootstrapping)
TO DETERMINE INPUT IMPEDANCE (Zi) AND OUTPUT
IMPEDANCE (Zo) (a) INPUT IMPEADNCE (Zi)
Fig 2.c Circuit to find the
input impedance
PROCEDURE
1. Connect the circuit as shown in the Fig 2.c to obtain input impedance and
set the input frequency at say 10kHz(center frequency).
2. Set the DRB value to minimum initially and note the corresponding output voltage.
Start increasing the resistance in the DRB from the minimum value until output
voltage becomes half. when the output voltage becomes half of the initial value, the
corresponding resistance in the DRB is the input impedance(Zi).

b) OUTPUT IMPEDANCE (Zo)
Fig 2.d - Circuit to find the output impedance
PROCEDURE
1. Connect the circuit as shown in the diagram 2.d to obtain output impedance and
set the input frequency at say 10kHz(center frequency).
2. Set the DRB value to maximum initially and note the corresponding output voltage.
Start decreasing the resistance in the DRB from the maximum value until output
voltage becomes half. When the output voltage becomes half of the initial value,
the corresponding resistance in the DRB is the output impedance (Zo). Repeat the
procedure for circuit diagram in Fig 1(A).b (with bootstrapping)
EXPECTED GRAPH OF FREQUENCY RESPONSE
Fig 2.e Frequency response curve.

Frequency
(Hz)
Output
Voltage
(VOP-P) Volts
Voltage Gain
V / V
Gain(dB) =20 log (Vo /
V )
TABULAR COLUMN:
o i i
Table 2.a. To record the observed values
RESULT
Midband Voltage Gain =
With bootstrapping Without bootstrapping
Input Resistance(Zi)
Output Resistance(Zo)
EXPECTED VIVA QUESTIONS
1. What is Darlington emitter follower?
2 .Why do you call it as Darlington emitter follower?
3. What is the difference between with and without bootstrapping?
4. Benefits of with and without bootstrapping?
5. What is the difference between Darlington emitter follower and FET amplifier?
6.Mention the application of emitter follower?
OBSERVATION AND WORK SHEET
SEMILOG GRAPH SHEET HAS TO BE INSERTED

EXPERIMENT NO.9
Aim : Simplification, realization of Boolean expressions using logic gates/Universal
gates.
a) Verification of Logic gates.
b) Realize the following expressions Using Logic gates and universal gates in
i) SOP form ii) POS form
Components required:-
SlNo NAME OF THE
COMPONENT
IC NUMBER QUANTITY
1
2
3
4
5
6
7
8
9
AND gate
OR gate
Not gate
EXOR gate
NAND gate
NOR gate
EX-NOR gate
Patch chords
Trainer Kit
7408
7432
7404
7486
7400
7402
4077
2
2
2
2
2
2
1
1) NOT GATE
2) OR GATE
2) AND GATE
SYMBOL TRUTH TABLE IC 7408
UNIVERSAL GATES
1) NAND GATE

2) NOR GATE
3) XOR GATE
4) EX-NOR GATE
IMPIMENTATION OF BASIC GATES USUNG UNIVERSAL GATES
NAND GATE AS
(a)
LOGICDIAGRAM TRUTH TABLE

(b) OR GATE
(c) NOT GATE
(d) NOR GATE
(e) EX-OR GATE
(f) EX-NOR GATE
NOR GATE AS

(a) AND GATE
LOGIC DIAGRAM TRUTH TABLE
(b) OR GATE
(c) NOT GATE
(d) NAND GATE
(e) EX-NOR GATE
(f) EX-OR GATE

Realize the fallowing expressions in
(1) SOP form (sum of product)
(2) POS form (product of sum)
SOP FORM
F(A,B,C,D) = ∑(5,7,9,11,13,15)
Simplification- SOP form using basic gates
Using NAND gates using NOR gates
POS FORM
F(A,B,C,D) =∏(0,1,2,3,4,6,8,10,12,14)
Simplification- POS form Using basic gates
Using NAND gates Using NOR gates

Procedure:
Truth table:
1.Place the Ic in the socket of the trainer
kit.*complex boolean Expression s are
simplified by using Kmaps.
2.make the connections as shown in the
circuit diagram.
3.Apply diff combinations of i/ps
according to the truth table verify the
o/p.
4.Repeat the above procedure for all the
circuit diagrams.
A B C D Y=(A+B)D
0 0 0 0 0
0 0 0 1 0
0 0 1 0 0
0 0 1 1 0
0 1 0 0 0
0 1 0 1 1
0 1 1 0 0
0 1 1 1 1
1 0 0 0 0
1 0 0 1 1
1 0 1 0 0
1 0 1 1 1
1 1 0 0 0
1 1 0 1 1
1 1 1 0 0
1 1 1 1 1

EXPERIMENT NO.10
Realization of half/Full adder and Half/Full Sub tractors using logic gates.
Aim: (a) realization of half /full adder and (b)half/full subtractor using logic gates
Components required :-
Sl.No NAME OF THE
COMPONENT
IC NUMBER QUANTITY
1
2
3
4
5
6
7
8
AND gate
OR gate
Not gate
EXOR gate
NAND gate
NOR gate
Patch chords
Trainer Kit
7408
7432
7404
7486
7400
7402
1
1
1
3
3
3
(a) HALF ADDER USING BASIC GATES
(b) HALF ADDER USING NAND GATES
FULL ADDER USING NAND GATES Truth table (Full adder)
USING NAND GATES
Circuit Diagram

USING NOR GATES
Procedure:-verify the truth table for half adder and full adder circuits using basic and universal gates.
HALF SUBTRACTOR
Truth Table Circuit Diagram
Using NAND gates
FULL SUBTRACTOR
Truth Table
A B Diff Barrow
0 0 0 0
0 1 1 1
1 0 1 0
1 1 0 0
A B Bin Diff Borrow
0 0 0 0 0
0 0 1 1 1

Logic Diagram
USING NANAD GATES
USING NOR GATES
0 1 0 1 1
0 1 1 0 1
1 0 0 1 0
1 0 1 0 0
1 1 0 0 0
1 1 1 1 1

Procedure:-verify the truth table for half subtractor and full subtractor circuits using basic and universal
gates.

EXPERIMENT NO11
REALIZATION OF PARALLEL ADDER/SUB TRACTORS USING 7483 CHIP- BCD TO
EXCESS-3 CODE
AIM: (1) REALISATION of Parallel adder/subtractor using 7483chip
(2) BCD to XS3 code conversion and vice versa
Components required
:-
Sl.No NAME OF THE
COMPONENT
IC NUMBER QUANTITY
1
2
3
4
EXOR gate
4 bit parallel
adder/subtractor
Patch chords
Trainer Kit
7486
7483
1
1
Pin diagram:
LOGIC diagram
Block Diagram

Procedure:
Make the connections as shown .
For addition ,make Cin=0 and apply the 4 bits as i/p for A and aply another set of A bits to B. Observe
the o/p at S3, S2 S1 S0 and carry generated at Cout.
Repeat the above steps for different inputs and tabulate the result.
3.For subtration Cin is made equal to 1 and A-B format is used.
First no
second no.
By Xor –ing the i/p bits of ‘B’ by 1 , is complement of ‘B’ is obtained. Further Cin ,which is 1 is added to
the LSB of the Xor –ed bits. This generates 2’s complement of B.
verify the difference and polarity of differences at S0, S1, S2, S3.and Cout.
If Cout is 0 , diff is –ve and diff is 2’s complement form.
If Cout is 1, diff is +ve .
Repeate the above steps for different inputs. And tabulate the result.
Readings:-
Cin A3 A2 A1 A0 B3 B2 B1 B0 Cout S3 S2 S1 S0
0 1 0 0 1 1 0 0 1 1 0 0 1 0
0 0 1 1 1 0 0 0 1 0 1 0 0 0
1 1 0 0 1 1 0 0 0 1 0 0 0 1
1 0 0 0 1 0 0 1 1 0 1 1 1 0
Truth table
BCD XS3
B4 B3 B2 B1 X4 X3 X2 X1
0 0 0 0 0 0 1 1
0 0 0 1 0 1 0 0
0 0 1 0 0 1 0 1
0 0 1 1 0 1 1 0
0 1 0 0 0 1 0 0
0 1 0 1 1 0 0 0
0 1 1 0 1 0 0 1
0 1 1 1 1 0 1 0
1 0 0 0 1 0 1 1

BCD to Ex-3
Ex-3 to BCD
Circuit Diagram
Procedure:
BCD to XS-3 code conversion and vice-versa can be implemented using Ic 7483 along with
7486 Xor gates. The four i/p bits of ‘B’ ie B3, B2, B1, B0, are fixed as 0011. cin =0, performs
addition and Cin =1 performs subtraction.
1 0 0 1 1 1 0 0
Truth table
XS3 BCD
X4 X3 X2 X1 B4 B3 B2 B1
0 0 1 1 0 0 0 0
0 1 0 0 0 0 0 1
0 1 0 1 0 0 1 0
0 1 1 0 0 0 1 1
0 1 0 0 0 1 0 0
1 0 0 0 0 1 0 1
1 0 0 1 0 1 1 0
1 0 1 0 0 1 1 1
1 0 1 1 1 0 0 0
1 1 0 0 1 0 0 1

 For BCD to xs –3 code conversion 3 has to be added to i/p bits of A there for Cin
=0.
 For Xs-3 to BCD code conversion ‘3’ has to be subtarcated from the i/p of A
therefor Cin =1.
 Verify the truth table.

Experinment -12
AIM : Realization of Binary to Gray code conversion and vice versa.
Sl.No NAME OF THE
COMPONENT
IC NUMBER QUANTITY
1
2
3
4
EXOR gate
NAND gate
Patch chords
Trainer Kit
7486
7400
1
4
Binary to Gray code converter
G3 = ∑(8,9,10,11,12,13,14,15) G2 = ∑(4,5,6,7,8,9,10,11)
BINARY GRAY CODE
B3 B2 B1 B0 G3 G2 G1 G0
0 0 0 0 0 0 0 0
0 0 0 1 0 0 0 1
0 0 1 0 0 0 1 1
0 0 1 1 0 0 1 0
0 1 0 0 0 1 1 0
0 1 0 1 0 1 1 1
0 1 1 0 0 1 0 1
0 1 1 1 0 1 0 0
1 0 0 0 1 1 0 0
1 0 0 1 1 1 0 1
1 0 1 0 1 1 1 1
1 0 1 1 1 1 1 0
1 1 0 0 1 0 1 0
1 1 0 1 1 0 1 1
1 1 1 0 1 0 0 1
1 1 1 1 1 0 0 0

G3=B3
G2 =
G1= ∑(2,3,4,5,10.,11,12,13) G0 = ∑(1,2,3,5,6,9,10,13,14)
USING XOR GATES ONLY
Using NAND Gates only

GRAY TO BINARY
GRAY CODE BINARY CODE
G3 G2 G1 G0 B3 B2 B1 B0
0 0 0 0 0 0 0 0
0 0 0 1 0 0 0 1
0 0 1 0 0 0 1 1
0 0 1 1 0 0 1 0
0 1 0 0 0 1 1 1
0 1 0 1 0 1 1 0
0 1 1 0 0 1 0 0
0 1 1 1 0 1 0 1
1 0 0 0 1 1 1 1
1 0 0 1 1 1 1 0
1 0 1 0 1 1 0 0
1 0 1 1 1 1 0 1
1 1 0 0 1 0 0 0
1 1 0 1 1 0 0 1
1 1 1 0 1 0 1 1
1 1 1 1 1 0 1 0

LOGIC DIAGRAM
USING NAND GATE

Procedure:-
1. place the Ic’s in the socket of the trainer kit.
2. make connections for the gate as shown in the circuit diagram.
3.Apply different combinations of the input according to the truth table and verify the
corresponding o/ps
shown on the truth table.

EXPERIMENT-13
Ring counter/Johnson counter
AIM –Design and testing of Ring counter/Johnson counter using IC-7495
RING COUNTER USING IC-7495
Sl.No NAME OF THE
COMPONENT
IC
NUMBER
QUANTITY
1
2
Ring Counter
NAND gate
Patch chords
Trainer Kit
7495
7400
1
1
TRUTH TABLE CIRCUIT DIAGRAM
CP QA QB QC QD
t0 1 0 0 0
t1 0 1 0 0
t2 0 0 1 0
t3 0 0 0 1
t4 1 0 0 0
Procedure-
(1). Rig up the circuit as shown in the diagram,DS is not given as input.
(2). Load data parallely with clkp and M=1
(3). Then make M=0,Clks-cp
(4). Verify the working of a ring counter.
JHONSON COUNTER USING
TRUTH TABLE CIRCUIT DIAGRAM

CP QA QB QC QD
t0 1 0 0 0
t1 1 1 0 0
t2 1 1 1 0
t3 1 1 1 1
t4 0 1 1 1
t5 0 0 1 1
t6 0 0 1 1
t7 0 0 0 1
t8 1 0 0 0
Procedure-
(1). Rig up the circuit as shown in the diagram,DS is not given as input.
(2). Load data parallely with clkp and M=1
(3). Then make M=0,Clks-cp
(4). Verify the whether the ckt works as a Jhonoson counter or twisted ring
counter.

EXPERIMENT NO.14
SEQUENCE GENERATOR
AIM: DESIGN A SEQUENCE GENERATOR
Sequence: 100010011010111
Design: There are 15 bits, so there will be 15 states s=15. So at least 4 flip-flops are required.
Sl.No NAME OF THE
COMPONENT
IC NUMBER QUANTITY
1
2
3
4
Shift register
Ex-OR
NANDgate(3i/ps)
Trainer Kit
Patch Chords
7495
7486
7410
1
1
1
TRUTH TABLE SIMPLFICATION
QA QB QC QD f
1 1 1 1 0
0 1 1 1 0
0 0 1 1 0
0 0 0 1 1
1 0 0 0 0
0 1 0 0 0
0 0 1 0 1
1 0 0 1 1
1 1 0 0 0
0 1 1 0 1
1 0 1 1 0
0 1 0 1 1
1 0 1 0 1
1 1 0 1 1
1 1 1 0 1
Procedure-
(1).The sequence is written such that no state repeats itself.The binary sequence is repeated
once in every 2N-1
clock cycles.
(2). The Expression for (QA, QB, QC, QD) is got using K-maps.
(3). Rig up the circuit as shown in the figure.
(4). Intially let M = 1,clkp = cp, the intial state (A, B, C, D – 1110/1111) is losded.
(5). Then make clks =Cp, M = 0, output is observed at MSB (QA).
Note:-When we observe the sequence, which is to be generated, the LSB is a 1, following bit
is a 0. If 0 has to be generated, then input to that particular D-FF must be a 0. There fore
f(QA, QB, QC, QD) has its first entery as a 0.

EXPERIMENT NO.15
COUNTERS
Aim -Realization of 3-bit counters as a sequential circuit and mod-N counter design
(7476,7490,74192,74193)
a) Asynchrones type
b) Synchronous type
Components required:-
Sl.No NAME OF THE
COMPONENT
IC
NUMBER
QUANTITY
1
2
3
4
5
6
JK flip flop
NANDgate(3 pin)
AND gate
OR gate
Decade Counter
Decade Up/down
Counter
MOD 16 counter
Patch chords
Trainer Kit
7476
7408
7432
7490
74192
74193
2
2
1
1
1
1
1
(A). ASYNCRONOUS COUNTERS
(a) Realization of 3-bit binary counters using IC7476(MOD-8)
PIN DIAGRAM
UP COUNT (MOD-8)
WAVE FORMS

TRUTH TABLE
DOWN COUNT
TRUTH TABLE
Number of
clock pulses
Flip Flop outputs
Qc Qb Qa
0 0 0 0
1 0 0 1
2 0 1 0
3 0 1 1
4 1 0 0
5 1 0 1
6 1 1 0
7 1 1 1
8 0 0 0
Number of clock
pulses
Flip Flop outputs
Qc Qb Qa

MOD-N COUNTER (UP COUNTER)
MOD-4 COUNTER –
In MOD-4 counter 1 0 0 (Qa, Qb, Qc) is Invalid state
TRUTH TABLE
Circuit diagram
0 1 1 1
1 1 1 0
2 1 0 1
3 1 0 0
4 0 1 1
5 0 1 0
6 0 0 1
7 0 0 0
8 1 1 1
Number of clock
pulses
Flip Flop outputs
Qc Qb Qa
0 0 0 0
1 0 0 1
`2 0 1 0
3 0 1 1

WAVE FORMS
MOD-6 COUNTER TRUTH TABLE
In MOD-6 counter 1 1 0 (QC, Qb, Qa) invalid state
Circuit diagram
WAVEFORMS

MOD-N COUNTERS (DOWN COUNTER)
MOD-4 COUNTER-
Invalid state is from 1 0 0 to 1 1 1. Then sequence is 011(C, B, A), 010,001,000,011
TRUTH TABLE
WAVEFORMS

MOD-7 COUNTER
In MOD-7 Counter the invalid state is 111, the data sequence will starts from 110 and should
count down to 000
Ie. 110- 101-100-011-010-001-000-110
TRUTH TABLE
CIRCUIT
DIAGRAM
WAVEFORMS

REALISATION OF 3 BIT UP/DOWN COUNTER
WAVEFORMS
SYNCHRONOUS COUNTERS
UP COUNTER
DESIGN AND REALIZATION OF 3 BIT SYNCHRONOUS COUNTER USING IC7476
Excitation table
Present
state
Next
State J K

output output
0 0 0 X
0 1 1 X
1 0 X 1
1 1 X 0
simplifications
PRESENT STATE NEXT STATE EXCITATION
Qc Qb Qa Qc Qb Qa JC KC JB KB JA KA
0 0 0 0 0 1 0 X 0 X 1 X
0 0 1 0 1 0 0 X 1 X X 1
0 1 0 0 1 1 0 X X 0 1 X
0 1 1 1 0 0 1 X X 1 X 1
1 0 0 1 0 1 X 0 0 X 1 X
1 0 1 1 1 0 X 0 1 X X 1
1 1 0 1 1 1 X 0 X 0 1 X
1 1 1 0 0 0 X 1 X 1 X 1

CIRCUIT DIAGRAM
WAVE FORMS
MOD-6 COUNTER
In MOD-6 counter invalid state is 110
simplifications
0 0 0 0 0 1 0 X 0 X 1 X
0 0 1 0 1 0 0 X 1 X X 1
0 1 0 0 1 1 0 X X 0 1 X
0 1 1 1 0 0 1 X X 1 X 1
1 0 0 1 0 1 X 0 0 X 1 X
1 0 1 0 0 0 X 1 0 X X 1

CIRCUIT DIAGRAM
WAVE FORMS
MOD-5 SYNCHRONOUS DOWN COUNTER
PRESENT STATE NEXT STATE FLIP-FLOPS
Qc Qb Qa Qc Qb Qa Jc Kc Jb Kb Ja Ka
1 0 0 0 1 1 X 1 1 X 1 X
0 1 1 0 1 0 0 X X 0 X 1
0 1 0 0 0 1 0 X X 1 1 X
0 0 1 0 0 0 0 X 0 X X 1

0 0 0 1 0 0 1 X 0 X 0 X
SIMPLIFICATION
CIRCUIT DIAGRAM
MOD-8 DOWN COUNTER
TRUTH TABLE
1 1 1 1 1 0 X 0 X 0 X 1
1 1 0 1 0 1 X 0 X 1 1 X
1 0 1 1 0 0 X 0 0 X X 1
1 0 0 0 1 1 X 1 1 X 1 X
0 1 1 0 1 0 0 X X 0 X 1

0 1 0 0 0 1 0 X X 1 1 X
0 0 1 0 0 0 0 X 0 X X 1
0 0 0 1 1 1 1 X 1 X 1 X
SIMPLIFICATIONS
CIRCUIT
DIAGRAM
MOD-N COUNTERS
To realize a MOD-N counter using IC-74193 with a given preset value, write down the
expected function table
Pin details of IC 74193(Synchronous counter)
[MOD-16 UP/DOWN COUNTER]

FUNCTION TABLE
Load Up Down Qd Qc Qb Qa
H X X X 0 0 0 0
L L X X D C B A
L H Cp H COUNT UP
L H H Cp COUNT DOWN
L H H H NO CHANGE
Design a counter which counts from (6-
12)
Invalid state 1101
WAVE FORMS

REALIZE A (15-6) COUNTER USING IC 74193
Invalid state---0101
Note:-Lo and Bo are used basically for cascading the counters
To realize a MOD-N counter using IC-74193
PIN DIAGRAM
INTERNAL DIAGRAM

Functional Table
R1 R2 S1 S2 Qa Qb Qc Qd
H H L X L L L L
H H X L L L L L
X L H H 1 0 0 1
L X L X MOD-2 COUNTER
X L X L MOD-5 COUNTER
7490 AS MOD-2 COUNTER

To realize a MOD-N counter using IC74192 with given preset value, write down the
expected function table
SYNCHRONOUS COUNTER
PIN DETAILS OF IC-74192[ MOD-10 UP/DOWN COUNTER]
MOD-6 UP COUNTER:Invalid state o110

MOD-9 DOWN COUNTER:
DESIGN A COUNTER WHICH CAN COUNT FROM 7 TO 9
NOTE After 1001, out put becomes 0000

VIVA QUESTION
1. Define minterm and maxterm?
2. Define minterm Canonical formula and maxterm canonical formula?
3. What does standard SOP and standard POS mean?
4. What are basic gates? disjunctive conjunctive.
5. What are universal gates? Why are they called so?
6. What does prime implicant, prime implicant and essential prime implicant mean?
7. What does the term subsume mean?
8. What are the different methods for solving Boolean expressions?
9. Among K-map, VEM technique & Quine-McCluskey, which is easier to use?
10. What is the other name given to K-map?
11. What is the other name given to VEM technique?
12. What is the other name given to Quine-McCluskey?
13. What is the difference between Combinational circuit and Sequential circuit?
14. What is a Parallel adder?
15. If 2 10-bit numbers have to be added using Parallel adder how many full adders are
required?
16. What are the disadvantages of parallel adder?
17. What is a look ahead carry adder?
18. What is a serial adder?
19. Is a serial adder a sequential circuit or a combinational circuit?
20. What is a BCD adder?
21. Which is better when compared to a binary adder and BCD adder?
22. If 2 10-bit numbers have to be added using serial adder, how many full adders are
required?
23. Which is fastest among parallel adder, serial adder and look ahead carry adder?
24. Which is called a reflecting code?
25. Is Gray code a weighted code?
26. Which is called a self complementing code?
27. Is EXESSES-3 code a weighted code?
28. What is an ASCII code?
29. What is an EBCDIC code?
30. What is parity checking?
31. What is the difference between encoder and priority encoder?

32. Why Enable input is generally an active LOW signal?
33. What is the difference between de-multiplexer and decoder?
34. What is the role of a multiplexer?
35. Is it possible to realize 4-bit Binary to Gray using a 1:16 de-multiplexer?
36. What are PLD‟s?
37. What is the difference between PAL and PLA?
38. What is the difference between RAM and ROM?
39. Can a PROM be reprogrammed?
40. What is the difference between Static RAM and Dynamic RAM?
41. Is it possible to realize comparator using Multiplexer?
42. What is the difference between LCD and LED?
43. What is the input voltage to a Common Anode LED?
44. Why is a low frequency signal required for an LCD?
45. Why an ordinary decoder is cannot be used with displays?
46. What does RBI, RBO and BI mean?
47. What does a basic bi-stable element mean?
48. What is the forbidden state in SR latch?
49. How contact de-bouncing can be eliminated using SR latch?
50. What is the difference between a latch and a flip flop?
51. What is meant by race around condition? How is it eliminated?
52. How to convert a JK flip flop to a D flip flop?
53. How to convert a JK flip flop to a T flip flop?
54. How to convert a T flip flop to a D flip flop?
55. How to convert a D flip flop to a T flip flop?
56. Why is a D flip flop called as a transparent flip flop?
57. What are the two asynchronous inputs? Why are they called so?
58. What is the difference between synchronous PRESET and asynchronous PRESET?
59. Design a traffic light system where in RED and GREEN do not glow simultaneously.
When RED glows GREEN is OFF and when RED glows GREEN is OFF.
60. What is MEALY sequential circuit?
61. What is MOORE sequential circuit?
62. Can a multiplexer be used to realize a JK flip flop?
63. What is the difference between a counter and a register?
64. What is the difference between a synchronous counter and asynchronous

counter?
64. What type of flip flop is used to design a asynchronous counter?
65. In a mod-16 counter, each of the flip flop used has a delay of 20ns. To get an
output at the MSB how long does it take in a synchronous counter and in a
asynchronous counter?
66. Why is an asynchronous counter called as a ripple counter?
67. Define modulus of a counter.
68. What type of counter is 7490?
69. What type of counter is 74192 and 74193?
70. What type of flip flop is used to design a synchronous counter?
71. What is data lockout in a counter?
72. How is data lockout overcome?
73. Is the output of a counter always a square wave?
74. What is the difference between a shift register and register?
75. Is a shift register a synchronous circuit or an asynchronous circuit?
76. If a ADC is interfaced to a shift register, what type of shifting is preferred?
77. What are the features of a shift register?
78. Do we have a register which can perform all the basic shifting operation?
79. Name a combinational circuit which works similar to a PISO?
80. Name a combinational circuit which works similar to a SIPO?
81. What is a ring counter and a Johnson counter?
82. Is the output of a ring counter a square wave?
83. What is the difference between a self starting ring counter and a counter
loaded initially with a parallel data?
84. What happens if 0000 is loaded into a ring counter, provided it is not self
starting?
85. If there are 6 states in a ring counter, then how many flip flops are required?
86. If there are 7 states in a Johnson counter, then how many flip flops are
required?
87. What is the use of ring counter and Johnson counter?

88. Is the ring counter a synchronous circuit or an asynchronous circuit?
89. What are the other names by which a Johnson counter is called?
90. Define SET-UP time and HOLD time?
91. If the maximum clock frequency of a JK flip flop is 45Mhz, what happens if 100
Mhz is given?
92. What is meant by active time or triggering time?
93. What is a sequence generator? What is its application?
94. What happens if the states repeat in a sequence generator?
95. Is the sequence generator periodic in nature?
96. What is the difference between a sequence generator and a ring counter?
97. What is the sequence generator made up of?
98. Is the sequence generator a synchronous circuit or an asynchronous circuit?
99. If the length of the sequence to be generated is specified, then what is the
name given to such sequence generator?
100. Explain the working of a 555 timer?
101. What is the purpose of a monostable and astable?
102. Why is a monostable called so?
103. Why is a astable called so?
104. What is the use of the two diodes in a astable circuit?
105. Is a trigger input required for a astable?
106. Why doesn‟t triggering pulse affect a monostable output when the capacitor is
still charging?
107. What is positive logic?
108. What is negative logic?
109. What are the output voltage ranges for a TTL gate?
110. What are the input voltage ranges for a TTL gate?
111. What is meant by current sourcing and current sinking?
112. Define Fan in?
113. Define Fan out?

114. Define Speed Power product?
115. In the basic circuit of transistor as an inverter, why CE configuration is used?
116. Why Diode-Transistor Logic is not used?
117. What is the advantage of Totempole configuration?
118. What is the disadvantage of Totempole configuration?
119. What is Open Collector configuration?
120. What are the advantages and disadvantages of Open Collector configuration?
121. What are Schottky diodes and transistors?
122. What are the different TTL families?
123. What does 74LS series mean?
124. If the input or output voltages fall in the intermediate range, what happens?
125. What is the difference between FET and MOSFET?
126. What is the difference between FET and BJT?
127. What are the advantages of using MOSFET?
128. Why p-MOS is not used?
129. Why n-MOS is used?
130. Why c-MOS is used?
131. What is the range of voltages that can be given as a input to MOS devices?
132. What is the output range of voltage for a MOS device?
133. What is the application of FET in digital circuits?
134. Compare the performance of different TTL sub families?
135. What is wired AND gate?
136. What is the input impedance of FET?
137. What is the input impedance of MOSFET?
138. What is enhancement MOSFET?
139. What is depletioin mode MOSFET?
140. What is inversion layer in a MOSFET?
141. Why MOS devices are preferred in digital circuits?
142. Give some application of MOS devices in digital circuits.

APPENDIX – I
MAXIMUM RATINGS OF COMMONLY USED TRANSISTORS
BC107
Specifications:
1. Type : Si – NPN
2. operating point temp : 65o to 200oC
3. IC(max) : 100mA
4. hfe (min) = 110 : 100
5. hfe (max) : 450
6. VCE (max) : 45V
7. Ptot(max) : 300mW
8. Category(typical use) : Audio, low power
9. Possible substitutes :BC182, BC547

DIODE

APPENDIX – II
COMPONENT VALUE IDENTIFICATION
RESISTOR VALUE IDENTIFICATION

STANDARD CAPACITOR VALUES

APPENDIX – III
SYMBOLS
VARIABLES

Gopi 1 lab manual

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