adc r2013 lab manual

EC6311 Analog and Digital Lab

Semester 03 Department of ECE Rajalakshmi Institute of Technology

RAJALAKSHMI INSTITUTE OF TECHNOLOGY

CHENNAI 600 124

DEPARTMENT

of ELECTRONICS AND COMMUNICATION ENGINEERING

Regulation 2013

Academic Year: 2014-2015

EC6311 ANALOG AND DIGITAL LABORATORY

STUDENT MANUAL

Faculty In-charge

Mr.S.Rajalingam /AP/ECE

Mr.L.Franklin Telfer/AP/ECE

Mr.V.S.Vignesh/AP/ECE

Lab Assistant

Ms.T.Jeevitha

www.Vidyarthiplus.com


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SYLLABUS

EC6311 ANALOG AND DIGITAL CIRCUITS LABORATORY

LIST OF EXPERIMENTS

LIST OF ANALOG EXPERIMENTS: 1. Frequency Response of CE / CB / CC amplifier

2. Frequency response of CS Amplifiers

3. Darlington Amplifier

4. Differential Amplifiers- Transfer characteristic.

5. CMRR Measurement

6. Cascode / Cascade amplifier

7. Determination of bandwidth of single stage and multistage amplifiers

8. Spice Simulation of Common Emitter and Common Source amplifiers

LIST OF DIGITAL EXPERIMENTS 9. Design and implementation of code converters using logic gates

(i) BCD to excess-3 code and vice versa (ii) Binary to gray and vice-versa

10. Design and implementation of 4 bit binary Adder/ Subtractor and BCD adder using IC 7483

11. Design and implementation of Multiplexer and De-multiplexer using logic gates

12. Design and implementation of encoder and decoder using logic gates

13. Construction and verification of 4 bit ripple counter and Mod-10 / Mod-12 Ripple counters

14. Design and implementation of 3-bit synchronous up/down counter

15. Implementation of SISO, SIPO, PISO and PIPO shift registers using Flip- flops



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SL NO LIST OF EXPERIMENTS

LIST OF ANALOG EXPERIMENTS

Introduction - Study of electronic components [ Active and Passive] - Study of Signal Generator, CRO, Bread board and Regulated Power supply - Study of transistor parameters using Transistor Data Sheets

1. Design and Analysis of Common Emitter Amplifier - To Determine a. DC characters tics b. AC characteristics c. Gain d. Bandwidth e. Gain- Bandwidth Product f. SPICE Simulation of Amplifier(Additional)

2.

Design and Analysis of Common Collector Amplifier - To Determine a. DC characteristics b. AC characteristics c. Gain d. Bandwidth e. Gain- Bandwidth Product f. SPICE Simulation of Amplifier(Additional)

3. Design and Analysis of Common Base Amplifier - To Determine a. DC characteristics b. AC characteristics c. Gain d. Bandwidth e. Gain- Bandwidth Product f. SPICE Simulation of Amplifier(Additional)

4.

Design and Analysis of Darlington Amplifier - To Determine a. DC characteristics b. AC characteristics c. Gain d. Bandwidth e. Gain- Bandwidth Product f. SPICE Simulation of Amplifier(Additional)

5. Design and Analysis of Common Source Amplifier - To Determine a. DC characteristics b. AC characteristics c. Gain d. Bandwidth e. Gain- Bandwidth Product f. SPICE Simulation of Amplifier (Additional)

6.

Design and Analysis of Cascade Amplifier - To Determine a. DC characteristics b. AC characteristics c. Gain d. Bandwidth e. Gain- Bandwidth Product f. SPICE Simulation of Amplifier (Additional)

7. Design and Analysis of Cascode Amplifier - To Determine a. DC characteristics b. AC characteristics c. Gain d. Bandwidth e. Gain- Bandwidth Product f. SPICE Simulation of Amplifier (Additional)

8.

Design and Analysis of Differential Amplifier - To Determine a. Transfer characteristics b. CMRR

9.

Pspice Simulation of Common Emitter Amplifier a. Gain b. Bandwidth

10. Pspice Simulation of Common source Amplifier a. Gain b. Bandwidth

LIST OF DIGITAL EXPERIMENTS

Introduction - Study of Digital IC's using IC data sheets - Study of IC trainer Kit

11. Design and implementation of code converters using logic gates (i) BCD to excess-3 code and vice versa (ii) Binary to gray and vice-versa

12. Design and implementation of 4 bit binary Adder/ Subtractor and BCD adder using IC 7483

13. Design and implementation of Multiplexer and De-multiplexer using logic gates

14. Design and implementation of encoder and decoder using logic gates

15. Construction and verification of 4 bit ripple counter and Mod-10 / Mod-12 Ripple counters

16. Design and implementation of 3-bit synchronous up/down counter

17. Implementation of SISO, SIPO, PISO and PIPO shift registers using Flip- flops



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INDEX

S.No Date Experiment Names Pg No Faculty Sign

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

13.

14.

15.

16.

17.



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Common Emitter Amplifier circuit diagram

CE Amplifier without Feedback :

CE Amplifier with Feedback:



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COMMON EMITTER AMPLIFIER

EXPERIMENT: 01 DATE:

1. OBJECTIVE:

To Design and Construct a Common Emitter Amplifier using voltage divider bias and to

determine its:

a. DC Characteristics

b. Maximum Signal Handling Capacity

c. Gain of the amplifier

d. Bandwidth of the amplifier

e. Gain -Bandwidth Product

2. REQUIREMENTS:

S.no Requirement Name Range Quantity

1

Components

Transistor [Active] BC 107 1

2 Resistor [Passive] 61kΩ, 10kΩ, 1kΩ,

4.7kΩ 1,1,1,2

3 Capacitor [Passive] 10µf, 100µf 2,1

4

Equipment

Signal Generator (0-3)MHz 1

5 CRO 30MHz 1

6 Regulated power

supply (0-30)V 1

7

Accessories

Bread Board - 1

8 Connecting Wires Single strand as required



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DESIGN PROCEDURE:

Given specifications:

VCC= 10V, IC=1.2mA, AV= 30, hFE= 100

(i) To calculate RC:

The voltage gain is given by,

AV= -hfe (RC|| RF) / hie

h ie = β re

re = 26mV / IE = 26mV / 1.2mA = 21.6Ω

hie = 150 x 21.6 =3.2KΏ

Apply KVL to output loop,

VCC= IC RC + VCE+ IE RE ----- (1)

Where VE = IE RE (IC= IE)

VE= VCC / 10= 1V

Therefore RE= 1/1.2x10-3=0.8K= 1KΏ

VCE= VCC/2= 5V

From equation (1), RC= ( Vcc - VCE - IE RE / IC ) = ________

(ii) To calculate R1&R2:

S=1+ (RB/RE)

Where RE = 1 KΏ and S = 9

RB= (S-1) RE= (R1 || R2) =1KΏ

RB=( R 1R2 ) /( R1+ R2) ------- (2)

VB= VBE + VE = 0.7+ 1= 1.7V

VB= VCC (R2 / R1+ R2 )------- (3)

Solving equation (2) & (3),

R1= ____ & R2= ______

(iii) Input coupling capacitor :

Xci= Rif / 10= 2.4 Ω (since XCi << Rif)

Ci = 1/ 2пfXCi = _____

(iv) Output coupling capacitor:

XCO= Rof /10= 5.2 Ω

CO = 1/ 2пfXCO = _____



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3. THEORY:

A common emitter amplifier is type of BJT amplifier which increases the voltage level of the

applied input signal Vin at output of collector.

The CE amplifier typically has a relatively high input resistance (1 - 10 KΩ) and a fairly high

output resistance. Therefore it is generally used to drive medium to high resistance loads. It is

typically used in applications where a small voltage signal needs to be amplified to a large

voltage signal like radio receivers.

The input signal Vin is applied to base emitter junction of the transistor and amplifier output

Vo is taken across collector terminal. Transistor is maintained at the active region by using the

resistors R1,R2 and Rc. A very small change in base current produces a much larger change in

collector current. The output Vo of the common emitter amplifier is 180 degrees out of phase

with the applied the input signal Vin.

4. PROCEDURE:

1. Connect the circuit as per the circuit diagram

2. Determine the Q-point of the CE amplifier using DC analysis.

3. Determine Maximum input voltage that can be applied to CE amplifier using AC analysis.

4. Set the input voltage Vin=V MSH /2 and vary the input signal frequency from 0Hz to 1MHz in

incremental steps and note down the corresponding output voltage Vo for at least 20

different values for the considered range.

5. The voltage gain is calculated as Av = 20log (V0/Vi) dB

6. Find the Bandwidth and Gain-Bandwidth Product from Semi-log graph taking

frequency on x-axis and gain in dB on y-axis.,

Bandwidth, BW = f2-f1

where f1 lower cut-off frequency

f2 upper cut-off frequency



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a. DC ANALYSIS:

It is the procedure to find the operating region of transistor

Steps:

i) Set Vin = 0 by reducing the amplitude of the input signal from signal

generator

ii) Open circuit the capacitors since it blocks DC voltage

iii) Set VCC= +10v and measure the voltage drop across the Resistor VRC, voltage

across Collector- Emitter Junction VCE and Voltage drop across base emitter

junction. VBE

iv) Find the Q-point of the transistor and draw the DC load line.

To verify dc condition

1. VBE : (forward bias)

2. VRC = ____________

3. VCE = _______ (REVERSE BIAS)

4. Ic( Ic = (Vcc – VCE ) / Rc) =________

Q point analysis:

It is the procedure to choose the opearating point of transistor

Q-point: ( ICQ =_____ ; VCEQ =______ )



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b. Maximum signal handling capacity :

It is the process to find the maximum input voltage that can be handled by the

amplifier, so that it amplifies the input signal without any distortion.

Procedure:

i. Apply input signal Vin = 20 mV of 1Khz frequency to the amplifier using the

signal generator between base emitter junction of the transistor. Find the

sinusoidal output using CRO across RL.

ii. By increasing the amplitude of the input signal find maximum input voltage

V MSH across VBE at which the sinusoidal signal gets distorted during the

process which can be seen in the CRO. The amplitude obtained at this point

is maximum voltage that can be applied to the transistor for efficient

operating of transistor.

V MSH = _________ volts

MODEL GRAPH:



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5. TABULATION [Without Feedback ] :

Input voltage (Vin=V MSH/2) =____________V

S. NO FREQUENCY [Hz] OUTPUT VOLTAGE [ VO]

in Volts

GAIN= 20 log Vo/Vin

dB

1. 0

2. 100

3. 500

4. 600

5. 800

6. 900

7. 1 KHz

8. 100 KHz

9. 500 KHz

10. 600 KHz

11. 700 KHz

12. 800 KHz

13. 900 KHz

14. 1 MHz

15. 1.1 MHz

16. 1.5 MHz



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With Feedback :

Input voltage (Vin=V MSH/2) =____________ V

S. NO FREQUENCY [Hz] OUTPUT VOLTAGE [

VO] in Volts

GAIN= 20 log ( vo/vin )

dB

1.

0

2. 100

3. 500

4. 600

5. 800

6. 900

7. 1 KHz

8. 100 KHz

9. 500 KHz

10. 600 KHz

11. 700 KHz

12. 800 KHz

13. 900 KHz

14. 1 MHz

15. 1.1 MHz

16. 1.5 MHz



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WORKSHEET



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6. RESULT:

INFERENCE:

The Common Emitter Amplifier was constructed and the following results were determined:

a) Gain of the amplifier :

b) Bandwidth of the amplifier :

c) Gain-Bandwidth product :

CONCLUSION:



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Common Collector Amplifier Circuit Diagram:

MODEL GRAPH:



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COMMON COLLECTOR AMPLIFIER


1. OBJECTIVE:

To Design and Construct a Common collector Amplifier and to determine its:





e. Gain -Bandwidth Product using frequency response curve

2. REQUIREMENTS:

S.No Requirement Name Range Quantity

1

Components Transistor [Active] BC 107 1

2 Resistor [Passive]

3 Capacitor [Passive]

4

Equipment Signal Generator 0-3MHz 1

5 CRO 0-30MHz 1

6 Regulated power supply 0-30 V 1

7

Accessories Bread Board - 1




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Design of Common collector amplifier:


VCC= 15V, IC=1.2mA, hie = 2.1kΩ hFE= 75 hib= 27.6Ω

(i) To calculate Zb ( Device input impedance )

Zb = hie + hfe ( RE || RL)

Assume RE = 4.7 KΩ and RL= 3.3 KΩ

Zb = 2.1k + 75 (4.7 K || 3.3 K) = __________

(ii) To calculate Zi ( Input Impedance )

Zi = R1 || R2 || Zb

Assume R1= R2= 10k Ω

Zi = _______

(iii) To Calculate Voltage gain Av :

Av = [ ( RE || RL ) / ( hib + ( RE || RL) ) ]

Av = _____

3. THEORY:



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A common collector amplifier is a unity gain BJT amplifier used for impedance matching and as

a buffer amplifier.

Circuit Operation : When a positive half-cycle of the input signal is applied to Base emitter

junction of transistor the forward bias voltage Vbe is increased, which in turn increases the base

current Ib of transistor. Since emitter current Ie is directly proportional to Ib the voltage drop across

the Emitter Ve= IeRe is increased, hence, output voltage Vo is increased, thus, we get positive half-

cycle of the output. It means that a positive-going input signal results in a positive going output

signal and, consequently, the input and output signals are in phase with each other. Similarly the

negative half cycle of input signal produces negative going output signal.

Characteristics of a CC Amplifier

1. high input impedance (20-500 K Ω)

2. low output impedance (50-1000 Ω)

3. high current gain of (1 + β) i.e. 50 – 500

4. voltage gain of less than 1 (unity)

5. power gain of 10 to 20 dB

6. no phase reversal of the input signal

4. PROCEDURE:







5. The voltage gain is calculated as Av = 20log (V0/Vin)




Where f1 - lower cut-off frequency

f2 - upper cut-off frequency



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a. DC ANALYSIS:


Steps:


generator




junction. VBE




2. VRC = ____________


4. Ic( Ic = (Vcc – VCE ) / Rc) =________

Q point analysis:


Q-point: ( ICQ =_____ ; VCEQ =______ )




Procedure:

i. Apply input signal Vin = 1 V of 1Khz frequency to the CC amplifier using the








V MSH = _________ volts



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5. TABULATION

Input voltage (Vin=V MSH /2) =____________ volts

S. NO FREQUENCY

[Hz]

OUTPUT VOLTAGE

[ VO] in Volts GAIN= 20 log vo/vin dB

1. 0

2. 100

3. 500

4. 600

5. 800

6. 900

7. 1 KHz

8. 100 KHz

9. 500 KHz

10. 600 KHz

11. 700 KHz

12. 800 KHz

13. 900 KHz

14. 1 MHz

15. 1.1 MHz

16. 1.5 MHz



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WORKSHEET



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6. RESULT:

INFERENCE:

The common collector amplifier was constructed and input resistance and gain were determined.

The results are found to be as given below

a) Gain of the amplifier (in dB) :

b) Bandwidth of the amplifier (in Hz) :

c) Gain-Bandwidth product (GBWP) :

CONCLUSION:



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Common Base Amplifier Circuit Diagram:

MODEL GRAPH:



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COMMON BASE AMPLIFIER

EXPERIMENT:03 DATE:

1. OBJECTIVE:

To Design and Construct a Common Base Amplifier and to determine its:






2. REQUIREMENTS:

S.No. Requirement Name Range Quantity

1 Components




4 Equipment

signal Generator (0-3)MHz 1

5 CRO 30MHz 1

6 Regulated power supply (0-30)V 1

7 Accessories

Bread Board - 1




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DESIGN PROCEDURE:

Given Transistor specifications:

hie = 2.1kΩ ; hfe = 75 ; hfb =0.987

i) To find Device input impedance :

hib = ( hie / (1+ hfe))

hib = ____

ii) To find Circuit input impedance (Zi) :

Zi = hib || Re ,Where Re= 2.2 kΩ

Zi = _____

iii) To find Circuit output impedance (Zo) :

Zo = Rc ; where Rc = 4.7 kΩ

iv) To find Voltage Gain (Av) :

Av = [ hfb (Rc || RL) / hib ]

Where RL= 10kΩ

Av = _____



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3. THEORY:

A common base amplifier is type of BJT amplifier which increases the voltage level of the


The Common base amplifier typically has good voltage gain and relatively high output

impedance. But the Common base amplifier unlike CE amplifier has very low input impedance which

makes it unsuitable for most voltage amplifier. It is typically used used as an active load for a

cascode amplifier and also as a current follower circuit.

Circuit Opeartion:

A positive-going signal voltage at the input of a CB pushes the transistor emitter in a positive

direction while the base voltage remains fixed, hence Vbe reduces. The reduction in VBE results in

reduction in VRC, consequently VCE increases. The rise in collector voltage effectively rises the output

voltage. The positive going pulse at the input produces a positive-going output, hence the there is no

phase shift from input to output in CB circuit. In the same way the negative-going input produces a

negative-going output.

4. PROCEDURE:


2. Determine the Q-point of the CB amplifier using DC analysis.

3. Determine Maximum input voltage that can be applied to CE amplifier using

AC analysis.


incremental steps and note down the corresponding output voltage Vo for atleast 20


5. The voltage gain is calculated as Av = 20log (V0/Vi)




where f1 lower cut-off frequency

f2 upper cut-off frequency



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a. DC ANALYSIS:


Steps:


generator




junction. VBE




2. VRC = ____________


4. Ic( Ic = (Vcc – VCE ) / Rc) =________

Q point analysis:


Q-point: ( ICQ =_____ ; VCEQ =______ )




Procedure:









V MSH = _________ volts



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5. TABULATION

Input voltage (Vin=V MSH / 2) =____________V

S. NO FREQUENCY

[Hz]

OUTPUT VOLTAGE

[ VO] in Volts

GAIN= 20 log Vo/ vin

dB

17. 0

18. 100

19. 500

20. 600

21. 800

22. 900

23. 1 KHz

24. 100 KHz

25. 500 KHz

26. 600 KHz

27. 700 KHz

28. 800 KHz

29. 900 KHz

30. 1 MHz

31. 1.1 MHz

32. 1.5 MHz



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WORKSHEET



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6. RESULT:

INFERENCE:

The Common base amplifier was constructed and input resistance and gain were determined. The

results are found to be as given below




CONCLUSION:



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Darlington Amplifier Circuit Diagram:

MODEL GRAPH:



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DARLINGTON AMPLIFIER

EXPERIMENT:04 DATE:

1. OBJECTIVE:

To Design and Construct a BJT amplifier using Darlington pair and to determine its:






2. REQUIREMENTS:


1 Components




4 Equipment


5 CRO 30MHz 1


7 Accessories

Bread Board - 1




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DESIGN PROCEDURE:


VCC= 12V, IC=1.2mA, AV= 30, f1 = 300 HZ, f2 = 500KHZ, hFE= 150


Assume R2 = 10KΩ and Ic = 1mA.

Since voltage amplification is done in the Darlington transistor amplifier circuit, we assume

equal drops across VCE and load resistance RC. The ICQ = 1mA is assumed.

Drop across Re is assumed to be 1V.

The drop across VCE with a supply of 1.2 V is given by

12 – 1 = 1V.

It is equal to VRC & VCE = 5.5V x RC

Therefore, Rc = 5.5 KΩ (4.7 KΩ) ; IC = 1mA

(ii) To calculate R1&R2:

S=1+ (RB/RE)

RB= (S-1) RE= (R1 || R2) =1KΏ

RB=( R 1R2 ) /( R1+ R2) ------- (2)

VB= VBE + VE = 0.7+ 1= 1.7V

VB= VCC (R2 / R1+ R2 )------- (3)

Solving equation (2) & (3),

Since R2=10kΩ , the other resistor is found to be, R1= 47kΩ

(iii) To Find Cin :

Cin = * 1 / 2πf1 (Zi / 10) ]

Where Zi = ( RB || hie ) = 1.1KΩ and

f1 = Lower cut-off frequency= 25HZ

= 57.9µF

(iv) To Find CO :

C0 = * 1 / 2πf2 ( (RC + RL) / 10) ]

Where RC = 9KΩ; RL = 90KΩ and

f2 = Upper cut-off frequency= 500KHZ

= 64 µF



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3. THEORY:

The Darlington transistor (often called a Darlington pair) is compound structure consisting of

two bipolar transistors connected in such a way that the First transistor does current

amplification of input signal and then it will be fed to the second transistor which performs

voltage amplification.

This configuration gives a much higher gain than each transistor taken separately and, in the

case of integrated devices, can take less space than two individual transistors because they can use

a shared collector. The Darlington amplifier typically has a relatively high input resistance (1 - 10 KΩ)

and a fairly high output resistance. Therefore it is generally used to drive medium to high resistance

loads. It is typically used in applications where a small voltage signal needs to be amplified to a large


4. PROCEDURE:


2. Determine the Q-point of the Darlington amplifier using DC analysis.

3. Determine Maximum input voltage that can be applied to Darlington amplifier using AC

analysis.








where f1 - lower cut-off frequency




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http://en.wikipedia.org/wiki/Bipolar_transistor

http://en.wikipedia.org/wiki/Gain



a. DC ANALYSIS:


Steps:


generator




junction. VBE




2. VRC = ____________


4. Ic( Ic = (Vcc – VCE ) / Rc) =________

Q point analysis:


Q-point: ( ICQ =_____ ; VCEQ =______ )




Procedure:









V MSH = _________ volts



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5. TABULATION


S. NO FREQUENCY

[Hz]

OUTPUT VOLTAGE

[ VO] in Volts

GAIN= 20 log Vo/Vin

dB

1. 0

2. 100

3. 500

4. 600

5. 800

6. 900

7. 1 KHz

8. 100 KHz

9. 500 KHz

10. 600 KHz

11. 700 KHz

12. 800 KHz

13. 900 KHz

14. 1 MHz

15. 1.1 MHz

16. 1.5 MHz



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WORKSHEET



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6. RESULT:

INFERENCE:

The Darlington amplifier was constructed and the results are found to be

a. Gain of the amplifier :

b. Bandwidth of the amplifier :

c. Gain-Bandwidth product :

CONCLUSION:



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Common Source Amplifier Circuit Diagram:

MODEL GRAPH:



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COMMON SOURCE AMPLIFIER

EXPERIMENT:05 DATE:

1. OBJECTIVE:

To Design and Construct a Common source amplifier using the bootstrapped gate resistance

and to determine its:






2. REQUIREMENTS:


1 Components

Transistor [Active] BFW10 1



4 Equipment


5 CRO 30MHz 1


7 Accessories

Bread Board - 1




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DESIGN ANALYSIS : Given :

VDD = 20 V, IDSS = 5mA, ID = 1.5 mA, i) To Find the voltage across the Gate-source region (VGS)

VGS = ID RS

Assume RS = 3.3KΩ,

VGS = 1.5mA x 3.3KΩ = _____

ii) To find Voltage Across Drain to Source (VDS)

VDS= VDD - ID ( RD + Rs) ; Where RD= 3.3 KΩ = 20V – 1mA ( 3.3 KΩ + 3.3 KΩ)

= ________

iii) To Find input impedance :

Zi = RG ; Assume RG = 1MΩ

iv) To Find output impedance :

ZO = RD ||



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3. THEORY

There are three basic types of FET amplifier or FET transistor namely common source

amplifier, common gate amplifier and source follower amplifier.

The common-source (CS) amplifier may be viewed as a transconductance amplifier or as a

voltage amplifier.

i) As a transconductance amplifier, the input voltage is seen as modulating the current going

to the load.

ii) 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.

However, the FET device's output resistance typically is not high enough for a reasonable

transconductance amplifier (ideally infinite), nor low enough for a decent voltage amplifier (ideally

zero). Another major drawback is the amplifier's 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 favorable output and frequency

characteristics

4. PROCEDURE:


2. Determine the Q-point of the CS amplifier using DC analysis.

3. Determine Maximum input voltage that can be applied to CE amplifier using

AC analysis.












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http://en.wikipedia.org/wiki/Ohm%27s_law

http://en.wikipedia.org/wiki/Electronic_amplifier#Input_and_output_variables



http://en.wikipedia.org/wiki/Common-drain

http://en.wikipedia.org/wiki/Common-gate



a. DC ANALYSIS:


Steps:


generator




junction. VBE



1. VGS : = ____________

2. VDS = ____________

3 ID = _______




Procedure:

i. Apply input signal Vin = 1 V of 1Khz frequency to the CS amplifier using the








V MSH = _________ volts



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5. TABULATION

Input voltage (Vin=V MSH/2) =____________V

S. NO FREQUENCY

[Hz]

OUTPUT VOLTAGE [

VO] in Volts

GAIN= 20 log Vo/Vin

dB

1. 0

2. 100

3. 500

4. 600

5. 800

6. 900

7. 1 KHz

8. 100 KHz

9. 500 KHz

10. 600 KHz

11. 700 KHz

12. 800 KHz

13. 900 KHz

14. 1 MHz

15. 1.1 MHz

16. 1.5 MHz



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6. RESULT:

INFERENCE:

The common Source amplifier was constructed and input resistance and gain were determined. The


a) Gain of the amplifier (in db) :

b) Bandwidth of the amplifier (in HZ) :

c) Gain-Bandwidth product (GBWP) :

CONCLUSION:



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Cascade amplifier Circuit Diagram:

MODEL GRAPH:



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CASCADE AMPLIFIER

EXPERIMENT:06 DATE:

1. OBJECTIVE:

To Design and Construct a Cascade Amplifier and to determine its:






2. REQUIREMENTS:


1 Components




4 Equipment


5 CRO 30MHz 1


7 Accessories

Bread Board - 1




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DESIGN PROCEDURE:


VCC= 14 V, IC1=1.2mA, RL = 40KΏ hFE= 100

(i) To calculate R5 :

Assume VE1 = 5V , VCE1 = VCE2 = 3V;

VB2 = VC1 = VE1 + VCE1 = 5V + 3V = 8V

VE2 = VB2 – VBE = 8V – 0.7V = 7.3V

VR5 = Vcc – VE2 – VCE2 = 14V – 7.3V – 3V = 3.7V

Choose R5 = RL / 10 = 40KΩ / 10 = 4KΩ ;

IC2 = ( VR5 / R5 ) = 3.7V / 3.9KΩ = 1000µA

(ii) To calculate R6 :

VR6 = VE2 / IC2 = 7.7KΩ;

IC2 = VE2 / R6 = 7.3V / 8.2 KΩ = 890µA

(iii) To calculate R1, R2 , R3 & R4:

Voltage across resistor R3 is given by

VR3 = Vcc – VC1 = 14V – 8V = 6V

R3 = VR3 / IC1 = 6V / 1mA = 6KΩ

R4 = VE1 / IC1 = 5V/ 1mA = 4.7KΩ

Voltage across resistor R2 is given by

VR2 = VE1 – VBE = 5V + 0.7V =5.7V

R2 = 10 R4 = 4.7 KΩ

VR1 = VCC – VB1 = 14V + 5.7V =8.3V

R1 = [ VR1 x R2 / VR2] = 68.4 KΩ



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3. THEORY:

A cascade is type of multistage amplifier where two or more single stage amplifiers are

connected serially. Many times the primary requirement of the amplifier cannot be achieved with

single stage amplifier, because Of the limitation of the transistor parameters. In such situations more

than one amplifier stages are cascaded such that input and output stages provide impedance

matching requirements with some amplification and remaining middle stages provide most of the

amplification. These types of amplifier circuits are employed in designing microphone and

loudspeaker.

4. PROCEDURE:


2. Determine the Q-point of the amplifier using DC analysis.

3. Determine Maximum input voltage that can be applied to amplifier using AC analysis.












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a. DC ANALYSIS:


Steps:

v) Set Vin = 0 by reducing the amplitude of the input signal from signal

generator

vi) Open circuit the capacitors since it blocks DC voltage

vii) Set VCC= +10v and measure the voltage drop across the Resistor VRC, voltage


junction. VBE

viii) Find the Q-point of the transistor and draw the DC load line.



2. VRC = ____________


4. Ic( Ic = (Vcc – VCE ) / Rc) =________

Q point analysis:


Q-point: ( ICQ =_____ ; VCEQ =______ )




Procedure:

iii. Apply input signal Vin = 20 mV of 1Khz frequency to the amplifier using the

signal generator between base emitter junction of the transistor.Find the


iv. By increasing the amplitude of the input signal find maximum input voltage


processwhich can be seen in the CRO. The amplitude obtained at this point



V MSH = _________ volts



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5. TABULATION

Input voltage (Vin=V MSH/2) =____________ volts

S. NO FREQUENCY

[Hz]

OUTPUT VOLTAGE

[ VO] in Volts

GAIN= 20 log Vo/Vin

dB

1. 0

2. 100

3. 500

4. 600

5. 800

6. 900

7. 1 KHz

8. 100 KHz

9. 500 KHz

10. 600 KHz

11. 700 KHz

12. 800 KHz

13. 900 KHz

14. 1 MHz

15. 1.1 MHz

16. 1.5 MHz



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6. RESULT:

INFERENCE:

The Cascade amplifier was constructed and input resistance and gain were determined. The results

are found to be as given below




CONCLUSION:



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Cascode amplifier Circuit Diagram:

MODEL GRAPH:



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CASCODE AMPLIFIER


1. OBJECTIVE:

To Design and Construct a Cascode Amplifier and to determine its:






2. REQUIREMENTS:


1 Components


2 Resistor [Passive] 61kΩ, 10kΩ, 1kΩ, 4.7kΩ

1,1,1,2

3 Capacitor [Passive] 10µf, 100µf 2,1

4 Equipment


5 CRO 30MHz 1


7 Accessories

Bread Board - 1




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DESIGN PROCEDURE:


VCC= 20V, IC =1.2mA, AV= 30, , RL = 90KΏ ;

Transistor Parameters: hFE= 50 , hie = 1.2KΏ and hib= 24Ώ


Rc = RL / 10 = 90kΏ / 10 = 9KΏ

(ii) To calculate RE:

Assume VCE1 = VCE2 = 3V, and VE = 5V;

The voltage drop across collector resistor is given by,

VRC = VCC - VCE1 - VCE2 – VE

= 20V – 3V – 3V- 5V

VRC = 9V

RE = VE / IE ; Where IE = Ic = 1.1mA

RE = 4.5K

(iii) To Calculate Bias Resistors R1, R2, R3 :

a. R3 = 10 RE = 47KΩ

b. Voltage Across the base of Transistor 1 is given by,

VB1 = VBE + VE

VB1 = 5V + 0.7V = 5.7V

I3 = (VB1/ R3) = 5.7V / 47KΏ = 121 µA

c. Voltage Across the base of Transistor 2 is given by

VB2 = VBE2 + VE + VBE2

= 5V + 3V + 0.7V

= 8.7V

d. Voltage across resistor R2 is given by

VR2 = VB2 - VB1

VR2 = 8.7V – 5.7V = 3V

e. The resistor R2 is given by R2 = (VR2 / I3 ) = (3V / 121µA)

R2 = 24.8KΩ

f. Resistor R1 = [VCC - VB2 / I3 ]

= [ 20V – 8.7V / 121 µA]

= 93.4 KΏ

Determination of Capacitor Values:

To Find C1 :

C1 = * 1 / 2πf1 (Zi / 10) ]

Where Zi = ( R3 || R2 ) = 1.1KΩ and


= 57.9µF

To Find C2 :

C2 = * 1 / 2πf1 (hie2 / 10) ] = 53 µF

Where hie2= 1.2 KΩ and f1 = Lower cut-off frequency= 25HZ;



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To Find C3 :

C3 = * 1 / 2πf1hib ]

Where hib= 24Ω and


= 256µF

To Find C4 :

C4 = * 1 / 2πf1 ( (RC + RL) / 10) ]

Where RC = 9KΩ; RL = 90KΩ and


= 0.64 µF

THEORY:

The cascode configuration has one of two configurations of multistage amplifier. In each case

the collector of the leading transistor is connected to the emitter of the following transistor. The

arrangement of the two transistors is shown in the circuit diagram. The cascode amplifier consists of

CE stage connected in series with CB stage. The arrangement provides a relatively high input

impedance with low voltage gain for the first stage to ensure the input miller capacitance is at a

minimum, whereas the following CB stage provides an excellent high frequency response.

Features:

1. It provides high voltage gain and has high input impedance.

2. It provides high stability and has high output impedance

PROCEDURE:





incremental steps and note down the corresponding output voltage Vo for atleast 20 different

values for the considered range.


6. Find the Bandwidth and Gain-Bandwidth Product from Semi-log graph taking frequency on x-

axis and gain in dB on y-axis., Bandwidth, BW = f2-f1

where f1 - lower cut-off frequency f2 - upper cut-off frequency



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a. DC ANALYSIS:


Steps:

ix) Set Vin = 0 by reducing the amplitude of the input signal from signal

generator

x) Open circuit the capacitors since it blocks DC voltage

xi) Set VCC= +10v and measure the voltage drop across the Resistor VRC, voltage


junction. VBE

xii) Find the Q-point of the transistor and draw the DC load line.



2. VRC = ____________


4. Ic( Ic = (Vcc – VCE ) / Rc) =________

Q point analysis:


Q-point: ( ICQ =_____ ; VCEQ =______ )




Procedure:

v. Apply input signal Vin = 20 mV of 1Khz frequency to the amplifier using the

signal generator between base emitter junction of the transistor.Find the


vi. By increasing the amplitude of the input signal find maximum input voltage


processwhich can be seen in the CRO. The amplitude obtained at this point



V MSH = _________ volts



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5. TABULATION


S. NO FREQUENCY

[Hz]

OUTPUT VOLTAGE

[ VO] in Volts

GAIN= 20 log Vo/Vin

dB

1. 0

2. 100

3. 500

4. 600

5. 800

6. 900

7. 1 KHz

8. 100 KHz

9. 500 KHz

10. 600 KHz

11. 700 KHz

12. 800 KHz

13. 900 KHz

14. 1 MHz

15. 1.1 MHz

16. 1.5 MHz



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6. RESULT:

INFERENCE:

The Cascode amplifier was constructed and input resistance and gain were determined. The results

are found to be as given below




CONCLUSION:



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Differential amplifier Circuit Diagram:

Common Mode :

Differential Mode :



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DIFFERENTIAL AMPLIFIER


1. OBJECTIVE:

To Design and Construct a Differential Amplifier using BJT and to determine its:

a. Transfer Characteristics

b. Gain of the amplifier in common mode

c. Gain of the amplifier in differential mode

d. CMRR (Common Mode Rejection Ratio)

2. REQUIREMENTS:

3. THEORY:

A differential amplifier is a type of electronic amplifier that amplifies the difference

between two voltages but does not amplify the particular voltages. The need for differential

amplifier arises in many physical measurements where response from D.C to many MHZ is

required. It is also used in input stage of integrated amplifier.


1 Components




4 Equipment


5 CRO 30MHz 1


7 Accessories

Bread Board - 1




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http://en.wikipedia.org/wiki/Electronic_amplifier



DESIGN PROCEDURE:


VCC= 12V, IC=1.2mA, VCE = 5V

Assume Q1 = Q2

Where Q1 = Transistor 1 and Q2 = Transistor 2.

The Collector resistance Rc1 is given by using KVL at the transistor 1

Vcc = IcRc1 – VCE – Ie1Re1

Rc1 = (Vcc + VCE + Ie1Re1 ) / Ic,

Where VCE = 5V and Ic = 1.2mA, Ie1 = Ic/10 and assume Re1 = 470Ω

Now Rc1 = 1kΩ .

MODEL GRAPH:

Differential amplifier – Transfer Characteristics:



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The output signal in differential amplifier is proportional to the difference between the two

input signals.

Vo = Ad (V1 – V2 ).

Where V1,V2 are the input voltages and Ad is the differential gain.

If V1 = V2, then output voltage is zero. A non zero output voltage is obtained if V1 and V2

are not equal.

i) The difference mode input voltage is defined as Vd = (V1-V2)

ii) The common mode input voltage is defined as the Vcm= (V1+V2)/2

iii) The CMRR is defined as the ratio of the differential gain Ad to common mode gain

Ac and is generally expressed in dB.

CMRR= 20 log10 ( Ad / Ac)

4. PROCEDURE:


2. Determine the Q-point of the Differential amplifier using DC analysis.

3. Determine Maximum input voltage that can be applied to amplifier using

AC analysis.

4. Determine the Transfer characteristics of Differential amplifier by plotting the graph for

normalized differential input voltage [ (Vb1 – Vb2) / VT ] vs. Normalized collector current [ Ic / Io].

5. Calculate the voltage gain of differential amplifier for differential mode

as Ad = 20log (V0/Vi) , Where Vi = V1 – V2

6. Calculate the voltage gain of differential amplifier for Common mode

as AC = 20log (V0/Vi) , Where Vi = (V1+ V2 / 2 )

7. Find the Common mode rejection ratio of differential amplifier using the formula given

below.

CMRR= 20 log10 ( Ad/Ac)

Where Ad- Differential mode gain in dB

Ac – Common Mode gain in dB



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a. DC ANALYSIS:


Steps:


generator




junction. VBE




2. VRC = ____________


4. Ic( Ic = (Vcc – VCE ) / Rc) =________

Q point analysis:

It is the procedure to choose the operating point of transistor

Q-point: ( ICQ =_____ ; VCEQ =______ )




Procedure:

vii. Apply input signal Vin = 20 mV of 1Khz frequency to the amplifier using the



viii. By increasing the amplitude of the input signal find maximum input voltage





V MSH = _________ volts



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5. TABULATION

a. Transfer Characteristics Calculation:

S.no Input Voltage

Vi = (Vb1 – Vb2) in Volts

Output Current

Ic2 in Ampere

1.

2.

3.

4.

5.

6.

b. CMRR Calculation:

To Find Differential Gain (Ad ) :

S. NO INPUT VOLTAGE

in volts

OUTPUT VOLTAGE [ VO] in

Volts

Differntial gain in dB

Ad = 20log (V0/Vi)

Where Vi = Vi1 – Vi2

33. Vi1

34. Vi2

To Find Common Mode Gain (AC ) :

S. NO INPUT VOLTAGE

in volts

OUTPUT VOLTAGE [ VO]

in Volts

Common mode gain in dB

AC = 20log (V0/Vi)

where Vi = (V1+ V2 / 2 )

1. Vi1

2. Vi2



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6. RESULT:

INFERENCE:

The Differential amplifier was constructed and input resistance and gain were determined. The


d) Trans-Conductance of Differential amplifier ( in millisiemens) :

e) Differential mode gain in dB :

f) Common Mode Gain in dB :

g) CMRR in dB :

CONCLUSION:



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SPICE SIMULATION USING PSPICE



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Common Emitter Amplifier circuit diagram

CE Amplifier without Feedback :

CE Amplifier with Feedback:



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COMMON EMITTER AMPLIFIER


1. OBJECTIVE:

To Design and Construct a Common Emitter Amplifier using Pspice simulation tool and to

determine its:

a. Gain of the amplifier

b. Bandwidth of the amplifier

c. Gain -Bandwidth Product

2. REQUIREMENTS:

THEORY:

A common emitter amplifer is type of BJT amplifier which increases the voltage level of the


The CE amplifier typically has a relatively high input resistance (1 - 10 KΩ) and a fairly high

output resistance. Therefore it is generally used to drive medium to high resistance loads. It is

typically used in applications where a small voltage signal needs to be amplified to a large


The input signal Vin is applied to base emitter junction of the transistor and amplifier output

Vo is taken across collector terminal. Transistor is maintained at the active region by using the

resistors R1,R2 and Rc. A very small change in base current produces a much larger change in

collector current. The output Vo of the common emitter amplifier is 180 degrees out of phase

with the applied the input signal Vin.

S.no Requirements Quantity

1 PC 1

2 Pspice Software -



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3. PROCEDURE:

1. Click on the start menu and select the pspice simulation software

2. Select the parts required for the circuit from the parts menu and place

them in the work space

3. Connect the parts using wires

4. Save the file and select the appropriate analysis

5. Simulate the circuit and observe the corresponding output waveforms

MODEL GRAPH:

4.RESULT:

INFERENCE:

The Common Emitter Amplifier was simulated and the following results were determined:




CONCLUSION:



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Common source Amplifier circuit diagram

CS Amplifier:



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COMMON SOURCE AMPLIFIER


1. OBJECTIVE:

To Design and Construct a Common Emitter Amplifier using Pspice simulation tool and to

determine its:

a. Gain of the amplifier

b. Bandwidth of the amplifier

c. Gain -Bandwidth Product

2. REQUIREMENTS:

THEORY:

There are three basic types of FET amplifier or FET transistor namely common source

amplifier, common gate amplifier and source follower amplifier.

The common-source (CS) amplifier may be viewed as a transconductance amplifier or as a

voltage amplifier.

i) As a transconductance amplifier, the input voltage is seen as modulating the current going

to the load.

ii) 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.

However, the FET device's output resistance typically is not high enough for a reasonable

transconductance amplifier (ideally infinite), nor low enough for a decent voltage amplifier (ideally

zero). Another major drawback is the amplifier's 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 favorable output and frequency

characteristics

S.no Requirements Quantity

1 PC 1

2 Pspice Software -



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http://en.wikipedia.org/wiki/Ohm%27s_law




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WORKSHEET



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3. PROCEDURE:

1. Click on the start menu and select the pspice simulation software

2. Select the parts required for the circuit from the parts menu and place

them in the work space

3. Connect the parts using wires

4. Save the file and select the appropriate analysis

5. Simulate the circuit and observe the corresponding output waveforms

MODEL GRAPH:

4.RESULT:

INFERENCE:

The Common Emitter Amplifier was simulated and the following results were determined:




CONCLUSION:



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DIGITAL EXPERIMENTS



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DESIGN AND IMPLEMENTATION OF CODE CONVERTERS

EXPERIMENT: DATE:

1. OBJECTIVE:

To design and verify the truth table of the following code converters

a. Binary to Gray converter

b. Gray to Binary converter &

c. BCD to Excess3 &

d. Excess3 to BCD.

2. REQUIREMENTS:

S. No. Components / Equipments Specifications Quantity

1.

2.

3.

Digital IC trainer

NOT, AND, OR, Ex-OR Gate

Connecting wires

---

IC7404,7408,7432,7486

---

1

1 in each

Sufficient

numbers

3. THEORY:

Binary to GRAY Converter:

By representing the ten decimal digits with a four bit Gray code, we have another form

of BCD code. The Gray code however can be extended to any number of bits and conversion

between binary code and Gray code is sometimes useful. The following rules apply for conversion:

1. The MSB in the Gray code is the same as the corresponding bit in the binary number.

2. Going from left to right, add each adjacent pair of binary bits to get the next Gray code bit.

Disregard carries.

GRAY to Binary Converter:

To convert from Gray code to binary code, A similar method is used, at there are some

differences. The following rules apply:

1. The MSB in the binary code is the same as the corresponding digit in the Gray code

2. Add each binary digit generated to the gray digit in the next adjacent position Disregard

carries.



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TRUTH TABLE FOR BINARY TO GRAY CODE CONVERTER:

| Binary input | Gray code output

B3 B2 B1 B0 G3 G2 G1 G0

0

0

0

0

0

0

0

0

1

1

1

1

1

1

1

1

0

0

0

0

1

1

1

1

0

0

0

0

1

1

1

1

0

0

1

1

0

0

1

1

0

0

1

1

0

0

1

1

0

1

0

1

0

1

0

1

0

1

0

1

0

1

0

1

0

0

0

0

0

0

0

0

1

1

1

1

1

1

1

1

0

0

0

0

1

1

1

1

1

1

1

1

0

0

0

0

0

0

1

1

1

1

0

0

0

0

1

1

1

1

0

0

0

1

1

0

0

1

1

0

0

1

1

0

0

1

1

0



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K-Map for G3:

G3 = B3

K-Map for G2:

K-Map for G1:



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K-Map for G0:

Binary to GRAY Logic Diagram :



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TRUTH TABLE FOR GRAY CODE TO BINARY CONVERTOR:

| Gray Code | Binary Code

G3 G2 G1 G0 B3 B2 B1 B0

0

0

0

0

0

0

0

0

1

1

1

1

1

1

1

1

0

0

0

0

1

1

1

1

1

1

1

1

0

0

0

0

0

0

1

1

1

1

0

0

0

0

1

1

1

1

0

0

0

1

1

0

0

1

1

0

0

1

1

0

0

1

1

0

0

0

0

0

0

0

0

0

1

1

1

1

1

1

1

1

0

0

0

0

1

1

1

1

0

0

0

0

1

1

1

1

0

0

1

1

0

0

1

1

0

0

1

1

0

0

1

1

0

1

0

1

0

1

0

1

0

1

0

1

0

1

0

1



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K-Map for B3:

B3 = G3

K-Map for B2:

K-Map for B1:



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K-Map for B0:

GRAY to Binary LOGIC DIAGRAM



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TRUTH TABLE FOR BCD TO EXCESS-3 CONVERTOR:

| BCD input | Excess – 3 output |

B3 B2 B1 B0 G3 G2 G1 G0

0

0

0

0

0

0

0

0

1

1

1

1

1

1

1

1

0

0

0

0

1

1

1

1

0

0

0

0

1

1

1

1

0

0

1

1

0

0

1

1

0

0

1

1

0

0

1

1

0

1

0

1

0

1

0

1

0

1

0

1

0

1

0

1

0

0

0

0

0

1

1

1

1

1

x

x

x

x

x

x

0

1

1

1

1

0

0

0

0

1

x

x

x

x

x

x

1

0

0

1

1

0

0

1

1

0

x

x

x

x

x

x

1

0

1

0

1

0

1

0

1

0

x

x

x

x

x

x



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K-Map for E3:

E3 = B3 + B2 (B0 + B1)

K-Map for E2:

K-Map for E1:



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K-Map for E0:

BCD TO EXCESS-3 Convertor Logic Diagram



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TRUTH TABLE FOR EXCESS-3 TO BCD CONVERTOR:

| Excess – 3 Input | BCD Output |

B3 B2 B1 B0 G3 G2 G1 G0

0

0

0

0

0

1

1

1

1

1

0

1

1

1

1

0

0

0

0

1

1

0

0

1

1

0

0

1

1

0

1

0

1

0

1

0

1

0

1

0

0

0

0

0

0

0

0

0

1

1

0

0

0

0

1

1

1

1

0

0

0

0

1

1

0

0

1

1

0

0

0

1

0

1

0

1

0

1

0

1

K-Map for A:

A = X1 X2 + X3 X4 X1



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K-Map for B:

K-Map for C:

K-Map for D:



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EXCESS-3 TO BCD convertor Logic diagram:

4. PROCEDURE:

1. Connections are given as per the circuit diagram (Binary to GRAY).

2. Switch on the power supply.

3. Verify the truth table given for different inputs.

4. Repeat the above procedures for other converters.



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5. Results:

INFERENCE:

Thus the truth tables for Binary to Gray, Gray to Binary and BCD to Excess3 converters were

verified.

CONCLUSION:



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DESIGN AND IMPLEMENTATION OF 4 BIT BINARY ADDER / SUBTRACTOR USING IC 7483


1. OBJECTIVE:

To study the 4 bit binary adder/subtractor using IC7483.

2. REQUIREMENTS:

S.No. Name of the apparatus Specifications Quantity

1 Digital Trainer kit - 1

2 OR gate IC 7432 1

3 AND gate IC 7408 1

4 Binary Adder / Subtractor IC 7483 2

5 Connecting wires - some

3. THEORY:

The full adder/sub tractors are capable of adding/subtracting only two single digit binary

numbers along with a carry input. But in practice we need to add/subtract binary numbers, which

are much longer than just one bit. To add/subtract two n-bit binary numbers we need to use the n-

bit parallel subtractor/adder.

Binary adder: IC type 7483 is a 4-bit binary parallel adder/subtractor .The two 4-bit input

binary numbers are A1 through A4 and B1 through B4. The sum is obtained from S1 through S4. C0 is

the input carry and C4 the output carry. Test the 4-bit binary adder 7483 by connecting the power

supply and ground terminals. Then connect the four A inputs to a fixed binary numbers such as 1001

and the B inputs and the input carry to five toggle switches. The five outputs are applied to indicator

lamps. Perform the addition of a few binary numbers and check that the output sum and output

carry give the proper values. Show that when the input carry is equal to 1, it adds 1 to the output

sum.

Binary subtractor : The subtraction of two binary numbers can be done by taking the 2’s

complement of the subtrahend and adding it to the minuend. The 2’s complement can be obtained

by taking the 1’s complement and adding. To perform A-B, we complement the four bits of B, add

them to the four bits of A, and add 1 through the input carry. The four XOR gates complement the

bits of B when the mode select M=1(because xx 0 ) and leave the bits of B unchanged when

M=0(because xx 0 ) .Thus , when the mode select M is equal to 1, the input carry C0 is equal 1

and the sum output is A plus the 2’s complement of B. when M is equal to 0, the input carry is equal

to 0 and the sum generates A+B.



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Functional symbol for IC 7483:

Operand1 Operand2

Pin Diagram of IC7483:

B3 B2 B1 B0 A3 A2 A1 A0

4 bit IC 7483

C4 C0

O/P



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Circuit Diagram for 4-bit Binary adder/subtractor:

4-BIT BINARY ADDER:



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4-BIT BINARY SUBTRACTOR



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TRUTH TABLE FOR BCD ADDER:

BCD SUM CARRY

S4 S3 S2 S1 C

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 0

0 1 1 0 0

0 1 1 1 0

1 0 0 0 0

1 0 0 1 0

1 0 1 0 1

1 0 1 1 1

1 1 0 0 1

1 1 0 1 1

1 1 1 0 1

1 1 1 1 1

K MAP

Y = S4 (S3 + S2)



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LOGIC DIAGRAM OF BCD ADDER

4. PROCEDURE:

1. Connections are given as per the circuit diagram. 2. Set mode M =0 such that the circuit will operate in addition mode. 3. Set the Value of inputs A as 1001 and B as 1001 note the sum and output carry. 4. Repeat the same step in step 3 by keeping M=1 such that circuit will operate in subtraction

mode.



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5. RESULTS:

INFERENCE:

Thus the 4 bit Binary Adder / Subtractor using IC7483 is been implemented for both addition

and subtraction and the corresponding truth tables are verified.

CONCLUSION:



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DESIGN AND IMPLEMENTATION

OF

MULTIPLEXER AND DEMULTIPLEXER USING LOGIC GATES


1. OBJECTIVE:

To design and implement multiplexer and demultiplexer using logic gates

2. REQUIREMENTS :

S.No. Name of the apparatus Specification Quantity

1 Digital Trainer kit -

2 OR gate IC7432 1

3 AND gate IC7411 1

4 NOT gate IC7404 1

5 Connecting wires -

3. THEORY:

Multiplexer:

It has a group of data inputs and a group of control inputs. The control inputs are

used to select one of the data inputs and connected to the output terminal. It selects one

information out of many information lines and directed to a single output line.

Demultiplexer:

Demultiplexers perform the opposite function of multiplexers. They transfer a small

number of information units (usually one unit) over a larger number of channels under the

control of selection signals. Fig shows a 1-line to 2-line Demultiplexer circuit. Construct this

circuit; connect an LED to each of the outputs D0 and D1. Set the select signal S to logic 1 or

logic 0, and toggle the input I between logic 1 and logic 0. Which output followed the input

when S = 1 and S = 0.



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4:1 MULTIPLEXER:

BLOCK DIAGRAM

Circuit Diagram:

Truth Table:

S1 S0 Y

0 0 I0

0 1 I1

1 0 I2

1 1 I3



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1:4 DEMULTIPLEXER:

BLOCK DIAGRAM

:

Circuit Diagram:

Truth Table:

Selection Lines OUTPUT V0

S1 S0

0 0 D0=Di

0 1 D1= Di

1 0 D2= Di

1 1 D3= Di



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TRUTH TABLE:

8X1 Multiplexer

LOGIC DIAGRAM FOR 8:1 MULTIPLEXER:

S0 S1 D0 D1 D2 D3 O/P

0 0 1 X X X 1

0 0 0 X X X 0

0 1 X 1 X X 1

0 1 X 0 X X 0

1 0 X X 1 X 1

1 0 X X 0 X 0

1 1 X X X 1 1

1 1 X X X 0 0



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1X8 De-Multiplexer:

LOGIC DIAGRAM FOR 1X8 DEMULTIPLEXER:

TRUTH TABLE

1:8 DEMULTIPLEXER:



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PIN DIAGRAM FOR IC 74150:

PIN DIAGRAM FOR IC 74154:



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4. PROCEDURE:

1. Connections are given as per in the circuit diagram. 2. Inputs are given through the logic switches. 3. Outputs are noted and verified with truth table

5. RESULTS

INFERENCE :

Thus the truth table of multiplexer and demultiplexer was studied and verified using logic

gates.

CONCLUSION:

6. VIVA QUESTIONS:

1. What is a multiplexer? 2. What are the applications of multiplexer? 3. What is the difference between multiplexer & demultiplexer? 4. In 2n: 1 multiplexer how many selection lines are used? 5. Draw a 2 to 1 multiplexer circuit 6. Draw a 1 to 2 demultiplexer circuit.



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DESIGN AND IMPLEMENTATION OF ENCODER

EXPERIMENT:12.a. DATE:

1. OBJECTIVE:

To construct and verify the 8 X 3 Encoder using logic gates.

2. REQUIREMENTS:

S. No Components / Equipments Specification Quantity

1. Digital IC trainer kit - 1

2. OR Gate IC7432 3

3. Connecting Wires - Sufficient Numbers

3. THEORY:

Digital Computers, Microprocessors and other digital systems are binary operated

whereas our language of communication is in decimal numbers and alphabetical characters

only. Therefore, the need arises for interfacing between digital system and human operators.

To accomplish this task, Encoder is used.



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Encoder

Lgic Diagram:

Truth Table:

Outputs:

A = D4 + D5 + D6 + D7

B = D2 + D3 + D6 + D7

C = D1 + D3 + D5 + D7

INPUT OUTPUT

D0 D1 D2 D3 D4 D5 D6 D7 A B C

1 0 0 0 0 0 0 0 0 0 0

0 1 0 0 0 0 0 0 0 0 1

0 0 1 0 0 0 0 0 0 1 0

0 0 0 1 0 0 0 0 0 1 1

0 0 0 0 1 0 0 0 1 0 0

0 0 0 0 0 1 0 0 1 0 1

0 0 0 0 0 0 1 0 1 1 0

0 0 0 0 0 0 0 1 1 1 1



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4. PROCEDURE:

1. Construct the circuit as per the diagram 2. Switch on the power supply. 3. Apply the necessary input and observe the outputs to verify the truth table.

5. RESULTS:

INFERENCE:

Thus an 8 x 3 encoder is constructed and verified.

CONCLUSION:

6. REVIEW QUESTIONS:

1. Draw the basic block diagram of a practical decoder.

2. What is the need for decoder?

3. Name the procedure involved in decoding.

4. Give some practical applications where decoding is necessary.

5. List the advantages of decoding.



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DESIGN AND IMPLEMENTATION OF DECODER

EXPERIMENT:12.b DATE:

1. OBJECTIVE :

To construct and verify the decoder / driver along with seven segment LED display unit and

verify the results.

2. REQUIREMENTS :

S. No. Components / Equipments Specification Quantity

1. Decoder IC 7447 1

2. Seven Segment Display Common Anode 1

3. Resistors 330Ω 7

4. IC Trainer Kit --- 1

5. Connecting Wires ---

Required

numbers

3. THEORY:

The 7-segment LED display is the most popular display device used in digital systems.

To use this device the data that is in the BCD form has to be changed suitably. For this

purpose a BCD to 7-segment decoder is required. The IC7447 is a BCD to 7-segment pattern

converter. . The 7447 converts the four input bits (BCD) to their corresponding 7-segment

codes. The outputs of the 7447 are connected to the 7-segment display.



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Connection Circuit Diagram:

4. PROCEDURE:

1. Connect the circuit as per circuit diagram. 2. Apply the inputs to the IC7447(A,B,C&D). 3. Observe the output and verify the result.

5. RESULTS:

INFERENCE :

A decoder / driver unit along with 7 segment display unit is constructed and the

results were verified.

CONCLUSION:

6. REVIEW QUESTIONS:

1. Draw the basic block diagram of a practical decoder.

2. What is the need for decoder?

3. Name the procedure involved in decoding.

4. Give some practical applications where decoding is necessary.

5. List the advantages of decoding.



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DESIGN AND IMPLEMENTATION OF SYNCHRONOUS COUNTER


1 . OBJECTIVE:

To construct and verify the synchronous up/down counters.

2. REQUIREMENTS:

S. No. Components / Equipments Specification Quantity

1.

2.

3.

Digital IC trainer kit

JK Flip-Flop, AND Gate

Connecting wires

----

IC 7473,7408

----

1

2,1

Sufficient Nos

3. THEORY:

Synchronous Counter

Clock input is applied simultaneously to all flip-flops. The output of the first FLIP-FLOP is

connected to the input of second FLIP-FLOP and so on.

Design of synchronous counter

Step 1: Find the number of flip-flops required. For an n-bit counter, n- flip-flops is

required.

Step 2: Write the count sequence in tabular form.

Step 3: Determine the flip-flop inputs, which must be present for the desired next State from the present state using excitation table of flip-flops.

Step 4: Prepare K-map for each flip-flop input in terms of flip-flop output as input

Variables. Simplify the K-map and obtain the minimized expressions.

Step 5: Connect the circuit using the flip-flops.



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CIRCUIT DIAGRAM:

Design of 3-bit synchronous up:

Design of 3-bit synchronous down counter:

Pin Diagram



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Truth Table:

3 Bit Synchronous UP

Counter

3 Bit Synchronous DOWN

Counter

Clock Q2 Q1 Q0 Clock Q2 Q1 Q0

0 0 0 0 0 1 1 1

1 0 0 1 1 1 1 0

2 0 1 0 2 1 0 1

3 0 1 1 3 1 0 0

4 1 0 0 4 0 1 1

5 1 0 1 5 0 1 0

6 1 1 0 6 0 0 1

7 1 1 1 7 0 0 0

4. PROCEDURE:

1. The connections are made as per the circuit diagram. 2. Switch on the power supply. 3. The input is given at the appropriate terminal and corresponding output is observed

and truth table is verified.

5. RESULTS:

INFERENCE:

Thus the counters were constructed and their truth tables verified.

CONCLUSION:

6. REVIEW QUESTIONS

1. Name any four flip-flop used to construct the counter. 2. Draw the basic block diagram of a practical 4-bit counter. 3. What is MOD 5 counter? 4. What is the need for counters?

5. Give some practical applications of counters.



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IMPLEMENTATION OF SISO, SIPO, PISO AND PIPO SHIFT REGISTERS


1. OBJECTIVE:

To implement the 4 bit shift register using flip flops and to study the operations in the

following modes.

(i) Serial in serial out (ii) Serial in parallel out (iii) Parallel in parallel out (iv) Parallel in serial out

2. REQUIREMENTS:

S.No. Name of the apparatus Range Quantity

1 Digital Trainer kit 1

2 D Flip Flop IC 7474 2

3 Connecting wires some

3. THEORY:

SHIFT REGISTER:

A register is a device capable of storing a bit. The data can be serial or parallel. The

register can convert a data from serial to parallel and vice versa shifting then digits to left and right is

the important aspect for arithmetic operations,

A register capable of shifting its binary information either to the right or to the left is called a shift register. An N bit shift register consists of N flip-flops and the gates that control the shift operation. A shift register can be used in four different configurations depending upon the way in which the data are entered into and taken out of it. These four configurations are:

a. Serial-input, Serial-output b. Parallel-input, Serial-output c. Serial-input, parallel-output d. Parallel-output, parallel-output

The serial input is a single line going to the input of the leftmost flip-flop of the register. The

serial output is a single line from the output of the rightmost flip-flop of the register, so that the bits stored in the register can come out through this line one at a time. The parallel output consists of N lines, one for each of the flip-flops in the register, so the

information stored in the register can be inspected through these lines all at once.



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PIN DIAGRAM:

Logic Diagram for Serial in Serial out:



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Logic Diagram for Serial in Parallel Out:

Logic Diagram for Parallel In Parallel Out

Logic Diagram for Parallel in Serial out



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TRUTH TABLE FOR PISO SHIFT REGISTER:

CLK Q3 Q2 Q1 Q0 O/P

0 1 0 0 1 1

1 0 0 0 0 0

2 0 0 0 0 0

3 0 0 0 0 1

4. PROCEDURE:

1. The flip-flop is connected using connecting wires as shown in the circuit. 2. The flip flop are then reset to zero internally with the help of reset to set inputs. 3. The bits are shifted in by giving suitable clock input. 4. Thus the truth table is then verified.

5. RESULTS:

INFERENCE:

Thus the operation of 4 bit shift register for SISO, SIPO, and PIPO was

studied and verified.

CONCLUSION:

.



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VIVA QUESTIONS

1. What do you understand by Operating point?

2. Why do we choose the Q point at the center of the load line?

3. Name the two techniques used in the stability of the q point .explain.?

4. Define stability factor &Give the expression for stability factor?

5. List out the different types of biasing.

6. What do you meant by thermal runway?

7. Why transistor is called as a current controlled device?

8. Define current amplification factor?

9. What are the requirements for biasing circuits?

10. When does a transistor act as a switch?

11. What is biasing?

12. What is an operating point?

13. What is d.c load line?

14. Explain about the various regions in a transistor?

15. Explain about the characteristics of a transistor?

16. Why the operating point is selected at the Centre of the active region?

17. What is an amplifier?

18. What is small signal amplifier?

19. What is a Darlington pair?

20. Define Common Mode Rejection Ratio.

21. What is meant by Differential Amplifier?

22. What do you mean by balanced and unbalanced output?

23. What are the methods of improving CMRR?

24. Give few applications of differential amplifier

25. What are the advantages of double tuned over single tuned?

26. List the four differential amplifier configurations

27. What are the advantages and disadvantages of single-stage amplifiers?

28. Why gain falls at HF and LF?

29. Why the gain remains constant at MF?

30. Explain the function of emitter bypass capacitor, Ce?

31. How the band width will effect as more number of stages are cascaded?

32. Define frequency response?

33. What is the phase difference between input and output waveforms of a CE amplifier?

34. What is early effect?

35. What is the difference between FET and BJT?

36. FET is unipolar or bipolar?

37. What are the advantages and disadvantages of common base amplifier?

38. Why the gain remains constant at MF?

39. Define frequency response?

40. What is the phase difference between input and output waveforms of a CB amplifier?

41. What are the applications of CB amplifier?

42. What is the difference between FET and BJT?



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43. FET is unipolar or bipolar?

44. Draw the symbol of FET?

45. What are the applications of FET?

46. FET is voltage controlled or current controlled?

47. Draw a hybrid model for a BJT. (2)

48. What is the relationship between bandwidth and rise time? (2)

49. What are the high frequency effects? (2)

50. If the rise time of a BJT is 35 nano seconds, what is the bandwidth that can be obtained

using this BJT? (2)

51. Explain the usefulness of the decibel unit. (2)

52. Define the term bandwidth of an amplifier? (2)

53. State various capacitances in the hybrid model? (2)

54. Define the term bandwidth of an amplifier? (2)

55. Why it is not possible to use the h- parameters at high frequencies? (2)

56. What do you mean by the half power or 3 db frequencies? (2)



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DATA SHEETS



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NPN general purpose transistors BC107; BC108; BC109

FEATURES

APPLICATIONS

amplification.

DESCRIPTION

NPN transistor in a TO-18; SOT18 metal package.

PNP complement: BC177.



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QUICK REFERENCE DATA



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adc r2013 lab manual

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