Electronics-I Laboratory Manual By .pdf

7/28/2019 Electronics-I Laboratory Manual By .pdf

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BAHIR DAR UNIVERSITY

FACULTY OF ENGINEERING

ELECRICAL & COMPUTER ENGINEERING DEPARTMENT

ELECTRONIC DEVICES AND CIRCUITS I

LABORATORY MANUAL

Lists of Experiments

Experiment I:- Semiconductor diode Experiment II:- Half-wave rectification Experiment III:- Full-wave rectification Experiment IV:- Diode Logic Gates Experiment V:- Zener Diode Experiment VI:- Limiter circuits Experiment VII:- Transistor familiarization Experiment VIII:- Common Emitter Connection

Endalew B.

Sep. 2008


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BDUEF, Electrical and Computer Engineering Department Page 1 of 19

EXPERIMENT I: SEMICONDUCTOR DIODE

I INTRODUCTION:

Semiconductor Diodes are two terminal, single PN junction devices, made from a

semiconductor that has been dopped in such a way that it changes from P back to N or from N

back to P.A typical diagram is shown in the figure below.

NType

PType

vol t agesour ce

Fig. 1-1 Function Diode

Diodes, unlike resistors and other circuit components, are unilateral devices---the amount of

current that flows across their terminals depends on the polarity and size of voltage applied.When a diode is connected so as to conduct, it is said to be forward biased otherwise reversebiased.

Schematically a diode is represented in a circuit by the symbol shown below.

Anode ( A)

P- t ype

Cat hode( K)

N- t ype

Fig. 1-2 Diode Symbol

Note: Physically a band locates the cathode terminal.

II OBJECTIVES:

-To determine the polarity of diode.

-To gain knowledge of V/I characteristics of a diode.

III COMPONENTS AND EQUIPMENT REQUIRED:

Type Quantity

Breadboard 1Resistors: 1K, 100R, 10K pot. 3

Diode 1

Power supply unit; (0-20V variable DC regulated) 1Multimeters or 2

High impedance, DC Voltmeter 1


3/19


IV PROCEDURE:

PART I Determining the diode polarity:

1. Construct the circuit of fig. 1-3 on the construction board.

DC V

NO DATA

D1

R1

1k

+ V110V

Fig. 1-3 Diode Test Circuit

2. Switch on the power supply (V1) and set its control to give 10V on the meter.

3. Measure the voltage drop across R1 (read the DC Voltmeter) and calculate the Id (Id = VR1/

R1 = [VR1] mA) record it in the data table 1-1.

Data Table 1-1

4. Turn off the power supply5. Reverse the diode polarity

6. Repeat step 2 and 3 in the same way

PART II Characteristics of forward biased diodes:


Fig. 1-4 Test Circuit

Circuit Id Current (mA)

Forward Biased

Reverse Biased

R11k

D1

DC VNO DATA

VR1

DC V

NO DATA

Vd

R2DC V

NO DATA

VS+

Vs 10k

0-20V


4/19


Note: Vd =Vs VR1 ; If= Id = VR1 / R1 =[VR1]mA

2. Turn the potentiometer R2 to zero.

3. Switch on the power supply and adjust it to supply 20V.4. Adjust the potentiometer to give a voltage of VS to 0 and measure VR1 and record it in

table 1-2

5. In the same way adjust Vs 0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9, 1, 2, 3, 4, 5, 6V and

measure VR1 for each and record it in data table 1-2.

6. Calculate Vd and If, and record in the data table 1-2

Data table 1-2

Vs (V) VR1 (V) Vd =Vs VR1 (V) Id = If (mA)

0

0.1

0.2

0.3

0.4

0.5

0.6

0.8

0.9

1

2

3

4

5

6

V. CONCLUSIVE QUESTIONS:

1. Plot Vd against If on a square paper in your report.

2. At what approximate value of Vd does the current Ifbegin to rise noticeably? Estimate fromthis value whether the diode is made of silicon or germanium.

3. Does Vd rise much above this value for larger values of If?

4. How accurate is the assumption of a diode as an Ideal Electronic Switch?5. In fig. 1-3, what is the function of the 4K7 resistor in series with the diode?

6. Provide a substantive conclusion to this experiment.


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EXPERIMENT II: HALF-WAVE RECTIFICATION

I INTRODUCTION:

Rectification is a process of changing a bi-directional ac into a pulsating dc. It is a

fundamental step in ac to dc conversion. Frequently, diodes are used to rectify an ac voltage

as shown below.

Fig. 2-1 Halfwave Rectifier Circuit

The above circuit allows only the positive half-cycle of the input voltage to reach at the outputand is called a half-wave rectifier circuit. The diode will conduct only when the P region has a

positive voltage applied to it relative to the N region. Hence the output voltage will appear

across R as frequently as the positive half-cycle of the input voltage Vin repeats and have the

same frequency as the input voltage. But such a rectifier circuit suffers from a large ripple.

If the diode in fig 2-1 is turned around, it will conduct during the negative portion of the inputsignal.

II OBJECTIVE:

-To recognize a half-wave rectified sinusoidal voltage.-To understand the term mean value as applied to a rectified waveform.-To understand the effect of a reservoir capacitor upon the rectified waveform and its mean

value.


Type Quantity

Bread board 1Resistor, 10K 1

Diode 1

Capaciotrs; 1uF, 22uF 2Power supply unit; A.C supply 50/60 Hz 1

Multimeters or D.C Voltmeter 50V 1

Oscilloscope 1

50 Hz


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IV PROCDRE:

PART I Simple Half-wave Rectification

1. Construct the circuit of fig 2-2 on the construction board.

Fig. 2-2 Half-wave Rectification

2. Adjust the signal generator to get an output of 10 Vpkusing an oscilloscope and give it toyour circuit.

3. With the oscilloscope d.c. coupled, adjust the time base and the Y-amplifier sensitivity to

obtain a steady trace of the waveform.4. Measure and record the time T and peak voltage Vpkof the trace. Note also the mean

voltage on the voltmeter.

5. Sketch the waveform and label it to show the periods, when the diode is conducting and

those when it is not. Time T depends on the frequency of your power supply.

PART II Effect of Reservoir Capacitor

1. Add a capacitor of 1uF to the circuit of fig. 2-2 in parallel with the load.

2. Observe the output waveform on the oscilloscope and note the value of the peak to peakvariations in voltage. Note also the new mean voltage on the voltmeter. Is the new mean

voltage greater or less than it was before.3. Now replace the 1uF capacitor by a much larger value of 22uF (make sure to connect the +

side of the capacitor to the diode cathode if an electrolytic type of capacitor is used).

Is the ripple now less or more than it was with the lower value capacitor?

Is the new mean rectified voltage now greater or less?

V CONCLUSIVE QUESTIONS:

PART I:

1. Why will Vpknot be exactly equal to the peak voltage of the alternating supply? How much

will it differ?

2. The mean voltage you obtain is positive relative to zero. How could you obtain a negativevoltage?

PART II:

1. Describe briefly the effect of the reservoir capacitor on the ripple and mean rectified

voltage?

2. How do you relate the value of the capacitor to the amount of ripple obtained?3. List several methods to reduce the ripple.

4. Summaries the things that you have learned from this experiment and give a reasonable

conclusion.

Osci l l oscope

common

Osci l l oscope50 Hz

R10K

Y i nput

+

-VL


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EXPERIMENT III: FULL-WAVE RECTIFICATION

I INTRODUCTION:

The large ripples available in the half-wave rectified ac voltage could be further reduced using

full-wave rectification. In this method, both the positive and negative half-cycles of the input

will appear at the output with the same polarity.

A full-wave rectifier circuit shown below is called a DIODE BRIDGE rectifier circuit.

Fig. 3-1 Diode Bridge Rectifier Circuit

In the above circuit, during the positive half cycle of the supply A is more positive than B.Diodes D1 and D2, therefore conduct while diodes D3 and D4 are reverse biased. During the

negative half cycle D3 and D4 conduct with the others reverse biased.

II OBJECTIVES:

-To recognize a full-wave rectified waveform with and without a reservoir capacitor.

-To understand the operation of a diode bridge circuit as a full wave rectifier and its

advantage over half-wave rectification.


Type Quantity

Breadboard 1Resistor, 10K 1

Diode 4

Capacitor, 1uF 1

Power supply unit; A.C supply 50/60 Hz 1Multimeter or D.C Voltmeter 50V 1

Oscilloscope 1

A

50 Hz

D1

D2 D3

D4

Load

B


8/19


IV PROCEDURE:

PART I A Bridge Rectifier with Resistive load:


Fig. 3-2 Test Circuit

2. Adjust the signal generator to give an output of 10 Vpkusing an oscilloscope and give it to

your circuit.

3. With the oscilloscope d.c coupled, adjust the controls to obtain a steady trace of the outputvoltage. Note the value of peak voltage Vpkand also the mean value of the output voltage

indicated on the voltmeter. Should Vpk be the same as it was for a half- wave rectifier? Doesyour observation confirm your answer?

PART II Effect of Reservoir Capacitor:

1. Add a capacitor of 1uF in parallel with the load resistor.

2. Note the mean value and peak-to-peak ripple amplitude of the rectified waveform.

3. Compare this figure with those obtained in expt. 2 for the same load and capacitor values.


PART I:

1. State briefly how the diode bridge works.

2. Why will Vpknot be exactly equal to the peak voltage of the alternating supply? How much

will it differ?

PART II:

1. Describe briefly the effect of the reservoir capacitor on the ripple and the mean rectified

voltage?2. How could the ripple further be reduced?

3. Summaries the things that you have learned from this experiment and give a reasonableconclusion.

D1

D2 D3

D4

50 Hz

10K

Osci l l oscope

Y i nput

Osci l l oscopeCommon

VL+

-


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EXPERIMENT IV: DIODE LOGIC GATES

I INTRODUCTION:

Diodes together with resistors can be used to implement digital logic functions. Figure below

shows two diode logic gates. To see how these circuits function, consider a positive logic

system in which voltage values close to 0V correspond to logic 0 (or Low) and voltage values

close to +5V correspond to logic 1 (or High)

Construct the circuit below and test the result

(a)OR Gate

(b) AND Gate

Fig. 4.1 Diode logic gates


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EXPERIMENT V: ZENER DIODE

I INTRODUCTION:

Diodes with a similar forward V/I characteristics as ordinary diodes but with a unique reverse

V/I characteristics are called Zener diodes. These devices are normally operated in the break

down region, where the voltage across the device remains substantially constant as the reverse

current varies over a large range as shown below.

Fig. 5-1

This ability of zener diode makes it an ideal voltage reference and it is widely implemented in

most demanding requirements, where a constant voltage is needed irrespective of changes insupply voltage, Vsand load.

II OBJECTIVES:

-To determine the constant voltage characteristics of reverse biased zener diodes.-To investigate the use of a zener diode in a simple voltage regulatory circuit.


Type Quantity

Breadboard 1

Resistors; 100R(1W), 100R(1W)220R(1/2W), 100R (2W) 1

Potentiometer,10K 1

Zener diode BZX85C10; Vbr=10V,Pmax=1W 1

Power supply unit; (0-20V variable d.c. regulated) 1Multimeters or 3

High impedance, d.c. Voltmeter 1

Ammeter, 100mA d.c. 1Ammeter, 100mA d.c. 1

VZ

IZ min

IZ max

I

V


11/19


IV PROCEDUER:

Part I: Zener diode reverse characteristics:


DC VNO DATA

+0- 20V

1k

Fig 5-2 Test Circuit.

2. Using the power supply variable control, set Vs to the values given in the data table 5-13. For each Vs setting record VR1, then calculate Vz= Vs - VR1 & Id = Iz = VR1 /1000=[VR1]mA

and record the result on the same data table.

Data Table 5-1

Vs (V) VR1 (V) Vz = Vs - VR1 (V) Id = [VR1]mA Pd =Vz Id (mW)

0

2

4

4.5

5.0

5.5

6.0

6.5

7.0

7.5

8.0

12

16

20

4. Calculate the power dissipated in the diode for each value of Vd and Id ; Enter it in the last

column of the above data table.5. From the result above, plot table Vd against Id on a graph paper.

VS

R1


12/19


Part II: A Simple Zener Diode Voltage Regulator with Variable load and fixed Supply.



2. Turn the potentiometer to maximum.

3. Turn on the variable voltage supply and adjust it to supply 20V. (Dont change this valueuntil the end of this expt.)

4. Measure the current through the diode Id , the output voltage Vout and the current throughthe load IL. Enter these in the data table 5-2.Is the diode in the break down region?

5. Increase the load IL by decreasing R, in steps of 10mA and record the corresponding I d , IL

and Vout in the data table.

6. Repeat this until the diode is out of the break down region (Id < 5 mA).

Data Table 5-2

Id IL Vout

Part III: A Simple Zener Diode Voltage Regulator with variable Supply and fixed Load.



DC

V

NO

DATA

DC V

NO DATA

+

0- 20VR2100

R1100

It

Vout

ILId

DC

V

NO

DATA

DC V

NO DATA

+

0- 300VR2100

R1100

It

Vout

ILId


13/19


2. Turn on the variable voltage supply and adjust it to supply 13V.

3. Measure the current through the diode Id , the voltage at the output Vout . Enter these in the

data table along with your supply setting Vs. Is the diode in the breakdown region?4. Increase the supply voltage Vs in steps of 1V and, measure and record IL, Vout and Vs in the

data table.

5. Repeat step 5 until the supply reaches 25V. (The current through the diode is less than its

maximum power dissipation).

Data Table 5-3

Vs Id IL Vout


PART I:

1. Describe the Zener Diode characteristics in your own words.2. Does the breakdown voltage obtained from the graph agrees with its nominal value?

If not, can you suggest a reason for the difference?

PART II:

3. Analyze the performance of the circuit for the given load variation. Is the diode doing itsjob well?

4. Is the series resistance Rs large enough to limit the current through the diode within its

rating?

5. How will the circuit react if the supply voltage starts to vary also?

PART III:

5. Analyze the performance of the circuit for the given supply voltage variation.

6. How will the circuit react if the load starts to vary also?


14/19


EXPERIMENT VI: LIMITER CIRCUITS

I INTRODUCTION:

The limiter acts a linear circuit, providing an output proportional to the input, Vo = K Vi. If

the input voltage exceeds the upper threshold, the output voltage is limited or clamped to the

upper limiting level L+. On the other hand, if the input voltage is reduced below the lower

limiting threshold, the output voltage is limited to the lower limiting level L-

Double limiter is limiter that works on both the positive and negative peaks of an input

waveform. Limiters therefore are sometimes referred to as Clippers.

1. Construct the circuit given below on the construction board

Fig 6-1 Clipper circuit

2. Connect the oscilloscope in the output side

3. Increase the amplitude (the voltage level) of the input signal and observe the output

wave shape


15/19


EXPERIMENT VII: TRANSISTOR FAMILARISATION

I INTRODUCTION:

Transistors are three terminal semiconductor devices containing two PN junctions. It can be

formed from a bar of material that has been doped in such a way that it changes from N to P

and back to N or from P to N and back to P, forming NPN and PNP types of transistorsrespectively (as shown below). In either case, junctions are created at each of the two

boundaries where the material changes from one type to the other, called Emitter to Base andCollector to Base junctions.

NPNPNP

Fig. 7-1 Two types of transistors

Each of the PN junctions in the diagram behaves individually like a simple diode but when

joined this way the behavior is very different. In normal use the emitterbase junction isforward biased and the collectorbase junction is reverse biased and normally you wouldexpect it to pass no current. But if the E-B junction is conducting forward current, this

influences the reverse biased C-B junction and causes it to pass almost as much reverse

current. The small difference current flows in the base circuit.

The ratio Ic/Ie is called hfb. Because Ic is almost as big as Ie , hfb is nearly 1 (e.g. 0.99). The

ratio Ic / Ib is usually called hfe.

Ie= Ib + Ic= Ic / hfb

Ic / Ib = hfe= hfb /(1- hfb )

If hfb = 0.99

hfe = 99

It is this large ratio between Ic and Ib that makes the transistor a useful amplifying devicewhen connected so that Ib is derived from an input and Ic provides an output.

II OBJECTIVES:

-To understand the basic construction of PNP and NPN transistor.

-To understand Junction biasing and magnitude of current flows.

PNP TRANSISTOR NPN TRANSISTOR

CONSTRUCTION

SYMBOLE

E E

E

CC

CC

B B

B B


16/19


III COMPONENTS AND EQUIMENT REQUIERED:

Type Quantity

Breadboard 1Resistors; 1K0, 2K2 pot. 2

Capacitor, 0.1uF 1

Transistor, BC107 1

Power supply unit; 0-5V, 0-15V variable d.c regulated 1

Multimeters or 2Microammeter 1

Voltmeter, 3V d.c. 1

IV PROCEDURE:


Fig. 7-2 Transistor Test Circuit.

2. Turn the potentiometer to zero and switch on both power supplies.

3. Slowly increase Veb by turning the potentiometer until Ic just begins to flow.

4. Connect the voltmeter temporarily between E and B on the transistor (3V d.c. range) andnote the value of Veb in data table. 7-1.

Data Table 7-1.

Ic (mA) Veb (V) I (uA) hfe= Ic/I Ie= Ic+I (mA) hfb= Ic/Ie

5. Remove the meter and then continue increasing Veb until Ic approximately equals 1mA and

record the values of Ic and Ib in the table.

6. Now, increase Veb until Ic approximately equals 10mA; again record Ic and Ib. Did yourreading of Veb confirm that forward biased E-B junction is acting like a simple diode?

7. For each set of readings calculate hfe , Ie and hfb as follows:

hfe= (Ic x1000)/Ib ; Ie= Ic+ (I b/1000) (mA) ; hfb= Ic/Ie


1. Do your results show that hfe andhfb increase, decrease or stay constant as Ic increases ?2. What is the use o the 0.1uF capacitor in the circuit of fig. 7-2?3. Summarize the things that you have learned from this experiment and come to a

conclusion.

-15V

0V

1K0

2K2

0.1uF

A

A


17/19


EXPERIMENT VIII: THE COMMON EMITER TRANSISTOR CIRCUIT

I INTRODUCTION:

Transistor junction can be biased in three different ways, called common base, common

collector and common emitter configuration. Figure 6-1 shows one of the three possible

ways to configure a transistor called common-emitter configuration.

++

Fig. 8-1 NPN transistor in common-emitter configuration.

In this configuration, the emitter is common to the input and output terminals.Ib is derived from input signal and Ic is used to generate an output. The ratio Ib/ Ic = hferepresents the gain of the transistor in terms of the current.

It is this large ratio between Ic and Ib that makes the transistor a useful amplifying device.

II OBJECTIVES:

-To be familiar with the common-emitter output (collector) characterstics.

-To provide an understanding of the meaning and importance of OPERATING POINT andLOAD LINE.

III COMPONENTS AND EQUIMENT REQUIERED:

Type Quantity

Breadboard 1Resistors; 22K, 10K pot. 1

Transistor, BC107 1

Power supply unit; 0-20V variable d.c. regulated and

+5V d.c.regulated 1Multimeters or 3

Microammeter 100uA 1

Milliammeter 10mA 1High resistance voltmeter, 10V d.c. 1

IC

IB

IE

VCEVBE


18/19


IV PROCEDURE:

Part I-To find how Ic is controlled by Ib and Vce:


Q1NPN

Fig. 8-2 Test Circuit.

2. Set Vce to 0.5V,then use the potentiometer to adjust I b to each value given in table 8-1.

Data Table 8-1.

Ib (uA) Ic (mA) for Vce ..(V)

0.5 1 2 5 10

0

10

2030

40

50

3. At each setting record Ic in the appropriate column. Repeat to other Vce values in the sameway.

4. Plot Ic against Vce for each value of Ib on a graph paper.

+5V

22K

10K

0-20V d.c.

variable

0V

VCE

A

A


19/19

Part II Constructing a load line:

Fig 8-3 shows a circuit in which the collector bias is applied through a resistor.

Vcc is the term now used for the bias voltage to distinguish it from Vce , because when thecurrent Ic is flowing these two will be different.

+

Fig. 8-3 Addition of a load resistor.

By ohms law

Vce = Vcc- Ic RTherefore, when Ic =0, Vce= VccAnd when Vce =0, Ic = Vcc /R

On your graph, plot these two points Vce and Ic =0 for:Ic =0, R=2K and Vcc=+10V

Plot two or three more points for these values of Ic such as 2 and 5mA using the same valuesof R and Vce and then join up the points by a line.


Part I:

1. What happens to Ic when Vce becomes less than 0.6V? What is the significance of this

value?2. What do you notice about the effect of Vce upon Ic when Vce is greater than about 1.0V?

Part II:

1. What is the importance of setting an operating point?

2. Summarize the things that you have learned from this experiment and come to a

conclusion.

VCC

VCEVBE

RIC

IB

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