This document contains instructions and procedures for experiments on basic electrical and mechanical engineering concepts. It includes 7 experiments: 1) Ohm's Law, 2) Characteristics of series and parallel DC circuits, 3) Kirchhoff's voltage laws, 4) Kirchhoff's current laws, 5) Analysis of a silicon diode characteristic curve, 6) Analysis of half-wave and full-wave rectifiers, and 7) Identification of diode schematic symbols and observation of normal operations in a diode circuit. The experiments are designed to investigate circuit laws and properties of series, parallel, and diode circuits through construction of circuits and measurement of voltages and currents. Procedures and observations are recorded in tables for analysis and verification of theoretical concepts.

1. PRACTICAL WORK BOOK
For Academic Session Fall 2020
BASIC ELECTRICAL & MECHANICAL ENGINEERING
CE-116
Name: ___________________________________________
Registration # _____________________________________
Class: ____________________________________________
Batch: ____________________________________________
Semester:b _________________________________________

2. 2
List of Experiments
S# Experiments
1. Ohm’s Law
2. Characteristics of a Series & Parallel DC Circuit
3. Kirchhoff’s Voltage Laws
4. Kirchhoff’s Current Laws
5. To analyze the characteristic curve of Silicon diode.
6. To analyze the Half-wave & Full Wave rectifier.

3. Experiment No. 02
Characteristics of a Series & Parallel DC Circuit
OBJECTIVE:
To investigate the characteristics of a series DC circuit.
APPARATUS:
 DMM
 DC Supply (20V)
 Resistors of 220Ω(R, R, Br), 330Ω (Or,Or, Br) & 470Ω(Y, Vio, Br).
THEORY:
In a series circuit, (Fig 3.1), the current is the same through all of the circuit elements.
The total Resistance RT = R1 + R2 + R3 .
By Ohm’s Law, the Current “I” is
I = E/ R T
Applying Kirchhoff’s Voltage Law around closed loop of Fig 3.1, we find.
E = V1 + V2 + V3
Where, V1= IR1 , V2= IR2 , V3= IR3
Note in Fig 2.1, that I is the same throughout the Circuit. The voltage divider rule states that the voltage
across an element or across a series combination of elements in a series circuit is equal to the resistance of
the element divided by total resistance of the series circuit and multiplied by the total impressed voltage.
For the elements of Fig 3.1
V1 =R1 E/R T , V2 =R2 E/R T , V3 =R3 E/ R T
V1= 220*20/1.24K=

4. 4

5. 5
PROCEDURE:
1- Construct the circuit shown in Fig 3.2.
2- Set the Dc supply to 20V by using DMM. Pick the resistances having values 220Ω, 330Ω &
470Ω. Also verify their resistance by using DMM.
3- Measure voltage across each resistor with DMM and record it in the Table (b).
4- Measure Current I delivered by source.
5- Shut down and disconnect the power supply. Then measure input resistance R T using DMM.
Record that value.
6- Now Calculate, Total current (using I= V/ RT ) and Total Resistance R T (R T = E/I).
7- Calculate V1,V2, V3 &V4 using voltage divider rule and measured resistance value.
8- Create an open circuit by removing R3 & measure all voltages and current I.
Note: Use measured value of resistance for all calculations.
OBSERVATIONS:
a) Resistors.
S No. Nominal
Values(Ω)
Measured
Value(Ω)
R T (Measured)
(Ω)
R T (Calculated)
(R T = E/I)(Ω)
1 220 1.24k 1.24k 20/16.129m=1.24k
2 220 1.24k
3 330 1.24k
4 470 1.24k

6. 6
b) Voltages.
S.No. Measured Value(V) Calculated Value (V) (VDR)
1 3.548 220*20/1.24=3.548
2 3.548 220*20/1.24=3.548
3 5.232 330*20/1.24=5.232
4 7.581 470*20/1.24=7.581
a) Current.
S. No. Measured Value of I (A) Calculated Value (A) (Ohm’s Law)
1 16.129m 16.129
2 16.129m 16.129
3 16.129m .129
4 16.129m 16.129
CALCULATIONS:
1.20/1.24=16.129
2. 20/1.24=16.129
3. 20/1.24=16.129
4. 20/1.24=16.129
Series circuit
When a number of electrical components are connected in series, the same
current flows through all the components of the circuit.
The applied voltage across a series circuit is equal to the sum total
of voltage drops across each component.
The voltage drop across individual components is directly proportional to
its resistance value.

7. 7
Answer the following:
Q-1) What can you deduce about the characteristics of a series circuit from observation
Table ‘b’ & ‘c’?
In series circuit we observation the same flow of current total voltage of
circuit is some of
Voltage applied to a series circuit is equal to the sum of the
individual voltage drops.
________________________________________________________________________________
Q-2) Viewing observation table ‘b’, comment on whether equal resistors in series have
Equal voltage drops across them?
Ans. In series circuit we observation that a cross current is same but voltage
drop across them is
Not same individual same will be create different voltage across each
resistance.
Q-3) Referring observation table (a). Would you recommend using measured rather
than
Color coded nominal values in the future? Why?
Ans.The actual (measured) resistance will vary from the nominal value due to
subtle mechanical and chemical differences that occur during manufacturing.

8. 8
The manufacturer specifies the maximum deviation from the nominal value as a
±percentage. This range of deviation is called the tolerance of
the resistor family.
Q-4) Referring to observation table’s b & c compare open circuit condition & normal
Circuit with reference to current & voltage values.
Ans. A Normal circuit implies that the two terminals are externally connected
with resistance R=0, the same as an ideal wire. This means there is
zero voltage difference for any current value. ... An open circuit implies that the
two terminals are points are externally disconnected, which is equivalent to a
resistance R=∞.
____________________________________________________________________________

9. Characteristics of a Parallel DC Circuit
OBJECTIVE:
To investigate the characteristics of parallel dc circuits
REQUIRED:
 12V DC Power Supply.
 DMM.
 470Ω (Y, Violet, Br).
 1KΩ (Br, Black, Red).
 1.8KΩ (R, Grey, Red).
THEORY:
In a parallel circuit (Fig4.1) the voltage across parallel elements is the same.
The total or equivalent resistance (RT) is given by.
1/R T =1/R 1+1/R 2+1/R3+ ------------------- +1/R N
If there are only two resistors in parallel, it is more convenient to use.
In any case, the total resistance will always be less than the resistance of the smallest resistor
Of the parallel network.
For the network of Fig 4.1. The currents are related by the following expression.
IT =I1 + I2 + I3+-----------+ IN
Applying current divider rule (CDR) & the network of Fig 4.2.
I1= R 2 I T and
R 1 + R 2
I2= R 1 I T
R 1 + R 2

10. 10

11. 11
For equal parallel resistors, the current divides equally and the total resistance is the value of One divided
by the ‘N’ number of equal parallel resistors, i.e:
R T =
𝑅
𝑁
For a parallel combination of N resistors, the current Ik through Rk is.
IK= IT X
1
Rk
1
R 1
+
1
R 2
+
1
R 3
+⋯
1
RN
PROCEDURE:
1- Construct the circuit shown in Fig 4.3.
2- Set the DC supply to 12V by using DMM.Pick the resistances values 470Ω, 1KΩ & 1.8KΩ. Also
verify their resistance by using DMM.
3- Measure voltage across each resistor with DMM and record it in the Table b.
4- Measure the currents IT , I1, I2, I3.
5- Shut down & disconnect the power supply. Then measure input resistance R T using DMM. Record
that value.
6- Now calculate respective voltages (using V=IR) and R T (using equivalent resistance Formula).
7- Calculate I1, I2 , I3 using CDR.
8- Create an open circuit by removing R2 and measure all voltages and currents.
Note: Use measured value of resistance for all calculations.

12. 12
OBSERVATION:
a) Resistors:
S.
No.
Nominal Values
(Ω)
Measured Value
(Ω)
R T(Measured)
(Ω)
R T(Calculated)
(Ω)
1 R1=470Ω 9.4k(Ω) 271.529 271.529
2 R2=1kΩ 5.4k(Ω)
3 R3=1.8kΩ 3k(Ω)
b) Voltages:
S.
No.
Measured Value
(V)
Calculated Value
(Ohm’s Law)
(V)
1 V1 20 20
2 V2 20 20
3 V3 20 20
c) Current:
S.No Measured Value
(A)
Calculated Value
(A) (CDR)
1 I1=3.55u I1=3.55u
2 I2=3.55 I2=3.55u
3 I3=11.113 I3=11.113u
4 IT=18.213A IT=18.213A

13. 13
CALCULATIONS:
Rt=1/R1+ 1/R2 + 1/R2 =271.529
I=v/R
Total current It=3.55+3.55+18.213=18.213
Conclusion
Parallel circuit
Voltage drops are the same across all the components connected in parallel.
Current through individual components connected in parallel is inversely
proportional to their resistances.
Total circuit current is the arithmetic sum of the currents passing through
individual components connected in parallel.
The reciprocal of equivalent resistance is equal to the sum of the
reciprocals of the resistances of individual components connected in
parallel.
Answer the following:
Q-1) How does the total resistance compare to that of the smallest of the
three parallel resistors?

14. 14
The total resistance in a parallel circuit is less than the smallest of
the individual resistances. Each resistor in parallel has the same
voltage of the source applied to it (voltage is constant in
a parallel circuit).
Q-2) Is IT under normal circuit condition less than IT for open circuit
condition?
A short circuit implies that the two terminals are
externally connected with resistance R=0 , the same as an ideal wire.
This means there is zero voltage difference for any current value.
(Note that real wires have non-zero resistance!)
An open circuit implies that the two terminals are points are
externally disconnected, which is equivalent to a resistance R=∞ .
This means that zero current can flow between the two terminals,
regardless of any voltage difference

15. 15
Experiment No. 03
Kirchhoff’s Voltage Laws
OBJECTIVE:
To verify experimentally Kirchhoff’s Voltage and Current Law:
REQUIRED:
 Resistors,
 DMM,
 DC power supply.
THEORY:
1) KIRCHOFF’S VOLTAGE LAW:
Explanation:
Consider the simple series circuit Fig. 5-1. Here we have numbered the points in the Circuit for
voltage reference.
As we are dealing with dc circuits, therefore we should carefully connect the voltmeter while
measuring voltage across supply or any of the resistances as shown in fig.5.2, keeping in mind the
similarity of polarities of voltage across the element and that of the connected probes of meter. In
such case,we will observe that,
This principle is known as Kirchhoff's Voltage Law,and it can be stated as such:
"The algebraic sum of all voltages in a loop must equal zero"

16. 16
OBSERVATIONS:
Table 5-1:
VT V1 V2 V3 Sum (V1+V2+V3)
-20v 5.98v 12.739v 1.274v -20+5.9+12.73+1.27=0

17. 17
CALCULATIONS:
Sum of voltage (-20+5.9+12.73+1.27=0)
Kirchhoff’s Voltage Laws the sum of current in close circuit is equal to
zero hence we prove that in KVL circuit that sum of current in close
circuit is zero.
Applied voltage V1+V2+V3+V4+(10+12+20+15=57
VS _V1_V2_V3=0 (100_15_20_v3=0) 25 –V3 =0 V3 =25
Answer the following:
Q-1) In fig.5-1 V1=10V, V2=12V, V3=20V, V4=15V. The applied voltage Vs must
then equal _57___________ V.
Q-2) In Fig.5-1,V1=15V,V2=20V and Vs = 100V. The Voltage V3 =
____25________V.

18. 18
Q-3) Is KVL verified practically as well as mathematically in the above
performed lab? If no, why?
Yes KVL is prove in practically performed as well as mathematically in above lab performed we
prove that sum ofclose circuit voltage is equal to zero.

19. 19
Experiment No. 04
Kirchhoff’s Current Laws
OBJECTIVE:
To verify experimentally Kirchhoff’s Voltage and Current Law:
REQUIRED:
 Resistors,
 DMM,
 DC power supply.
THEORY:
2) KIRCHHOFF’S CURRENT LAW:
Let's take a closer look at the circuit given in Fig 5-3:
Then, according to Kirchoff’s Current Law:
"The algebraic sum of all currents entering and exiting a node must equal zero"
Mathematically, we can express this general relationship as such:
That is, if we assign a mathematical sign (polarity) to each current,denoting whether they enter
(+) or exit (-) a node, we can add them together to arrive at a total of zero, guaranteed.
Note:
Whether negative or positive denotes current entering or exiting is entirely arbitrary, so long as they
are opposite signs for opposite directions and we stay consistent in our notation, KCL will work.
PROCEDURE:
a) For KVL
1. Construct circuit of fig. 5-1 using the values R1 , R2, R3 as shown in the figure 5-1.
2. Adjust the output of the power supply so that Vs = 20V. Measure and record this voltage in
table. 5-1, also measure and record the voltages V1, V2, V3 and enter the sum in the same
table.
b) For KCL:
1- Connect the circuit of Fig with Vs =5 V.
2- Measure and record in currents IR1,IR2,IR3 and Itotal in Table

20. 20
Itotal IR1 IR2 IR3 Sum (IR1+IR2+IR3)
-17.177mA 3.553A 6.061A 11.113A 17
CALCULATIONS:
-17.177 + 3.553 +6.061+11.113=0

21. 21
Conclusion
In this lab we prove the Kcl the interring current
and exited current is equal to zero we absorbed
that the current interring is positive and exited
current is negative there sum is equal to zero.

22. 22
Answer the following:
Q-1) In fig 5-3, if IR1=5A,IR2=2A and IR3=1A, then Itotal should be
equal to _______8______ A.
Q-2) Is KCL verified practically as well as mathematically in the above
performed lab? If no, why?
According to Kcl the sum of current interring current is positive and
exited current is negative so in above equation we prove practically and
so mathematically sum of entering current and exit current is equal to
zero.

23. 23
LAB SESSION 05
OBJECTIVES
 Identify diode schematic symbols
 Describe Silicon based diode operating characteristics.
 Identify diode construction characteristics.
 Observe normal operations in a diode circuit.
APPARATUS
 Power Supply.
 Multimeter
 Oscilloscope.
 Wire Stripper
 1N4007 Diode
 Graph Paper
THEORY
Although the diode is a simple device, it forms the basis for an entire branch of
electronics. Transistors, integrated circuits, and microprocessors are all based on its theory
and technology. In today’s world, semiconductors are found all around us. Cars, telephones,
consumer electronics, and more depend upon solid state devices for proper operation.
PN Junction
Now, we are ready to build a diode. To do this, we need two blocks of material, one N
type and one P type.

24. 24
The resulting block of material is a diode. At the instant the two blocks are fused, their point of
contact becomes the PN junction. Some of the electrons on the N side are attracted to the P side, while
at the same time, an equal number of hole charges are attracted to the N side.
As a result, the PN junction becomes electrically neutral. The barrier in Figure 3 is greatly
exaggerated. In some semiconductor devices, the PN junction barrier may only be a few atoms thick.
The PN junction is an electrical condition, rather than a physical one.The junction has no charge; it is
depleted of charges. Thus, another name for it is the depletion zone. Because of the existence of the
depletion zone, there is no static current flow from the N material to the P material.
The diode consists of two parts or elements, the N material and the P material. Their proper
names are cathode and anode. The cathode is the N material and the anode is the P material. Electron
current flow is from the cathode to the anode. Figure 4 illustrates a PN junction diode.
Bias
Average DC level of current to set operating characteristics.
There are two types of bias in semiconductors, forward and reverse. Forward bias will
eliminate the depletion zone and cause a diode to pass current. Reverse bias will increase the
size of the depletion zone and in turn, block current. Figure 6 and 7 illustrates forward and
reverse bias.

25. 25
A diode is biased by placing a difference in potential across it. Figure 7 illustrates a
forward biased diode. Because of the positive potential applied to the anode and the negative
potential applied to the cathode, the depletion zone disappears. Current flows from the
negative terminal of the battery through the N region, across the non-existent depletion zone,
and through the P region to the positive terminal of the battery. It takes a specific value of
voltage for a diode to begin conduction. Approximately .3 volts across a germanium diode or
.7 volts across a silicon diode are necessary to provide forward bias and conduction. A
germanium diode requires a lower voltage due to its higher atomic number, which makes it
more unstable. Silicon is used far more extensively than germanium in solid state devices
because of its stability.
FIGURE 6 REVERSED BIASED DIODE FIGURE 7 FORWARD BIASED DIODE
Reverse bias is accomplished by applying a positive potential to the cathode and a negative
potential to the anode as shown in Figure 6. The positive potential on the cathode will attract electrons
from the depletion zone. At the same time, the negative potential on the anode will attract holes. The
net result is that the depletion zone will increase in size.
A forward biased diode will conduct, with only a small voltage drop over it. The
voltage drop for a forward biased germanium diode is .3 volts, while .7 volts is normal for a
silicon diode. We can say that a forward biased conducting diode is almost a short. A
reversed biased diode will not conduct. Therefore, it can be considered an open circuit. We
call a reversed biased diode cut off. Cut off refers to the current flow through the diode being
blocked, or cut off.
Diode characteristics
The diode consists of two elements, the anode and the cathode. The anode
corresponds to the P material and the cathode to the N material. Current flow is from the
cathode to the anode.

26. 26
Figure 8 illustrates a forward biased diode with current flow and the diode elements labeled. The
graph in Figure 9 depicts current flow through a diode with different values of forward and reverse bias.
PROCEDURE
PART 1. Diode Test
a. Diode testing Scale
The diode-testing scale of a DMM can be used to determine the operating condition of a
diode. With one polarity, the DMM should provide “offset voltage” of the diode, while the
reverse connection should result is an “OL” response to support the open-circuit
approximation.
Using the connections shown in fig1.2, the constant-current source of about 2
mA internal to the meter will forward bias the junction, and a voltage about 0.7 V
(700mV) will be obtained for silicon and 0.3 V (300mV) for germanium. If the leads
are reserved, an OL indication will be obtained.
TEST SILICON GERMANIUM
FORWARD RESISTANCE 0.3 0.7

27. 27
REVERSE RESISTANCE 0 0
PART 2. FORWARD-BIAS DIODE CHARACTERISTICS
 The connections are made as shown in the circuit diagram.
 The power supply is switched ON
 Vary the voltages and note the corresponding Ammeter &Voltmeter readings
 Draw VI characteristics for forward bias and reverse bias in one graph
PART 3. REVERSE-BIAS DIODE CHARACTERISTICS
 The connections are made as shown in the circuit diagram.
 The power supply is Switched ON.
 Vary the Voltages and note the corresponding Ammeter & Voltmeter Readings
Forward Bias
Reverse Bias

28. 28

29. 29
OBSERVATION: (FORWARD BIAS)
Sr.
No.
Supply voltage (Volt) Diode voltage (VD)
(V)
Diode current (ID)
(mA)
1 0 0 0
2 0.2 0.194 6109NA
3 0.4 0.399 444NA
4 0.6 0.420 59958NA
5 0.8 0.0196 780359NA
6 1 0.1313 868699NA
7 2 0.035 964672NA
8 4 2.986 1014NA
9 6 4.962 1038NA
10 8 6.947 1053NA
11 10 8.935 1065NA
12 15 0.0139 1085NA
13 20 0.0189 11NA

30. 30
OBSERVATION: (REVERSE BIAS)
Sr.
No.
Supply voltage (Volt) Diode voltage (VD)
(V)
Diode current (ID)
(mA)
1 0 0 0
2 0.2 0.0126 199987NA
3 0.4 0.058 399942NA
4 0.6 0.2304 59977NA
5 0.8 0.877465 799123NA
6 1 3.281 996719NA
7 2 0.3175 1683NA
8 4 2.039 1961NA
9 6 2.039 1961NA
10 8 5.88 212NA
11 10 7.837 2163NA
12 15 0.0128 2237NA
13 20 0.0178 2286NA

31. 31
CONCLUSION
In this lab we absorbedthat During forward bias, the diode conducts
voltage-enhancing current. During reversing bias, the diode does not
conduct an increase in voltage (break down usually results in damage
of diode).

32. 32
LAB SESSION 06
OBJECTIVE
 To study the half-wave rectifier.
APPARATUS
 Transformer/ Function Generator
 Digital Multi-meter (DMM)
 Oscilloscope
 Diode : Silicon (D1N4002 or equivalent)
 Resistors: 2.2kΩ, 3.3kΩ
BASIC THEORY
The rectifier circuit converts the AC voltage furnished by the utilities company into the DC
voltage required to operate electronic equipment. Many common electrical products use voltages
provided by a rectifier. Without the rectifier to convert the AC voltage to the DC voltage required to
operate these electrical units, it would be virtually impossible to have the conveniences that we enjoy
today. A television without a rectifier would require several extremely large batteries. These batteries
would have to be large because of the current that is required. In other words, a television without the
rectifier would be so large that it would occupy an entire room. The rectifier is the heart of the
electronic unit.
Introduction
A rectifier system can be divided into five sections, each performing a separate function. Figure
1 is a block diagram of a rectifier system. This lesson deals with the input, rectifier, and filter sections.
Input Block
The input block consists of a transformer, normally a power transformer that receives
the AC input signal from some power source. The transformer transfers the electrical energy
received to the rectifier section by electromagnetic induction or mutual inductance. The
transformer performs the transfer of energy without any change in frequency, but it is able to

33. 33
change the voltage and current from the input source to the voltage and current required by
the rectifier section. The phase relationship of the current in the secondary of the transformer
is dependent upon the phase of the voltage in the primary winding and the direction of the
winding in the secondary. If the secondary windings are wound in the same direction as the
primary windings, the phase between the input signal and the output signal will be the same.
If the secondary windings are wound in the opposite direction of the primary windings, the
phase between the input signal and the output signal will be 180 degrees out of phase. The
schematic drawings of a transformer indicate the phase relationship between the primary and
secondary with the use of dots. The dots on a schematic diagram indicate which windings are
in phase. Figure 2 illustrates this relationship.
RectifierBlock
The rectifier circuit is the most important part in the rectifier system. The rectifier circuit
converts the AC waveform from the input block into a pulsating DC waveform. One of several
different rectifier circuits may be utilized to perform this function. These circuits are the half-wave
rectifier, the full-wave rectifier, the full-wave bridge rectifier, and the voltage doubler.
Half-Wave Rectifier
Figure 3 shows the schematic diagram for a half-wave rectifier. The half-wave rectifier is the
simplest type of rectifier; it consists of only one diode. For explanation purposes, a load resistance
must be placed in the circuit to complete the path for current flow and to develop the output signal.
The half-wave rectifier in Figure 3 is a positive half-wave rectifier. It is called a positive half-
wave rectifier because it only uses the positive portion of the input sine wave and produces a positive
pulsating DC signal. During the positive alternation of the input voltage, the positive alternation of the
sine wave causes the anode of the diode to become positive with respect to the cathode. The diode is
now forward-biased and will conduct. Current will flow from the negative side of the transformer
secondary, through the load resistor, through the diode, to the positive side of the transformer

34. 34
secondary. This path for current flow will exist during the complete positive alternation of the input
waveform because the diode will remain forward-biased as long as the positive signal is applied to the
anode. The resulting output of the rectifier will be developed across the load resistor and will be a
positive pulse very similar to the positive alternation of the input waveform. Figure 4 illustrates the
output waveform across the load resistor. During the negative alternation of the input sine wave, the
anode is negative with respect to the cathode and the diode will become reverse-biased. As long as this
condition exists, no current will flow in the circuit and an output signal cannot be developed across the
load. The circuit gives the appearance of producing a series of positive pulses. A negative half-wave
rectifier operates very similar to a positive half-wave rectifier, except the output will be a series of
negative pulses. (Refer to Figure 5.)
During the positive alternation, the diode is reverse-biased; no current will flow through the
circuit, and no signal will be developed across the load. This condition will exist any time a positive
alternation is present on the cathode. When the negative alternation is present on the cathode, the diode
is forward-biased; current flows from the negative side of the secondary through the diode, through the
load resistance, to the positive side of the secondary. This condition allows a negative pulse to be
developed across the load resistance and continues until the negative cycle is removed from the
cathode. The output of a negative half-wave rectifier will be a series of negative pulses. The amplitude
of the output is approximately the same as the peak voltage of the input signal if measured with the
oscilloscope. If a multimeter is used to measure the pulsating DC voltage, it will indicate the average
voltage. The average voltage of a sine wave is zero volts; however, if the negative portion of a sine
wave is clipped off, the average value changes to some positive value. Since the waveform swings
positive but never goes negative, the average voltage will be positive. To determine the average value
of a pulsating DC signal using a half-wave rectifier, multiply the peak voltage by .318.
PROCEDURE
Half Wave Rectification
1. Construct the circuit of Fig. 3.1. Set the supply to 6 V p-p sinusoidal wave with the frequency of

35. 35
1 kHz. Put the oscilloscope probes at function generator and sketch the input
2. Put the oscilloscope probes across the resistor and sketch the output waveform obtained.
3. Observe the waveform and measure the voltages (Vp-p) across resistor.
4. Now With your DMM, measure the dc voltage (Vdc) across the resistor.
5. Connect the Oscilloscope to display both the input voltage and the voltage across the load.
6. Compare the two waveforms and determine at which time the diode conducts.
7. Reverse the diode of circuit of Fig. 3.1. Sketch the output waveform across the resistor
RESULT

36. 36
ReversedBias Diode
The Full-Wave Bridge Rectifier
A basic full-wave bridge rectifier is illustrated in Figure 1.
A full wave bridge rectifier has one advantage over the conventional full-wave rectifier: the
amplitude of the output signal. The frequency of the positive pulses will be the same in either rectifier.
When the output signal is taken from a bridge rectifier, it is taken across the entire potential of the
transformer; thus, the output signal will be twice the amplitude of a conventional full-wave rectifier.
For the first half cycle of a bridge rectifier, refer to Figure 2.
During the first half cycle of the input signal, a positive potential is felt at Point A and a
negative potential is felt at Point B. Under this condition, a positive potential is felt on the anode of
CR3 and on the cathode of CR4. CR3 will be forward-biased, while CR4 will be reverse-biased. Also,

37. 37
a negative potential will be placed on the cathode of CR1 and the anode of CR2. CR1 will be forward-
biased, while CR2 will be reverse biased. With CR1 and CR3 forward-biased, a path for current flow
has been developed. The current will flow from the lower side of the transformer to Point D. CR1 is
forward-biased, so current will flow through CR1 to Point E, from Point E to the bottom of the load
resistor, and up to Point F. R3 is forward-biased, so current will flow through CR3, to Point C, and to
Point A. The difference of potential across the secondary of the transformer causes the current to flow.
Diodes CR1 and CR3 are forward-biased, so very little resistance is offered to the current flow by
these components. Also, the resistance of the transformer is very small, so approximately all the
applied potential will be developed across the load resistor. If the potential from Point A to Point B of
the transformer is 24 volts, the output developed across the load resistor will be a positive pulse
approximately 24 volts in amplitude.
When the next alternation of the input is felt (Figure 3), the potential across the transformer
reverses polarity. Now, a negative potential is felt at Point A and a positive potential is felt at Point B.
With a negative felt at Point C, CR4 will have a negative on the cathode and CR3 will have a negative
on the anode. A positive at Point D will be felt on the anode of CR2 and the cathode of CR1. CR4 and
CR2 will be forward-biased and will create a path for current flow. CR1 and CR3 will be reverse-
biased, so no current will flow. The path for current flow is from Point A to Point C, through CR4 to
Point E, to the lower side of the load resistor, through the load resistor to Point F, through CR2 to Point
D, and to the lower side of T1. Current flows because of the full potential being present across the
entire transformer; therefore, the current through the load resistor will develop the complete voltage
potential. The frequency of the output pulses will be twice that of the input pulses because both cycles
of the input AC voltage are being used to produce an output. The average value Vm of the rectified
voltage is:
PROCEDURE:
Full-Wave Rectification

38. 38
a) Construct the circuit of Fig. 3.2. Set the supply to 6 V p-p with the frequency of 1 kHz. Put the
oscilloscope probes at function generator (trainer board) and sketch the input waveform
obtained.
b) Put the oscilloscope probes across the resistor and sketch the output waveform obtained.
Measure and record the DC level of the output voltage using the DMM.
Figure 3.2
c) Replace diodes D3 and D4 of circuit of Fig. 3.2 by 2.2 kΩ. Draw the output waveform across
the resistor. Measure and record the DC level of the output voltage.
d) What is the major effect of replacing the two diodes (D3 and D4) with resistors?
OBSERVATION
Results and Calculations
1. Input waveform, Vi
2. Output waveform, Vo

39. 39
DC level of Vo (measured) = __________

40. 40
Half Wave Rectification
Result
Input wave form and out put wave form

41. 41