2.4.2 Rectification - WJEC

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Topic
2.4.
2



Rectification
.


1

Learning Objectives:



At the end of this topic you will be able to;




draw and understand the use of diodes in half wave and full wave
bridge rectifiers;



calculate the peak value of the output voltage of half wave and full
wave rectifiers given the rms inp
ut voltage.



Module ET
2


Electronic Circuits and Components
.


2

Rectification
.


In the previous section we discussed the use of alternating current to form
the basis of a power supply for electronic components
.

In that section we
learnt the difference between peak and rms voltage. At the end of the
sectio
n we posed an interesting problem which was that if an alternating
current supply is to be used to power modern electronic circuits then we must
have a way of changing a.c. into d.c.


There are a couple of stages in the conversion process, and we will con
sider
the first of these in this topic


the process of rectification. To achieve this
we will need to use one of the components we met in Topic 2.3
, and this is the
silicon diode.












You should remember from our previous work that the diode has t
he following
characteristic, i.e. only allows current to flow in one direction.







Cath
ode

Anode

Conventional current flows in this direction.


Topic
2.4.
2



Rectification
.


3

Now we will consider what happens when an a.c. source is applied to a silicon
diode.




The graph from the oscilloscope below shows the effect of diode on the a.c.
v
oltage. The
blue

trace, shows the output from the
step down transformer or
a.c. voltage source, and the
red

trace shows the output after the diode.



There are a couple of things to notice from the graph:


i.

T
he

negative part of the a.c. graph has be
en removed
.

ii.

The voltage across the resistor is now a variable voltage d.c. signal.

iii.

The peak voltage across the resistor is 0.7V less than peak of the input
signal due to the voltage drop across the diode.

i

i
i

i
ii

This symbol represents a
transformer


a device for

changing
the dangerous high voltage

mains a.c. into low voltage a.c.


Module ET
2


Electronic Circuits and Components
.


4

The process of changing a.c. into d.c. is called
re
ctification
.
The graph
shows
that we have created a variable voltage d.c. output from an a.c. source.

Unfortunately this method of rectification wastes 50% of the energy from
the a.c. source because the negative half cycle is completely block
ed

from
the l
oad

resistor

by the diode.

This particular circuit is called a
half
-
wave
rectifier
.


A much

improved version involves
3

extra diodes arranged in an
un
usual
pattern

called a
bridge rectifier

as shown below.











Consider the flow of current during ea
ch half cycle of an a.c. input using the
diagrams below:



First Half Cycle






Second Half Cycle


A careful examination of the current flowing through the
load
resistor, shows
that current flows in both half cycles of the a.c. input. The current also
flows
in the same direction, i.e.
we have achieved a

variable voltage d.c. output once
again.

Output

Voltage

Input

Voltage

+




_



_





+





Topic
2.4.
2



Rectification
.


5

If we now consider
the
bridge rectifier in a circuit, and monitor the output
across the
load
resistor as before then the circuit and oscilloscope trace will
look
like those shown below.





The
blue

trace shows the output of the
transformer or
a.c. source, and the
red

trace shows the voltage across the resistor. There are a couple of things
to notice from the graph:


i.

The negative part of the a.c. graph has been f
lipped to provide a second
positive pulse within the same cycle
, called
full
-
wave rectification
.

ii.

The voltage across the resistor is now a variable voltage d.c. signal.

iii.

The peak voltage across the resistor is 1.4V less than peak of the input
signal due to t
he voltage drop across the two conducting diodes in the
bridge rectifier.


Module ET
2


Electronic Circuits and Components
.


6

The process of rectification is the first stage of converting an a.c. source
into a d.c. source suitable for operating electronic circuits
. The output
produced by the half
-
wave recti
fier and full
-
wave rectifier are both
unsuitable for electronic circuits because of the ‘pulsing’ nature of the
output.


We can use the work from our previous topic to determine the peak value of
any rms a.c. input voltage. The peak output voltage will th
en depend on
whether the rectification method is
half
-
wave

or
full
-
wave
.


i.

If
half
-
wave

rectification is used then the peak output voltage value
will be 0.7V less than the peak a.c. voltage due to a single diode being
used.

ii.

If
full
-
wave

rectification is us
ed then the peak output voltage
value
will
be 1.4V
less
than the peak a.c. voltage due to two 0.7V diode drops in
the bridge rectifier.


Clearly we have not
achieved
a suitable d.c. supply for electronic circuits yet,
but we have completed everything need
ed for this particular section


let’s
look at a couple of examples before moving on.


Example:

A 6V rms a.c. source is half wave rectified, and connected to a
1kΩ
load
resistor.



i)

Calculate the peak
value

of the output voltage.



ii)

Draw a sketch graph of the input voltage and
output voltage:






Label all important values.


Topic
2.4.
2



Rectification
.


7

Solution
:


i)

Input Voltage = 6V rms





Peak voltage =






Peak Output voltage =



ii)

Draw a sketch graph of the input voltage and output voltage on
the grid below:




Now here’s
a cou
ple for you to do
.


Student Exercise 1.


1
.

A 10V rms a.c. source

from a transformer

is half
-
wave

rectified, and
connected to a 2.2
k
Ω resistor.



i)

Draw a circuit diagram of this arrangement.


-
8.5V


Voltage

time

8.5V

7.8V


Module ET
2


Electronic Circuits and Components
.


8


ii)

Calculate the peak value of the output voltage.




.............................................................................................................................




.......
......................................................................................................................




.............................................................................................................................




.....
........................................................................................................................



i
i
i)

Draw a sketch graph of the input voltage and output voltage on
the grid below
,

label all important values
:
























2.

A 12.75V rms a.c. source
from a transformer
is full
-
wave rectified, and
connected to a 3.9kΩ
load
resistor.



i)

Draw a circuit diagram of this arrangement.


time

Voltage


Topic
2.4.
2



Rectification
.


9


ii)

Calculate the peak value of the output voltage.




...............................
..............................................................................................




.............................................................................................................................




.............................
................................................................................................




.............................................................................................................................



iii)

Draw a sketch graph of
the input voltage and output voltage on
the grid below,

label all important values:

























No examination style questions have been set in this topic as they are
integral to longer questions on power supplies, which we are not yet in a

position to answer, so time to move on to topic 2.4.
3



Capacitive Smoothing
.



time

Voltage


Module ET
2


Electronic Circuits and Components
.


10

Solutions to Student Exercise 1.


1.

i)





ii)

Input Voltage = 10V rms





Peak voltage =






Peak Output voltage =



ii
i)






14.1
V

13.4
V

-
14.1
V


time

Voltage


Topic
2.4.
2



Rectification
.


11

2.

i)





ii)

Input Voltage = 12.75V rms





Peak voltage =






Peak Output voltage =



iii)







1
8
V

16.6
V

-
1
8
V


time

Voltage


Module ET
2


Electronic Circuits and Components
.


12

Self Evaluation Review



Learning Objectives

My personal review of these objectives:







Draw and understand the use of
diodes in half wave and full wave
bridge rectifiers;




Calculate the peak value of the
output voltage of half wave and full
wave rectifiers given the rms input
voltage.






Targets:

1.

…………………
…………
…………………………………………………………………………………





………………………………………………………………………………………………………………


2.

………………………………………………………………………………………………………………


……………………………………………………

………………………………………………………