Topic 2.4.2 – Rectification.
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 input voltage.
1
Module ET2
Electronic Circuits and Components.
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
section 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 consider
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.
Anode
Cathode
Conventional current flows in this direction.
You should remember from our previous work that the diode has the following
characteristic, i.e. only allows current to flow in one direction.
Silicon Diode Characteristic
18
16
14
12
Current (mA)
10
8
6
4
2
0
-5
2
-4
-3
-2
-1
0
-2
Voltage Applied (V)
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Topic 2.4.2 – Rectification.
Now we will consider what happens when an a.c. source is applied to a silicon
diode.
This symbol represents a
transformer – a device for changing
the dangerous high voltage
mains a.c. into low voltage a.c.
The graph from the oscilloscope below shows the effect of diode on the a.c.
voltage. 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.
ii
iii
i
There are a couple of things to notice from the graph:
i. The negative part of the a.c. graph has been 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.
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Module ET2
Electronic Circuits and Components.
The process of changing a.c. into d.c. is called rectification. 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 blocked from
the load resistor by the diode. This particular circuit is called a half-wave
rectifier.
A much improved version involves 3 extra diodes arranged in an unusual
pattern called a bridge rectifier as shown below.
Input
Voltage
Output
Voltage
Consider the flow of current during each 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.
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Topic 2.4.2 – Rectification.
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 flipped 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 the voltage drop across the two conducting diodes in the
bridge rectifier.
5
Module ET2
Electronic Circuits and Components.
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 rectifier 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 then 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 used 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 needed 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.
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Topic 2.4.2 – Rectification.
Solution:
i)
Input Voltage = 6V rms
Peak voltage =
2  Vrms  1.414  6  8.48V  8.5V
Peak Output voltage = 8.48  0.7  7.78V  7.8V
ii)
Draw a sketch graph of the input voltage and output voltage on
the grid below:
Voltage
8.5V
7.8V
time
-8.5V
Now here’s a couple 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 2.2kΩ resistor.
i)
Draw a circuit diagram of this arrangement.
7
Module ET2
Electronic Circuits and Components.
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:
Voltage
time
2.
A 12.75V rms a.c. source from a transformer is full-wave rectified, and
connected to a 3.9kΩ load resistor.
i)
8
Draw a circuit diagram of this arrangement.
Topic 2.4.2 – Rectification.
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:
Voltage
time
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.
9
Module ET2
Electronic Circuits and Components.
Solutions to Student Exercise 1.
1.
i)
ii)
Input Voltage = 10V rms
Peak voltage =
2  Vrms  1.414  10  14.14V  14.1V
Peak Output voltage = 14.14  0.7  13.44V  13.4V
iii)
Voltage
14.1V
13.4V
time
-14.1V
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Topic 2.4.2 – Rectification.
2.
i)
ii)
Input Voltage = 12.75V rms
Peak voltage =
2  Vrms  1.414  12.75  18.0285V  18V
Peak Output voltage = 18  1.4  16.6V
iii)
Voltage
18V
16.6V
time
-18V
11
Module ET2
Electronic Circuits and Components.
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.
………………………………………………………………………………………………………………
………………………………………………………………………………………………………………
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