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.
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.
You should remember from our previous work that the diode has the following characteristic, i.e. only allows current to flow in one direction.
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. 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.
There are a couple of things to notice from the graph:
- Thenegative part of the a.c. graph has been removed.
- The voltage across the resistor is now a variable voltage d.c. signal.
- The peak voltage across the resistor is 0.7V less than peak of the input signal due to the voltage drop across the diode.
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 resistorby 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.
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.
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:
- 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.
- The voltage across the resistor is now a variable voltage d.c. signal.
- 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.
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.
- 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.
- 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.
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 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.
ii)Calculate the peak value of the output voltage.
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iii)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.
ii)Calculate the peak value of the output voltage.
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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.
Solutions to Student Exercise 1.
1.i)
ii)Input Voltage = 10V rms
Peak voltage =
Peak Output voltage =
iii)
2.i)
ii)Input Voltage = 12.75V rms
Peak voltage =
Peak Output voltage =
iii)
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.………………………………………………………………………………………………………………
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