Direct Current
(DC): It is characterised by a
uniform direction of flow and a steady Voltage level (amount) with respect to
time.
Figure 4.1: Shows direct current with Amplitude of Voltage
with respect to time.
Shown in the figure 4.1 is the direction of DC current with
its voltage amplitude sketched with respect to time. We can observe that the
voltage level is steady at any given time with a direct current flow. We can recall that current just a the
movement of electrons through a conductor, in DC, the electrons flow steadily
in a single direction (forward).
Alternating Current
(AC): On the contrary, AC is an electric current that reverses its
direction many at regular intervals with respect to time. The flow of charges
(i.e. electrons) periodically reverses direction. Also its voltage level
(amplitude) fluctuates from Peak positive
(+Vp) to zero to peak negative (-Vp).
(a) (b) (c)
Figure 4.2: AC Sine
Waveform.
Figure 4.2 (a) is an example of a typical sine waveform of
an AC, 4.2(b) depicts how the current in the AC changes direction with respect
to time, while figure 4.2(c) illustrates how the voltage amplitude changes from
+Vp to -Vp.
For an AC voltage;
Vin=
Vp Sin (ωt)
Vin is the
input voltage
Vp is the peak
voltage
ω = 2πf
The DC component of the signal Vdc = Vav = 0
(Since we have equal voltage amplitude pulses in the positive
and negative portion of the wave).
But the root-mean-square
voltage can be given by;
This is equal to the DC Voltage that delivers the same average power to a resistor as the periodic Voltage of an AC.
Recently, most electronics appliances we have today are
comprises of semiconductor devices (e.g. diodes, transistors, Field Effect
Transistors (FETs), e.t.c) which a preferred direction for current flow (DC)
but, alternating
current (AC) is supplied in the mains of our homes because it is easier
to transmit over a long distance and redistribute. These necessitate a means of
converting the supplied AC to DC in most of these devices for them to work
appropriately. The process of converting AC to DC is known as rectification and it is achieved in
electronics with an appropriate use of a diode.
The Transformer
A transformer is a very common magnetic structure found in
many applications. It is used to connect AC circuits to each other. It couples two
circuits together magnetically rather than through any direct connection. Its
main use is to raise or lower voltage and current between one circuit and the
other.
A transformer is a necessary component in all power supply;
it is applied in small power supply
circuit for example in you small devices like in figure 4.3, it also find
application in Electrical power system transmission and distribution, as shown
in figure 4.4.
Figure 4.3: Transformer in small power supply
Figure 4.4: Transformer application in electrical
power transmission and distribution
An Ideal Transformer consists of two
conducting coils wound on a common core, made of high grade iron with no
electrical connection between the coils; they are connected to each other through
magnetic
flux.
The arrangement of primary and secondary windings on the
transformer core is shown in figure 4.5 below. The voltage, current and flux
due to the current in the primary winding is also shown.
Figure 4.5
The voltage relationship in a transformer is given by;
V1 - is the voltage across the primary winding of
the transformer
V2
– is the voltage output from the secondary winding of the transformer
N1 - is the number of turns in the primary
winding
N2
– is the number of turns in the secondary winding
a = is
the ratio of N1:N2
This ration a, determines the amount of voltage change form
the primary to the secondary winding of the transformer.
In summary, an Ideal transformer divides a sinusoidal input
voltage by a factor of a and multiplies a sinusoidal
current by a to obtain secondary voltage and current.
The voltage and current are expressed in terms of their RMS
values.
The equivalent circuit of an ideal transformer can be drawn
as follows
Figure 4.6: Equivalent Circuit of an Ideal Transformer
If a <1, i.e. N1<N2
The output voltage is greater than the input voltage and the
transformer is called a step-up transformer
If a>1, i.e. N1>N2;
The output voltage is smaller than the input voltage and the
transformer is called a step-down transformer.
If a =1, i.e. N1=N2;
The output voltage is the same as the input voltage and the
transformer is called an isolation transformer. This is
applied in a very useful application where two circuits need to be electrically isolated from each other.
(a) (b) (c)
Figure 4.7: a: Step-up transformer, b: Step-down
transformer, c: Isolation transformer
There are some application in which the secondary winding is
tapped at two different points, giving rise to two output circuits, for example
in center-tapped
transformer which splits the secondary voltage into two equal voltages
as in figure 4.8.
Figure 4.8: Center-tapped transformer
Sample Question:
A transformer is required to deliver 1A current at 12V from
a 220V rms Supply voltage. The number of turns in the primary is 2000.
i. How many turns are
required in the secondary winding?
ii. What is the current
in the primary winding?
Half Wave Rectifier
Now, we’ve come to the most popular application of diodes. Rectification can be simply defined as
the conversion of alternating current
(AC) to direct current DC). The simplest kind of rectifier is the half wave rectifier. In this, only one
half of an AC waveform is allowed to pass through the load. We are familiar
with the fact that a diode only allow
current to flow when it is forward biased, if we forward bias a diode with an
AC, considering the waveform in figure 4.9 (a), when the AC voltage is in the portion
in blue color the polarity is reversed and the diode connection is reversed i.e. reversed bias. At this time there
is no current flow and the voltage amplitude will be zero since V=IR, in the output we can see only the black colored potion of the waveform. If the diode is reversed in the connection as in figure 4.9(b), only the blue portion of the waveform will be passed for the same reason as in figure 4.9(a)
Figure 4.9: Half wave rectifier
Sample Question 2:
Assume a 40:1 transformer with a 240V r.m.s input of 50Hz, determine
the peak voltage, the dc voltage and the voltage at the load.
Let's Go To the Virtual Laboratory and demonstrate Half-Wave Rectifier:
Lab 1
In the Lab 1, we can see the circuit contains a function generator used to generate AC of 220V, 6Hz sent as input to a transformer of ration 1:0.05 (shown as 50m). when we set the voltage VM1 as input to our oscilloscope with 10 Volts/div and 5ms/division. We observed the simulated output of the oscilloscope,
we can see that the peak voltage of the sine wave is around 1.1division of the oscilloscope, therefore the input Voltage from the secondary of the transformer is given by; 1.1 x 10 = 11V.
Proves:
Now in the Lab 2, we include a diode in the circuit and observe the waveform of the input voltage.
Lab 2
In the Lab 2, we have included the diode in the circuit diagram, observing the output voltage waveform in the simulated oscilloscope, we can see that the negative portion of the waveform has be removed, diode only passes the positive portion of the sine wave. It has been half rectified.
Let's Go To the Virtual Laboratory and demonstrate Half-Wave Rectifier:
Lab 1
In the Lab 1, we can see the circuit contains a function generator used to generate AC of 220V, 6Hz sent as input to a transformer of ration 1:0.05 (shown as 50m). when we set the voltage VM1 as input to our oscilloscope with 10 Volts/div and 5ms/division. We observed the simulated output of the oscilloscope,
we can see that the peak voltage of the sine wave is around 1.1division of the oscilloscope, therefore the input Voltage from the secondary of the transformer is given by; 1.1 x 10 = 11V.
Proves:
Now in the Lab 2, we include a diode in the circuit and observe the waveform of the input voltage.
Lab 2
In the Lab 2, we have included the diode in the circuit diagram, observing the output voltage waveform in the simulated oscilloscope, we can see that the negative portion of the waveform has be removed, diode only passes the positive portion of the sine wave. It has been half rectified.
Lab 3
we can observe that the diode now passes only the negative portion of the waveform, this obviates our earlier discussions, in this orientation of the diode, it is the negative portion of the input voltage that forward biases the diode. Thus passes only the negative portion of the waveform as shown in Lab 3.
See Simulation Video for half-wave rectifier Lab:
Full Wave rectifier using;
1. Center-tapped
Transformer:
Figure 4.10: Full
wave rectifier circuit with center tapped transformer
In this circuit, two diodes which are connected in such a
way that only one of them will conduct (i.e. forward biased) for each half of
the waveform cycle. A center tapped transformer has its secondary winding split
into two equal halves with common center taped connection.
The circuit configuration makes one of the diode to be
forward biased in each half of the cycle.
With a full wave rectifier, we have the following relations;
Lab 4: Full wave rectifier with center tapped transformerIn the Lab 4, we can see that the output voltage has a peak amplitude of about half the input at the secondary turn of the transformer. This is because the center tapped has halved the input into two equals, while the flow of current is in such a way that one diode conducts (i.e. forward biased) at a time.
See Simulation Lab Video for full wave rectifier with center tapped transformer:
2. Bridge Rectifier
Figure 4.11: Bridge
Rectifier
In the bridge rectifier circuit of figure 11, four diodes are configured in the design so that, two of the diodes will conduct electricity for each half of the AC waveform. figure 11(a) and 11(b) depict how current flows in each half of the AC waveform.
In the bridge rectifier circuit of figure 11, four diodes are configured in the design so that, two of the diodes will conduct electricity for each half of the AC waveform. figure 11(a) and 11(b) depict how current flows in each half of the AC waveform.
The advantage of a bridge rectifier is that, there is no
centre-tap and the entire secondary voltage can be used.
Lab 5: Bridge Rectifier Virtual Lab Simulation
See Lab 5 Video:
Sample Question 3:
Assume a 20:1 primary to secondary turns ratio transformer,
with a 220V rms input of 50Hz.
i.
Determine its secondary output.
ii.
The peak voltage, Vp and the effective DC
voltage Vdc
iii.
Also determine the actual voltage across the load using
a center-tap and a bridge rectifier.
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Electronic Components
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