DC Power Supplies

Rectification, filtering and regulation

EE320 CBU SE 1
Half-Wave Rectification

EE320 CBU SE 2
The Half-Wave Rectifier circuit
• Figure a shows a half-wave rectifier circuit.
The ac source produces a sinusoidal voltage.
Assuming an ideal diode, the positive half-
cycle of source voltage will forward-bias the
diode.
• Since the switch is closed, as shown in Fig. b,
the positive half-cycle of source voltage will
appear across the load resistor. On the
negative half-cycle, the diode is reverse
biased.
• In this case, the ideal diode will appear as an
open switch, as shown in Fig. c, and no
voltage appears across the load resistor.

EE320 CBU SE 3
Waveform Analysis

EE320 CBU SE 4
• Figure (a) shows a graphical representation of the input voltage waveform. It
is a sine wave with an instantaneous value of vin and a peak value of Vp(in).
A pure sinusoid like this has an average value of zero over one cycle because
each instantaneous voltage has an equal and opposite voltage half a cycle
later. If you measure this voltage with a dc voltmeter, you will get a reading
of zero because a dc voltmeter indicates the average value.
• In the half-wave rectifier of Figure (b), the diode is conducting during the
positive half-cycles but is nonconducting during the negative half-cycles.
Because of this, the circuit clips off the negative half-cycles, as shown in
Figure (c). We call a waveform like this a half-wave signal. This half-wave
voltage produces a unidirectional load current. This means that it flows in
only one direction. If the diode were reversed, it would become forward
biased when the input voltage was negative. As a result, the output pulses
would be negative.

EE320 CBU SE 5
• This is shown in Figure (d). Notice how the negative peaks are offset
from the positive peaks and follow the negative alternations of the
input voltage.
• A half-wave signal like the one in Figure (c) is a pulsating dc voltage
that increases to a maximum, decreases to zero, and then remains at
zero during the negative half-cycle. This is not the kind of dc voltage
we need for electronics equipment.
• What we need is a constant voltage, the same as that you get from a
battery. We therefore need to filter out the ripples.
Ideal half wave: Vp(out) = Vp(in)

EE320 CBU SE 6
DC-value of a half-wave signal
• The DC value is basically the average value of the signal. Take a graph
for example, the average is basically the area under the graph over
the entire time period.
• The area under a curve is found by finding the definite integral over a
period of time T.
• Let us find the integral of the half-wave rectified output shown in
figure c. This will give us the DC value which can be measured by a DC
voltmeter.

EE320 CBU SE 7
Derivation….
1 $%
𝑉!" = & 𝑉& 𝑆𝑖𝑛 𝜔𝑡 𝑑𝑡
2𝜋 #

𝑉&
𝑉!" = − cos 𝜋 − (− cos 0)
2𝜋

𝑉&
𝑉!" =
𝜋

Hence the dc value of a half-wave rectified output is 0.318Vp

EE320 CBU SE 8
Frequency and practical diode outputs
• The input frequency is the same as the output frequency.
• If we consider the voltage drop of the silicon diode at 0.7V
• 0.7 V will be clipped from the output waveform as we shall see from
diode clippers.
• We expect Vp(out) to be equal to Vp(in) – 0.7V
Let us see try a multisim example

EE320 CBU SE 9
Multisim Example
• The figure shows a half-wave rectifier that you can build on the lab
bench or on a computer screen with Multisim. An oscilloscope is
across the 1 kΩ. Set the oscilloscope’s vertical input coupling switch
or setting to dc. This will show us the half-wave load voltage. Also, a
multimeter is across the 1 kΩ to read the dc load voltage.
• Calculate the theoretical values of peak load voltage and the dc load
voltage. Then, compare these values to the readings on the
oscilloscope and the multimeter.
• Connect the simulation in multisim as shown in the next slide.

EE320 CBU SE 10
EE320 CBU SE 11
A few things to note;
• You know that the rms value of a sine wave is 0.707Vp.
• You then calculate the Vp from the give rms value of 10V
• From the ideal diode you will find that the dc value is 4.49V
• For a practical diode you will need to consider the voltage drop of
0.7V from Vp.
• You should be able to find a Vdc of 4.27V
• Try it out in calculation and also in the simulator then compare the
results.

EE320 CBU SE 12
The Transformer
• Line voltage is too high for most of the circuits used in electronics
equipment. This is why a transformer is commonly used in the power-
supply section of almost all electronics equipment. The transformer
steps the line voltage down to safer and lower levels that are more
suitable for use with diodes, transistors, and other semiconductor
devices.
• I expect that you might have come across transformer function at this
level but we will go through a few basic concepts. Consider the figure:

EE320 CBU SE 13
• In the figure we see that the line voltage is applied to the primary
winding of the transformer.
• The conversion from primary voltage to secondary voltage is based on
the turns ratio N1/N2
• Notice the two dots on the figure. If the dots are both on top as in the
figure, it means the primary and secondary voltages are in phase. If
the input has a positive peak then the output has a positive peak.
• If the dot appears on the ground side of the secondary then the
voltage will be 180 degrees out of phase.
'$
• From your experience, 𝑉$ = '( 𝑉1
• This equation is used for all kinds of voltages, but since ac sources are
almost always rms, we will use it for rms voltages.

EE320 CBU SE 14
Let us try an example….
What is the peak load voltage and dc load voltage shown in the figure?

EE320 CBU SE 15
Hints.
• First find V2 using the turns ratio (24V)
• Then find the peak voltage Vp from the rms value (34V)
• With an ideal diode, the peak load voltage is Vp(out) = Vp
• With a practical diode the peak load voltage is Vp(out) = Vp – 0.7V

)!
• The dc load voltage is %
(10.8)

EE320 CBU SE 16
Summary
• So far we have learnt how to use a single diode in a half-wave rectifier
giving us a dc content of 0.318Vp
• How about if could increase the dc content level?
• We also revised the operation of the transformer.

EE320 CBU SE 17
Full-Wave Rectifiers
Improving the DC content
Full-Wave Rectifier operation
• In the figure, the full-wave rectifier
acts like two back-to-back half-wave
rectifiers. Notice the grounded
center tap on the secondary of the
transformer.
• Because of the center tap, each HW
rectifier has a voltage of half the
secondary voltage.
• D1 conducts on the positive half
cycle and D2 conducts on the
negative half cycle.
• Hence, we get a rectified load
through both half cycles.
FW Rectifier Operation
cont.
• Let us analyze figure b and c. During the positive
half cycle, D1 will be forward biased and
conducting allowing a positive voltage drop across
the load RL.
• During the negative half cycle D1 will be open and
D2 will be forward biased and conducting again
enabling a positive voltage drop across the load.
• During both half-cycles, the load voltage has the
same polarity and the load current is in the same
direction.
 FW rectifier cont.
• The circuit is known as a full-wave rectifier because it has changed the AC input voltage to a
pulsating DC output voltage present in both positive and negative cycles. We will now look at some
of the properties of this pulsating output voltage.
Properties of FW rectifier output
a) DC or average value
We can find the DC content in the output waveform by integrating it
over the period.
1 %
𝐹𝑢𝑙𝑙 − 𝑤𝑎𝑣𝑒 𝑉!" = & 𝑉𝑝 𝑆𝑖𝑛𝑤𝑡 𝑑𝑡
𝜋 #
𝑉&
= −𝐶𝑜𝑠 𝜋 + 𝐶𝑜𝑠 0
𝜋
2𝑉&
= ≈ 0.636 𝑉𝑝
𝜋
We see that the DC content has increased from about 36.3% to 63.6%
of the peak voltage Vp.
Properties of FW rectifier output cont…
(b) Output frequency
If we look and compare the HW output waveform to the FW output
waveform we notice something unusual. The period is halved in the FW
output. This means if the period of a HW is T then for a FW it is 0.5T.
What does this mean for the output frequency? Since frequency is the
inverse of period, we expect the frequency to double in FW rectifier
applications. Thus a full-wave has twice as many cycles as a sine-wave
input has.
𝑓*+, = 2 𝑓-.
FW rectifier with practical diode
• With practical applications we
expect the 0.7 V to be
subtracted from the peak output
voltage. Then the rest follows as
normal.
• Let us try this in multisim.
Connect the circuit as shown in
the figure.
• Set the oscilloscope such that
channel 1 shows the primary
voltage and Channel 2 shows
the voltage across the load.
Multisim output • We can compare with the following
theoretical analysis:
#!"# "'$
• 𝑉!" = $.&$& = $.&$& = 170 𝑉
($
• 𝑉!' = 𝑉 = 17 𝑉
(% !"
• The full-wave rectifier acts like two back-to-
back half-wave rectifiers. Because of the
center tap, the input voltage to each half-
wave rectifier is only half the secondary
voltage:
• 𝑉! )* = 0.5 𝑉!' = 8.5 𝑉
• For ideal diode 𝑉!+,- = 8.5 𝑉
• For practical diodes 𝑉!+,- = 8.5 − 0.7 =
7.8 𝑉
• Depending on what you are analysing you can
go ahead and find the DC load voltage by
multiplying each term by 0.636.
THE BRIDGE RECTIFIER

• Consider the bridge rectifier

shown in the figure.
• The bridge rectifier is similar to
a full-wave rectifier because it
produces a full-wave output
voltage.
• Diodes D1 and D2 conduct on
the positive half-cycle, and D3
and D4 conduct on the negative
half-cycle. As a result, the
rectified load current flows
during both half-cycles.
Bridge rectifier cont….
• During the positive half
cycle, diodes D1 and D2 are
forward–biased and
conducting, producing a
positive load voltage.
• You can visualize D2 to be
shorted and we notice that
the circuit becomes a half-
wave rectifier.
Bridge rectifier cont….
• During the positive half
cycle, diodes D3 and D3 are
forward–biased and
conducting, producing a
positive load voltage.
• You can visualize D3 to be
shorted and we notice that
the circuit becomes a half-
wave rectifier.
 Bridge rectifier
cont….
• We observe that the bridge rectifier also acts
like 2 back-to-back HW rectifiers.
• During both half-cycles, the load voltage has
the same polarity and the load current is in the
same direction. The circuit has changed the ac
input voltage to the pulsating dc output voltage
shown in figure.
• One advantage of the bridge rectifier over the
FW rectifier is that the entire secondary voltage
can be used as input to the rectifier.
• Which means for the same output DC voltage
a bridge rectifier can use a transformer with a
lower turns ratio than that of FW.
• This makes the design lighter and cost less.
DC Value and Output Frequency
• We notice that the bridge and the FW rectifier give the same full-wave
unidirectional output. The DC component and the frequency are the same.
#$!
• Therefore; 𝑉!" = and 𝑓&'( = 2𝑓)*
%
• The average value is 63.6 percent of the peak value, and the output
frequency is 120 Hz, given a line frequency of 60 Hz.
• One advantage of a bridge rectifier is that all the secondary voltage is used
as the input to the rectifier.
• Given the same transformer, we get twice as much peak voltage and twice
as much dc voltage with a bridge rectifier as with a full-wave rectifier.
• Doubling the dc output voltage compensates for having to use two extra
diodes.
 Considering the practical diode…
• If we have to consider the 0.7V voltage drop across the diode, we get
twice the diode voltage drop for a bridge rectifier unlike the center-
tapped FW.
• Therefore the Vp(out)= Vp(in) – 1.4V
• In your own time try out the multisim representation given below:
It is the same supply and transformer as we
used in the center-tapped FW.
What do you now get for the Vdc??
Is it almost twice??
What do you get for Vp(out) for an ideal
diode?
What about Vp(out) for a practical diode?
What do you get if you use a 5:1 turns ratio
instead? What can you deduce from this?
Summary
Rectifier Filters
Ripple reduction towards pure DC

EE 220 CBU SE 33
Why filter?
• At this stage we now how to produce a unidirectional power supply
from a sinusoidal input.
• However, although the rectifier output has some level of DC content,
the existence of ripples in the output makes in unsuitable for use in
most DC equipment.
• The goal is to have a constant DC source similar to that provided by a
DC battery, not with fluctuations introduced by the ripple content.
• So we go a step further and introduce filters to remove most of the
ripple content.

EE 220 CBU SE 34
The Choke-Input
Filter
• This used to be a commonly used filter but
not anymore due to its cost, bulk and weight.
However we will still look at it briefly.
• The figures show a typical choke input filter
and its equivalent ac circuit.
• The ac source induces a current in the
inductor, capacitor and resistor.
• This current depends on the inductive and
capacitive reactance and resistance.

EE 220 CBU SE 35
LC (Choke input) filter characteristics…..
• Inductive and capacitive reactances are given by the following
formulae respectively:
1
𝑋8 = 2𝜋𝑓𝐿 𝑎𝑛𝑑 𝑋" =
2𝜋𝑓𝐶
• We notice that the primary characteristic of an inductor is to oppose
a change in current. Therefore, a choke significantly reduced the ac
load current to a very small value.
• The first requirement to a properly designed LC filter is to ensure Xc is
much smaller than 𝑅8 this will enable us to ignore the effects of it on
the load.
EE 220 CBU SE 36
LC (Choke input) filter characteristics…..
• Once this requirement is met we achieve the ac equivalent circuit
shown previously.
• The other requirement is that 𝑋L should be much larger than XC. This
allows the AC output voltage to approach zero and input frequency.
• This means at DC when the frequency is 0Hz, the choke assumes a
short circuit and the capacitor assumes an open circuit hence passing
the DC current with minimal loss.
• Let us perform a mathematical analysis to see how much AC content
is removed by this design.

EE 220 CBU SE 37
LC filter design analysis…..
• Take. 𝑉-. = 15 𝑉 , 𝑋8 = 10 𝐾Ω. 𝑎𝑛𝑑 𝑋9 = 100 Ω, we can use
voltage divider rule to find the output voltage across the load.

100
𝑉*+, ≈ × 15 𝑉 = 0.15𝑉
10000
• In the above example we notice that the AC content is reduced by a
factor of 100.
• How then can we make use of this design in reducing the AC content
in the rectifier output?

EE 220 CBU SE 38
Filtering the output of a rectifier
• The output of a rectifier at this stage
can be looked at as having two
sources: A DC component (average
value) and an AC source (the
fluctuating part).
• We can thus use the superposition
theorem to find effect of a choke input
filter. This is shown in the figure.
• Almost all the DC component is
passed on to the load resistor and
almost all of the AC component is
blocked.
• In this way, we get an almost perfect
DC voltage. What remains is a small
AC ripple. A ripple is the small AC
component remaining in the output.

EE 220 CBU SE 39
The problem with a choke input filter
• A power supply is the circuit inside electronics equipment that converts the
ac input voltage to an almost perfect dc output voltage. It includes a
rectifier and a filter.
• The trend nowadays is toward low-voltage, high-current power supplies.
• Because line frequency is only 50 Hz, large inductances have to be used to
get enough reactance for adequate filtering.
• But large inductors have large winding resistances, which create a serious
design problem with large load currents. In other words, too much dc
voltage is dropped across the choke resistance. This also makes them too
bulky for use in modern electronic systems.
• However, the choke input filter is still useful for use in switching regulators
using in computers. They support much higher frequencies than 50Hz. They
usually range beyond 20kHz.
EE 220 CBU SE 40
Capacitor Input filter
• This produces a DC output
voltage equal to peak value of
the rectified voltage. This one is
most commonly used one.
• Let us understand a simple circuit
with a diode and capacitor to get
the picture of how the capacitor
input filter works.

EE 220 CBU SE 41
Sequence of operation……
• Initially, the capacitor is uncharged. During the first quarter-cycle of input sine
wave, the diode is forward biased. Since it ideally acts like a closed switch, the
capacitor charges, and its voltage equals the source voltage at each instant of the
first quarter-cycle.
• The charging continues until the input reaches its maximum value. At this point,
the capacitor voltage equals Vp.
• After the input voltage reaches the peak, it starts to decrease. As soon as the
input voltage is less than Vp, the diode turns off. In this case, it acts like the open
switch.
• During the remaining cycles, the capacitor stays fully charged and the diode
remains open. For a load less situation, this means the output remains constant
at Vp.
• This peak voltage is constant, the perfect dc voltage we need for electronics
equipment. There’s only one problem: There is no load resistor.

EE 220 CBU SE 42
What happens when we
add the load resistor?
• For the capacitor-input filter to be useful, we
need to connect a load resistor across the
capacitor, as shown in figure a.
• As long as the RLC time constant is much
greater than the period, the capacitor
remains almost fully charged and the load
voltage is approximately Vp.
• The only deviation from a perfect dc voltage
is the small ripple seen in figure b. The
smaller the peak-to-peak value of this ripple,
the more closely the output approaches a
perfect dc voltage.

EE 220 CBU SE 43
Effect of load resistor continued….
• Between peaks, the diode is off and the capacitor discharges through
the load resistor. In other words, the capacitor supplies the load
current.
• Since the capacitor discharges only slightly between peaks, the peak-
to-peak ripple is small.
• When the next peak arrives, the diode conducts briefly and recharges
the capacitor to the peak value.

EE 220 CBU SE 44
 Ripple from a full-wave rectified output
• When we connect a full-wave or bridge rectifier, the peak to peak ripple is cut in half.
• When the full-wave voltage is applied to the RC circuit, the capacitor discharges for only half as long.
• Therefore, the peak to peak ripple is reduced to half what we get with the half wave. This is shown in the
figure.

EE 220 CBU SE 45
Determining the ripple content…
• We are going to estimate the peak to peak ripple of any capacitor
input filter:
𝐼
𝑉: =
𝑓𝐶
Where 𝑉: = peak to peak ripple voltage
𝐼 = 𝑑𝑐 𝑙𝑜𝑎𝑑 𝑐𝑢𝑟𝑟𝑒𝑛𝑡
𝑓 = 𝑟𝑖𝑝𝑝𝑙𝑒 𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦 𝑡𝑤𝑖𝑐𝑒 𝑓𝑜𝑟 𝑎 𝑓𝑢𝑙𝑙 − 𝑤𝑎𝑣𝑒 𝑟𝑒𝑐𝑡𝑖𝑓𝑖𝑒𝑟
𝐶 = 𝑐𝑎𝑝𝑎𝑐𝑖𝑡𝑎𝑛𝑐𝑒
• For this course, it is sufficient just to use the approximation. In
industrial practice and projects you can use multisim to get a more
accurate answer.
EE 220 CBU SE 46
Tutorial Question……
• What is the DC load voltage and ripple in the figure?

EE 220 CBU SE 47
Tutorial solution…….
• First we find the rms secondary voltage
120
𝑉# = = 24 𝑉
5
• Next we find the peak secondary voltage by converting from the rms value
24
𝑉+ = = 34 V
0.707
Assuming an ideal diode and small ripple, the DC load voltage is
𝑉, = 34 𝑉
• Next we need to calculate the ripple that requires the load current first.
-
We then go ahead and use our formula. ./ you can try that!
• A final point about the circuit. The plus and minus signs on the filter
capacitor indicates a polarized capacitor, one whose plus side must be
connected to the positive rectifier output. Always watch for this polarity!
EE 220 CBU SE 48
Self Exercise….
• What is the DC load voltage and ripple in the following figure?

EE 220 CBU SE 49
Self Exercise….
• What is the DC load voltage and ripple in the following figure?

EE 220 CBU SE 50
Some thinking…..
• Of the three configurations we have just looked at, which one would
be the best choice for use generally?

EE 220 CBU SE 51
Lab exercise
Play with the following circuit is multisim. Observe the DC load voltage
and ripple content. Compare with calculations

EE 220 CBU SE 52
Peak Inverse Voltage
• The peak inverse voltage (PIV) is the maximum voltage across a
reverse bias diode of a rectifier.
• This voltage must be less than the breakdown voltage of the diode;
otherwise, the diode will be destroyed.
• It mainly depends on the type of rectifier and filter. Worse cases occur
with capacitor filters as they add extra reverse bias voltage.
• PIV on spec sheets has different names including PRV, VB, VBR, VR(max)
etc.

EE 220 CBU SE 53
PIVs for different rectifier configurations
(a) HW with Capacitor input filter
𝑃𝐼𝑉 = 2𝑉𝑝
(b) Full-wave with capacitor input filter
𝑃𝐼𝑉 = 𝑉;
(c) Bridge rectifier with capacitor input
filter
𝑃𝐼𝑉 = 𝑉&
In this case the bridge rectifier has the
lowest PIV. The full wave rectifier needs
twice the secondary voltage for the same
load voltage.
EE 220 CBU SE 54
Exercise
• What is the PIV of a bridge rectifier with capacitor input filter that has
a turns ratio of 8:1? The AC supply is 120V 60Hz.
• If the diodes being used have a breakdown voltage of 50 V. Is it safe to
use it in such a circuit?

EE 220 CBU SE 55
SUMMARY TABLES……

EE 220 CBU SE 56
General block diagram of a power supply…

EE 220 CBU SE 57
Voltage Regulation
The Zener Regulator
The Zener diode
• Unlike other diodes, the Zener diode is optimized to work in the
reverse bias region or the breakdown region
• The Zener is the what makes voltage regulation possible.
• A voltage regulator ensures the load voltage is almost constant
despite large changes in line voltage and load resistance.
• To better understand this, we have to look at the current-voltage
graph of the Zener diode.
Zener schematic symbols and graph
Let us analyse the graph……
• Manufactures vary the breakdown voltage from 2 to over a 1000V.
• It can work in any mode, forward, leakage and breakdown
• The voltage is almost constant at breakdown, from the graph we can
see that the reverse current is varying from IZT to IZM.
• The Zener diode gets destroyed only when the reverse current is
greater than IZM.
• The slope of the graph in the breakdown region means it is the
inverse of the Zener resistance.
• Having an almost vertical graph means an increase in reverse current
will only cause a slight increase in reverse voltage.
The Zener Regulator
• A regulator, the Zener diode maintains a constant output voltage even
as current through it changes.
• For normal operation you have to reverse bias the Zener diode and
ensure that the source voltage is greater than the breakdown voltage
Vz.
• A series resistance Rs is always used to limit the Zener current to less
than its maximum current rating. Again this is to prevent burning out
the Zener diode.
• How does this look like in a circuit? We will have a look at it…..
 The Zener Regulator circuit

The first circuit shows how to bias the Zener diode into the breakdown region. The second one shows a voltage regulator
circuit after rectification.
Because of the series resistor, the output is less than the output of the power supply but varies fairly constant.
Ideal Zener diode
• The current Is can be calculated via Ohms law.
𝑉< − 𝑉=
𝐼< =
𝑅<
• In this course, we will consider an ideal Zener diode. Which means Vz
will be constant even though the current changes.
• We ideally will ignore the Zener resistance. This means in our analysis
we can replace the Zener with a voltage source Vz.
• This is true as long as the diode remains within the breakdown region.
• Let us try an example
Example
• Suppose the Zener diode in figure (a) has a breakdown voltage of 10 V. What
are the minimum and maximum Zener currents? What is the Zener current
Is if Vin is 30V?
Working area
Note that:
• In a voltage regulator like in our example, the output voltage is held
constant at 10 V, despite the change in source voltage from 20 to 40
V. The larger source voltage produces more zener current, but the
output voltage holds rock-solid at 10 V.
What happens when we add a load?

The Zener diode operates in the breakdown region and holds the load
voltage constant.
AGAIN HERE, EVEN IF THE SOURCE VOLTAGE CHANGES OR THE LOAD
RESISTANCE VARIES, THE LOAD VOLTAGE WILL REMAIN FIXED AND
EQUAL TO THE ZENER VOLTAGE.
Ensuring breakdown operations in loaded
situations
• To push the Zener into breakdown, we have to reverse bias it. To do
this the EMF forcing the reverse-bias condition has to be greater than
Vz.
• In our loaded circuit the EMF, responsible for this is the Thevenin
voltage:
𝑅8
𝑉,> = 𝑉<
𝑅< + 𝑅8
Therefore, the Thevenin voltage has to be greater than Vz to push it
into breakdown operation.
In examples and questions for this course, always assume the Zener is
in breakdown unless otherwise stated.
Series Current
• Using ohms law, the current through the series resistor is
𝑉< − 𝑉@
𝐼? =
𝑅?

How about the current through the load?

Ideally, the load voltage equals the Zener voltage because the load
resistor is in parallel with the Zener diode. 𝑉8 = 𝑉@
)"
Therefore: 𝐼8 =
:"
How about the current going through the
Zener diode?
• This one is easy, you just use Kirchhoff’s current law:

𝐼< = 𝐼= + 𝐼8
• To find the current in the Zener diode simply make Iz subject
Summary of Steps
•Calculate the series current
•Calculate the load voltage
•Calculate the load current
•Calculate the Zener current
Exercise
• Analyse the circuit and determine if the Zener is in breakdown
• Calculate the Zener current?
What does the following circuit do?