chapter-5 diode circuits and power supplies
TRANSCRIPT
Diode Circuits and Power Supplies 1
Muhammed Riyas A.M,Assistant Professor,Dept. of ECE,MCET Pathanamthitta
Chapter-5
Diode Circuits and Power Supplies
Contents
Diode circuits: Series and parallel diode circuits, Clippers, Clampers, Voltage multipliers
Rectifiers: Half-wave and full wave (including bridge) rectifiers, Derivation of Vrms, Vdc, ripple factor,
peak inverse voltage, rectification efficiency in each case, capacitor filter, working and design of a
simple zener voltage regulator.
Power Supplies: Block diagram description of a DC Power supply, Principle of SMPS.
5.1 Introduction
Diode circuits to be considered perform functions such as rectification, clipping. and clamping.
These functions are possible only because of the nonlinear properties of the pn junction diode. The
conversion of an ac voltage to a dc voltage, such as for a dc power supply, is called rectification. Clipper
diode circuits clip portions of a signal that are above or below some reference level. Clamper circuits shift
the entire signal by some dc value, Zener diodes, which operate in the reverse-bias breakdown region, have
the advantage that the voltage across the diode in this region is nearly constant over a wide range of
currents. Such diodes are used in voltage reference or voltage regulator circuits.
5.2 Clippers
There are a variety of diode networks called clippers that have the ability to “clip” off a portion of the
input signal without distorting the remaining part of the alternating waveform. Depending on the orientation
of the diode, the positive or negative region of the input signal is “clipped” off. Often, dc battery is also used to
fix the clipping level. The input waveform can be clipped at different levels by simply changing the battery
voltage and by interchanging the position of various elements.
Clipper circuits are used in radars and digital computers etc. when it is desired to remove signal
voltages above or below a specified voltage level. Another application is in radio receivers for communication
circuits where noise pulses that rise well above the signal amplitude are clipped down to the desired level.
There are two general categories of clippers: series and parallel. The series configuration is defined as
one where the diode is in series with the load, while the parallel variety has the diode in a branch parallel to
the load. Here we assume that the diode is ideal. This simply means that the turn-on voltage of the diode is
any voltage greater than 0 V, and the diode resistance is 0 Ω.
5.2.1 Series Clippers
When the diode is connected in series with the load to form a clipper, the configuration is called
series clippers. Series clippers can be used to clip either the positive or negative portions of the input signal.
Diode Circuits and Power Supplies 2
Muhammed Riyas A.M,Assistant Professor,Dept. of ECE,MCET Pathanamthitta
Series Negative Clipper
During the positive half cycle of the input signal, the diode is turned on and it acts as closed switch. Hence no voltage drops across the diode. All the input voltage appears across the load, and hence the ouptut signal voltage is same as input voltage.
During negative half cycle of the input voltage, the diode is reverse biased and acts as open switch. In such a case, no current flows through the diode and the load, and hence the output voltage across the load is zero.
Fig: Series Negative Clipper
Biased Series Negative Clippers
If we want to change/adjust the clipping level of AC voltage, then external biasing voltage must be
used. The figure given below shows a biased (series) clipper. A biasing voltage is connected in series with the
diode as shown in the figure.
Diode Circuits and Power Supplies 3
Muhammed Riyas A.M,Assistant Professor,Dept. of ECE,MCET Pathanamthitta
Fig: Biased Series Negative Clipper
When the input signal is positive, it tries to establish the current through the diode. However, the
biasing voltage V opposes the flow of current and tries to keep the diode in “off” state. As soon as the signal
voltage rises above voltage V, the diode starts to conduct and the current flows through the load. The output
appears across the load as soon as the current flows though it. Maximum value of output voltage is Vm – V.
Series Positive Clipper
Positive clippers are used to clip positive portions of the input signal and allow the negative portions
of the signal to pass through .That means,during the positive half cycle of the input signal, diode is reverse
biased and acts as open switch. In such a case, no current flows through the diode and the load, and hence the
output voltage across the load is zero.
Diode Circuits and Power Supplies 4
Muhammed Riyas A.M,Assistant Professor,Dept. of ECE,MCET Pathanamthitta
Fig: Series Positive Clipper
During negative half cycle of the input voltage, the diode is turned on and it acts as closed switch.
Hence no voltage drops across the diode. All the input voltage appears across the load, and hence the ouptut
signal voltage is same as input voltage.
Biased Series Positive Clippers
When the input signal is positive, the diode is reverse biased, it is considered to be in “off” state and it
acts as an open switch. As the signal voltage goes positive, it aids the battery voltage in reverse biasing the
diode and consequently the reverse bias voltage across the diode increases. For the complete positive half
cycle, the diode stays in reverse bias and acts as open switch.
Now consider the instant when the signal voltage goes negative. As soon as the signal voltage goes
negative, it starts to oppose the battery voltage and tries to forward bias the diode. However at this point, the
signal voltage is not sufficient to forward bias the diode. Once the signal voltage is greater than the battery
voltage (we assume Vm to be greater than battery voltage V) it forward biases the diode. The diode is now in
the “on” state and output voltage is developed acoss the load. Minimum value of output voltage is -(Vm – V).
Diode Circuits and Power Supplies 5
Muhammed Riyas A.M,Assistant Professor,Dept. of ECE,MCET Pathanamthitta
Fig: Biased Series Positive Clipper
In all the above discussions, note that the diode is considered to be ideal one. In a practical diode, the
breakdown voltage will exist (0.7 V for silicon and 0.3 V for Germanium).
5.2.2 Parallel Clippers
In parallel or shunt clipping circuits, the diode is connected in parallel to the output. In this discussions, the
diode is considered to be a practical one. In a practical diode, the breakdown voltage will exist (0.7 V for silicon
and 0.3 V for Germanium).
Parallel Positive Clipper
In this diode clipping circuit, the diode is forward biased during the positive half cycle of the sinusoidal input
waveform. For the diode to become forward biased, it must have the input voltage magnitude greater than
+0.7 volts (0.3 volts for a germanium diode).
When this happens the diodes begins to conduct and all positive half cycles are bypassed through the diode to
the ground. Thus the output voltage which is taken across the diode can never exceed 0.7 volts during the
positive half cycle. During the negative half cycle, the diode is reverse biased and does not conduct. As a result
input voltage will appears at the output. In this way positive clipper passes only negative going half cycle of the
input to the output.
Diode Circuits and Power Supplies 6
Muhammed Riyas A.M,Assistant Professor,Dept. of ECE,MCET Pathanamthitta
Fig: Parallel Positive Clipper
Parallel Negative Clipper
The diode is forward biased during the negative half cycle of the sinusoidal waveform and limits or clips it to -
0.7 volts while allowing the positive half cycle to pass unaltered when reverse biased. As the diode limits the
negative half cycle of the input voltage it is therefore called a negative clipper circuit.
Fig:Parallel Negative Clipper
Combinational Clipper
Diode Circuits and Power Supplies 7
Muhammed Riyas A.M,Assistant Professor,Dept. of ECE,MCET Pathanamthitta
If we connected two diodes in inverse parallel as shown, then both the positive and negative half cycles would
be clipped as diode D1 clips the negative half cycle of the sinusoidal input waveform while diode D2 clips the
positive half cycle.
For ideal diodes the output waveform above would be zero. However, due to the forward bias voltage drop
across the diodes the actual clipping point occurs at +0.7 volts and -0.7 volts respectively. But we can increase
this ±0.7V threshold to any value we want up to the maximum value, (Vm) of the sinusoidal waveform either by
connecting together more diodes in series creating multiples of 0.7 volts, or by adding a voltage bias to the
diodes.
Fig: Combinational clipper
Biased Diode Clipping Circuits
To produce diode clipping circuits for voltage waveforms at different levels, a bias voltage, V is added
in series with the diode as shown. The voltage across the series combination must be greater
than V + 0.7V before the diode becomes sufficiently forward biased to conduct. For example, if the V level is
set at 4.0 volts, then the sinusoidal voltage at the diode’s anode terminal must be greater than 4.0 + 0.7 = 4.7
volts for it to become forward biased.
Biased Positive Parallel Clippers
Here the positive terminal of the external bias is connected to the cathode of the diode. Therefore
the diode is reverse biased until the voltage across it is greater than V+.7V (here the voltage level is 4+.7= 4.7
Volt).When the input signal voltage level is greater than 4.7 volt, the diode becomes forward biased and begin
to conduct. Any voltage levels above 4.7 volt are clipped off.
Diode Circuits and Power Supplies 8
Muhammed Riyas A.M,Assistant Professor,Dept. of ECE,MCET Pathanamthitta
Fig: Biased Positive Parallel Clipper
Biased Negative Parallel Clippers
Here the negative terminal of the external bias is connected to the anode of the diode. Therefore the
diode is reverse biased until the voltage across it is less than –V-.7V (here the voltage level is -4-.7= -4.7
Volt).When the input signal voltage level is less than 4.7 volt, the diode becomes forward biased and begin to
conduct. Any voltage levels below -4.7 volt are clipped off.
Fig: Biased Negative Parallel clipper
Biased Combinational Clippers
During the positive half cycle of the input signal,first branch is effective and others remain open.
When the voltage of the positive half cycle reaches +4.7 V, diode D1 conducts and limits the waveform at +4.7
Diode Circuits and Power Supplies 9
Muhammed Riyas A.M,Assistant Professor,Dept. of ECE,MCET Pathanamthitta
V. Diode D2 does not conduct during this period. During the negative half cycle of the input, second branch will
effective and first branch is open. When the voltage of the negative half cycle reaches -4.7 V,
diode D2 conducts and limits the waveform at -4.7 V. Diode D1 does not conduct during this period. Therefore,
all positive voltages above +4.7 V and negative voltages below –4.7 V are automatically clipped.
Fig: Biased combinational Clipper
The advantage of biased diode clipping circuits is that it prevents the output signal from exceeding preset
voltage limits for both half cycles of the input waveform, which could be an input from a noisy sensor or the
positive and negative supply rails of a power supply.
If the diode clipping levels are set too low or the input waveform is too great then the elimination of both
waveform peaks could end up with a square-wave shaped waveform.
Drawbacks of Series and Parallel Diode Clippers
In series clippers, when the diode is in ‘OFF’ position, there will be no transmission of input signal to
output. But in case of high frequency signals transmission occurs through diode capacitance which is
undesirable. This is the drawback of using diode as a series element in such clippers.
In shunt clippers, when diode is in the ‘off condition, transmission of input signal should take place to
output. But in case of high frequency input signals, diode capacitance affects the circuit operation
adversely and the signal gets attenuated (that is, it passes through diode capacitance to ground).
Diode Circuits and Power Supplies 10
Muhammed Riyas A.M,Assistant Professor,Dept. of ECE,MCET Pathanamthitta
5.3 Clampers
The clamping network is one that will “clamp” a signal to a different dc level. The network must have
a capacitor, a diode, and a resistive element, but it can also employ an independent dc supply to introduce an
additional shift. The magnitude of R and C must be chosen such that the time constant τ= RC is large enough
to ensure that the voltage across the capacitor does not discharge significantly during the interval the diode is
non conducting.
Clamping circuits are often used in TV receivers as dc restorers. The incoming composite video signal
is normally processed through capacitively-coupled amplifiers which eliminate the dc component thereby
losing the black and white reference levels and the blanking level. These reference levels have to be restored
before applying the video signal to the picture tube.
Note that voltage drop across the diode (.7 volt) is considered for representing the output wave forms
of the following clamper circuits and it is avoided in the discussions.
Positive Clamper
Let the input signal be Vm sint. During the negative half cycle of the input sine wave, the diode
conducts and capacitor charges to Vm (peak value of the input signal) with positive polarity at the right side of
the capacitor.Here we take the Vm as 5 volt. During the positive half cycle of the input signal, the capacitor
can’t discharge since the diode does not conduct. Thus the capacitor acts as a DC source of Vm volts connected
in series with the input signal source. Then the output voltage can be expressed as Vo= Vm+Vm sint.
Fig: Positive Clamper
Negative Clamper
Diode Circuits and Power Supplies 11
Muhammed Riyas A.M,Assistant Professor,Dept. of ECE,MCET Pathanamthitta
During the positive half cycle of the input sine wave, the diode conducts and capacitor charges to Vm
with negative polarity at the right side of the capacitor. During the negative half cycle of the input signal, the
capacitor can’t discharge since the diode does not conduct. Thus the capacitor acts as a DC source of -Vm volts
connected in series with the input signal source. Then the output voltage can be expressed as Vo= -Vm+Vm
sint.
Fig: Negative Clamper
Biased Positive Clamper
Here we take a 3 volt battery as DC source connected in series with the diode in such a way that
positive terminal of the battery is connected to the anode terminal of the diode. During the negative half cycle
of the input sine wave, the diode conducts and capacitor charges through diode and the DC source till (Vm +3)
volts with positive polarity at the right side of the capacitor. The charging of the capacitor is limited to (Vm +3)
volts due to the presence of the DC source. The output is then represented as Vo= (Vm +3)+ Vm sint.
Diode Circuits and Power Supplies 12
Muhammed Riyas A.M,Assistant Professor,Dept. of ECE,MCET Pathanamthitta
Fig: Biased Positive Clamper
Biased Negative Clamper
Here we take a 3 volt battery as DC source connected in series with the diode in such a way that
negative terminal of the battery is connected to the cathode terminal of the diode. During the positive half
cycle of the input sine wave, the diode conducts and capacitor charges through diode and the DC source till
(Vm +3) volts with negative polarity at the right side of the capacitor. The charging of the capacitor is extended
up to (-Vm-3) volts due to the presence of the DC source. The output is then represented as Vo= -Vm -3+ Vm
sint = -(Vm +3)+ Vm sint.
Fig: Biased Negative Clamper
Diode Circuits and Power Supplies 13
Muhammed Riyas A.M,Assistant Professor,Dept. of ECE,MCET Pathanamthitta
5.4 Voltage Multipliers
Voltage multipliers are AC-to-DC power conversion devices, comprised of diodes and capacitors, that
produce a high potential DC voltage from a lower voltage AC source. Voltage multipliers are similar in many
ways to rectifiers in that they convert AC-to-DC voltages for use in many electrical and electronic circuit
applications such as in microwave ovens, strong electric field coils for cathode-ray tubes, electrostatic and high
voltage test equipment, etc, where it is necessary to have a very high DC voltage generated from a relatively
low AC supply.
Generally, the DC output voltage (Vdc) of a rectifier circuit is limited by the peak value of its sinusoidal
input voltage. But by using combinations of rectifier diodes and capacitors together we can effectively multiply
this input peak voltage to give a DC output equal to some odd or even multiple of the peak voltage value of the
AC input voltage.
Voltage multiplier circuits are classified as voltage doubler’s, tripler’s, or quadrupler’s, etc, depending
on the ratio of the output voltage to the input voltage. In theory any desired amount of voltage multiplication
can be obtained and a cascade of “N” doublers, would produce an output voltage of 2N.Vp volts.
Half-Wave Voltage Doubler
The circuit of a half-wave voltage doubler is given in figure shown below. During the positive half cycle of the
ac input, voltage, diode D1 being forward biased conducts (diode D2 does not conduct because it is reverse-
biased) and charges capacitor C1 upto peak values of secondary voltage Vm with the polarity, as marked in
figure shown below.
Fig: Half Wave Voltage Doubler
During the negative half-cycle of the input voltage diode D2 gets forward biased and conducts charging
capacitor C2. For the negative half cycle, the lower end of the transformer secondary is positive while upper
end is negative. The polarity of the capacitor C2 has also been marked in the figure. Now starting from the
Diode Circuits and Power Supplies 14
Muhammed Riyas A.M,Assistant Professor,Dept. of ECE,MCET Pathanamthitta
bottom of the transformer secondary and moving clockwise and applying Kirchoff’s voltage law to the outer
loop we have
−푉 − 푉 + 푉 = 0
Or
푉 = 푉 + 푉 = 푉 + 푉 = 2푉 = Twice the peak value of the transformer secondary voltage. (Since Vc1 = Vm)
During the next positive half-.cycle diode D2 is reverse-biased and so acts as an open and capacitor
C2 discharges through the load If there is no load across the capacitor, C2 both capacitors stay charged – C1 to
Vm and C2 to 2Vm. If, as expected there is a load connected to the output terminals of the voltage doubler, the
capacitor C2 discharges a little bit and consequently the voltage across capacitor C2 drops slightly. The
capacitor C2 gets recharged again in the next half-cycle. The ripple frequency in this case will be the signal
frequency (that is, 50 Hz for supply mains.)
Full-Wave Voltage Doubler
The circuit diagram for a full-wave voltage doubler is given in the figure shown below. During the
positive cycle of the ac input voltage, diode D1 gets forward biased and so conducts charging the capacitor
C1 to a peak voltage Vm with polarity indicated in the figure, while diode D2 is reverse-biased and does not
conduct.During the negative half-cycle, diode D2 being forward biased conducts and charges the capacitor
C2 with polarity shown in the figure while diode D1 does not conduct. With no load connected to the output
terminals, the output voltage will be equal to sum of voltages across capacitors C1 and C2 that is, VC1 + VC2 or
(Vm + Vm) or 2 Vm. When the load is connected to the output terminals, the output voltage VL will be somewhat
less than 2 Vm.
Fig: Full Wave Voltage Doubler
Comparison between Half wave and Full Wave Voltage Doublers
Diode Circuits and Power Supplies 15
Muhammed Riyas A.M,Assistant Professor,Dept. of ECE,MCET Pathanamthitta
Half-Wave voltage doubler
Availability of common point between input and output lines for earthing is available
Ripple content is high
Ripple frequency is the supply frequency
Voltage regulation is poor
Maximum voltage across each capacitor is 2 Vm
PIV rating of each diode is 2 Vm
Full-Wave voltage doubler
Availability of common point between input and output lines for earthing is not available.
Ripple content is low.
Ripple frequency is twice the supply frequency.
Voltage regulation is better than that of half-wave voltage doubler.
Maximum Voltage Across Each Capacitor is Vm.
PIV rating of each diode is 2 Vm.
Voltage Tripler and Quadruples
The half-wave voltage doubler, shown in the earlier figure can be extended to provide any multiple of
the peak input voltage (that is, 3 Vm, 4 Vm or 5 Vm), as illustrated in the figure shown below. It is obvious from
the pattern of the circuit connections how additional diodes and capacitors are to be connected to provide
output voltage, 5,6,7 or 8 times the peak input voltage from a supply transformer of rating only Vm, and each
diode in the circuit of PIV rating 2 Vm. If load is small and the capacitors have little leakage, extremely high dc
voltages can be obtained from such a circuit using many sections to step-up the dc voltage.
In operation capacitor C1 is charged through diode Dl to a peak value of transformer secondary
voltage, Vm during first positive half-cycle of the ac input voltage. During the negative half cycle capacitor C2 is
charged to twice the peak voltage 2 Vm developed by the sum of voltages across capacitor C1 and the
transformer secondary. During the second positive half-cycle, diode D3 conducts and the voltage across
capacitor C2 charges the capacitor C3 to the same 2 Vm peak voltage. During the negative half-cycle diodes
D2 and D4 conduct allowing capacitor C3 to charge capacitor C4to peak voltage 2 Vm. From the fogure shown
below it is obvious that the voltage across capacitor C2 is 2 Vm, across capacitors C1 and C3 it is 3 Vm and across
capacitors C2 and C4 it is 4 Vm.
If additional diodes (each diode of PIV rating 2 Vm) and capacitors (each capacitor of voltage rating 2
Vm) are used, each capacitor will be charged to 2 Vm. Measuring from the top of the transformer secondary
winding (figure below) will give odd multiples of Vm at the output, while measuring from the bottom of
transformer secondary winding will give even multiples of the peak voltage, Vm.
Diode Circuits and Power Supplies 16
Muhammed Riyas A.M,Assistant Professor,Dept. of ECE,MCET Pathanamthitta
Fig: Voltage Tripler and Voltage Quadruplar
5.5 DC Power Supply:
The electric energy available in our country is in the form of AC voltage 230V, 50 Hz. But most of
electronic devices operate in DC power supplies. Almost all basic household electronic circuits need an
unregulated AC to be converted to constant DC. All devices will have a certain power supply limit and the
electronic circuits inside these devices must be able to supply a constant DC voltage within this limit. That
is, all the active and passive electronic devices will have a certain DC operating point (Q-point or
Quiescent point), and this point must be achieved by the source of DC power. The DC power supply is
practically converted to each and every stage in an electronic system. Thus a common requirement for all
this phases will be the DC power supply. Regulated DC power supply is an electronic circuit that is
designed to provide a constant dc voltage of predetermined value across load terminals irrespective of ac
mains fluctuations or load variations. The block diagram of a regulated DC power supply is shown in
figure.
Fig:Block Diagram of a Power Supply
The first block in the DC power supply is a transformer. A transformer is a device which transforms
high voltage AC into low voltage AC or vice versa. Our goal is to convert high voltage AC into low voltage DC. So
Diode Circuits and Power Supplies 17
Muhammed Riyas A.M,Assistant Professor,Dept. of ECE,MCET Pathanamthitta
there is absolutely no reason to use step-up transformer. The transformer that is used in power supply is step-
down transformer, which steps down the input AC voltage. The main reason why we use transformer in the
system are as follows.
We want to reduce the voltage level which we get from the AC mains. Transformer can do the job of
reducing the voltage level in a simple and efficient manner.
The diodes used in the rectifier block cannot handle such a high level of voltage from the AC mains. So the
voltage is first stepped down by the transformer and the reduced voltage is applied to the rectifier section.
The output of the transformer is given to the input of the rectifier.
The rectifier converts the input sinusoidal voltage to a unipolar output or pulsating DC. Full wave rectifiers
are the most commonly used rectifiers in power supply. The output of the rectifier is given to input of the filter
circuit. The output after being processed by full wave rectifier is not a pure DC. The output is a pulsating DC.
The output contains large fluctuations in voltages. The power supply that we intend to design must not have
any variation in output voltage. The voltage that we get from full wave rectifier fluctuates between 0 V and
Vpeak, and hence it contains AC components. These AC components needs to be filtered out so as to obtain DC
voltage. The filter circuit reduces the variation in magnitude (ripples) of the rectifier output. However, the
output from the filter is still not a pure DC but filters removes the AC component in the voltage to a
considerable extent. This increases the average DC value of the output voltage.
The output of the filter is given to the voltage regulator circuit for stabilizing the magnitude of the DC output
voltage against the variations caused by changes in the load current and input voltage.
5.5.1 Characteristics of a power supply
There are various factors that determine the quality of the power supply like the load voltage, load current,
voltage regulation, source regulation, output impedance, ripple rejection, and so on. Some of the
characteristics are briefly explained below:
Load Regulation – The load regulation or load effect is the change in regulated output voltage when the
load current changes from minimum to maximum value.
퐿표푎푑푅푒푔푢푙푎푡푖표푛 = 푉 − 푉
푉 – Load Voltage at no load
푉 – Load voltage at full load.
From the above equation we can understand that when 푉 occurs the load resistance is infinite, that is,
the out terminals are open circuited. 푉 occurs when the load resistance is of the minimum value
where voltage regulation is lost.
%퐿표푎푑푅푒푔푢푙푎푡푖표푛 = 푉 − 푉 /푉 ∗ 100
Diode Circuits and Power Supplies 18
Muhammed Riyas A.M,Assistant Professor,Dept. of ECE,MCET Pathanamthitta
Minimum Load Resistance – The load resistance at which a power supply delivers its full-load rated current
at rated voltage is referred to as minimum load resistance.
푀푖푛푖푚푢푚퐿표푎푑푅푒푠푖푠푡푎푛푐푒 = 푉 퐼⁄
The value of 퐼 , full load current should never increase than that mentioned in the data sheet of the
power supply.
Source/Line Regulation – In the block diagram, the input line voltage has a nominal value of 230 Volts but
in practice, here are considerable variations in ac supply mains voltage. Since this ac supply mains voltage is
the input to the ordinary power supply, the filtered output of the bridge rectifier is almost directly
proportional to the ac mains voltage.
The source regulation is defined as the change in regulated output voltage for a specified rage of lie voltage.
Output Impedance – A regulated power supply is a very stiff dc voltage source. This means that the output
resistance is very small. Even though the external load resistance is varied, almost no change is seen in the
load voltage. An ideal voltage source has an output impedance of zero.
Ripple Rejection – Voltage regulators stabilize the output voltage against variations in input voltage. Ripple is
equivalent to a periodic variation in the input voltage. Thus, a voltage regulator attenuates the ripple that
comes in with the unregulated input voltage. Since a voltage regulator uses negative feedback, the distortion is
reduced by the same factor as the gain.
5.6 Rectifiers:
Diode is a unidirectional device. It provides high resistance in one direction and a low resistance in the
other direction. Therefore it can be used as a rectifier for converting AC to DC.There are two types of rectifiers.
Half wave rectifier
Full wave rectifier
5.6.1 Half wave rectifier:
Half wave rectifier is the simplest form of the rectifier. The circuit arrangement is shown in fig(a).the
primary side is connected to the power mains(230 V,50 Hz).An AC voltage induces across the secondary
winding, which is equal to less than or greater than the primary voltage depending upon the turns ratio of
the transformer. This will be the input of the rectifier.
Diode Circuits and Power Supplies 19
Muhammed Riyas A.M,Assistant Professor,Dept. of ECE,MCET Pathanamthitta
When the transformer secondary voltage is on its positive half cycle, the diode is forward biased and
current flows through the load RL, developing a voltage across it. When the negative voltage is on its
negative half cycle, the diode is reverse biased and no current flows through the RL.Thus there will be no
voltage across the output terminal during negative half cycle. The fig (b) shows the input and output wave
forms of a half wave rectifier. Here only positive voltages are obtained at the output and all negative half
cycles are suppressed. Thus we get a pulsating DC at the output.
Peak Inverse Voltage (PIV)
In the rectifier circuit, during negative half cycle of the secondary voltage, the diode is reverse biased.
As there is no voltage across the load resistor RL during this half cycle. The whole secondary voltage will
come across the diode. When the secondary voltage reaches its maximum value Vm in the negative half
cycle, the voltage across the diode in reverse bias is also maximum. This maximum voltage is known as
peak inverse voltage (PIV).It is the maximum voltage the diode must withstand during the reverse bias
half cycle of the input. Therefore the diode is used in the rectifier circuit must have the higher PIV rating
than this voltage.
Diode Circuits and Power Supplies 20
Muhammed Riyas A.M,Assistant Professor,Dept. of ECE,MCET Pathanamthitta
Disadvantages:
Low output because only one half cycle delivers the output. AC components are more in the output.
5.6.1.1 Performance of Half wave Rectifier:
DC Output:
The DC output voltage or a current of the rectifier is the average value of the output voltage or current.
Ie,
The average value of a sine wave over one complete cycle is zero because it is symmetrical. But in the case of a half wave rectifier output, it consists of only alternate cycles as shown in output wave form. Hence there will be some DC current flowing through the RL.
Let,
Alternating voltage appearing across the secondary winding of the transformer V = V Sinωt
Current flowing through the load resistor i = I Sinωt
Diode forward resistance r = 0
From wave form, it is clear that current flowing through the load only in the period 0 to 휋.
i.e., i = I Sinωt ∶ for0 ≤ ωt ≤ π
i = 0 ∶ forπ ≤ ωt ≤ 2π Here area = ∫ ImSinωtd(ωt) = I [−Cosωt]
= 2 I
Time period = 2 π
Now, I = 2I /2π
I = I /π Voltage across RL V = I R
= I /πR
While diode forward resistance r is taken into account
I = V (R + r )⁄
DC voltage across RL can be written as
V = V R /π(R + r )
I = Areaunderonecompletecycle
Timeperiod
Diode Circuits and Power Supplies 21
Muhammed Riyas A.M,Assistant Professor,Dept. of ECE,MCET Pathanamthitta
= V π(1 + r /R )⁄
Since, r ≪R
RMS Value:
Root mean square (RMS) is the effective value of the current flowing through the load and is given by
In the case of half wave rectifier,
Hence,
This is the rms value of the total current which include DC value and AC components. In the output of the rectifier, the instantaneous value of AC fluctuation is the difference of the instantaneous total value and the DC value.
Instantaneous AC value is given by
i = i − I
Here the rms value of AC component is given by,
i = I − I
V = V π⁄
퐼 =퐴푟푒푎푢푛푑푒푟푡ℎ푒푠푞푢푎푟푒푑푟푒푐푡푖푓푖푒푑푤푎푣푒표푣푒푟푎푐푦푐푙푒
푇푖푚푒푝푒푟푖표푑
I = 1
2π iπ
d(ωt)
=1
2π I Sin ωtd(ωt)
=I2π
(1− Cos2ωt)
2d(ωt)
=I
2π× 2ωt − Sin2ωt
2π0
Irms = Im / 2
Vrms = Vm / 2
Diode Circuits and Power Supplies 22
Muhammed Riyas A.M,Assistant Professor,Dept. of ECE,MCET Pathanamthitta
Ripple Factor:
The purpose of the rectifier is to convert AC voltage to DC. But not type of rectifier converts AC to pure DC. It produces pulsating DC. This residual pulsation is called ripple. Ripple factor indicates the effectiveness of the rectifier in converting AC to perfect DC. It is the ratio of the ripple voltage to the DC voltage.
= V /V
Or 훾 = I /I
= I − I /I
= I I⁄ − 1
= (퐼 /2)/(퐼 /휋) − 1
= (1.57) − 1
= 1.21
Efficiency:
Efficiency of the rectifier is a measure of conversion of AC power to the useful DC output power. It is expressed as the ratio of DC output power to AC input power.
i.e., 휂 = 푃 푃⁄
Where 푃 is the power delivered to load and 푃 is the AC input power from the secondary winding of the transformer.
Here 푃 = 퐼 × 푅
= (퐼 휋⁄ ) × 푅
푃 = 퐼 (푅 + 푟 )
= (퐼 2⁄ ) (푅 + 푟 )
후 = 퐑퐢퐩퐩퐥퐞퐯퐨퐥퐭퐚퐠퐞퐃퐂퐯퐨퐥퐭퐚퐠퐞
=rmsvalueoftheACcomponent
DCcomponent
Diode Circuits and Power Supplies 23
Muhammed Riyas A.M,Assistant Professor,Dept. of ECE,MCET Pathanamthitta
Therefore,
휂 =(퐼 휋⁄ ) × 푅
(퐼 2⁄ ) (푅 + 푟 )
= . 406푅(푅 + 푟 )
= . 406(1 + 푟 /푅 )
휂 = .406표푟40.6%
Transformer Utilization Factor:
For designing a power supply using transformer, transformer rating must be known. In order to calculate the
transformer utilization factor, the power supplied by the transformer must be arranged in terms of the d.c
power.
푉 × 퐼 = 푃
TUF is defined as the ratio of d.c output power to a.c power supplied to it by the secondary winding,
i.e,푇푈퐹 = (푟푎푡푒푑)
As the rating of the transformer is different than the actual power delivered by the secondary,the TUF is
entirely different quantity than efficiency.
In case of a single diode half wave rectifier,the rated voltage of the secondary=Vm/√2 but the actual r.m.s
value of the current flowing through the secondary winding =Im/2 and not Im/√2.
Thus
푇푈퐹 =(퐼푚 휋⁄ )2 × 푅퐿
푉√2
× 퐼2
=2√2휋
퐼 푅푉
푉 = 퐼 (푟 + 푅 )
푇푈퐹 =2√2휋
퐼 푅퐼 (푟 + 푅 )
Hence,
푇푈퐹 =2√2휋2 = .287
Regulation of Half Wave Rectifier
Regulation is defined as the variation of output DC voltage with the changes in the output DC current.
%푅푒푔푢푙푎푡푖표푛 =푉 − 푉
푉 × 100%
Diode Circuits and Power Supplies 24
Muhammed Riyas A.M,Assistant Professor,Dept. of ECE,MCET Pathanamthitta
Where, 푉 = no load voltage.
푉 = full load voltage.
An ideal power supply should have a 0% regulation. The output voltage should not vary with load current IL.
For a half wave rectifier, the percentage regulation is given by,
%푅푒푔푢푙푎푡푖표푛 =
푉휋 − 푉
휋 − 푉휋푅 푅 + 푅
푉휋 − 푉
휋푅 푅 + 푅
Where,
푉 =푉휋
푉 = 푉 − 퐼
%푅푒푔푢푙푎푡푖표푛 =푉휋푉휋 ⎣⎢⎢⎢⎡ 1− 1 +
푅 + 푅푅
1−푅 + 푅푅 ⎦
⎥⎥⎥⎤
%푅푒푔푢푙푎푡푖표푛 = −푅 + 푅푅
Form Factor:
Form factor is the ratio of rms value to the average value.
퐹표푟푚푓푎푐푡표푟 =푟푚푠푣푎푙푢푒
푎푣푒푟푎푔푒푣푎푙푢푒
푓표푟푚푓푎푐푡표푟 = 푉 2⁄푉 휋⁄
= 휋 2 = 1.57⁄
Peak Factor:
Peak factor is the ratio of peak value to rms value.
푃푒푎푘푓푎푐푡표푟 =푝푒푎푘푣푎푙푢푒푟푚푠푣푎푙푢푒
푓표푟푚푓푎푐푡표푟 = 푉푉 2⁄
= 2
5.6.2 Full Wave Rectifiers:
In a full wave rectifier, current flows through the load during both half cycles of the input AC supply.
Thus in full wave rectifier, alternate half cycle of the input AC are inverted to get a unidirectional output
current. Full wave rectifiers are of two types,
Centre tap rectifier
Diode Circuits and Power Supplies 25
Muhammed Riyas A.M,Assistant Professor,Dept. of ECE,MCET Pathanamthitta
Bridge rectifier
5.6.2.1 Centre Tap Rectifier:
The circuit arrangement of a centre-tap full wave rectifier is shown in fig(a).during positive half cycle
of the secondary voltage, the diode D1 is forward biased and D2 is reverse biased. The current flows through
D1, RL and upper half of the secondary winding. During negative half cycle of the secondary voltage, current
flows through D2, RL and lower half of the secondary winding. Here current through RL during both half cycle of
input AC in the same direction. Therefore the output voltage taken across RL will be DC. The input and output
wave forms of this rectifier are shown in fig (b).
Peak Inverse Voltage (PIV)
In the centre tap full wave rectifier, during positive half cycle of the secondary voltage diode D1 conducts
and when secondary voltage attains its maximum value Vm, a voltage equal to Vm will develop across RL with
the polarity marked in fig(c).The diode D2 at that instant is reverse biased and the voltage across it will be the
sum of the voltages developed across the lower winding of the secondary (Vm) and the voltage developed
across RL=Vm. Therefore peak inverse voltage across the diode in this case will be Vm+Vm=2Vm.
Diode Circuits and Power Supplies 26
Muhammed Riyas A.M,Assistant Professor,Dept. of ECE,MCET Pathanamthitta
i.e, PIV=2 Vm
Advantages:
High efficiency
Low ripple factor
Disadvantages:
Requires centre tap rectifier with secondary voltage twice as required for half wave or bridge rectifier.
Diodes of high PIV (2V) rating are to be used.
5.6.2.2 Bridge Rectifier:
The circuit arrangement of the bridge rectifier is show in fig (a).It contains four diodes, but avoids the
need for a centre-taped transformer. During the positive half cycle of the secondary voltage, diode D2 and D3
will be forward biased. At the same time diodes D1 and D4 are reverse biased. There fore diodes D2 and D3
conduct and current flows through D2, RL, D3 and transformer secondary. During the negative half cycle of the
secondary voltage diodes D1 and D4 will be forward biased and D2 and D3 will be reverse biased. There fore the
current flows through D1, RL ,D4 and transformer secondary. Here in both cases the current flow through load
resistor RL is in the same direction. Hence DC voltage is obtained at the output. The wave forms are shown in
fig (b).
Diode Circuits and Power Supplies 27
Muhammed Riyas A.M,Assistant Professor,Dept. of ECE,MCET Pathanamthitta
Peak inverse voltage (PIV)
When the secondary voltage is at its positive maximum value Vm diodes D2 and D3 are forward biased
and conduct. Conducting diodes have zero resistance and zero voltage drop across them, the points A and C in
fig (a) are in same potential. Similarly points B and D .Therefore the reverse voltage coming across diodes D1
and D4 will be equal to the potential difference between the points A and D will be equal to Vm.
Advantages:
Does not require centre tap rectifier.
Diodes of low PIV rating can be used.
Disadvantages:
Needs four diodes.
5.6.2.2 Performance of Full Wave Rectifier:
DC output:
In a full wave rectifier, it utilizes both half cycles of the input AC voltage. Therefore the DC voltage or average voltage available in the output of a full wave rectifier will be double the DC voltage available in the half wave rectifier.
i.e. 2× Vdc of the half wave rectifier
Diode Circuits and Power Supplies 28
Muhammed Riyas A.M,Assistant Professor,Dept. of ECE,MCET Pathanamthitta
푉 = 2(푉 휋⁄ )
퐼 = 2(퐼 휋⁄ )
RMS Value:
When writing the rms value of the full wave rectifier both half cycles are to be considered. Hence the rms value of the current is given by,
Hence,
Ripple Factor:
훾 = I − I /I
= (I I⁄ ) − 1
= 퐼 /√2 /(2퐼 /휋) − 1
훾 = .482
I = 1π iπ
d(ωt)
=1π I Sin ωtd(ωt)
=Iπ
(1− Cos2ωt)2
d(ωt)
=I2π
ωt − Sin2ωt
2π0
Irms = Im / √2
Vrms = Vm / √2
=I π
2π
Diode Circuits and Power Supplies 29
Muhammed Riyas A.M,Assistant Professor,Dept. of ECE,MCET Pathanamthitta
Efficiency:
휂 = 푃 푃⁄
푃 = 퐼 × 푅
= (2퐼 휋⁄ ) × 푅
푃 = 퐼 (푅 + 푟 )
= 퐼 √2⁄ (푅 + 푟 )
휂 =(2 퐼 휋⁄ ) × 푅
퐼 √2⁄ (푅 + 푟 )
= . 812(1 + 푟 /푅 )
( Where 푅 ≫ 푟 )
휂 = .812표푟81.2%
TUF of Centre Tap Rectifier The two diode full-wave rectifier can be thought of equivalent to two half wave rectifiers feeding to a common load. Hence TUF with respect to two half wave secondary can be written as
TUF (secondary) = 2 TUF of half wave = 2× .287
=.574
푇푈퐹(푃푟푖푚푎푟푦) =(2퐼푚 휋⁄ )2 × 푅퐿
푉√2
× 퐼√2
=8퐼휋
퐼 푅푉 퐼
=8휋
= .812
Thus,
푇푈퐹 = (푇푈퐹 + 푇푈퐹 )/2
=. 812 + .574
2 = .693
TUF of Bridge Rectifier
푇푈퐹 = 푃 푃⁄
Diode Circuits and Power Supplies 30
Muhammed Riyas A.M,Assistant Professor,Dept. of ECE,MCET Pathanamthitta
푇푈퐹 =(2퐼푚 휋⁄ )2 × 푅퐿
푉√2
× 퐼√2
=8퐼휋
퐼 푅푉 퐼
=8휋
= .812
Regulation of Full Wave Rectifier
Regulation of output signal is defined as the variation of output DC voltage with the changes in the DC load current.
%푅푒푔푢푙푎푡푖표푛 =푉 − 푉
푉 × 100%
Where, 푉 = no load voltage.
푉 = full load voltage.
An ideal power supply should have a 0% regulation. The output voltage should not vary with load current IL. The % regulation is given by,
%푅푒푔푢푙푎푡푖표푛 =
2푉휋 − 2푉
휋 − 2푉휋푅 푅 + 푅
2푉휋 − 2푉
휋푅 푅 + 푅
Where,
푉 =2푉휋
푉 =2푉휋 − 퐼
%푅푒푔푢푙푎푡푖표푛 =2푉휋
2푉휋 ⎣
⎢⎢⎢⎡ 1− 1 +
푅 + 푅푅
1−푅 + 푅푅 ⎦
⎥⎥⎥⎤
%푅푒푔푢푙푎푡푖표푛 = −푅 + 푅푅
Form Factor:
퐹표푟푚푓푎푐푡표푟 =푟푚푠푣푎푙푢푒
푎푣푒푟푎푔푒푣푎푙푢푒
푓표푟푚푓푎푐푡표푟 = (푉 √2)⁄(2푉 휋⁄ )
= 휋 2√2 = 1.11⁄
Peak Factor:
Diode Circuits and Power Supplies 31
Muhammed Riyas A.M,Assistant Professor,Dept. of ECE,MCET Pathanamthitta
푃푒푎푘푓푎푐푡표푟 =푝푒푎푘푣푎푙푢푒푟푚푠푣푎푙푢푒
푃푒푎푘푓푎푐푡표푟 = 푉(푉 √2)⁄
= √2
5.6.Problems:
1. In a half wave rectifier the load resistance RL=1KΩ the forward resistance of the diode rd=100Ω, input
alternating voltage (Vm) =325 volt. Find (a) Peak value, average value and rms value of the output current
(b) Efficiency of the rectifier.
Solution
(a) Peak value
퐼 = 푉 (푟 + 푅 )⁄
= 325 (100 + 1000)⁄
퐼 = 295.45푚퐴
Average value
퐼 = 퐼 휋⁄
= 295.45 휋⁄
퐼 = 94.046푚퐴
Rms value
퐼 = 퐼 /2
= 295.45 2⁄
= 147.725푚퐴
(b) Efficiency
휂 = 푃 푃⁄
푃 = 퐼 × 푅
= (94.046 × 10 ) × 1000
푃 = 8.845푊
푃 = 퐼 (푅 + 푟 )
= (147.725 × 10 ) (100 + 1000)
= 24푊
휂 = 푃 푃⁄
= 8.845/24
휂 = .3685표푟36.85%
Diode Circuits and Power Supplies 32
Muhammed Riyas A.M,Assistant Professor,Dept. of ECE,MCET Pathanamthitta
2. In a full wave centre tap rectifier, the load resistance used is of 2KΩ,the forward resistance of each diode is
400Ω,the voltage across the half of the secondary winding is 240 Sin50t.Find (a) Im, (b) Idc (c) Irms (d) ripple
factor (e) PIV.
Solution
(a) 푰풎
퐼 = 푉 (푟 + 푅 )⁄
= 240 (400 + 2000)⁄
= .1퐴푚푝
(b) 푰풅풄
퐼 = 2 퐼 휋⁄
= (2 × .1) 휋⁄
= .636퐴푚푝 = 63.6푚퐴
(c) 푰풓풎풔
퐼 = 퐼 /√2
= . 1 √2⁄
= .070퐴푚푝 = 70.72푚퐴
(d) Ripple factor
훾 = (퐼 퐼⁄ ) − 1
= (. 07072/.0632) − 1
= .501표푟50.1%
(e) PIV
푃퐼푉 = 2푉 = 2 × 240 = 480 V
5.7 Filter Circuits:
The out put voltage of both half wave and full wave rectifier is pulsating and not pure DC.If such a
pulsating DC is given to any equipment, the working of the equipment will be disturbed. This disturbance of
ripples can be avoided by using filter circuits. Many types of passive filters are in use such as,
Shunt capacitor filter.
Series inductor filter.
Chock input LC filter.
Π- Section Filter.
5.7.1 Shunt Capacitor Filter:
This type of filter consists of a large value capacitor connected across the load resistance RL as shown
in fig (a).This capacitor offers low reactance path to AC components and very high impedance to DC.So that AC
components in the rectifier output find low reactance path through the capacitor and only a very small part
flows through RL, producing a small ripple voltage at the output as shown in fig (B).
Diode Circuits and Power Supplies 33
Muhammed Riyas A.M,Assistant Professor,Dept. of ECE,MCET Pathanamthitta
When the diode conducts, the capacitor charges up to peak voltage Vm (point B in figure).From this
point onwards the rectifier voltage starts to fall, but the charged capacitor tries to send current back through
the diodes. Thus the diodes are reverse biased and become open circuited. The capacitor then starts to
discharge through the load. This action prevents the load voltage from falling to zero during no conduction
period of the diode. The capacitor continues to discharge until the source voltage becomes more than the
voltage across the capacitor(at point C).Then the diode begins to conduct and the capacitor again charges to
Vm (portion CD in figure).During this time, the rectifier delivers the charging current as well as load current.
Thus the capacitor removes the AC components and also improves the output voltage. The discharging rate of
the capacitor depends upon the time constant C.RL. The ripple factor in the capacitor filter is given by,
휸 = ퟏퟒ√ퟑ풇푪푹푳
Half Wave Rectifier with Capacitor Filter:
Diode Circuits and Power Supplies 34
Muhammed Riyas A.M,Assistant Professor,Dept. of ECE,MCET Pathanamthitta
Full Wave Centre-tap Rectifier with Capacitor Filter:
Full Wave Bridge Rectifier with Capacitor Filter:
Diode Circuits and Power Supplies 35
Muhammed Riyas A.M,Assistant Professor,Dept. of ECE,MCET Pathanamthitta
5.7.2 Ripple factor of the full wave rectifier with a capacitor filter
From the figure, it is obvious that discharging of the capacitor is given by,
푄 = 퐼 × 푇
The charge acquired by capacitor is,
푄 = 푉 × 퐶
Diode Circuits and Power Supplies 36
Muhammed Riyas A.M,Assistant Professor,Dept. of ECE,MCET Pathanamthitta
We know that the charge acquired by the capacitor during the charging is equal to the charge lost during
discharge.
푄 = 푄
푉 × 퐶 = 퐼 × 푇
푉 × 퐶 = × 푇
푉 × 퐶 = × ---------------------(1)
For a full wave rectifier the capacitor charges and discharges the charge carries in both the half cycles and the
total time period is equal to charging and discharging of charges in positive cycle plus charging and discharging
of charges in negative cycle.
T = (T1 + T2) |positive cycle + (T1 + T2) |negative cycle
Since Tdischarge >> Tcharge , charging time Tcharge can be neglected.
Therefore, 푇 = -----------------------(2)
Substituting equation (2) in equation (1)
Vrp-p = ×
For the ripple waveform,
Vr rms = √
Vrp-p = 2√3 Vr rms ------------------------(3)
Substituting equation (3) in equation (2)
2√3 Vr rms = ×
Vr rms = √
But ripple factor is given by,
γ = = √
5.8 Zener Voltage Regulator:
The function of regulator is to provide an output voltage V0 that is as constant as possible in spite of
the ripple in source voltage, Vs and the variations in the load current IL. Two parameters used to measure the
performance of voltage regulator are line regulation and load regulation.
Line regulation is the ratio of change in output voltage to the change in source voltage.
퐿푖푛푒푟푒푔푢푙푎푡푖표푛 =∆푉∆푉
Load regulation is defined as the ratio of change in output voltage to the change in load current.
퐿표푎푑푟푒푔푢푙푎푡푖표푛 =∆푉∆퐼
Diode Circuits and Power Supplies 37
Muhammed Riyas A.M,Assistant Professor,Dept. of ECE,MCET Pathanamthitta
Working of Zener Voltage regulator
The figure shows the basic circuit arrangement for a zener voltage regulator. Here the zener diode
with a break down voltage Vz is connected in reverse bias across the load. A resistor Rs is also connected in
series. If a voltage Vs is applied to the circuit, until the voltage across the load is less than the break down
voltage Vz the zener diode does not conduct and no current flows through the zener diode. At this time the
same voltage Vs will appear across the load. When the input voltage Vs is more than Vz, the zener diode will
conduct and a current Iz will flow through it. This excess current will increase the drop across Rs and limits the
zener current from exceeding the maximum rated value.Thus the current ‘I’ from the DC supply splits to the
load and to the zener diode.
At this time when the zener diode conducts the voltage across it is equal to its break down voltage Vz
and remains fairly constant even when the current through it vary considerably. Thus a constant voltage is
maintained across the load.
A figure of merit for a voltage regulator is called the percent regulation, and is defined as
% Regulation = ( ) ( )
( )× 100
where VL(nom) is the nominal value of the output voltage.
As the percent regulation approaches zero percent, the circuit approaches that of an ideal voltage regulator.
Analysis and Design
We first determine the proper input series resistance Rs. This resistance limits the current through the
Zener diode and drops the excess voltage between Vs and VZ.
Rs = =
Diode Circuits and Power Supplies 38
Muhammed Riyas A.M,Assistant Professor,Dept. of ECE,MCET Pathanamthitta
which assumes that the Zener resistance is zero for the ideal diode. Solving this equation for the diode
current. Iz, we get
Iz = − 퐼
where IL = Vz / RL , and the variables are the input voltage source VS and the load current IL.
For proper operation of this circuit, the diode must remain in the breakdown region and the power
dissipation in the diode must not exceed its rated value. In other words:
1. The current in the diode is a minimum, Iz(min), when the load current is a maximum, IL(max), and the
source voltage is a minimum, VS(min).
2. The current in the diode is a maximum, Iz(max), when the load current is a minimum, IL(min), and
the source voltage is a maximum, VPS(max).
Inserting these two specifications into the previous equation, we obtain
Rs = ( )
( ) ( )
and
Rs = ( )
( ) ( )
Equating these two expressions, we then obtain
[푉 ( ) − 푉 ][퐼 ( ) + 퐼 ( )] =[푉 ( ) − 푉 ][퐼 ( ) + 퐼 ( )]
Reasonably, we can assume that we know the range of input voltage, the range of output load current, and
the Zener voltage. The previous equation then contains two unknowns Iz(min) and Iz(max). Further, as a
minimum requirement, we can set the minimum Zener current to be one-tenth the maximum Zener current,
or IZ(min) =0.1 IZ(max). We can then solve for IZ(max), using the previous equation, as follows:
Iz(max) = ( ) ( ) ( )[ ( ) ]
( ) . . ( )
Diode Circuits and Power Supplies 39
Muhammed Riyas A.M,Assistant Professor,Dept. of ECE,MCET Pathanamthitta
Use the maximum current obtained from the above equation, we can determine the maximum required
power rating of the Zener diode. Then we can determine the required value of the input resistance using
one of the previous equations.
Limitations
A basic zener voltage regulator has the following draw backs :
(i) It has low efficiency for heavy load currents. It is because if the load current is large, there will be
considerable power loss in the series limiting resistance.
(ii) The output voltage slightly changes due to zener impedance as Vout = VZ + IZ ZZ. Changes in load current
produce changes in zener current. Consequently, the output voltage also changes. Therefore, the use of this
circuit is limited to only such applications where variations in load current and input voltage are small.
5.8.1 Problems:
1. The zener diode shown in figure has a fixed voltage drop of 18V across it so long as the zener current is
maintained between 200mA and 2Amp.(i) find the value of R so that the load voltage remains 18V as
input voltage is free to vary from 22V to 28V.(ii) Find the maximum power dissipated by the zener diode.
Solution
Minimum voltage across R=22-18=4 V.
Current through R, when 퐼 ( ) flows in 푅 is
퐼 ( ) + 퐼 ( ) =18푉18Ω + 200푚퐴 = 1200푚퐴
Therefore,
(i) 푅 =
= 3.33Ω
(ii) Maximum power dissipated = 푉 × 퐼 ( )
Given, 퐼 ( ) = 2퐴푚푝
Thus, Maximum power dissipated = 18푉 × 2퐴푚푝 = 36푊
2. For the circuit shown in figure, find the maximum and minimum value of zener diode current.
Diode Circuits and Power Supplies 40
Muhammed Riyas A.M,Assistant Professor,Dept. of ECE,MCET Pathanamthitta
Solution
The zener diode will conduct maximum current, when the input voltage is maximum. i.e. 120 V.
The total current is the current through 5KΩ resistor and is
퐼 =120− 50
5퐾Ω = 14푚퐴
Load Current 퐼 = Ω
= 5푚퐴
Maximum zener current = 14− 5 = 9푚퐴
For minimum input voltage, zener current will be minimum. Therefore,
퐼 =80− 50
5퐾Ω = 6푚퐴
Minimum zener current = 6− 5 = 1푚퐴
3. Design a zener voltage regulator circuit that will maintain an output voltage of 20V across 1KΩ load when
the input voltage is 30-50V.Assume the zener knee current to be negligible compared to the load current.
Find the maximum power of the diode.
Solution
Load current 퐼 = Ω
= 20푚퐴;푅 = ( )
= .5퐾Ω = 500Ω
Maximum current through 푅 is
퐼 =50− 20. 5퐾Ω = 60푚퐴
We know that 퐼 = 퐼 + 퐼 . So,
퐼 = 퐼 − 퐼 = (60 − 20)푚퐴 = 40푚퐴
Maximum power rating of the zener diode is
40푚퐴 × 20푉 = 800푚푊 = .8푊
Diode Circuits and Power Supplies 41
Muhammed Riyas A.M,Assistant Professor,Dept. of ECE,MCET Pathanamthitta
5.9 Switching Mode Power Supply (SMPS)
SMPS stands for switch mode power supply. In such a device, power handling electronic components
are continuously switching "on" and "off" with high frequency in order to provide the transfer of electric
energy via energy storage components (inductors and capacitors). By varying duty cycle, frequency or a
relative phase of these transitions an average value of output voltage or current is controlled. The operating
frequency range of commercial SMPS units varies typically from 50 kHz to several MHz.
Basic Concept
The basic concept behind a switch mode power supply or SMPS is the fact that the regulation is undertaken by
using a switching regulator. This uses a series switching element that turns the current supply to a smoothing
capacitor on and off.
Fig: Switching regulator concept used in SMPS
The time the series element is turned on is controlled by the voltage on the capacitor. If it is higher than
required, the series switching element is turned off, if it is lower than required, it is turned on. In this way the
voltage on the smoothing or reservoir capacitor is maintained at the required level.
Block Diagram of SMPS
Fig: Block Diagram of SMPS
Diode Circuits and Power Supplies 42
Muhammed Riyas A.M,Assistant Professor,Dept. of ECE,MCET Pathanamthitta
If the SMPS has an AC input, then the first stage is to convert the input to DC by using a rectifier. A
SMPS with a DC input does not require this stage. In some power supplies, the rectifier circuit can be
configured as a voltage doubler by the addition of a switch operated either manually or automatically. This
feature permits operation from power sources that are normally at 115 V or at 230 V. The rectifier produces an
unregulated DC voltage which is then sent to a large filter capacitor.
The inverter stage converts DC, whether directly from the input or from the rectifier stage described above, to
AC by running it through a power oscillator, whose output transformer is very small with few windings at a
frequency of tens or hundreds of kilohertz. The frequency is usually chosen to be above 20 kHz, to make it
inaudible to humans. The switching is implemented as a multistage (to achieve high gain) MOSFET amplifier.
If the output is required to be isolated from the input, as is usually the case in mains power supplies, the
inverted AC is used to drive the primary winding of a high-frequency transformer. This converts the voltage up
or down to the required output level on its secondary winding. The output transformer in the block diagram
serves this purpose.
If a DC output is required, the AC output from the transformer is rectified. For output voltages above ten volts
or so, ordinary silicon diodes are commonly used. For lower voltages, Schottky diodes are commonly used as
the rectifier elements; they have the advantages of faster recovery times than silicon diodes and a lower
voltage drop when conducting.
The rectified output is then smoothed by a filter consisting of inductors and capacitors. For higher switching
frequencies, components with lower capacitance and inductance are needed.
A feedback circuit monitors the output voltage and compares it with a reference voltage, as shown in the block
diagram above. Depending on design and safety requirements, the controller may contain an isolation
mechanism (such as an opto-coupler) to isolate it from the DC output. Switching supplies in computers, TVs
and VCRs have these opto-couplers to tightly control the output voltage.
Advantages of SMPS
High efficiency: The switching action means the series regulator element is either on or off and
therefore little energy is dissipated as heat and very high efficiency levels can be achieved.
Compact: As a result of the high efficiency and low levels of heat dissipation, the switch mode power
supplies can be made more compact.
Flexible technology: Switch mode power supply technology can be sued to provide high efficiency
voltage conversions in voltage step up or "Boost" applications or step down "Buck" applications.
Disadvantages of SMPS
Diode Circuits and Power Supplies 43
Muhammed Riyas A.M,Assistant Professor,Dept. of ECE,MCET Pathanamthitta
Noise: The transient spikes that occur from the switching action on switch mode power supplies are
one of the largest problems. The spikes can migrate into all areas of the circuits that the SMPSs power
if the spikes are not properly filtered. Additionally the spikes or transients can cause electromagnetic
or RF interference which can affect other nearby items of electronic equipment, particularly if they
receive radio signals.
External components: While it is possible to design a switch mode regulator using a single integrated
circuit, external components are typically required. The most obvious is the reservoir capacitor, but
filter components are also needed. In some designs the series switch element may be incorporated
within the integrated circuit, but where any current is consumed, the series switch will be an external
component. These components all require space, and add to the cost.
Expert design required: It is often possible to put together a switch mode power supply that works.
To ensure that it performs to the required specification can be more difficult. Ensuring the ripple and
interference levels are maintained can be particularly tricky.
Costs: Careful consideration of the costs of a switch mode power supply must be made before
designing or using one. Beyond the basic power supply, additional filtering may be required and this
can add to the cost.
Despite the disadvantages, switch mode power supply technology is the major form of power supply
technology used for a whole variety of applications especially those included in computers. For applications
where very low noise are required, linear regulator technology is still widely used.
Review Questions
Short Answer Questions
1. Explain any two clamper circuits.
2. A HWR having resistive load of 1000? , it rectifies an ac voltage of 325V peak value and the diode has
a forward resistance of 100?. Calculate a) peak, average and rms value of current , and b) efficiency.
3. Define line regulation and load regulation of a voltage regulator.
4. What are the advantages of SMPS?
5. Explain the operation of voltage multiplier.
Essay Questions
1. Explain any four parallel clipper circuits.
2. Derive the expressions for the rectification efficiency, ripple factor, transformer utilization factor,
form factor and peak factor of full wave rectifier.
3. Explain the full-wave bridge rectifier with capacitor filter. Also derive the expression for ripple factor.
4. Explain the operation of a zener diode voltage regulator.
5. With neat block diagram,explain the operation of SMPS.