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2.1 Power Supply
Power supply is a group of circuits that convert the standard ac voltage (120 V, 60
Hz) provided by the wall outlet to constant dc voltage.
The voltage produced is used to power all types of electronic circuits including:
Consumer electronics (ex: radio, television, DVD, etc.)
Computers
Industrial controllers
Most laboratory instrumentation systems and equipment
The dc voltage level required depends on the application, but most applications
require relatively low voltage.
There are two basic types of power supplies: a linear power supply and a switching
power supply. These components are described as follows:
1. A linear power supply is one that provides a constant current path between its
input and its load.
2. A switching power supply provides an intermittent current path between its
input and its output.
(a) Complete power supply with transformer, rectifier, filter, and regulator
(b) Half-wave rectifier
Fig. 2-1: Block diagram of a dc power supply with a load and rectifier.
2.2 Transformer
A transformer is a device that changes ac electric
power at one voltage level to ac electric power at
another voltage level through the action of a
magnetic field.

Simple transformer consist of:
1. Primary winding (input winding)
2. Secondary winding (output winding)
3. Magnetic core
If the secondary has more turns than the primary,
the output voltage across the secondary will be
higher and the current will be smaller. If the
secondary has fewer turns than the primary, the
output voltage across the secondary will be lower
and the current will be higher.
The core has a function to concentrate the
magnetic flux.
Fig.2-2: The general arrangement
of a transformer
There are three types of transformers: step-up, step-down, and isolation. These
components are described as follows:
1. The step-up transformer provides a secondary voltage that is greater than the
primary voltage. Ex: a step-up transformer may provides a 240 Vac output with a 120
Vac input.
2. The step-down transformer provides a secondary voltage that is less than the
primary voltage. Ex: a step-down transformer may provides a 30 Vac output with a
120 Vac input.
3. An isolation transformer provides an output voltage that is equal to the input voltage.
This type of transformer is used to isolate the power supply electrically from the ac
power line.
+
NP
1:2 NS
120 Vac
+
240 Vac
-
-
+
NP 4:1 NS
120 Vac
+
30 Vac
-
-
+
NP 1:1 NS
120 Vac
120 Vac
-
-
Step-up
Step-down
Isolation
(a)
(b)
(c)
Fig.2-3
+
The turns ratio of a transformer is equal to the voltage ratio of the component and since,
the voltage ratio is the inverse of the current ratio. By formula:
N sec Vsec I pri


N pri V pri I sec
(2-1)
where
NSec = the number of turns in the secondary
NPri = the number of turns in the primary
VSec = the secondary voltage
VPri = the primary voltage
ISec = the secondary current
IPri = the primary current
By the equation (2-1) can be stated that:
Step-down transformer secondary current is greater than its primary current
(ISec > IPri).
Step-up transformer secondary current is less than its primary current (IPri > ISec).
2.3 Half-Wave Rectifiers
A rectifier is a diode circuits that converts the ac input voltage to a pulsating dc
voltage.
There are three basic types of rectifier circuits:
1. Half-wave rectifier
2. Full-wave rectifier
3. Bridge rectifier
The half-wave rectifier is simply a diode that is placed in series between a
transformer (or ac line input) and its load.
When a half-wave rectifier is positioned as shown in Fig.2-4 (a), it eliminates the
negative alternation of the input. And when positioned as shown in Fig.2-4 (b), it
eliminates the positive alternation of the input.
The direction of the diode determines whether the output from the rectifier is
positive or negative:
1. When the diode points toward the load (RL), the output from the rectifier will be
positive.
2. When the diode points toward the source, the output from the rectifier will be
negative.
Fig.2-4: Half-wave rectifiers.
2.3.1 Load Voltage and Load Current
When we take the value of VF (barrier potential) into account, the peak out voltage
(peak load voltage) is found as
V p ( out)  V p (sec)  VF
(2-2)
Vp(sec) is the peak secondary voltage which is the peak input voltage (Vp(in)) of the
transformer and is found as
V p (sec)
Nsec

V p ( pri)
N pri
where
Nsec/Npri = the ratio of transformer secondary turns to primary turns
Vp(pri) = the peak primary voltage of a transformer
(2-3)
The source voltage are given
as rms values as follows:
Vrms
Vp 
0.707
(2-4)
Fig.2-5: Ideal half-wave
rectifier operation.
2.3.2 Average Voltage and Current
The value of average voltage (VAVG) for an ac (or other) waveform is the value that
would be measured with a dc voltmeter. For a half-wave rectifier, VAVG is found as:
VAVG 
Vp

where Vp = the peak value of the voltage.
Fig.2-6 : Average value of the half-wave rectified signal.
(2-5)
We can also convert a peak current to an average current. The value of the average
current (IAVG) for an ac waveform is the value that would be measured with a dc
ammeter. The value of IAVG can be calculated in one of two ways:
1. We can determine the value of VAVG and then use Ohm’s law as follows:
I AVG
VAVG

RL
(2-6)
2. We can calculate the value of Ip. Then, we can convert this peak value to average
form using equations similar to those we used to convert Vp to VAVG. The current
forms of these equations are:
I AVG 
Ip

(2-7)
2.3.3 Peak Inverse Voltage (PIV)
The maximum reverse bias that will be applied to a diode in a given circuit is called the
peak inverse voltage (PIV) of the circuit. For the half-wave rectifier, the PIV is found as
PIV  V p (sec)
(2-8)
When the diode is reverse biased (Fig.2-5 (b)), no voltage is dropped across the load.
Therefore, all of Vsec is dropped across the diode.
Ex. 2-1:
The fuse shown in Fig. 2-7 is used to limit the current in the primary of the transformer.
Assuming that the fuse limits the value of Ipri to 1 A, what is the limit on the value of the
secondary current if the transformer has a turns ratio of 1:4? Ans.: 250 mA
Ex. 2-2:
A circuit like the one shown in Fig.2-7 has a turns ratio of 1:5 and a fuse that limits the
primary current to 750 mA. Calculate the maximum allowable value of Isec. Ans.: 150
mA
Ex. 2-3:
Determine the peak load voltage for the circuit shown in Fig.2-7. The ratio of secondary
turns to primary turns is 5:1. Ans.: 33.3 V
Ex. 2-4:
A half-wave rectifier has values of Npri = 10, Nsec = 1, and Vp(pri) = 180 V. What is the
peak load voltage for the circuit? Ans.: 17.3 V
Ex. 2-5:
Using the results of ex. 2-3 and 2-4, what are the peak load current, the value of VAVG
and the value of IAVG for the circuit if the load current of the circuit is 120 Ω?
F1
1A
Figure 2-7.
2.4 Full-Wave Rectifiers
There are two types of full-wave rectifiers: center-tapped and bridge full-wave rectifier.
2.4.1 The Center-Tapped Full-Wave Rectifier
A center-tapped rectifier is a type of rectifier that uses two diodes connected to the
center of the secondary winding of a center-tapped transformer, as shown in Fig. 2-8.
The input voltage is coupled through the transformer to the center-tapped secondary.
Half of the total secondary voltage appears between the center tap and each end of
the secondary winding as shown.
Fig.2-8: A center-tapped full-wave rectifier.
The number of positive alternations
that make up the full-wave rectified
voltage is twice that of the half-wave
voltage for the same time interval, as
illustrated by Fig. 2-9. Therefore, the
average load voltage for the full-wave
rectifier is found as:
VAVG 
2V p

(2-9)
Input signal
Half-wave
rectifier output
Full-wave
rectifier output
to
Fig. 2-9: Typical rectifier waveforms.

V p (sec)
2

V p (sec)
2
Fig. 2-30: The operation of the center-tapped full-wave rectifier during the one complete
cycle of the input signal.
During a positive half-cycle of the input voltage, the polarities of the secondary voltages
are shown in Fig. 2-30 (a). This condition causes diode D1 is forward-biased, and diode
D2 is reverse-biased. The current path is through D1 and the load resistor RL, as
indicated.
During a negative half-cycle of the input voltage, the polarities of the secondary
voltages are shown in Fig. 2-30 (b). This condition reverse-biases diode D1, and forwardbiases diode D2. The current path is through D2 and RL, as indicated.
Because the current through the load resistor is in the same direction, so the output
voltage always has the same polarity.
Assuming the diode is an ideal model, the peak output voltage (value of peak load
voltage, Vp(load) ) can be approximated as:
V p ( out) 
V p (sec)
2
(2-10)
The load voltage is approximately equal to half the secondary voltage because
the transformer is center tapped. The voltage from either end of a center-tapped
transformer to the center tap is always half the total secondary voltage.
Effect of the Turns Ratio on the Output Voltage
Using the practical diode model, in the any case, the peak load voltage for a full-wave
rectifier is always one-half of the peak secondary voltage less the diode drop (due to the
barrier potential):
V p (load) 
V p (sec)
2
 0.7 V
(2-11)
If the transformer’s turns ratio is 1, the output voltage equals half the peak primary
voltage less the voltage drop, as shown in Fig. 2-31.
Fig.2-31: Center-tapped full-wave rectifier
with a transformer turns ratio of 1.
V p (load) 
V p ( pri)
2
 0.7 V
(2-12)
In order to obtain an output voltage with a peak equal to the input peak, a steptransformer with a turns ratio of n = 2 must be used, as shown in Fig. 2-32. Meaning
that the total peak secondary voltage is twice the peak primary voltage (Vp(sec) =
2Vp(pri)), so the voltage across each half of the secondary is equal to Vp(pri).
Fig.2-32: Center-tapped full-wave rectifier with a transformer turns ratio of 2.
V p (load)  V p ( pri)  0.7 V
(2-13)
Peak Inverse Voltage
When one of the diodes in a full-wave rectifier is reverse-biased as shown in Fig. 2-33
where D2 is assumed to be reverse-biased, the maximum anode voltage of D1 is
Vm ( anode) 
V p (sec)
2
(2-14)
and the maximum anode voltage of D2 is
Vm ( anode)  
V p (sec)
2
(2-15)
Fig.2-33: Full-wave rectifier PIV (D1 is forward-biased and D2 is reverse-biased.
Since D1 is assumed to be forward-biased, its cathode is at the same voltage as its
anode minus diode drop; this is also the voltage on the cathode of D2. By formula,
Vm( cathode) 
V p (sec)
2
 0.7 V
(2-16)
Thus, the peak inverse voltage across diode D2 is
 V p (sec)
  V p (sec) 
  V p (sec)  0.7V
PIV  
 0.7V    
2 
 2
 
(2-17)
Because the peak load voltage supplied by the full-wave rectifier is approximately half
the secondary voltage. By formula,
V p (load)  V p ( out) 
V p (sec)
2
or
V p (sec)  2V p ( out)
(2-18)
Then, by substitution, the peak inverse voltage across either diode is
PIV  2V p ( out)  0.7 V
(2-19)
2.4.2 The Bridge Full-Wave Rectifier
A bridge rectifier is a type of rectifier that uses four diodes connected as shown in Fig.
2-34.
The bridge rectifier is the most commonly used full-wave rectifier for several reasons:
1. It does not require the use of a center-tapped transformer and therefore can be
coupled directly to the ac power line if desired.
2. When connected to a transformer with the same secondary voltage, it produces
nearly twice the peak output voltage of the conventional full-wave rectifier. This
results in a higher dc output voltage from the supply.
 V p (sec)
(a) During the positive half-cycle of the input, D1 and D2 are forward-biased and conduct current.
D3 and D4 are reverse-biased.
 V p (sec)
(b) During the negative half-cycle of the input, D3 and D4 are forward-biased and conduct
current. D1 and D2 are reverse-biased.
Fig.2-34: Bridge rectifier operation.
If the center-tapped full-wave rectifier produces its output by alternating conduction
between two diodes, the bridge full-wave rectifier alternates conduction between two
diode pairs.
When the input half-cycle is positive, the transformer has the polarity as shown in Fig.
2-34(a), causing diodes D1 and D2 are forward-biased and conduct the current in the
direction shown. A voltage is developed across RL that looks like the positive half of the
input cycle. During the time, diodes D3 and D4 are reverse-biased.
When the input half-cycle is negative, the transformer has the polarity as shown in Fig.
2-34(b). Diodes D3 and D4 now are forward-biased and conduct the current in the same
direction through RL. During the time, diodes D1 and D2 are reverse-biased. A full-wave
rectified output voltage appears across RL as a result of this action.
Bridge Output Voltage
Assuming the diodes in the bridge as shown in Fig. 2-35 to be ideal (in other word, the
diode drops are neglected), the rectifier has a peak output voltage of
V p ( out)  V p (sec)
(2-20)
Fig. 2-35: Bridge operation during a positive half-cycle of the
primary and secondary voltages for a ideal diodes.
If the voltage drops across the two conducting diodes as shown in Fig. 2-36 are taken
into account, the output voltage is
V p ( out)  V p (sec)  1.4V
(2-21)
The 1.4 V in the equation represents the sum of the diode voltage drops. Note that in
the bridge rectifier, two diodes are always in series with the load resistor during both
the positive and negative half-cycles.
Fig. 2-36: Bridge operation during a positive half-cycle of the primary
and secondary voltages for a practical diodes.
Peak Inverse Voltage
When Vsec has the polarity as shown in Fig. 2-37(a), D1 and D2 are forward-biased that
the current flows through D1 and D2 as shorts (ideal model). As shown, D3 and D4 are
connected across the transformer secondary and thus, D3 and D4 have a peak inverse
voltage equal to the peak secondary voltage. Since the output voltage is ideally equal
to the secondary voltage,
PIV  V p (out)
(2-22)
If the voltage drops of the forward-biased diodes are included as shown in Fig. 236(b), the peak inverse voltage across each reverse-biased diode in terms of Vp(out)
is
PIV  V p ( out)  0.7 V
(2-23)
Fig.2-37: Peak inverse voltages across diode D3 and D4 in a bridge rectifier during the
positive half-cycle of the secondary voltage.
Fig.2-37: Peak inverse voltages across diode D3 and D4 in a bridge rectifier during the
positive half-cycle of the secondary voltage.
2.5 Filters and Regulators
Filter is a circuit implemented with capacitor that follows the rectifier in a power
supply.
Filters are used to reduce the fluctuation in the rectified output voltage and produces
a constant dc output voltage as shown in Fig. 2-38.
The small amount of fluctuation in the filter output voltage is called ripple.
Fig.2-38: The effects of filtering on the output of a half-wave rectifier.
The amount of ripple in the output voltage from a given filter depends on rectifier used,
filter component values, and load resistance.
Too much ripple in the output can have different adverse effects, depending on the
application of the power supply.
For example:
In an audio amplifier, excessive ripple can produce an “annoying hum” at 60 or
120 Hz, depending on the type rectifier used.
In video circuits, excessive ripple can produce “video hum bars” in the picture.
In the digital circuits, excessive ripple can result in “erroneous outputs” from logic
gates.
Therefore, filtering is necessary to provide power and biasing for proper operation of
electronic circuits.
2.5.1 Capacitive Filter
Capacitive filter is simply a capacitor
connected in parallel with the load
resistance or connected from the rectifier
output to ground, as shown in Fig.2-39.
During the positive first quarter-cycle of
the input, the diode is forward-biased,
allowing the capacitor charges rapidly, as
illustrated in Fig.2-39(a).
When the input begins to go negative,
the diode is reverse-biased, and the
capacitor slowly discharges through the
load resistance (Fig.2-39(b)). As the
output from the rectifier drops below the
charged voltage of the capacitor, the
capacitor acts as the voltage source for
the load.
During first quarter of the next cycle, as
illustrated in part (c), the diode will again
become forward-biased when the input
voltage exceeds the capacitor voltage.
Fig.2-39: Basic capacitive filter. Current
indicates charging or discharging of the
capacitor
Capacitor
Ripple Voltage
Ripple voltage is the fluctuation in the capacitor voltage due to the difference between
the charge and discharge times.
The difference between the charge and discharge times is caused by two distinct RC
time constant in the circuit. One time constant is found as:
  RC
(2-24)
where R and C are the total circuit resistance and capacitance, respectively.
Since it takes five time constants for a capacitor to charge or discharge fully, this time
period (T) can be found as:
T  5RC  5
(2-25)
For example, refer to Fig. 2-40(a), the capacitor charges through the diode.
Assuming that diode has a forward resistance of 5 Ω, so the time constant for the
circuit is found as:
  (5 )(100 F )  500 s
and the total capacitor charge time is found as:
T  (5)(500 s)  2.5 ms
The discharge path for the capacitor is through the resistor as shown in Fig. 2-40(b).
For this circuit, the time constant is found as:
  (1 k)(100 F )  100 ms
and the total capacitor discharge time is found as:
T  (5)(100 ms)  500 ms
(a) Charge circuit
(b) Discharge circuit
Fig.2-40: The basic capacitive filter.
The shorter time between peaks of the rectifier output voltage causes the capacitor
reduces the more fluctuation. Therefore, the full-wave rectified voltage has a smaller
ripple than the half-wave rectified voltage for the same load resistance and capacitor
values (Fig. 2-41).
Fig.2-41: Comparison of ripple voltages for half-wave and full-wave rectified voltages with
the same filter capacitor and load and derived from the same sinusoidal input voltage.
Ripple Factor
The ripple factor (r) is an indication of the effectiveness of the filter and defined as:
r
Vr ( pp)
VDC
(2-26)
where, Vr(pp) is the peak-to-peak ripple voltage and VDC is the dc (average) value of
the filter’s output voltage (Fig.2-42).
Fig.2-42: Vr and VDC determine the ripple factor.
The lower the ripple factor, the better the filter.
The ripple factor (or the amplitude of the ripple voltage) at the output of a filter can be
lowered by increasing the value of filter capacitance or increasing the load resistance,
as illustrated in Fig. 2-43:
Fig. 2-43.
For a full-wave rectifier with a capacitive filter, approximations for the peak-to-peak
ripple voltage, Vr(pp), and dc value of the filter output voltage, VDC, are given in the
following equations:
Vr ( pp)
VDC
 1 
V p ( rect )
 
 fRLC 

1 
V p ( rect )
 1 
 2 fRLC 
Vp(rect) = the unfiltered peak rectified voltage.
(2-27)
(2-28)
2.5.2 Surge Current
Before the switch in Fig.2-44 is closed, the filter capacitor is uncharged. At the instant
the switch is closed, voltage is connected to the bridge and the uncharged capacitor
acts as a short circuit. As a result, the current is initially limited only by the resistance of
the transformer secondary and the bulk resistance of the diode. Since these resistances
are usually very low, the initial current tends to be extremely high. This high initial
current is referred to as surge current.
Fig.2-44: Surge current in a capacitive filter.
The value of surge current can be calculated as follows:
I surge 
V p (sec)
RW  RB
where, Vp(sec) = the peak secondary voltage
RW = the resistance of the secondary windings
RB = the diode bulk resistance
(2-29)
2.5.3 Voltage Regulators
A voltage regulator is connected to the output of a filtered rectifier and maintains a
constant output voltage (or current) despite changes in the input, the load current, or
the temperature.
The regulator reduces the ripple to negligible amount.
Most regulators are integrated circuits and have three terminals: an input terminal, an
output terminal, and a reference (or adjust) terminal.
Three-terminal regulators designed for fixed output voltages require only external
capacitors to complete the regulation portion of the power supply, as shown in Fig.2-45.
Fig.2-45: A voltage regulator with input and output capacitors.
Fig.2-46: A basic +5 V regulated power supply
2.5.4 Percent Regulation
The regulation can be stated in a percentage in terms of input (line) regulation or load
regulation.
Line regulation specifies how much change occurs in the output voltage for a given
change in the input voltage. It is mathematically defined as a ratio of a change in output
voltage for a corresponding change in the input voltage expressed as a percentage.
 VOUT
Line regulation  
 VIN

 x100%

(2-30)
Load regulation specifies how much change occurs in the output voltage over a certain
range of load current values, usually from minimum current (no load, NL) to maximum
current (full load, FL). It can be mathematically determined with the following formula:
 VNL  VFL 
 x100%
Load regulation  
 VFL 
(2-31)
2.6 Clippers
Clipper is a diode circuit that is used to limit or clip the positive part of the input voltage.
Clipper is often referred to as a limiter.
2.6.1 Series Clippers
Each series clipper contains a diode that is positioned in series with a load resistor, as
shown in Fig.2-47 (a) and (b).
+
-
-
+
+
+
I=0
RL
-
RL
-
(a) Negative series clipper
(b) Positive series clipper
Fig.2-47: Two basic clipper configurations.
The operating principles of the series clipper are as follows:
1. When the diode in a negative series clipper is forward biased by the input signal, it
conducts, and the load voltage is found as
VL  Vin  0.7 V
(2-32)
2. When the diode in the negative series clipper is reverse biased by the input signal,
it does not conduct. Therefore,
Vdiode  Vin
(2-33)
VL  0V
(2-34)
and
A negative series clipper and its associated waveforms are shown in Fig.2-48(a).
3. The positive series clipper operates in the same fashion. The only differences are:
a. The output voltage polarities are reversed.
b. The current directions through the circuit are reversed.
A positive series clipper and its associated waveforms are shown in Fig.2-48(b).
Fig.2-48: Series clipper operation
Comparing Fig.2-4 and 2-49, we can see that the operation of the series clipper is
identical to that of the half-wave rectifier.
2.6.2 Shunt Clippers
Each shunt clipper contains a diode that is positioned in parallel with a load resistor, as shown
in Fig.2-49 (a) and (c).
The operating principles of the shunt clipper are
as follows:
As the input signal goes positive (Fig.2-49(a)),
the diode is forward biased and conducts
current. With the diode conducting, the voltage
across the diode equals the diode forward
voltage (VF). Since the diode and load are in
parallel, the output voltage also equals the
diode forward voltage. By formula:
Vout  VF  0.7 V
(a) Positive shunt clipper. The diode is forward-biased
during the positive alternation (above 0.7 V).
(2-35)
-
When the input signal goes back below 0.7 V
(Fig.2-49(b)), the diode becomes reverse
biased and acts as an open circuit. The output
voltage looks like the negative part of the input
voltage, but with a magnitude determined by
the voltage divider formed by R1 and RL, as
follows:
+
(b) Positive shunt clipper. The diode is reverse-biased
during the negative alternation (below 0.7 V).
Fig.2-49: Two basic shunt clipper configurations.
Vout
 RL 
Vin
 
 R1  RL 
(2-36)
If R1 is small compared to RL, then
Vout  Vin
(2-37)
If the diode is turned around, as in Fig. 2-50, the negative part of the input signal is
clipped off. When the diode is forward-biased during the negative part of the input
signal, the voltage across the diode is held to - 0.7 V. When the input signal goes
above -0.7 V, the diode is no longer forward-biased; and the output voltage is
proportional to the input voltage.
Fig.2-50: Negative shunt
clipper. The diode is
forward-biased during the
negative alternation and
reverse-biased during the
positive alternation.
2.6.3 Biased Clippers/Limiters
A biased clipper uses a dc bias
source, VBias, in series with the diode
to limit an output voltage to certain
level, as shown in Fig. 2-51.
The positive-biased clipper (Fig.251(a)) clips the input signal at the
values of VBias + 0.7 V before the
diode will become forward-biased
and conduct. Once the diode starts to
conduct, the voltage at point A is
limited to VBias + 0.7 V so that all
input signal above this level is
clipped off.
The negative-biased clipper (Fig.251(b)) works in the same fashion, but
it clips the input signal at the values
of -VBias - 0.7 V.
(a) A positive clipper or limiter
(b) A negative clipper or limiter
Fig.2-51: Biased shunt clipper or limiter
By turning the diode around, the
positive clipper can be modified to
limit the output voltage to the
portion of the input voltage
waveform above VBias - 0.7 V, as
shown in Fig.2-52(a).
Similarly, the negative clipper can
be modified to limit the output
voltage to the portion of the input
voltage waveform below –VBias +
0.7 V, as shown in part (b).
Fig. 2-52.
2.6.4 Voltage-Divider Bias
The bias voltage sources can be replaced by a resistive voltage divider that derives the
desired bias voltage from the dc supply voltage, as illustrated in Fig.2-53.
(a) A positive clipper
(b) A negative clipper
(b) A variable positive clipper
Fig. 2–53: Diode clippers implemented with voltage–divider bias.
The bias voltage is calculated by using the voltage-divider formula as follows:
 R3 
VSupply
VBias  
 R2  R3 
(2-38)
2.7 Clampers (DC Restorers)
Clamper is a diode circuit designed to shift a waveform either above or below a given
reference voltage without distorting the waveform.
There are two types of clampers: the positive clamper and the negative clamper.
1. A positive clamper shifts its input waveform so that the negative peak of the
waveform is equal to the clamper dc reference voltage.
For example: Fig. 2-54 shows what happens when a 20 Vpp sin wave is applied to a
positive clamper with a dc reference of 0 V. The input and output waveforms have the
value of 20 Vpp. However, the clamper output waveform has the positive peak of +20 V and
the negative peak of 0 V. The positive clamper has shifted the entire waveform so that
its negative peak is equal to the circuit’s dc reference voltage.
Fig.2-54: The input/output characteristics of the positive clamper circuit
2. A negative clamper shifts its input waveform so that the positive peak of the waveform
is equal to the clamper dc reference voltage.
For example: Fig. 2-55 shows what happens when a 20 Vpp sin wave is applied to
a negative clamper with a dc reference of 0 V. In this case, The clamper output
waveform has the positive peak of 0 V and the negative peak of –20 V. The
negative clamper has shifted the entire waveform so that its negative peak is
equal to the circuit’s dc reference voltage.
Fig.2-55: The input/output characteristics of the negative clamper circuit
Clamper Operation
The clamper is similar to a shunt
clipper; the difference is added
capacitor in the clamper, as
illustrated in Fig. 2-56.
For the circuit in Fig. 2-56(a), the
diode is forward biased and it
charges the capacitor. Thus, the
charging time constant is found as:
  RDC
(a) Capacitor charge circuit
(2-39)
+
where RD is bulk resistance of
the diode and C is capacitance
of the capacitor. The total
charge time is:
TCh arg e  5 RD C
−
Reverse
-biased
(b) Capacitor discharge circuit
(2-40)
Fig.2-56: Clamper charge and discharge
When the diode is reverse biased, the capacitor starts to discharge through the
resistor, as shown in Fig. 2-56(b). Therefore, the discharge time constant is found as:
  RLC
(2-41)
and the total discharge time is found as:
TDischarg e  5RLC
(2-42)
The effect of the clamping action is shown in Fig. 2-56. The capacitor retains a charge
approximately equal to the input peak less the diode drop so that it acts as a battery.
Fig.2-57: Positive
clamper operation
If the diode is turned around, a negative dc voltage is added to the input voltage to
produce the output voltage as shown in Fig. 2-57:
Fig.2-57: Negative clamper operation
Ex. 2-6:
Assume the circuit in Fig.2-40 has an input of 170 Vpk, a turns ratio of 2:1, and values
of RW = 0.5 Ω and RB = 8 Ω. What is the initial surge current for the circuit?
Ex. 2-7:
A negative shunt clipper has a values of RL = 510 Ω and R1 = 100 Ω. If the input
voltage is +15 Vpk, what is the peak load voltage?
2.8 Voltage Multipliers
A voltage multiplier is a circuit providing a dc output voltage that is a multiple of its
peak input voltage.
When a voltage multiplier increases a peak input voltage by a given factor, the peak
input current is decreased by approximately the same factor.
The typical application of a voltage multiplier is to supply the high-voltage, low-current
input required to operate the cathode-ray tube in a television and to operate the
particle accelerators.
There are several types of the voltage multiplier:
Voltage doublers : half-wave voltage doubler and full-wave voltage doubler.
Voltage tripler
Voltage quadrupler
2.8.1 Voltage Doubler
Voltage doubler is a voltage multiplier with a multiplication factor of two.
Half-Wave Voltage Doubler
Half-wave voltage doubler consist of two diodes and two capacitors, as shown in Fig.2-58.
Fig.2–58: Half-wave voltage doubler operation. Vp is the peak secondary voltage.
During the positive half-cycle of the secondary voltage (in Fig.2-58(a)), diode D1 is
forward-biased and D2 is reverse-biased. Capacitor C1 charges until its plate-to-plate
voltage equals the peak value of the secondary voltage less the diode drop.
During the negative half-cycle, D2 is forward-biased and D1 is reverse-biased (in part b).
AC voltage source and C1 charge C2 until its plate-to-plate voltage is approximately 2 Vp.
Applying Kirchhoff’s law around the loop as shown in part (b), the voltage across C2 is
VC1  VC 2  V p  0
VC 2  V p  VC1
Neglecting the diode drop of D2, VC1 = Vp. Therefore,
VC 2  V p  V p  2V p
Input and output waveforms for a half-wave voltage doubler are shown in Fig. 2-59.
Fig. 2-59.
Full-Wave Voltage Doubler
The full-wave voltage doubler closly resembles the half-wave voltage doubler, as shown
in Fig.2-60.
Fig.2–60: Full-wave voltage doubler operation.
During the positive half-cycle of the secondary voltage, D1 is forward-biased and C1
charges until its plate-to-plate voltage is approximately equal to Vp.
During the negative half-cycle, D2 is forward-biased and C2 charges until its plate-toplate voltage is approximately equal to Vp. Since C1 and C2 are in series, the total output
voltage across the two capacitors is 2 Vp.
2.8.2 Voltage Tripler
Voltage tripler is variation on the half-wave voltage doubler. It is created by the addition
of another diode-capacitor section to the half-wave voltage doubler, as shown in Fig.261.
Fig. 2–61: Voltage tripler.
On the positive half-cycle of the secondary voltage, C1 charges through D1 until its
plate-to-plate voltage is approximately equal to Vp.
During the negative half-cycle, C2 charges through D2 until its plate-to-plate voltage is
approximately equal to 2Vp.
During next positive half-cycle, C3 charges through D3 until its plate-to-plate voltage is
approximately equal to 2Vp. The voltage across C1 and C3 add up to 3Vp.
2.8.3 Voltage Quadrupler
Voltage quadrupler is also variation on the half-wave voltage doubler. It is produced by
the addition of still another diode-capacitor section, as shown in Fig.2-62.
Fig. 2–62: Voltage quadrupler.
C4 charges through D4 until its plate-to-plate voltage is approximately equal to 2Vp
on a negative half-cycle. The voltage across C1 and C3 now add up to 4Vp.
Summary:
Half-wave voltage doubler provides a dc output voltage that is approximately twice its
peak input voltage.
Full-wave voltage doubler provides a dc output voltage that is approximately twice its
peak input voltage.
Voltage tripler produces a dc output voltage that is approximately three times its peak
input voltage.
Voltage quadrupler produces a dc output voltage that is approximately four times its
peak input voltage.
2.9 The Diode Datasheet
Datasheet gives detailed information on a device as maximum ratings, electrical
characteristics, mechanical data, and graphs of various parameter. Thus, it can be used
properly in a given application.
Diode datasheet is commonly divided into two categories:
1. Data given in table form: (a) absolute maximum ratings, (b) thermal characteristics
and (c) electrical characteristics.
2. Characteristics shown in graphical form: (a) forward current derating curve, (b)
forward characteristics curve, (c) nonrepetitive surge current and (d) reverse
characteristics.
Fig. 2–63: A selection of rectifier diodes based on maximum ratings of IO, IFSM, and VRRM.
Fig. 2–63: A selection of rectifier diodes based on maximum ratings of IO, IFSM, and VRRM.
2.10 Troubleshooting
Power supply faults may occur in the transformer, rectifier, filter, or regulator.
2.10.1 Transformer Faults
The transformer in a power supply develop one of several possible faults:
1. A shorted primary or secondary winding.
2. An open primary or secondary winding.
3. A short between the primary and secondary winding and the transformer frame.
In most cases, a shorted primary or secondary winding will cause the fuse to blow.
2.10.2 Rectifier Faults
Half-wave rectifier is the easiest rectifier to troubleshoot. If the diode in the rectifier
shorts, the output will be a sine wave. If the diode opens, the resulting symptom is an
output of 0 V, as illustrated in Fig. 2-64.
Fig. 2–64: The effect of an open diode in a half-wave rectifier is an output of 0 V.
In full-wave rectifier, a shorted diode will cause the power supply fuse to blow. Fullwave center-tapped rectifier is shown in Fig.2-65. If either of two diodes is open, the
output voltage will have twice the normal ripple voltage and the ripple frequency will go
from 120 to 60 Hz.
Fig. 2–66: Effect of an open diode in a bridge rectifier.
The symptoms for open diode in the bridge rectifier are the same as those for the
center-tapped full-wave rectifiers, as shown in Fig.2-66. In the case, we have more
diodes that must be tested.
Fig. 2–66: Effect of an open diode in a bridge rectifier.
2.10.3 Filter Faults
Capacitors store an electrical charge, and they can retain that charge even after the
power switch has been turned off or the ac plug has been disconnected. This is a
danger to the technician and also, the test equipment can be damaged.
As a safety precaution, capacitors should be discharged by shorting the terminals
with a shorting tool before taking any measurements.
Three types of defects of a filter capacitor are illustrated in Fig.2-67:
If the filter capacitor for a full-wave rectifier opens, the output is a full-wave
rectified voltage.
If the filter capacitor shorts, the fuse will blow and the output is 0 V. When the
filter capacitor shorts, this effectively shorts out the transformer secondary,
causing excessive secondary current.
If the filter capacitor leaks, the leakage resistance will reduce the time constant
and allow the capacitor to discharge more rapidly than the normal. This results in
an increase in the ripple voltage on the output. This fault is rare.
Fig. 2–67: Effects of a faulty filter capacitor.