1

Semiconductor Diodes and Application

Introduction
The diode is a two terminal device which behaves as one way device by offering low
resistance when forward biased and high resistance or almost open switch during reverse
biased condition. A practical diode offers some voltage drop 0.7 V for silicon diode and
0.3 V for Germanium diode. Diodes have many applications like clippers, clampers,
Regulators, Rectifiers etc. Low current diodes are used in switching, high current diodes are
used in Rectifiers.

PN-junction Diode
PN-junction is formed when a single crystal of semiconductor is added with an acceptor
impurity on one side and donor impurity on the other side and is as shown as in Figure
below. A small quantity of trivalent impurity is added to n-type silicon material and
pentavalent impurity is added to p-type silicon material.
connecting leads
P-Type N-Type

junction

Figure: PN- junction

The left side material is a p-type semiconductor having acceptor ions and positively charged
holes, the right material is n-type having positive donor ions and free electrons. Since
n-type has high concentration of electrons and p-type has high concentration of holes and
there exists a concentration gradient across the junction. Due to this, charge carriers move
from high concentration area towards low concentration area to achieve uniform distribution
of charge. This is shown in Figure below.
Migration hole Migration electron

P N

Figure: Charge accumulation in PN- junction

In p-type excess holes move towards n-side similarly electrons from n-side move
2
towards p-side, this process is called diffusion and diffusion of charge carries take place on
either side of the junction. This diffusion of charge carriers takes place in neighborhood of
the junction immediately after the junction is formed, and the rest of the material will be at
equilibrium under no bias condition.
When the migrating electrons diffuse into p-type and recombines with the acceptor
atoms on p-side, the acceptor ions accepts this additional electrons and becomes negatively
charged immobile ions, and the hole disappears and free electron becomes valence electron.
Similarly when hole diffuses into n-side they recombine with donor atom, this donor atom
accepts additional hole and they become positively charged immobile ion and electron
disappears. These ions are covalent bonded and hence cannot move around freely.
paragraph ka matlab how depletion region
p-n junction form??
ptype negative kyun show ho rha P-Type N-Type dekho because of tranfer of electron and
hai because during diffusion holes on opposite sides and hence become
electrons from n side moves immobile ions and due to these ions large
towards p side and hence full the positive charge form on N side and large
holes so holes ke karam jo wo negative charge appear on p side because of
positive tha wo an electron ke aane this n side positive repel the pside majority
se negative ho jayega and same for holes and same for ntype also which creates
n side also like extra lectrons are the depletion region
present hence holes jo migrate hue
n side me wo positive create kr Depletion width(1 × 10–6 m)
denge on n side

Figure: Creation of depletion region

After diffusion, negative ions are formed on the p-side and positive ions are formed
on the n-side closer to the junction as shown in Figure below. If the doping density is same
on both sides then large positive charge gets accumulated on n-side and large negative
charge gets accumulated on p-side of the junction, thus these charges at junction repel and
do not allow further migration of carriers from one side to the other side of the junction
Thus the uncovered ion in the neighborhood of the junction is depleted of mobile charges and is
called depletion region, or the space charge region or the transition region. Figure shows creation of
depletion region.
PN-Junction diode: A pn-junction provided with copper wire connecting copper lead
becomes an electronic device known as a Diode.
There are three types of diodes, classified according to their current or reverse voltage
carrying capacity, they are as follows:
Low current diodes:
The low current diode is as shown in Figure below. The structure of these diode is 3 mm
long. It allows a maximum forward current of 100 mA and can withstand reverse voltage
of 75 V.
Coloured band

Anode Cathode
Low current diodes:
Medium current diodes are as shown in Figure above. These diodes allow a forward current of
400 mA and survive a reverse voltage of 200 V.
High current diodes or Power diodes:
Power diodes are shown in Figure below. Since they are power diodes lot of heat is generated.
They are capable to carry forward current of many amperes and can tolerate several hundreds of
reverse voltage. Anode Cathode
Anode Cathode

Figure : Medium current diodes Figure: High current diodes

Characteristics and Parameters of PN-junction

Characteristics
PN-junction works under following conditions:
a) No bias, b) Forward bias and c) Reverse bias
Biasing: Biasing is connecting a p-n junction to an external d.c. voltage.

a) No Bias Condition: No Bias Condition is as shown in Figure below.

Under no bias condition, the positive charge on p-side repel the holes to cross from p to n
side, negative charge on n-side repel free electrons to enter from n to p side. Thus, a barrier is
setup against further movement of charge carriers, this is called potential barrier or junction
barrier. The potential barrier is of the order of 0.7V for silicon and 0.3V for germanium.
Barrier
Electric field E
E
P-Type N-Type

Metal contacts

(a)
d 2V D
=
dX 2 E
Charge density 'p'
(b)
I = 0 mA Distance from junction

V = fedx Vo Potential barrier for holes

(c)
Distance from junction
Potential barrier for electrons Eo
(d)
Distance from junction

+ –
V = 0 V (no bias)

Figure: A pn junction under no-bias condition

Since the potential barrier of electron is inverted compared to potential barrier of holes in
Figure, this is due to the charge on an electron is negative.
Similarly the potential barrier against flow of holes from p-side across the junction is
as shown in Figure, and the potential is positive due to charge on hole is positive. Due to
the presence of potential barrier (cut-in voltage) which in turn prevents further movement of
majority carriers across the junction. But the barrier promotes the minority carriers in n-
type (holes) that finds a path to pass directly into p-type material, due to negative in the
p-type near the junction. Similarly the minority carriers (electrons) in p-side pass directly
into n-type due to positive potential in n-type. Thus in the absence of an applied bias the net
flow of charge in anyone direction for a semiconductor is zero.
b) Forward Bias: The forward bias condition is as shown in Figure. When an external
voltage is applied to the junction is in such a direction that it cancels the potential
barrier, thus permitting current flow, is called forward biasing. To apply forward bias,
connect positive terminal of the battery to p-type and negative terminal to n-type.The
applied forward potential establishes the electric field which acts against the field
due to potential barrier. Therefore the resultant field is weakened and the barrier height
is reduced at the junction as shown in Figure below. Since the potential barrier voltage
is very small, a small forward voltage is sufficient to completely eliminate the
barrier.once the potential barrier is eliminated by the forward voltage, junction resistance
becomes almost zero and a low resistance path is established for the entire circuit.
Therefore current flows in the circuit. This is called forward current. Figure below
shows forward VI characteristics of pn junction. Cut in or knee voltage is the
minimum voltage below which the current through the diode is zero. For silicon diode
cut in voltage is 0.7 V and for Germanium diode it is 0.3 V.
P-Type N-Type

holes IP
Io
electrons

+
Reduced –
barrier
Conventional current
V
Figure Forward biasing of p-n junction potential (VB)
I

100
Forward current (mA)

F
or F
80 G or
er sil
60 m ic
an on
40 iu (S
m i)
20 di di
od od
e e V
0 0.2 0.4 0.6 0.8 1 Forward bias voltage
Figure Forward VI characteristics of pn junction

c) Reverse Biasing: The reverse bias condition is as shown in Figure below . The
condition under reverse bias is explained below.
When the external voltage applied to the junction is in such a direction, that the
potential barrier is increased, then it is called reverse biasing. To apply reverse bias, connect
negative terminal of the battery to p-type and positive terminal to n-type. The applied reverse
voltage establishes an electric field which acts in the same direction as the field due to
potential barrier. Therefore the resultant field at the junction is strengthened and the barrier
height is increased as shown in Figure below.The increased potential barrier prevents the
flow of charge carriers across the junction. Thus a high resistance path is established for the
entire circuit and hence current does not flow. Figure below shows reverse VI characteristics
of Silicon and Germanium diode.
Increased depletion
P-Type layer N-Type

Increased Barrier potential

t
V
– +

Figure : Reverse biasing of p-n junction

Reverse breakdown
For Silicon diode voltage For Germanium diode
(Reverse voltage)
VR –100 –80 –60 –40 –20 –50 –25
VR
–50 –0.5
Reverse breakdown
voltage –100 –1

–150 –1.5

–200 –2
IR (nA) (Reverse current) IR (μA)

Figure: reverse VI characteristics of Silicon and Germanium diode

Diode Parameters
⯈ Forward voltage drop (VF): It is the maximum forward voltage drop for a given
forward current, at a given device temperature.
 ⯈ Reverse saturation Drop (IR): Whenever pn junction is reverse biased a small current
flows due to the minority charge carriers across the junction, this is known as reverse
saturation current.
⯈ Reverse breakdown voltage (VBR): When the reverse bias is applied to a PN junction
and is increased, a point is reached when the junction breaks down and reverse current
shoots up to a value limited only by the external resistance connected in series with
the junction. This critical value of the voltage is known as the reverse breakdown
voltage.
⯈ Maximum forward current (IF(max)): It is the maximum current that may be passed
continously through the diode.
⯈ Static Resistance: The static resistance of a diode is the resistance offered by a
forward biased diode at a particular point on its V-I characteristics.
⯈ Dynamic resistance (rd) of the diode is the resistance offered to changing levels of
forward current. The dynamic resistance is also known as the incremental resistance or
AC resistance and is the reciprocal of the slope of the forward characteristics beyond
the knee voltage as shown in the Figure below.
IF

∆IF

∆VF

VF

Figure : Determination of diode dynamic resistance from the

forward characteristics
∆VF
rd =
∆I F

Where rd is the dynamic resistance

The dynamic resistance can also be calculated from the equation

26 mV
rd =
IF
Diode Approximations
a) Ideal Diode Characteristics
The cut-in voltage is zero. Since for an ideal diode, there is no barrier potential, any small
forward bias voltage causes conduction through the device. It has zero forward drop infinite
reverse resistance The diode readily conducts when forward biased and it blocks conduction
when reverse biased. The reverse saturation current is zero. This is as shown in Figure. But
Ideal diode does not exist to match this analysis.

IF

VR VF
IR
Ideal diode Forward condition Reverse condition

Figure: Ideal Diode Figure: Ideal diode characteristics

b) Second Approximation
Diodes are to be assumed nearly ideal for situations which require exact values of load
current and load voltage but Ideal diodes does not exist in practical situations. Hence we
assume the second Approximation model. In this we consider the diode voltage drop of
silicon as 0.7 V and Germanium as 0.3 V. Constant voltage source of 0.7 V or 0.3 V is
assumed to be in series with Ideal Diode as shown in Figure.

A K

VK Ideal diode

Figure: Second Approximation circuit

IF IF
0.3 V 0.7 V

VR VF VR VF
IR IR

Figure: Approximate Figure:

Approximate characteristics characteristics for
a for a silicon diode
germanium diode

c) The Third Approximation: Piecewise Linear Characteristics

Figure shows Third approximation. Since the forward characteristics
of diode is not a vertical straight line. A straight-line approximation, called the
piecewise linear characteristic, may be employed. To construct the piecewise
linear characteristic, VF is first. marked on the horizontal axis, as shown in
Figure. Then, starting at VF, a straight line is drawn with a slope equal to the diode dynamic
resistance. I (mA) (V + ∆V )
F F B
200
160
120
80
40
VK 0 A
Rf
–VF 0
0.2 0.4 0.6 (VF)
A K
Ideal diode VF

Figure: Third Approximation circuit Figure: Piecewise Linear

Characteristics of a diode
Example 1:
A silicon diode is used in the circuit shown in Figure. Calculate the diode current.

IF
+
RF IFVF
E
5 kΩ –
20 V
+
VF
–

Figure
Given: E = 20V, RF = 5 kΩ, IF = ?
E1 = IF R1 + VF
E − VF 20 V − 0.7 V 19.3
IF = = = = 3.86 mA
R1 5 k 5 10 3

Example 2:
Construct the piecewise-linear characteristic for a silicon diode that has a 0.4 Ω. Dynamic
resistance and a 200 mA maximum forward current.
Given: VF = 0.7 V
IF(max) = 200 mA,
 rd = 0.4 Ω
VF = IF  rd
= 200 mA  0.4 Ω
= 0.08 V

Plot point B (on Figure) at: IF = 200 mA and VF = (0.7V + 0.08 V). Draw the
characteristic through points A and B.
Plot point A on the horizontal axis at:

mA
B
200
180
160 0.7 + 0.08
140
120 (V + ∆VF)
100
80
60
40
20
0 A
0 0.2 0.4 0.7 0.8

Figure

Diode DC Equivalent Circuit

The equivalent for a device for a device is a circuit representing its internal behavior.
In case of a diode the circuit is made up of a number of compounds, such as resistors and
voltage cells. V VF
F
rd Ideal diode

(a) Basic dc equivalent circuit (b) Complete dc equivalent circuit

Figure: Diode equivalent circuit

An exact equivalent circuit for the diode includes the dynamic resistance (rd) in series
with diode forward drop VF as shown in the Figure. The small variations in VF due to change
in forward current is considered. With rd included in the equivalent circuit represents a diode
with piecewise characteristics. The circuit in Figure above is called as piecewise linear
equivalent circuit.
 particular biasing
DC Load Line condition
It is a graphical analysis of a diode circuit, giving precise levels of diode current and voltage. It
is a straight line that illustrates all dc conditions that could exists within the diode circuit.
Consider the diode circuit shown in Figure below. Applying the KVL we get,
E = V = IF R1 + VF
When IF = 0, in eqn. 1 we get V = VF and
When VF = 0 in eqn. 1 we get V = IF R or IF = V/R
Plotting these two conditions as shown in Figure, that is identifying point A equal to
V/R and point B equal to VF and drawing line AB which represents the dc load line.
(mA)
IF
+ 50 B
RF IFVF IF
E 40 Q
100 Ω
–
5V 30
+ dc load line
DF VF 20
–
10
(V)
0 A
0
1 2 3 4 5 6
VF

Figure Diode-resistor series circuit (b) Plotting the dc load line on the diode
characteristics
The Q point: It is the point of intersection of the diode forward characteristics with
the load line. There is only one point on the dc load line where the diode voltage and current
are compatible with the circuit conditions. The Q-point or Quiescent point or operations
point given the operating conditions of the diode.
DC load line analysis
DC load line analysis is defined as graphical analysis of DC load line which gives
precise values of forward voltage and current. Figure above shows graphical representations
of dc load line drawn on the diode forward characteristics. It is a straight line that demonstrates
all dc conditions that could exist within the circuit.

Example 3:
Draw the dc load line for the circuit in Figure The diode forward characteristic is given in
Figure
 mA
IF

100 Ω R1
10 V

VF

0.7VF

Given: E = V = 10 V R1 = 1000 Ω, VF = 0.7 V

Substitute IF = 0 in equation E = V = IFR1 + VF (1)
E = (IF R1) + VF = 0 + V
VF = E = 10 V
or,
Plot point A on the diode characteristic at, IF = 0, and VF = 5 V
Now substitute VF = 0 into Eq. 1,
E = (IF R1) + 0
E 10 V = 10 mA
giving, IF = =
R1 1000 
(mA)

12
10
IF 8
6 Q-point
4
2
0 (V)
0.7 10
VF

Figure

Plot point B on the diode characteristic at,

IF = 10 mA, and V1 = 0
Draw the dc load line through points A and B.
Temperature Dependence of VI-Characteristics
Power Dissipation: The power dissipated in a diode is simply calculated as the product of
terminal voltage of the diode multiplied by the current flowing through the diode.
PD = VF IF, where PD is the power dissipated in the diode.
Device manufacturer specify maximum power dissipation for each type of diode. If
the specified value of power is exceeded at higher temperature, the diode will get destroyed
either by becoming open or short due to overheating.

De-rating of Diodes
The power dissipation in a diode is calculated as the product of voltage across the
device and the current through the device.
i.e. PD = VF IF in watts
The diode manufacturer specifies in the data sheet, the maximum power that can be
dissipated by the device under normal operation. If the specified level is exceeds, then device
is over heated and may get damaged. A typical de-rating graph is shown in Figure.
The maximum power that can be dissipated is specified for an ambient temperature
of 25°C, if the temperature of the device exceeds maximum power dissipation then it must
be derated. The maximum power dissipation for any temperature can be read from the graph
provided by the manufacturer, and is shown in Figure below. The derating factor defines
the slope of power dissipation versus temperature and is shown in Figure below. The derated
power at P2 can be calculated as follows:
P2 = (P1 at T1) - [DT  (derating factor)]
(mW)
ΔP
100 derating factor =
ΔT
80
P 60
∆P
40
∆T
20
0
0 °C
10 20 30 40 50 60 70 80 90 100

Figure: Typical diode de-rating graph

Example 1:
Calculate the maximum current than can flow through IN 759 as shown in Figure below at
device temperature 60° and 80°, PD = 250 mW.
13

RS

+
V 30 V D1
6V
– IN759

Figure

Given: P1 = PD = 250 mW, Vz = 6 V, V = 30 V

PD 250´10 -3
Iz max = = = 41.66 mA
Vz 6V
at 80° C
P2 = (P1 at T1) – (80° C – 60° C) (derating factor)
P2 = 250 mW – 20° C × 2.2 mW /°C

P2 = 250 mW – 44 mW = 206 mW

P2 206x10 -3
Iz max = = = 34.33 mA
Vz 6

Example 2:
A germanium diode with a 600 mW power dissipation at 26° has a derating factor of
4 mW / °C. Find maximum forward current at 26° C and 45° C temperature.

Given: P1 =PD = 600 mW, derating factors = 4 mW/°C

P1 600´10 -3
IF = = =2 V F = 0.3 V for Ge
VF 0.3 A

P2 = (P1 at T1) – ∆T × (Derating factor)

P2 = (600 mW) – (45° C – 26° C)  4 mW / °C

P2 = 0.6 – 0.076 = 0.524 W = 524 mW

P 524x10 -3
IF = V F = 1.746 A.
0.3
14

Diode ac Equivalent Circuit

Junction Capacitance
The junction between p and n region of a diode is depleted of charge carriers. It results in a
capacitor. The p type material acts as positive metallic plate, n type material acts as negative
metallic plate and junction acts as dielectric medium. Hence the diode exhibits the behavior
of parallel plate capacitor.
There are two types of capacitances exists in the diode when it is subjected to
alternating voltage. They are as follows:
a) Diffusion capacitance: (capacitance of the forward biased junction) (Cd)
b) Depletion capacitance: or transition capacitance (capacitance of the reverse
biased junction) (Cj)
a) Diffusion capacitance or Storage Capacitance
When a pn-junction is forward biased, the majority carriers on p-side which are holes
diffuse into n-side. Similarly, from n-side the majority carriers which are electrons diffuse
into p-side resulting in decrease of depletion region. If the applied voltage increases then the
concentration of carriers increases, resulting in capacitance.
Cd

+ –

IF

Figure: Diffussion capacitances

A diffusion capacitance is present at a forward biased Figure shows capacitance under
forward bias. Therefore the diffusion capacitance is defined as the rate of change of carriers
with external applied ac voltage.
dQ
that is, Cd = ––
dV
The diffusion capacitance is proportional to forward current IF and the practical value
varies from nF to pF.
b) Depletion capacitance or Transition capacitance
This is the capacitance exhibited when diode is reverse biased. When pn-junction
is reverse biased, the majority carriers move away from the junction resulting in wider
depletion region. The p-region and the n-region acts as the metal plates of capacitor with the
depletion layer acting as dielectric material resulting in capacitance and this capacitance is
called “Depletion capacitance”. If the reverse voltage, increases, the width of the depletion
layer also increases, hence capacitance decreases. This capacitance is also called junction
capacitance or transition capacitance. Figure below shows depletion capacitance under
reverse
15
bias condition of diode.
Cpn

+ –

Figure A depletion layer capacitance occurs at a reverse-biased diode

AC Equivalent Circuit of a Diode

The ac equivalent circuit is the modification of the dc equivalent circuit discussed earlier
and is as shown in the Figure
+ – + –

RF VF rd

Cpd Cd
(a) Equivalent circuit for a reverse-biased diode (b) Equivalent circuit for a
forward-biased diode
Figure: Diode capacitance equivalent circuits
When the pn-junction is reverse biased, the equivalent circuit consists of a voltage
cell representing cut-in voltage in series with dynamic resistance rd. The whole circuit is a
combination of the above parameters connected in parallel with diffusion capacitance and is
as shown in Figure. Similarly the equivalent circuit of a pn-junction under reverse bias is
represented by the reverse bias resistance Rr in parallel combination with depletion
capacitances shown in Figure .

Reverse Recovery Time

The presence of the junction capacitances, affects the switching characteristics, of the
diode. Most diode switch quickly into forward biased condition, however, there is longer
turn off time due to junction diffusion capacitance.
Figure below illustrates the effect of a pulse on the diode forward current. When the
pulse switches from positive to negative, the diode conducts in the reverse direction instead
of switching off. The reverse current (Ir) initially equals the forward current (IF), then it
gradually decreases towards zero. The high level of reverse current occurs because, at the
instant of reverse bias there are charge carriers crossing the junction depletion region and
these must be removed.
Typical values of reverse recovery time for switching diodes ranges from
4ns to 50ns. Figure b e l o w shows that, to keep the diode current to a minimum, the
fall time (tf) of the applied voltage pulse to the must be much larger than the diode
reverse recovery time.
16

V VF Forward bias
Forward bias
VF
t
t
Reverse bias
VR Reverse bias
Vr
tf

I
IF
IF
IR
IR
t

IR Reverse saturation (b)

current
Trr
(a) tf >> trr
tf = 10 trr
(a) Effect of instantaneous reverse bias (b) Effect of tf >> trr

Figure: Diode reverse recovery time

The reverse recovery time (trr), is the time required for the current to decrease to the
reverse saturation level.
Typically, tr (min) = 10 trr

Diode Applications
Diode has numerous applications in this the main applications is in rectifiers and regulators.
The component which converts a sinusoidal AC signal waveform into single polarity half
cycle is diode. One important application of diode is rectification. Process of conversion of
AC to DC is called rectification. Any electrical device used for rectification offers a low
resistance to the current in one direction but a very high resistance to current in opposite
direction is called rectifier.

Rectifiers
Rectifiers are the devices which converts input AC to output DC.
There are two types of rectifiers.
1. Half Wave Rectifier.
2. Full Wave Rectifier.
17

Half Wave Rectifier

VS D
A

Vi R Vo

B
Figure
Vi
Vi

t
t
D D
ON OFF

VS I R Vo VS RL Vo

Figure Figure

Figure
(a) Halfwave rectifier circuit (b) Halfwave rectifiers during positive half cycle
(c) Half wave rectifiers during negative half cycle
In almost all the rectifier circuits, transformers are generally used for following purposes.
i. Either to step up or step down the input voltage.
ii. To provide better isolation between AC supply and rectifier circuit.
Operation: When AC input is applied to primary of the transformer, it is mutually coupled
across secondary. During positive half cycle of Vi (0 to π) point A is positive and point B is
negative. Since anode is positive, diode is forward biased and acts as short circuit. Current
flows through RL and Vo follows the input. This is shown in Figure.
 Vo = Vm sin wt for 0  wt  π

During negative half cycle of input Vi (π to 2π) point B is positive where as A is

negative hence diode D is reverse biased and acts as open circuit as a result no current flows
through RL. Therefore no output voltage flows.
Vo = 0 for π  wt  2π
18

Vi Vm

0 t
π 2π 3π

IL
Im

0 t
π 2π 3π
Vo

t
0 π 2π 3π

Figure: Half wave rectifier output waveform

Load current and load voltage waveforms are shown in the Figure,
Vi = Vm sin wt = Vm sin θ
During positive half cycle
i = Im sin θ for 0  θ  π
i = 0 for π  θ  2π
The maximum load current is given by
Vm
Im =
19
R f + Rs + RL

where Rf is forward resistance of diode.

Rs is transformer secondary winding resistance.
RL is load resistance.
20
21
22

half wave
23
24

the maximum reverse voltage the diode can withstand is

25
Full Wave Rectifiers
In a full wave rectifier the current flows through the load for the entire cycle of input. The full wave
rectifiers are classified into two types.
1. Centre tapped full Wave Rectifier. 2. Bridge Rectifier.
Centre Tapped Full Wave Rectifier

Full Wave rectifier consists of two diodes D1 and D2 connected across centre tap secondary
and load RL.Let the instantaneous sinusoidal voltage V = Vm t.
Transformer is splitted into two halves. One half of split voltage appears in series with
D1 and RL, but the other half appears across D2 and RL. Secondary of transformer is split
into two halveπs equally due to equal number of turns across point ‘0’. Hence the name
centre tapped.
Operation
The centred tapped full wave rectifier is as shown in Figure . During the positive half cycle
from 0 to π of AC input V = vm sin ωt the point ‘A’ is positive with respect to ‘0’, similarly
the point ‘B’ receives an inverted sinusoidal waveform with respect to point ‘A’.
D1
A
+

Vm
RL
0 –
AC i/p + E
C Vo
Vm + –
–
B
D2
Figure : Centre tapped full wave rectifier
Diode D1 anode is positive with respect to point ‘0’ and the diode D1 is forward biased
and acts as short circuit or closed switch which provides a very low resistance path for the
current to flow through diode D1 and RL. The direction of current flowing through RL as
indicated in Figure. D -ON
A 1

I1

RL
0
AC i/p E
F
Vo
+ –

B D2-OFF
Figure : Circuit during positive half cycle
Diode D2 receives opposite polarity and is reverse biased or acts as open switch.
Similarly during negative half cycle from p to 2p point ‘B’ is positive with respect to
‘0’ hence D2 is forward biased acts as a closed switch and allows the current to flow
through RL. But the point ‘A’ is negative hence the diode D1 is reverse biased and
acts as open switch to current flow. The operation is shown in Figure.
26

A D1-OFF

I1

RL
AC i/p E
F
Vo
+ –

B D2-ON

Figure : Circuit during negative

half cycle

Vm
Vi

0 t
π 2π 3π

IL
Im

0 t
π 2π 3π
Vo
Vm

0 t
π 2π 3π

Figure: Input and output waveforms

27

Peak Inverse Voltage (PIV).

It is defined as the maximum voltage that appears across diode during non-conducting
condition under reverse bias.
For a centre tapped circuit, if any one diode does not conduct then voltage appearing
28
across it is entire secondary voltage Vm + Vm = 2Vm. Thus diode used in center tapped rectifier
should have high PIV ratings.

Advantages
1. Lower ripple r = 48.2 %.
2. Efficiency is high h = 81.2 %.
3. Requires only two diodes.
4. Used only for 10 ω voltage applications.
Disadvantages
1. Lower DC output since each half cycle utilize half the transformer voltage.
2. High PIV (2 Vm) is required.
3. Very difficult to locate exact centre point of secondary.
4. Cost of the transformer is very high.

Example 1:
A FWR with transformer rating 230 V, 40 – 0 – 40 V uses a load resistance RL of
3 kΩ the diode forward resistance. RF = 10 Ω and RR = ∞. Find
a) Maximum value of current in diodes when conducting.
b) DC value of current through RL.
c) Output DC voltage.
d) PIV across non conducting diode.
Given: Vrms = 40 V, RL = 3 kΩ, RF = 10 Ω, RR = ∞

Vm = 40  2 = 56.568 V

Im = Vm 56.568 = 18.79 mA.

=
(R F + R L ) (10 + 3000)

DC value of current through RL is Idc.

-3
2´18.79´10
Idc = 2 I m = 37.58 mA
p p
Output DC voltage = Idc  R=
L

Vdc = 37.58  10–3  3000 = 112.74 v

PIV across non conducting diodes

–2 Vm = –2  56.568 = –113.136 V
29

Example 2:

A Full Wave Rectifier supplier power is 3 kΩ load. The a.c voltage applied to the diodes is
180 – 0 – 180 V. If diode resistance is 10 Ω. Determine.
a) Average load current b) Average load voltage
c) RMS value of ripple voltage d) Rectification efficiency
Given: RL = 3 kΩ, Vrms = 180 V, RF = 10 Ω

a) Vm = 2  180 = 254.55 V
Vm =
I = 254.55 = 84.5 mA.
m (R F + R L ) (10 + 3000)

2Im 2´84.5´10 -3
= = 53.79 mA
Idc =
p p

b) Average load voltage = Idc  RL = 53.79  10–3  3000 = 161.37 V

c) Ripple factor is given by

rms value of ripple voltage
r=
value of DC component of voltage

rms value of ripple voltage

0.48 =
Vdc

rms value of ripple voltage = 0.48  Vdc

= 0.48  161.37 = 77.45 V.

d) Rectification efficiency

Bridge Rectifier
Due to the disadvantage of high PIV and usage of centre tap transformer in centre tapped
full wave rectifier the bridge rectifier can be used. In Bridge rectifier four diodes will be
used and two diodes will be conducting in each cycle.
30

Operation:
Figure below shows the bridge rectifier circuit. During positive half cycle point K is
positive with respect to point L. Hence D2 and D4 are forward biased and acts like
a closed switch but where as diode D1 and D3 are reverse biased and acts as open
switch hence D2 and D4 conduct and allows the current flows through RL as shown in
Figure below.
K A
D1 D2

AC i/p D B
Vi

D4 D3 RL Vo

L C

Figure : Bridge rectifier circuit

During negative half cycle L is positive with respect to K. Hence diodes D1 and D3
are forward biased and acts as closed switch hence allows the current to flow through RL in
the same direction but D2 and D4 are under reverse bias they acts as open switch as shown
in Figure below.
K A

D2-ON

VS D B

D4-ON
RL Vo

L C

Figure : Bridge rectifier circuit when D2 and D4 is ON

Expression for Idc, Vdc, Irms, Vrms and h remains as that of centre tapped full wave
31

rectifier except ‘Rf ‘ . Here for each half cycle two diodes will be conducting hence
Rf + Rf = 2 Rf.
Figure below shows the input and output wave form of bridge rectifier.
K A

D1-ON

VS B
D

D3-ON RL Vo

L C

Figure : Bridge rectifier circuit when D1 and D3 is on

Vi Vm

0 π t
2π 3π

IL
Im

0 t
π 2π 3π
Vo
Vm

0 t
π 2π 3π

Figure : Input and output waveforms

Advantages:
1) The transformer which is connected to bridge rectifier has secondary voltage peak of
Vm which is sufficient enough to get output peak of Vm.
2) The transformer is not bulky when compared to centre tap FWR as centre tap
transformer is not required.
3) PIV is Vm.
Disadvantages:
1) Additional two diodes are required when compared to centre tap FWR.
2) Regulation and efficiency is slightly poor when compared to centre tap full wave
rectifier since it involves the potential drop across 2 diodes in each half cycle.
32

Table: : Comparison of Half wave rectifier, centre tap rectifier and bridge rectifier

Parameter H.W.R F.W.R

Centre tap Bridge

Idc Im /  2 Im /  2 Im / 

Irms Im / 2 Im / 2 Im / 2
Ripple factor 1.21 0.48 0.48
0.406 RL 0.812 RL 0.812 RL
Efficiency
(R S + R f + R L (R S + R f + R L (R S + 2R f + R L )
) )
Centre tap
Not required Required Not required
transformer
No. of diodes One Two Four
PIV Vm. 2Vm. Vm.

Formula’s to Remember

HALF WAVE RECTIFIER

1) Maximum load current, I = Vm

m
R f + Rs + RL
Rs → Transformer secondary winding resistance.
Im
2) Idc = .
p
3) Vdc = Idc RL
Im
4) Vdc = . RL Maximum load current → Im
p
Vm
5) Vdc = . Maximum load voltage → Vm
p
I m , I m = Vm
6) Irms =
2 RL
Im
7) Vrms = Irms . RL  Vrms = . RL.
2
2 - I2
I ac I rms dc
8) Ripple factor, r = =
I dc I dc
33

Pdc
9) efficiency h =
Pac
2
Pdc = I dc . R L
P = I 2 . (R + R + R )
ac rms f s L

Rf V No load -Vfull load

10) % of Regulation = ×100 ; % of Regulation =
RL Vfull load
Pdc
11) Transformer utilization factor = .
Vm I m
2 2
12) PIV = Vm.
13) V2 N2
=
V1 N1
Full Wave rectifier :
1) I = Vm
m
R s + R + RL
f
2) IDC
2 Im
= .
p

3) Vdc = Idc . RL

2 Vm
4) Vdc = .
p
Im
5) Irms = .
2
Vrms = Irms . RL

Im
Vrms = . RL
2
V m . RL
Vrms =
2(RL + R s + R f )

6) Ripple factor.
I
r = ac =
I dc
34

7) Transformer utilization factor (TUF).

I 2 .R
dc L
TUF = .
V rms . I rms

8) PIV = 2 Vm.
Bridge Rectifier :
same as Full Wave Rectifier
except PIV = Vm.
For Full wave Rectifier with capacitor filter:
Vr, rms 1
= r =
Vdc 4 3. f . R L
35

Filters
The output of rectifier is pulsating that is it contains AC and DC components. The AC
component is undesirable and should be avoided reaching the load. Filter circuit is used
between rectifier output and load in order to remove AC component so that only DC
component passes to the load.

Rectifier Filter Load

Figure : Filters in Rectifier

The filter may be defined as, it is a device, which removes AC component of the
rectifier output but allows the DC component to reach the load or it is a circuit, which
converts pulsating output of the rectifier into a steady DC level.
Filter should be installed between the rectifier between the rectifier and the load as
shown in the above Figure. A filter circuit is generally a combination of inductors
(L) and capacitors (C). The filtering action of L and C depends upon the basic electrical
principles. A capacitor passes AC but does not allow DC. Inductor opposes AC but allows
DC to pass through it.
The reactance of capacitor and inductor are given by
1
X = and XL = 2p fL respectively.
C
2 p fC
For capacitor at f = 0 (dc), XC =  (offers high resistance to DC)
For inductor at f = , finite high value of f (ac), XL =  (offers very high reactance to AC).

Types of filters
1. Capacitor 2. Inductor
3. LC-filter 4. CLC filter or p section of capacitor input filter
Out of all these ‘four types’ capacitor filter is commonly used because of its low cost,
small size and little weight.
36
Rectifier With Capacitor Filter

The capacitor in both the circuits are placed across rectifier output ‘ RL’ i.e. load.
The pulsating DC of the rectifier is applied across the capacitor. As rectifier voltage
increases it charges the capacitor towards Vm because initially capacitor acts as a short.
At the end of quarter cycle (p/2) it is charged to peak value Vm of the rectifier voltage
as shown in waveform. Once capacitor voltage reaches Vm diode is reverse biased. Where
anode of diode is positive but less than Vm hence diode is reverse biased and stops
conducting. Now capacitor discharges through the load and voltage across it decreases.
Capacitor discharges until the input voltage is less than capacitor voltage. Once the input
voltage is greater than capacitor voltage the diode is forward biased and capacitor charges
to Vm and same cycle repeats. Figure below shows full wave rectifier which conduct for full
cycle and output waveform is as shown in Figure below.
VS D

Vin C RL V
o

Figure : Half wave rectifier with capacitive filter

T2 T1
C≠0 C
Vrpp

Vdc
0 C=0
t
π 2π 3π

Figure : Output waveform of Half wave rectifier

D1 D2

Vo
RL

D4 D3

Figure : Full wave rectifier with capacitive filter

37

T1 T2
A C
Vrpp
B
Vdc
t
π/2 π 2π 3π

(d)

Figure : Output waveform of FWR with capacitor

The expression for Ripple factor (r) is given by the equation
1
r= for Half wave rectifier ...(2.22)
2 3 fCRL

1
r= for Full wave rectifier ...(2.23)
4 3 fCRL

Where,
f is Frequency,
C is Capacitance and RL is Load resistance
Expression for Ripple Factor
The presence of AC component in DC output of Rectifier known as Ripple. Value of load
current ‘Idc’ is the average capacitor discharge current over an interval of T2. Amount of
charge lost by capacitor during this interval is
Qdischarge = Idc . T2
This charge is replaced during short interval T1. During which the voltage across the
capacitor changes by an amount equal to the peak to peak voltage of ripple, ‘Vrpp’.
Using the relation
Q = CV [... C = Q / V]
Qcharge = C Vrpp

Qcharge = Qdischarge

C Vrpp = Idc  T2
I dc ×T2
Vrpp =
C
Let us assume the load resistance is very high so that ripples are so small and the time
for recharging the capacitor is very small compound with the time with which it discharges
(i. e. T2).
Now, Let us consider half wave rectifier with C-filter.
38

T1 + T2 = T
T1  T2
T = T2

I dc ×T2
Vrpp =
C

I dc ×T I dc
= =
C fC

V = V rpp r is ripple voltage.

r, rms
2 3
[... rms value of a ’r or sawtooth type of waveform is

V rpp
= ]
2 3

Vdc
Idc =
RL
Vdc
2 3 Vr, rms
= Vrrms = Vac
Cf R L
Vr, rms 1
Ripple factor γ = =
Vdc 2 3 f RL C
1
The ripple factor for Half Wave Rectifier =
2 3 R L Cf

In FWR with C - filter the only changes i.e., discharge time

T T
T1 = or T2 =
2 2
T 1
T2 @ =
2 2f
Vr, rms 1
 = rγ =
Vdc 4 3 f RL C

⯈ DC output voltage (VDC)

Vdc = Vm – V rpp
2
39

I dc Vdc
Vdc = Vm – I dc =
2 fC RL

Vdc = Vm – Vdc
2f R L C

For a Full Wave Rectifier.

I dc
Vdc = V m -
4 fC
Choke filter consists of an inductor connected in series with rectifier output circuit and a
capacitor connected in parallel with the load resistor. It is also called L-section filter because
the inductor and capacitor are connected in the shape of inverted L. The output pulsating DC
voltage from a rectifier circuit passes through the inductor or choke coil.
Figure below shows Choke capacitor filter circuit for full wave rectifier with
waveforms. It filters the rectifier in much more effective way, using an LC Choke. The
inductor of the filter attempts to keep the current constant while the capacitor attempts to
keep the output voltage steady.

L
Full wave +
C RL
rectified input

Vo

Full wave rectifier output

t
Full wave rectifier output
with capacitor filter
t
Full wave rectifier output
with choke capacitor filter
t

Figure : Choke capacitor filter circuit for full wave rectifier with waveforms
40

The series inductor (L1) and the capacitor (C1) form a voltage divider for the ac
component (ripple) of the applied input voltage. This is shown in Figure Above. As far as the
ripple component is concerned, the inductor offers a high impedance (Z) and the capacitor
offers a low impedance. As a result, the ripple component is greatly attenuated (reduced).
Since the inductance of the filter choke opposes changes in the value of the current flowing
through it, the average value of the voltage produced across the capacitor contains a much
smaller value of ripple component as compared with the value of ripple produced across the
coil. For large DC power output capacitor filter is not much effective .in order to reduce
the ripples choke capacitor filter i.e., LC filter is used. In low power electronic circuits non
linearity of choke is not used.

Zener Diode Voltage Regulator

Diodes which have adequate power dissipation capabilities to operate in the breakdown
region are commonly called zener diodes.
Avalanche multiplication and zener breakdown are the two process, which produce the
breakdown region in the reverse bias.
Avalanche Multiplication it is seen in normal diode
During reverse bias the depletion region widens in ordinary diode. Carriers which are
generated thermally will get sufficient energy from applied potential. The crystal atom are
collided by these carriers and imparts energy to disturb the covalent bond. Because of these
actions a new electron hole pair is generated and these electrons also gets energy from
applied field. Through continuos collision by these electron with another atom more number
of electron hole pair is generated. This action continuos and produce additional carriers.
This cumulative process is known as avalanche multiplication. This results in large reverse
current and diode will be in breakdown region.

Zener breakdown it is seen in zener diode

When a diode is heavily doped, there will be a narrow depletion layer due to this the
electric field across depletion layer will be very high. Because of the presence of this
electric field across the junction a strong force may be applied on bound electrons by
the field to tear it from covalent bonding. The new hole pair created raises the reverse
current. This process is known as zener breakdown.
Zener diode
A heavily doped diode, which has a sharp breakdown voltage is known as a zener
diode. Figure below shows the symbol of zener diode.

Symbol
Anode Cathode
A K

Figure : Zener diode symbol

41

Due to heavy doping the electric field in the junction increases during reverse bias.
Hence covalent bonds breaks producing a large number of electron hole combinations. This
inturn increases reverse current. The diode possess some resistance called zener dynamic
resistance. Figure below shows the zener diode characteristics.
IF
Forward current

Reverse voltage
VZ V 0
VR BR VF
Forward voltage

IZ
Reverse current
IR

Figure: Zener Diode characteristics

Voltage does not change once it goes into saturation which means ‘Vz’ remains constant
even if current ‘Iz’ increases.
Izmin : Minimum current to maintain zener in breakdown region if Iz < Izmin the zener will
not provide constant voltage.
Izmax : It is the maximum zener current that would flow through it without destroying it. It
is given by
P
Izmax = z .
Vz

Zener equivalent circuit

The equivalent circuit of zener diode is as shown in the Figure below. Zener
works like a battery ‘Vz’ provided zener operates in the breakdown region ‘Rz’ represents
small dynamic resistance of the diode.

RZ

+
VZ
–

Figure : Zener equivalent circuit

42

Zener Voltage Regulator

A zener diode can be used as a regulator because once the zener diode enters into
Breakdown region it maintains a constant output voltage even though there is change in
current through it. For zener diode to be a voltage regulator it should be operated in Reverse
bias mode by making the supply voltage greater than the zener breakdown voltage Vz. The
Figure below shows a simplified Zener diode voltage regulator circuit. In order to limit the
current through zener diode series resistor Rs is used.
The three main conditions for zener diode to work as regulator is a follows:
⯈ The zener diode must be always reverse biased.
⯈ Supply voltage must be greater than zener breakdown voltage.
⯈ Load current must be less than maximum zener current.

Rs

Is Iz
+
Vs Vo = Vz
– RL

Figure : Simplified Zener diode voltage regulator circuit

Case i) Regulation obtained with varying input voltage (V) Line regulator
Consider zener diode in reverse bias condition if input voltage Vi increases more than
zener voltage than output voltage remains constant. Since zener diode is in Break down
region. As Vi increases input current Is also increases inturn zener current Iz also increases.
Figure shows regulator with variable input voltage.
Rs

Iz
Vin RL Vo = Vz

Figure : Regulator obtained with variable input voltage

From the circuit shown in Figure above, we have,

Vo = Vz

Vo V
 IL = = z
RL RL
43

Since Rz, RL is constant then IL is also constant. Hence the current Iz also increases.
But Iz should be in the range from Izmin to Izmax.
If Vin decreases then zener current decreases maintaining constant load current IL.
Hence output remains constant untill Iz is in the range of Izmin to Izmax. Thus irrespective of
changes in input voltage or line voltage output voltage remains constant.
DVo for 10% change in Vin
Percentage of regulation = ...(2.24)
Vo

Case ii): Regulation obtained with varying load (RL)

Consider a zener diode in breakdown region since input voltage Vi is greater than zener
break down voltage but it is kept constant.
As Vin is constant Vo = Vz
The current through Rs is Is which is given as,
V -V z
Is = Iz + IL ; I s = in = constant = I z + I L
RS
Figure below shows variable load condition.
R

Is IL
Iz
Vin RL Vo = Vz
constant variable
resistance

Figure: Variable load condition

If load resistance (RL) is increased the current through the load decreases, to keep
Is constant Iz increases. But Iz should be between Izmin and Izmax to keep output voltage Vo
constant. Similarly if RL decreases to keep Is constant Iz decreases. Since IL increases as
there is less resistance at the output. Until Iz is between Izmin and Izmax the output voltage
remains constant. Thus irrespective of changes in load resistance (RL) The output voltage
remains constant.
The percentage regulation is given by the formula,
∆Vo for ∆I L max
Percentage of regulation = ×100 ...(2.25)
Vo
44
45
46

Review Questions
2.1 Explain the theory of PN junction.
2.2 With neat sketch, explain the formation of depletion region in unbiased p-n junction.
2.3 Explain the different types of diode approximations.
2.4 Draw and explain V-I characteristics of PN junction diode.
2.5 What is rectifier? Draw the circuit for half-wave rectifier and explain its working.
Derive the expression for Idc, h, PIV, RMS value of voltage.
2.6 Draw the circuit for Full-wave rectifier and explain its working. Derive the expression
for Idc, h, PIV, Ripple factor, RMS value of voltage.
2.7 Define Ripple factor show that for Half Wave Rectifier, Ripple factor is 1.21.
2.8 Explain the avalanche and zener breakdown with the help of VI characteristics.
2.9 Differentiate between zener breakdown and Avalanche break down.
2.10 Draw the bridge rectifier circuit and explain its operation with waveforms. Show that
ripple factor is 0.48.
2.11 With relevant waveforms, derive expression for Idc, Irms and ripple factor of a Full
Wave Rectifier.
2.12 Give the advantages and disadvantages of bridge rectifier over centre tap Full Wave
Rectifier.
2.13 Explain how a zener diode can be used as voltage Regulator.
2.14 Explain Regulator Performance.