5 Applications of Diodes

In the forgoing chapter of this book, the details of the semiconductors and
different types of diodes have been discussed. The applications of the semiconductor
diodes such as rectifiers, filters circuits, voltage multiplier circuits, clipping, clamping
Log antilog circuits will be discussed in this chapter. Zener diode as the voltage regulator
circuit will also be discussed in this chapter.

5.1 Rectifier Circuits: Many electronic equipments work with d.c. power supply.
The line voltage available from the power plugs is a.c. voltage (220 Volts & 50Hz
frequency). A circuit which converts a.c. voltage to d.c. voltage is required. Such a circuit
is called the rectifier circuit. The nonlinear elements like vacuum diodes or
semiconductor diodes may be used for the rectification of the supply. There are two types
of rectifier circuits (i) Half Wave Rectifier & (ii) Full Wave Rectifier. In the following
articles these circuits will be discussed in detail using the semiconductor diodes.

5.1.1 Half Wave Rectifier : Consider a circuit shown in the figure (5.1). In this
circuit a sinusoidal voltage E s = E m sin ω .t , obtained from the transformer, is applied to
a series combination of the diode D1 and a load resistance RL. Em is the peak voltage of
the signal and ω = 2 π f is the frequency of the a.c. mains in radians and f is the mains
frequency in Hz. The transformer used here may be step up or step down transformer
which is chosen according to the required d.c. supply. If the a.c. mains supply (220 volts,
50 Hz) is to be rectified, then it may directly be applied to the series combination of diode
and load resistance.

Fig. 5.1

To discuss its operation, assume that the peak value of voltage Em is large
enough than the cut in voltage of the diode, so that the diode behaves like an ideal diode.
During the positive half cycle of the input wave Es, the diode is in forward bias and
works as a closed switch. Therefore, applied voltage appears across the load resistance
RL. However, during the negative half cycle of the input wave, the diode is in reverse bias
and it acts as an open switch. In this case, no current flows through the circuit and voltage
across the load resistance RL is zero. The input, output waves of the circuit is shown in
the fig. (5.2).
 Fig. 5.2

From this figure it is clear that the current in the load resistance flows only
during positive half cycle of the input wave and is zero during the negative half cycle,
i.e., the current is unidirectional. Hence the circuit is called a half wave rectifier.
Analysis: The input a.c. voltage is given by E s = E m sin ω .t
Since the current in the load resistance is unidirectional, so we take the average value of
the voltage across the resistance. It may be given as:
T
1 2 E 2π
E d .c . = ∫ E m sin ω .t .dt = m since T =
T 0 π ω
The d.c. current flowing through the load resistance RL is given by:
E d .c . Em
I d .c . = =
RL πR L
Im
= Em
(where I m =is the peak current)
π RL
The output voltage across the load resistance may be given by the Fourier analysis as:
Em Em 2E 2E 2 Em
Eout = + sin ωt − m cos 2ωt − m cos 4ωt − cos 6ωt − ........
π 2 3π 15π 31π
------ (5.1)
From equation (5.1), it may be seen that the first term of the equation is the same as Ed.c.
as calculated above. In addition to this term there are other terms of frequency ω and its
higher harmonics. So the output voltage across the load resistance has the required
voltage (Ed.c) and other unwanted components called the ripple.
The root mean square value of the output voltage may be given by:
 T
2
1
= ∫E
2 2
E r .m .s . s . dt
T 0
T
2 2
1 E
= ∫E
2
m sin ω t .dt =
2 m
T 0 4
Em
or E r .m .s . = ------ (5.2)
2
The total power delivered to the load resistance is:
E r2. m . s . E m2
Pin = = ------ (5.3)
RL 4RL
The d.c. power delivered to the load is:
E d2 . c . E m2
Pd . c . = = 2 (Putting E d .c. =
Em
RL π RL π
)

Therefore, power delivered in load resistance due to unwanted components i.e. ripple
may be given as:
Pripple = Pin – Pd.c.
E r2. m . s . E d2 . c .
= − ------ (5.4)
RL RL
E a2 . c .
The Pripple is equal to . The equation (5.4) may be rewritten as:
RL
E a2.c . E2 E2 E a .c . E r2. m . s .
= r . m . s . − d .c . or γ = = ( − 1)
RL RL RL E d .c . E d2. c .
Where γ is the ripple factor and is defined as the ratio of the a.c. components to the d.c.
components at the output.
Putting the values of Ed.c. and Er.m.s.respectively we get:

π2
γ = ( − 1) = 1 .21
4
From this equation, it is clear that γ is more than 1, which indicates that at the output, the
unwanted a.c. components (ripple) are more than the wanted d.c. components. Hence the
half wave rectifier being a poor circuit for rectification is not of much practical use.

Rectifier Efficiency: It is useful to define the quantity called the rectifier efficiency,
which is defined as:
Rectifier Efficiency η = (d.c. power input to load) / (Input power delivered to load)
 Pd .c .
= x 100 %
P in
( E m2 π 2 R L )
= x100 = 42 x100 = 40.6 %
π
2
(Em
)
4RL
This means, for half wave rectifier, under ideal conditions, only 40.6 % of the a.c. input
power is converted into d.c. power in the load resistance.

Example 5.1 A 12 volt a.c. from the secondary of a transformer is applied to the input of
a half wave rectifier circuit having 10KΩ load resistance. If the diode is ideal, find:
(i) Peak value of the a.c. signal
(ii) D.C. output voltage
(iii) Peak value of the current through the load resistance
(iv) Average value of the current through the load resistance
(v) PIV of the diode
Solution: Given Er.m.s. =12 Volts , RL = 10 KΩ
(i) Em = 2 Er .m.s. = 1.414 x12 = 16.97volts
Em 16.97
(ii) E d .c = .= = 5.4volts
π 3.14
E 16.97
(iii) Im = m = = 1.7 mA
RL 10000
E 5.4
(iv) I d .c . = d .c . = = 540µA
RL 10000
(v) PIV of the diode = Em = 16.97 Volts

Example 5.2 A P – N junction diode having forward resistance Rf = 25Ω, is used in half
wave rectifier circuit. The input applied signal is given by Es = 25Sin(100πt ) . If the load
resistance RL is 500Ω, then calculate Im, Ir.m.s., Ed.c., d.c. current following through the
load resistance and the rectifier efficiency.
Solution: The peak value of the input a.c. is given Em = 25 Volts
Em 25
Im = = = 0 .0476 A = 47 .6 mA
R f + RL ( 25 + 500 )
I m 47.6
I r .m . s . = = = 23.8mA
2 2
E m xRL 25 x500
E d .c . = = = 7.58volts
( R f + RL )π 525 x3.14
E 7.85
I d .c . = d .c . = = 15.17 mA
RL 500
 Pd .c .
Rectifier Efficiency η= x100
Pin
Pd .c. = I d2.c. RL = (15.17 ) 500 = 0.115watt
2

Pin = I r2.m.s. x(RL + R f ) = (23.8) 2 x525 = 0.2974watt

Pd .c . 0.115
η= = x100 = 38.67%
Pin 0.2974

5.1.2 Full Wave Rectifier: Figure (5.3) shows the circuit diagram of full wave
rectifier. It consists of a centre tapped transformer and two diodes D1 and D2 in addition
to the load resistance RL. The centre tapped transformer converts a sinusoidal signal into
two equal sinusoidal signals which are 1800 out of phase. In order to understand the
operation of this circuit, we consider the diodes are ideal. The two signals ES1 and ES2
obtained from the transformer are being applied to the two diodes D1 and D2. During first
half cycle of the input wave (ES), point A is positive and B is negative with respect to. the
common terminal. The diode D1, therefore, conducts and D2 is in reverse bias. This
causes Ib1 current to flow through the diode D1 and load resistance RL. During the other
half cycle, diode D2 conducts and diode D1 is in reverse bias. This caused Ib2 current to
flow through the diode D2 and load resistance RL.

Fig 5.3

Fig. (5.4) shows the input and output waves of the full wave rectifier circuit,
Fig. 5.4
From these wave shapes, it is clear that the direction of current in the load resistance is
same for both the half cycles of the output wave. Hence, the circuit is called as full wave
rectifier.
Analysis: In this circuit the centre tapped terminal of the transformer is used as the
common terminal of two voltages ES1 and ES2 which are 180 out of phase.
Let E S 1 = − E S 1 = E m Sin ω t
The average value (d.c.) of the voltage across the load resistance for the full wave
rectifier is given by:

1  2 
T
T
E d .c . =
T  ∫0
E S 1 dt + ∫T S 2 
E dt
 2 

1  2 
T
T
ω ω 
T  ∫0 ∫T m
= E m sin t .dt − E sin t .dt

 2 
2E
which may be simplified as = m
.
π
The r.m.s. value of the voltage across the load resistance is given by:

1  2 2 
T
T
E r2.m .s . =
T  ∫0
+ ∫ S 2 
2
E S1 .dt E .dt
 T
2 
1 2 
T
T E m2
=  ∫ E m2 sin 2 ω t.dt + ∫ E m2 sin 2 ω t .dt  =
T 0  2
 T
2 
Em
or E r .m .s . =
2
From the Fourier analysis the voltage across the load resistance is given by:
2Em 4Em 4Em 4Em
EL = − cos 2ω .t − cos 4ω .t − cos 6ω t − ... ------ (5.5)
π 3π 15π 31π
It is clear from this equation that first term of the output voltage across the load resistance
is the same as that of Ed.c., calculated above. Further the lowest frequency term (ripple) in
the full wave rectifier circuit is twice the frequency of supply signal.
The ripple factor γ is calculated as:
 E r2. m . s .  E m2 2 
γ = ( − 1) =  − 1 
E d2 . c .  4Em π
2 2


π 2
= ( − 1) = 0 . 48 ------ (5.6)
8
It clearly indicates that γ being less that unity has the d.c. components (desired) in the
output more than the a.c. components (undesired).
Rectifier Efficiency: This is calculated as:
Pd . c .
η = x 100 %
Pin

( 4 E m2 π 2 R L ) 8
= x100 = 2 x100 = 81.2 %
(Em 2
) π
2RL
Thus for full wave rectifier 81.2 % of the a.c. input power is converted to d.c. output
power.
5.2 Peak Inverse Voltage (PIV): Peak inverse voltage of a rectifier diode in a
circuit is the maximum reverse voltage across the diode. For the rectifier circuit PIV of
the diodes used should be less than the break down voltage of the diode. It can be seen
from the circuit of half wave rectifier that PIV of the diode used is equal to peak voltage
of the applied input a.c. i.e. +Em.. But in case of full wave rectifier PIV of each diode
used is twice the peak voltage of the applied a.c. (+2Em). It is clear from the figure (5.5)
of the full wave rectifier that during +ve half cycle of the input wave, the diode is in
forward bias and the voltage at the point C may thus reaches to +Em and at point B it is -
Em. So the total reverse voltage across the diode D2 is 2Em. Similarly it can be seen that
during –ve half cycle PIV of the diode is also +2Em. From the above discussion it is clear
that in full wave rectifier circuit, we should use the diodes whose break down voltage in
greater than +2Em (i.e. double the peak value of the input a.c. voltage).

Fig. 5.5
Disadvantage of Full Wave Rectifier: The full wave rectifier circuit has the following
disadvantages:
 (i) As discussed above, the PIV of the diode used in the circuit is
+2Em. So such diodes whose break down voltage is greater than
2Em is to be chosen for this purpose.
(ii) In this circuit, especially designed transformer called centre tapped
transformer is to be used. To rectify the a.c. mains voltage a center
tapped transformer of 220-0-220 volts is to be used. However, in
case of half wave rectifier circuit, transformer is not to be used if
an a.c. main is to be rectified.
(iii) Since centre terminal of the centre tapped transformer is used as a
common terminal (or ground terminal) –ve d.c. output voltage may
not be obtained by simply reversing the output terminals; but will
be obtained by reversing the diode connections.
Example 5.3 A full wave rectifier circuit has the input signal E s = 100 Sin(100π .t ) , RL
= 900 Ω and each diode has the forward resistance of 100 Ω. Calculate
(i) Peak value of the current through the load resistance,
(ii) D.C. load current,
(iii) D.C. output voltage
(iv) Rectifier Efficiency
Solution: Given Em = 100 Volts, RL = 900 Ω, Rf = 100Ω
Em 100
(i) Im = = = 100mA
RL + R f 900 + 100
2I m 2 x100
(ii) I d .c . = =
= 63.7 mA
π 3.14
(iii) E d .c . = I d .c. R L = 63 .7 mAx 900 = 57 .3 Volts
P
(iv) Rectifier Efficiency η = d .c . x100
Pin
Pd .c. = I d2.c. RL = (63.7 ) 900 = 3.65watt
2

100 x100 x1000 x10 −6

x(R )= I m2
Pin = I 2
r . m. s . L + Rf x1000 = = 5watt
2 2 2
Pd .c . 3.65
η= = x100 = 73%
Pin 5

Example 5.4 A 15 – 0 – 15 volts transformer is used for full wave rectifier circuit. Each
diode has a forward resistance of 10Ω. The load resistance is 6000 Ω. Find Ed.c., Id.c.,
Ir.m.s., and rectifier efficiency η.
Solution: Rf = 10 Ω, RL = 600 Ω , Er.m.s. = 15 Volts
E m = 15 x 2 = 21.21volts
2Em 2 x 21.21x600
E d .c . = .R L = = 41.72volts
( R f + RL ) (10 + 600)
 E d .c. 41.72
I d .c . = = = 69.5mA
RL 600
Em 21.21
I r .m . s . = = = 24.6mA
2 ( R f + R L ) 1.414 x610
0.812 0.812
η= .R L =
( R f + RL ) Rf
(1 + )
RL
0.812 0.812 x30 x100
= x100 = = 78.6%
20 31
(1 + )
600

5.3 Bridge Rectifier: The disadvantages that we have in case of full wave
rectifier will be removed in the Bridge Rectifier. The basic circuit diagram of bridge
rectifier is shown in Fig. (5.6). This circuit has four diodes connected in the four arms of
a bridge, hence the name bridge rectifier. To understand the working of this circuit let us
consider that during +ve half cycle of the input ES, the point A is +ve with respect to B.
In this condition, diodes D1 and D3 conducts and diodes D2 and D4 are in reverse bias.
The current flows in the load resistance RL, in the direction of arrow as shown in fig.
(5.6a). During the next half cycle the point B is +ve with respect to A and thus diodes D2
and D4 conducts and diodes D1 and D3 goes in reverse bias. This also causes the flow of
current through the load resistance RL in the same direction which is shown by dotted
lines in the figure 5.6(b).

Fig. 5.6
 Fig. 5.6(a) Fig. 5.6(b)

The output wave form of the bridge rectifier is essentially the same as in the case of full
wave rectifier.
Advantages of Bridge Rectifier:
(i) No especially designed center tapped transformer is required in the
bridge rectifier circuit.
(ii) There is no common terminal (common between input and output),
so –ve supply may be obtained just by reversing the output
terminals.
(iii) The peak inverse voltage across any of the four diodes in the
bridge rectifier circuit is equal to Em and not +2Em as in the case of
full wave rectifier.
(iv) For the same output voltage, the transformer secondary line – to –
line voltage, in the bridge rectifier should be one half of that used
for full wave rectifier. Thus bridge rectifier supply large amount of
d.c. power.
Example 5.5 The forward resistance of each diode used in bridge rectifier is 10 Ω. The
a.c. voltage used to the input is 230 volts and load current is 1K Ω. Find Ed.c., Id.c., Im, a.c.
current to the load Ir.m.s. and PIV of the diodes used in the circuit.
Solution: Er.m.s. = 230 Volts Rf = 10Ω RL = 1000 Ω
E m = 230 2 = 325.22 Volts
2 E m xRL 2 x325.22 x1000
E d .c . = = = 203.1 Volts
(2 R f + RL )π (20 + 1000) x3.14
( Here 2Rf is used in place of Rf as two diodes go in forward bias at a time)
E 203.1
I d .c. = d .c. = = 203.1mA
RL 1000
Em 325.22
Im = = = 319mA
(2 R f + RL ) (20 + 1000)
 PIV of the diodes = Em = 325.22 Volts

5.4 Filter Circuits: In the preceding section we have studied the rectifier
circuits. The rectified output contains a large amount of unwanted a.c. components
(ripples) in addition to the d.c. voltage. The ripple can be eliminated or considerably
reduced by using a filter circuit between the output of the rectifier and the load resistance.
We shall now study the different commonly used filter circuits.

5.4.1 Half Wave Rectifier with Shunt Capacitor Filter: Figure (5.7)
shows the circuit diagram of a half wave rectifier with shunt capacitor filter. In this
circuit a capacitor C is connected in parallel (in shunt) with RL hence it is called shunt
capacitor filter. The operation of this circuit may be explained as follows:

Fig. 5.7

During the voltage rise of +ve half cycle of the input wave Es, the diode conducts
and the output follows as the input voltage at point A. The capacitor C starts charging and
thus it charges to the maximum of peak voltage Em of the input wave Es. Further during
the decrease of the input wave from Em to zero of the same half cycle, the diode D is in
reverse bias. The output seems to be disconnected from the point A and the capacitor C
will now start discharging through the load resistance RL. The RLC time constant is
chosen large enough compared to half the time period of the input wave, so that the
capacitor C is not completely discharged. As soon as the input voltage again becomes
greater than the output voltage (during the next +ve half cycle) the diode again conducts
and capacitor C starts charging further. In this way the cycle repeats. The input and
output wave with shunt capacitor filter is shown in Fig. (5.8).
 Fig. 5.8

The discharging path will depend on the RLC time constant. Since it is large, the
discharging path will therefore, seem to be almost linear.

Calculation of ripple factor: Given input signal is E S = E m sin ω .t

The total change in the output (ripple voltage) is shown in Fig. (5.8). The average value
of this triangular wave is approximately half the total change in the output wave (ER),
ER
which is given by: E d .c . = E m − . ----- (5.7)
2
For the good filtering action the discharge period T2 should be very much greater than the
charging period T1 i.e.
T2 >> T1 ≈ T (say) where T is the time period of the one complete cycle.
For simplicity of analysis the ripple voltage is approximated the exact triangular wave as
shown in Fig.(5.9).
 Fig. 5.9

The equation of the Triangular wave is given by:

E = A1 + A2 .t ------ (5.8)
where A1 and A2 are constants which may be obtained from the boundary conditions:
ER ER
E= at t = o and E=− at t = T ------(5.9)
2 2
ER ER
so we get from equations (5.8) & (5.9) A1 = & A2 = −
2 T
ER E
The wave equation (5.8) will become E = − R .t
2 T
The r.m.s. value of the ripple voltage is given by:
T
1
E r2. m . s . = ∫E
2
dt
T 0
2 T 2
E 1 t 
=
T
R
∫0  2 − T  .dt
E R2
T
1 t2 t E R2
= ∫0 ( 4 + T 2 − T ). dt =
T 12
ER
or E r .m .s . = ------ (5.10)
2 3
During the discharge period T2, the capacitor loses the charge Id.c..T2 at a constant rate.
Hence the change in capacitor voltage ER is calculated as:
ER = (Loss of charge)/(capacity of the capacitor)
I d . c . .T 2
= ------ (5.11)
C
1
It has already been assumed that T 2 = T = for good filtering action. Where f is the
f
frequency of the wave in Hz.
I
So E R = d .c . ------ (5.12)
f .C
From the equations (5.7) & (5.12) we get :
I d .c .
E d .c . = E m −
2 f .C
1
The term has the dimension of resistance, which represents the source resistance
2. f .C
of d.c. supply Ed.c.; and may assumed to be an open circuit voltage. For having the low
ripple voltage and ensure good voltage regulation the capacitor C must be large enough.
The ripple factor γ is given by:
 γ = (r.m.s. value of a.c. output)/(d.c. output voltage)
ER I d .c .
= (since E R = )
2 3 . E d .c . f .C
1 E d .c.
= (where = RL ) ----- (5.13)
2 3 . f .C . R L I d .c .
From this equation it is clear that we may get small ripple content at high load resistance.
At no load ( RL → ∞ ) the output voltage will be equal to Em , the open circuit voltage of
the d.c. supply; and we thus say that this circuit behaves like a peak detector. This filter is
suitable for low current.
The PIV of the diode used in half wave rectifier with shunt capacitor filter is
approximately 2Em since the point D (in fig. 5.7), the cathode of the diode is at about +Em
potential and anode goes to –Em with respect to common point. Hence the total reverse
voltage across the diode is 2Em.

Example 5.6 A transformer whose secondary winding is rated at 12 volts r.m.s., is used
for half rectifier with shunt capacitor filter. If the value of capacitor C = 100 µF and load
resistance RL = 1KΩ and frequency f = 50 Hz, determine Ed.c. and Id.c., peak to peak ripple
voltage at the output, peak forward current in the diode. How do the above change if C is
increased to 1000 µF? What should be voltage ratings of the capacitor, if 25% variation is
allowed in the input a.c. voltage.
Solution: C = 100 µF RL = 1000 Ω E r.m.s. =12 volts f = 50 Hz
E
E m = 12 2 = 12 x1.414 = 16.97 Volts, I d .c . = d .c .
RL
I d .c . E d .c .
E d .c . = E m − = Em −
2 f .C 2 f .C . R L
1 Em
or E d .c . (1 + ) = Em and E d .c . =
2 f .C.RL  1 
1 + 
 2. f .C.RL 
16.97 16.97
Ed .c. = = = 15.43 Volts
1 1.1
1+
2 x50 x100 x10 −6 x1000
E 15.43
I d .c . = d .c . = = 15.43mA
RL 1000
Peak to peak ripple voltage at the output ER is given by:
I 15 . 43 x 10 − 3
E R = d .c . = = 3 . 086 Volts
f .C 50 x 100 x 10 − 6
E 16.97
Peak forward current in the diode I m = m = = 16.97mA
RL 1000
Now the value of C is increased to 1000 µF so put C = 1000 µF in the above calculations,
we get:
 16.97 16.97
Ed .c. = = = 16.8 Volts
1 1.1
1+
2 x50 x1000 x10 −6 x1000
E 16.8
I d .c . = d .c . = = 16.8mA
RL 1000
I d .c . 16 . 8 x 10 − 3
ER = = = 0 . 336 Volts
f .C 50 x 1000 x 10 − 6
E 16.97
Im = m = = 16.97mA
RL 1000
If 25% increase in the a.c. signal then peak voltage also increases in the same ratio i.e.
E m = 16.97 + 0.25 x16.97 = 21.2 Volts
So the voltage rating of the capacitor should be greater than 21.2 Volts may be taken of
25 Volts or 50 volts.
5.4.2 Full Wave Rectifier with Shunt Capacitor Filter: The circuit diagram
of this filter is shown in figure (5.10). The working of this circuit is similar to that of the
half wave rectifier with shunt capacitor filter; the only difference between the two is that
the circuit works for the half cycles. The input – output wave shaped is given in figure (
5.11).

Fig. 5.10
 Fig. 5.11

The ripple factor for the full wave rectifier with shunt capacitor filter may also be
calculated in the similar fashion as for half wave case simply by replacing T2 ≈ T 2 .
Since in this case, the discharge period T2 must be greater than T/2. So T2 is assumed to
be equal to T/2.
The ripple factor γ is given by:
1
γ = ------ (5.14)
4 3. f .C.RL

5.4.3 Percentage Regulation: The percentage regulation of a d.c. power supply

may be defined as:
Percentage regulation = (No load voltage – Full load voltage) / Full load voltage
If the supply designed by the rectifier and filter, then it can be represented by the model
of the battery, i. e., a voltage source Ei ( open circuit voltage) and a source resistance R0
in series with it as given in figure (5.12).

Fig. 5.12

From this figure we get E i = R0 I L + E 0 and E0 = RL I L

or E i − E 0 = R0 I L
EI = No load voltage E0 = Full load voltage
 R0 I L R
So Percentage regulation = x100% = 0 x100%
RL I L RL
Example 5.7 A full wave rectifier is supplied with an a.c. signal of 50 – 0 – 50 volts
r.m.s. and 50 Hz frequency. A 100 µF capacitor serves as a filter and takes 300 mA load
current. What is the d.c. load voltage? Also find the ripple factor.
Solution: Er.m.s. = 50 Volts, f = 50 Hz C = 100 µF Id.c. = 300 mA
E m = 50 x 2 = 50 x1.414 = 70.7 Volts
For full wave rectifier with shunt capacitor filter is given by:
I d .c .
E d .c . = E m −
4 f .C
300 x10 −3
= 70.7 − = 70.7 − 15 = 55.7 Volts
4 x50 x100 x10 −6
E d .c. 55.7
RL = = = 185.7Ω
I d .c . 300 x10 −3
1 1
γ = = −6
4 3 . f .C .R L 4 x1 .73 x 50 x100 x10 x185 .7
= 0.156 = 15.6%
Example 5.8 It is required to design full wave rectifier with shunt capacitor filter which
is capable of supplying 20 volts d.c. at no load. The regulation of this supply is required
to be less than 10% for a full load current of 1 ampere. The maximum ripple is to be less
than 3 volts (peak to peak). Find (i) the required secondary rating of the transformer,
(ii) the value and voltage rating of the capacitor,
(iii) the peak forward current and PIV ratings of the diodes.
Solution: No load voltage Ed.c. = 20 Volts % regulation = 10%
Percentage regulation = (No load voltage – Full load voltage) x100/ Full load voltage
10 = (20 – Full load voltage)x100/Full load voltage
or Full load voltage = 18.18 volts

The full load voltage should be greater than 18.18 volts say 18.5 Volts

Ripple voltage ER = 3 Volts (peak to peak)

I d .c . 1 amp
ER = = = 3 volts
2 . f .C 2 x 50 xC
1
or C= = 0.003333F = 3333µF
300
E 18.5
RL = d .c. = = 18.5Ω
I d .c . 1amp
I d .c .
We know I d .c . R L = E m −
4 f .C
 Em 1
I d .c . = or E m = I d .c . ( R L + )
1 4. f .C
( RL + )
4. f .C
300
Em = 1(18.5 + ) = 20 volts
4 x50 x1
Em 20
Er.m.s. = = = 14.14 volts = 15 (say)
2 1.414
(i) So secondary of the transformer should be rated as 15 – 0 – 15 volts and current 1amp.
(ii) The value of capacitor C= 3333 µF and voltage ratings should greater than 20 volts.
Em 20
(iii) Peak forward current Im =
= = 1.08amp
RL 18.5
PIV of the diodes = 2Em= 40 volts

5.4.4 Series Inductor Filter: The circuit diagram of the series inductor filter is
shown in figure (5.13). Here an inductance is connected in series with the load resistance.
Therefore, it is known as series inductor filter. The signal EI has the d.c. components
along with the higher harmonics of input frequency ω . EI is given by:

2Em 4Em 4Em

EI = − cos 2ω .t − cos 4ω .t − ......
π 3π 15 π

Fig. 5.13

Inductance L is connected to eliminate or reduce the ripple content at the output.

4Em
Out of the ripple contents, the second harmonic term . cos ω .t is the only
3π
effective term. Other terms may be neglected. The inductance offers zero reactance for
d.c. components and thus d.c. voltage appears at the output terminals. However, it offers
very high reactance for the second and other higher harmonic terms. Thus the inductance
attenuates the a.c. components to appear at the output terminals.
Now the ripple for this type of filter is calculated as below:
Peak value of the a.c. components to appear at the output is
4Em RL
E a .c . = . ) where X L = 2 j ω . L , is the inductive
3π ( RL + X L
reactance at second harmonic frequency 2 ω . The root mean square value of this ripple
content is :
4Em RL
E r .m. s . =
3 2π ( R L + X L )
4 Em RL 4Em
= =
[ ]
E r .m .s.
3 2π R L + 4ω L
2 2
 4ω 2 L2 
3 2π  1 + 
 R L2 
4ω 2 L2 2 E m .R L
If >> 1 then E r .m . s . =
RL2 3 . 2π .ω .L
2Em
E d .c . =
π
E r .m .s . 2 E m R L 3 2πω 1 RL
The ripple factor γ = = = . ------ (5.15)
E d .c . 2Em 3 2 ω.L
π
From this equation, it is clear that the ripple factor depends on both the load resistance
and the magnitude of the inductances. The ripple factor decrease with the increase of
inductance; and also it is smaller for smaller value of RL. That is the inductor filter is
suitable for higher value of load current.

Example 5.9 A full wave rectifier with series inductor filter has the load resistance of
1KΩ and inductor of 25 Henry. The peak value of the applied a.c. signal is 50 volts and
frequency 50Hz. Calculate d.c. output voltage, d.c. current to the load resistance and the
ripple factor. Diodes used are ideal.
Solution: L = 25 Henry Em = 50 Volts RL = 1000 Ω f = 50 Hz
2 E m 2 x50
E d .c . = = = 31.85 Volts
π 3.14
E 31.85
I d .c . = d .c . = = 31.85mA
RL 1000
1 RL RL 1000
Ripple factor γ = . = = = 0.75
3 2 ω.L 3 2 x 2 xπxf 3x1.414 x 2 x3.14 x50

5.4.5 L –Section ( or L-C) Filter: It has been discussed earlier that the shunt
capacitor filter is used for small value of load currents and inductor filter is suitable
where we wish to draw large amount of currents. The series combination of inductor and
capacitor used for the filtering action is suitable for all values of the currents. Figure
(5.14) shows the circuit diagram of L-section or L-C filter with full wave rectifier.

Fig. 5.14

In this circuit, the values of inductance L and capacitance C are chosen so that XL
(inductive reactance) is much greater than XC (capacitive reactance) at the ripple
frequency. In this way the inductor attenuates the ripple and capacitor bypasses it.
The voltage EI available at the input of the L-section filter is given by:
2Em 4Em 4Em
EI = − cos 2 ω .t − cos 4 ω .t − ......
π 3π 15 π
Higher Harmonic terms (having frequency more than 2 ω ) may be neglected. The
effective ripple frequency may be assumed as 2 ω .
2Em
E d .c . =
π
The a.c. components (ripple content) at the output terminals may be given by:
4 E m .( X C R L )
E a .c . = cos 2 ω .t
3π .( X L + X C R L )
It is further assumed that XL (inductive reactance at 2 ω frequency) is large enough than
the parallel combination of load resistance RL and XC (capacitive reactance at 2 ω ripple
frequency) also RL>> XC.
4 E m .X C
So Ea .c . = cos 2 ω .t .
3π . X L
4 Em X C
The r.m.s. value of a.c. component is E r . m. s . =
3 2π X L
(capacitive reactance at 2 ω ripple frequency)
1
Putting XC =
2ω.C
and XL = 2 ω
L (inductive reactance at 2 ω
ripple frequency)
4 Em Em
we get E r .m. s . = =
3 2πω 2 .2ωL.2ωC 3 2πω 2 L.C
 E r .m.s E m (3 2πω 2 LC )
The ripple factor γ is given as: γ = =
E d .c . 2Em π
1
or γ = ------ (5.16)
6 2ω 2 L.C
It is clear from this equation that the ripple factor is independent of load resistance. So
this filter circuit is suitable for all values of currents.

5.4.6 π -Section Filter: A more common filter is the π -Section Filter shown in
figure (5.15). It has a capacitor filter followed by L-section filter and is used to provide

Fig. 5.15
higher d.c. output voltage with low ripple. It used for medium load currents. The
capacitors C1, C2 and inductance L form a π − type network, hence it is known as π -
Section filter. Here the capacitor C1 bypasses the ripple frequency, the inductance L
attenuates the ripple and C2 further bypasses it. So it is assumed that pure d.c. is available
at the output.
The r.m.s. value of a.c. components (ripple content) across the capacitor C1 is given by:
I d .c. π . I d .c .
E r . m. s . = = where ω = 2πf
4 3 fC1 2 3ω C 1
The r.m.s. value of the a.c. voltage across the parallel combination of C2 and RL may be
calculated as:
E r .m.s. ( X C 2 R L ) I d . c .π . X C 2
E r' .m . s . = ≅ (since X L >> X C 2 and
( X L + X C2 R L ) 2 3ω C 1 X L
RL >> X C2 )
(capacitive reactance at 2 ω ripple frequency)
1
Putting XC =
2ω.C
and XL = 2 ω L (inductive reactance at 2 ω ripple frequency)
π .I d .c.
We get E r .m.s. ≅
'

8 3ω 3 C1C 2 L
d.c. voltage across the load resistance Ed.c. = Id.c. RL
The ripple factor is given by:
 E r' .m.s. π .I d .c .
γ = =
E d .c . 8 3ω 3 .C1 .C 2 .L.I d .c. R L
π
or γ = ------ (5.17)
8 3ω .C1 .C 2 .L.R L
3

the ripple factor is inversely proportional to the load resistance.

We can, of course, use more than one section of filters depending upon the extent of
ripple reduction as shown in figure (5.16). Each section acts like a voltage divider; the
overall attenuation equals the product of the individual attenuation.

Fig. 5.16

Example 5.10 A full wave rectifier with π − section filter has the following circuit
elements. C1 = C2 =20 µF, L = 20H, RL = 5KΩ, peak value of the a.c. signal 200 volts
and frequency = 50 Hz. Find the ripple factor of this circuit.
π
Solution: The ripple factor γ is given by γ =
8 3ω 3 .C1 .C 2 .L.R L
3.14
or γ =
8 x 3 x(2 x3.14 x50) x 20 x 20 x10 −6 x10 −6 x 20 x5000
3

= 0.002 = 0.02%

5.5 Voltage Multiplier Circuits: Some times high voltage / low current d.c.
supply is required in electronic circuits. Such a supply is needed for accelerating the
electrons in a cathode ray tube of CRO. Voltage multiplier circuits are used to design
such a supply, it gives a d.c. output which is a multiple of the peak value of the input a.c.
signal applied to the circuit.

5.5.1 Half Wave Voltage Doubler: As the name indicates it gives d.c. output
which is almost double the peak value of the input signal. It is also known as the cascade
voltage doubler. The basic circuit diagram of the half wave voltage doubler is shown in
figure (5.17a).

It works as follows. During the negative half cycle of the input sinusoidal wave,
diode D1 conducts and the capacitor C1 charges to the maximum of the peak value of the
input wave (Em) with the polarity of the capacitor as shown in figure (5.17). The diode D2
will be in the reverse bias. Now during the positive half cycle of the input wave, diode D1
will
 Fig. 5.17

be in reverse bias and diode D2 conducts as the voltage across the diode D1 is positive
and its magnitude will be approximately equal to double the peak value of the input
sinusoidal wave. The capacitor C2 will charge to steady voltage +2Em for the number of
cycles of operations. So the output voltage E0 which is equal to the double of the peak
value of the input will be available across the load resistance RL. Hence this circuit is
known as voltage doubler. The PIV ratings of the diodes used in this circuit, is 2Em. Half
wave voltage doubler has very poor regulation and its ripple content is also high as ripple
frequency is equal to the frequency of the input signal.
5.5.2 Full Wave Voltage Doubler: The basic circuit diagram of the full wave
voltage doubler is shown in figure (5.18). It works as follows.

Fig. 5.18

During the first positive half cycle of the input sinusoidal signal of the wave the diode D1
will conduct and capacitor C1 charges to the maximum of the peak value Em of the wave,
with the polarity of the capacitor C1. In this case, the diode D2 will be in reverse bias.
Now during the next negative half cycle of the input wave, diode D1 will be in reverse
bias and diode D2 will conduct and capacitor C2 will charge to Em. The total voltage
across the load resistance RL will be just double of the peak value of the input wave i.e.
+2Em. The PIV ratings of the diodes used in this circuit, is 2Em. The ripple frequency in
this circuit is equal to twice the frequency of the input signal (second harmonic terms).

5.5.3 Half Wave Voltage Multiplier: The half wave voltage doubler discussed
above can be extended to obtain any multiple of the peak value of the input wave i.e.
3Em, 4Em, 5Em, 6Em etc. The circuit thus obtained may be called as voltage multiplier
which is shown in figure (5.19).
 The circuit may be explained on the same lines as the half wave voltage doubler.
The voltage obtained across A & B terminals is equal to 2Em, and the voltage across A &

Fig. 5.19

C terminals is equal to 4Em (Quadrupler), and across A & D is 6Em. The voltage across
the terminals E & F is 3Em (Tripler), and the voltage across E & G terminals is 5Em.

Example 5.11 What will be voltage at the output of half wave voltage doubler, if the
voltage at the secondary of the transformer is 25 volts? What will be PIV of the diodes
used?
Solution: Er.m.s. = 25 volts,
Peak value Em = 2 xE r .m.s. = 1.414 x 25 = 35.35 volts
Output voltage of the doubler = 2.Em = 2x35.35 = 70.70 volts
PIV of the diodes used = 2.Em =70.70 volts

5.6 Clipper Circuits: The clipping circuits clip off the unwanted portion of the
signal without distorting the remaining part of the applied signal. The clipping circuits
basically are of two types (i) Series Clipper and (ii) Shunt Clipper. In series clippers the
diodes are connected in series with the load resistance and hence the name series clipper.
In shunt clippers the diodes are connected in parallel or in shunt with the load resistance.
These clippers may further be classified as (i) unbiased and (ii) biased clippers. In
unbiased clippers no additional battery is applied to the diodes. However, in case of
biased clippers, the diodes are biased with the additional battery of desired magnitude.
The classification of clipper circuits is shown in figure (5.20). Now we shall
discuss each of these circuits in detail.
 Clippers

Series Clippers Shunt Clippers

Unbiased Biased
Unbiased Biased
Clippers Clippers
Clippers Clippers

Positive Negative Positive Negative Positive Negative Positive Negative

clippers clippers clippers clippers clippers clippers clippers clippers

Fig. 5.20

5.6.1 Unbiased Positive Series Clipper: This circuit is basically the half wave
rectifier as shown in figure (5.21). If the input wave sine, triangular or square wave is
applied to the input of the circuit, we get zero output for the positive half cycle of the
input wave; and for the negative half cycle output is the same as the input. This is
because that the diode goes in reverse bias during positive half cycle of the input wave
and during the negative half cycle the diode is forward bias and output follows the input.
The input output wave shapes are shown in figure (5.22). From the wave shapes it is clear
that this circuit clip off the positive half cycle of the input wave hence the name positive
clipper.

Fig. 5.21

Fig. 5.22
5.6.2 Unbiased Negative Series Clipper: If the connections of the diode are
reversed as shown in figure (5.23), then the circuit is known as negative unbiased series
clipper circuit. It clips off the negative half cycle of the input wave; and the output is the
same as the input for negative half cycle. The input output wave shapes are shown in
figure (5.24).

Fig. 5.23

Fig 5.24

5.6.3 Biased Positive Series Clipper: The circuit diagram of this clipper is shown
in figure (5.25). To explain the working of the circuit let us assume the diode and battery

Fig. 5.25

are ideal. During the positive half cycle of the input, the point A is positive with respect
to point B and the diode will remain in reverse bias till the cathode of the diode is more
negative than E1. During the next half cycle of the input wave the diode will remain in
reverse bias and the output E0 is zero. So when the input is positive and beyond E1, the
output follows the input; and when the input is between zero and E1 then the output E0 is
zero. It is, therefore, concluded that this circuit has clipped off the portion between zero
and +E1 of the input wave. Hence the name biased positive clipper. The input output
wave shapes are given in figure (5.26). It is worth while to mention that the input should
be greater than the magnitude of the battery otherwise the clipping will not take place.
 Fig. 5.26

5.6.4 Biased Negative Series Clipper: The circuit diagram of this clipper is
shown in figure (5.27). To explain the working of the circuit let us assume the diode and

Fig. 5.27
battery are ideal. During the positive half cycle of the input, the point A is positive with
respect to point B and the diode will be in reverse bias. During the next half cycle of the
input wave the diode will remain in reverse bias till the anode of the diode is more
positive than E2 and the output E0 will follow the input beyond E2. So when the input is
positive, the output is zero and when the input is negative and beyond E1 then the output
follows the input. It is, therefore, concluded that this circuit has clipped off the portion
between zero and – E2 of the input wave. Hence the name biased negative clipper. The
input output wave shapes are given in figure (5.28). It is worth while to mention that the
input should be greater than the magnitude of the battery otherwise the clipping will not
take place.

Fig. 5.28
Further if both the circuits of figures (5.25) & (5.27) are combined as shown in figure
(5.29), then the clipper will be both sided biased clipper.
 Fig. 5.29

It is very easy to explain the circuit on the similar lines as discussed above. In this
circuit the diode D1 conducts when input is positive and beyond E1. The output thus
follows the input, clipping off the portion between 0 and E1. The diode D2 will conduct
when the input is negative and beyond E2. The output follows the input, clipping off the
portion between 0 and – E2. The input output wave shapes are shown in figure (5.30).

Fig. 5.30

In the clipping circuits discussed above, we have considered the ideal diodes and
ideal batteries. Practically it is sometimes important to consider the knee voltage or cut –
in voltage Vγ of the diode, its resistance rd and also the internal resistance of the battery
rb. In this situation the Vγ potential will be added with the voltage of the battery that is
clipping voltage will be E1+ Vγ and E2 + Vγ in double sided series clipper. The transfer
characteristics which is the input output relationship, will have less than unity slope at the
RL
points (E1 + Vγ ) and – (E2 + Vγ ). The value of slope is given by . This
RL + rd + rb
will result a distortion in the out put wave forms.

Example 5.12 Draw the transfer characteristics of the series clipper circuit shown in
the figure (5.31a). The values of the circuit parameters are given as: E1 =1.5 V, E2 = 2.0
V, RL = 200 Ω, diode resistance of each diode rd=20 Ω and internal resistance of each
battery rb = 10 Ω. The cut – in voltage of each diode Vγ = 0.7 V.
 Fig. 5.31a

Fig. 5.31b

Solution: The break points are (E1 + Vγ ) = 1.5 + 0.7 = 2.2 V

– (E2 + Vγ ) = 2.0 + 0.7 = – 2.7 V
RL 200
The slope at these points is = = 0.87
RL + rd + rb 200 + 20 + 10
The transfer characteristics are shown in figure (5.31b).

5.6.5 Unbiased Positive Shunt Clipper: The circuit diagram for unbiased

Fig. 5.32

positive shunt clipper is shown in figure (5.32). Its working may very easily be explained
as follows. During the positive half cycle of the input wave, the diode is forward biased
and will act as an on switch. The output will be zero for the positive half cycle. However,
for the negative half cycle diode is in reverse bias and will act as open switch. The output
will therefore, follows the input. So during positive half cycle no output is obtained
clipping off this cycle of the input wave. The output is taken in parallel with the diode
hence called unbiased positive shunt clipper. The input output wave shapes are shown in
figure (5.33).

Fig. 5.33

5.6.6 Unbiased Negative Shunt Clipper: In this circuit the connections of the
diode are reversed as shown in figure (5.34a). This circuit clip off the negative half cycle
of the input wave and positive half cycle is obtained without any distortion. The input
output wave shapes are shown in figure (5.34 b)
5.6.7 Biased Positive Shunt Clipper: The circuit diagram of the biased positive

clipper is shown in figure (5.35a). It work as follows, during the positive half cycle of the
input wave, the diode will be in reverse bias and behaves as an open switch till input is
less than or equal to the magnitude of the battery E1. In this condition, output follows the
input. When the input becomes more positive than E1, diode conducts and works as an on
switch. The output will equal to E1. During the next half cycle the diode will be in reverse
bias and output will follow the input. So it is concluded that the portion of the input
beyond E1, in positive half cycle is clipped off. The input output wave shapes is shown in
figure (5.35 b).

5.6.8 Biased Negative Shunt Clipper: The circuit diagram of the biased negative
clipper is shown in figure (5.36 a). In this circuit the connections of the diode and
battery are reversed. It work as follows, during the positive half cycle of the input wave,
the diode will be in reverse bias and behaves as an open switch and the output follows the
input. During the next half cycle of the input wave the diode will be in reverse bias till
the input is less negative than E2.When the input becomes more negative than E2, diode
conducts and works as an on switch. The output will equal to E2. So it is concluded that
the portion of the input beyond E2, in negative half cycle is clipped off. The input output
wave shapes is shown in figure (5.36 b).

Further both the circuits of figures (5.35 a) & (5.36 a) may be combined to get both sided

biased shunt clipper as shown in figure (5.37 a). It may be explained that the portion
beyond E1 in the positive half cycle of the input, is clipped off and in the negative half
cycle the portion beyond E2 is clipped off. The input output wave shapes are shown in
figure (5.37 b).

Example 5.13: The diodes connected in the circuit of figure (5.37 a) are not ideal but
have some finite forward resistance rd. Draw its transfer characteristics and the input
output wave forms.
Solution: When the input signal is lying between E1 and E2, both the diodes will be in
reverse bias and the slope will be unity at the origin. When the input is beyond E1 or E2,
the slope at these points E1 and E2 will not be zero but will have the slope equal to
rd
which is small positive. So the output will be slightly distorted at the points E1 &
R + rd
E2. The transfer characteristics and its input output wave shapes are given in figure (5.38)
 Fig. 5.38

5.7 Clamping Circuits: The clamping circuit is used to clamp a signal to a desired
d.c. level. The sinusoidal, square or triangular signal normally swings symmetrically
about the X- axis with equal magnitude on both sides as shown in figure (5.39). The
signal in this condition is said to have the zero average value or its d.c. level is zero. If the
signal is lifted upward to touch the negative peak points of the signal to X- axis, the
signal is said to have positively clamped at zero d.c. level. If on the contrary the signal is
lifted downward to touch the positive peak points of the signal to X- axis, the signal is
said to have negatively clamped at zero d.c. level (ref fig. 5.39). The signal may further
be lifted upward or down ward to a desired positive or negative voltage (E) as shown in
figure (5.39). In this condition the signal is said to have clamped at positive E volts or
negative E volts.

Fig. 5.39
The clamping circuit shown in figure (5.40a) consists of a diode, a capacitance
and a resistance. To the input of this circuit we apply a signal whose d.c. level is zero i.e.
which swings both sides of X – axis symmetrically. During the first negative half cycle of
the input wave Ei, diode conducts and it behaves as a closed switch. The capacitor C
charges to the peak value of the negative swing (EN) with the polarity shown in figure
(5.40a). The values of the capacitor C and resistance R are so chosen so that the RC time
constant is large enough so that the capacitor is not immediately discharged. Now during
the next positive half cycle of the input wave, diode is in reverse bias and behaves as an
open switch. The voltage at the output will be equal to the magnitude (EN+EP).The cycle
repeats and we get the output as shown in figure (5.40b). It is clear from this figure that
the output wave touches the negative peak of the signal to the X-axis hence it is
positively clamped at zero d.c. level.

If the diode connections are reversed as shown in figure (5.41a), the circuit may easily be
explained that the signal is negatively clamped at zero d.c. level (ref. 5.41 b).
Now to clamp the signal at some positive E volt, then a battery of E volt may be
introduced is series with the diode as shown in figure (5.42 a). Input output signals are
shown in figure (5.42 b).

Fig. 5.42 b
Fig. 5.42 a
Similarly, to clamp the signal at some negative E volt, then a battery of E volt may be
introduced in series with the diode as shown in figure (5.43 a). Input output signals are
shown in figure (5.43 b).

5.8 Log and anti log circuit: The signal processing operations such as logarithm
and antilogarithm can be performed using the p – n junction diode characteristics. As it is
well known that the current flowing through the diode is given by
I = I s (eV / VT − 1) ------- (5.18)
where VT is the thermal voltage and Is is the reverse saturation current which are
constants at the particular temperature and semiconductor material. If V >> VT, this
equation will of the form
I = I s eV /VT ------ (5.19)
i.e. if a voltage V is applied across a diode as shown in figure (5.44a), the current
flowing through the diode will be proportional to the antilog function

Fig. 5.44 a Fig. 5.44 b

of this voltage V. This circuit may be called as the antilog circuit. Figure (5.43b) shows
the antilog function of the input voltage V.
Taking the log on both sides of equation (5.19), we get:
 I  V I 
ln  = or V = VT ln  ------ (5.20)
 I s  VT  Is 
from this equation it is clear that if a known current I from a current source

Fig. 5.45a Fig. 5.45 b

is passed through the diode (fig. 5.45 a), then the voltage across the diode will be
proportional to the logarithm of current I. hence this circuit may be called as log circuit.
The logarithm function is shown in figure (5.45 b).

5.9 Zener Diode as Voltage Regulator: The heavily doped P N junction diodes
which work in reverse bias and operate in the break down region, known as Zener diodes
have already been discussed in the preceding chapter. Here we shall discuss its most
common application as a voltage regulator which provides a constant voltage from a
source whose voltage may vary over a sufficient range. A simple voltage regulator circuit
using Zener diode is shown in figure (5.46).
In this circuit a series combination of Zener diode and a resistance Rs, is connected across
the unregulated supply Ei. The regulated output voltage E0 is available across the parallel
combination of Zener diode and the load resistance RL. The polarity of the Zener diode is
such that it is biased in reverse bias. The Zener diodes of different break down voltages

Fig. 5.46

are available in the market, so the diode of required break down voltage Ez is chosen in
the circuit. If the circuit elements are such that the voltage across the Zener diode is less
than the Zener break down voltage Ez, the diode will behave like an off switch.
 Ei RL
The output voltage E0 will be given by E 0 = . The Zener diodes are never used
Rs + R L
in this state.
If the voltage across the Zener diode is more than or equal to the Zener break down
voltage Ez, the diode is in on state and acts as voltage source of voltage Ez. The output
voltage E0 will be equal to Ez. This is the required state for the voltage regulation.
The current drawn Is from the unregulated supply is given by Is = Iz + IL
E E − E0
or Iz = Is – IL where I L = i and Is = i
RL Rs
The power dissipation of the Zener diode is given by: Pz = Ez.Iz.
As discussed above the Zener diode is used in the break down region. The output voltage
will be equal to the Zener voltage. This circuit is commonly used as voltage regulator or
as a fixed reference source. The regulator circuit gives the constant voltage irrespective of
the change in the load resistance RL or the change in the unregulated supply Ei. We shall
now find the minimum value of load resistance RL and the minimum value of the Ei. If
the load resistance RL or unregulated supply is too small the diode will not be in the break
down region or the Zener will be off.
Ei xRL
We know E0 = E z =
( Rs + RL )
or ( Rs + R L ) E z = Ei R L
or RL ( Ei − E z ) = Rs E z
Rs E z
or RL min = ,
( Ei − E z )
so any load resistance greater than this value will ensure the Zener diode is in the break
down region or on state.
E0 Ez
The maximum value of RL may also be calculated. I L max = =
RL R L max
Ei − E z E s
or I L max = = = Is
Rs Rs
The Zener current is minimum, as Is = Iz + IL .
ILmin may also be given by: I L min = I s − I z max
Ez
so R L max =
I L max
The minimum value of the unregulated supply is to be calculated as follows:
Ei xRL
We know E0 = E z =
( Rs + RL )
( R + Rs ) xE z
and Ei min = L
RL
The maximum value of Ei is limited by the maximum Zener current Izmax. Since
Izmax = Is – IL .
Now Ei = I.Rs+ E0
As E0 = Ez is constant, the input voltage will be maximum when Is is maximum, so
Eimax = Ismax.Rs + Ez
It is worth while to mention the following points regarding the Zener regulator
circuits.
1. More than one Zener diodes may be connected in series, and the
output voltage will be sum of all the break down voltages of the Zener
diodes. Let three Zener diodes Z1, Z2, Z3 are connected in series as shown
in figure (5.47), whose values are Ez1 = 6 volts, Ez2 = 12 volts, Ez3 = 15

Fig. 5.47
volts. The output voltage will be 33 volts, provided the circuit parameters
are properly chosen, so that the voltage across the load resistance is more
than 33 volts, when diodes are assumed in off state.
2. The Zener diodes are never connected in parallel, as in the break down
region the Zener diodes behave like ideal voltage source.
3. The Zener diodes may be connected in series back to back as shown in
figure (5.48a). This circuit will behave like the shunt clipper circuit. Input

Fig. 5.48 a Fig. 5.48 b

signal connected to this circuit is a varying signal whose peak value

should be greater than break down voltage of each diode. During the
positive half cycle of the input signal the diode Z1 will be in forward and
will work like an ordinary diode. The Zener diode Z2 will be in reverse
bias till the input is less than its break down voltage, in this condition the
 output follows the input. But when the input is greater than the break
down voltage of Z2, a constant voltage equal to Ez2 will be maintained
across the output. Thus it clips off the portion of the input signal beyond
Ez2. Similarly it can be shown that this circuit clips off the portion of the
negative half cycle beyond Ez1, as the process is reverse in this case. The
input output wave shapes are shown in figure (5.48 b).

Example 5.14 The Zener diode regulator circuit shown in figure (5.49) has the following
parameters. Ei = 20 volts, Rs = 1KΩ, Ez = 8.2 volts and RL = 2KΩ. Find E0, Es, Iz. If the
maximum wattage of the Zener diode is 35 mW, then suggest whether this diode will
work or not.

Fig. 5.49
Solution: If the Zener diode is in the off state, the voltage across the load resistance RL
20Vx 2 KΩ
is given by: E= = 13.33Volts
(1 + 2) KΩ
This voltage is greater than the break down voltage of the Zener diode, the diode will go
in the on state and out put voltage will be equal to the Ez.
So E0 = Ez = 8.2 Volts
and voltage across Rs is Es = 20 – 8.2 = 11.8 volts
11.8Volts
The current following through the resistance Rs is Is = = 11.8mA
1KΩ
8.2Volts
The current following through the resistance RL is IL = = 4.1mA
2 KΩ
The current following through the Zener diode is
I z = I s − I L = 11.8 − 4.1 = 7.7mA
Power of the Zener diode Pz =Ez x Iz =7.7 mA x 8.2 volts = 63.14 mW
This is the required power of the Zener diode. So we have to use the Zener diode whose
maximum power is greater than 63.14 mW. But in the given problem, the required power
of the Zener diode is less than the given power 35 mW. So the Zener diode of 35 mW
power will not work. The Zener diode of higher than 63.14 mW should be used.

Example 5.15 The Zener diode regulator circuit shown in figure (5.50) has the
following parameters. Ei = 35 volts, Rs = 1.2 KΩ, Ez = 12 volts and maximum Zener
current is 10mA. Find the range of RL so that E0 remains to be constant to 12 volts.
 Fig. 5.50

Solution: The minimum value of RL is given by:

Rs E z 1200 x12
RL min = = = 626Ω
( Ei − E z ) 35 − 12
The current flowing through the resistance Rs is then given by:
E − E z (35 − 12 )volts
Is = i = = 19.2mA
Rs 1.2 KΩ
The minimum value of IL is I L min . = I s − I z max = 19 .2 − 10 = 9 .2 mA
Ez 12 volts
The maximum value of RL is R L max = = = 1 .3 K Ω
I L min . 9.2 mA

Problems:

1. Draw a circuit diagram of a full wave rectifier along with a series inductor
(choke) filter and obtain a formula for the ripple factor.
2. Give the circuit diagram and describe the working of a full wave rectifier with
shunt capacitor filter. Why for very low ripple the output current should be very
small?
3. Explain, giving the circuit diagram, the working of a full wave rectifier. Obtain
the expressions for average, r.m.s. value of the output voltage, efficiency and
ripple factor.
4. Drive the expressions for (a) average and (b) r.m.s. value of the output voltage of
a half wave rectifier. Find the expression of rectifier efficiency and ripple factor
for half wave rectifier.
5. Describe the circuit diagram and explain the working of half wave rectifier.
Explain the ripple voltage and ripple factor for the same.
6. Give the circuit diagram of a full wave rectifier with an L-Section filter and
explain its working. Find the expression for the ripple factor.
7. Give the circuit diagram of a full wave rectifier with π - section filter and explain
its working. Find the expression for the ripple factor.
8. What do you mean by the peak inverse voltage of the diode? Show that when a
capacitor is connected across the load resistance of a half wave rectifier circuit,
and then the peak inverse voltage of the diode is approximately twice the peak
voltage of the input signal.