College of Computer Science

Department of Computer Engineering

CCE121: Semiconductors

Chapter 2: Diodes applications

This lectures notes are extracted from the text book: "Electronic
Devices and Circuit Theory", 11th edition 2013, by Robert L.
Boylestad and Louis Nashelsky.
2020-2021
 Load-Line Analysis (graphical solution)

The analysis of diode can follow one of two paths: using the
actual characteristics or applying an approximate model for
the device.
Load Line Analysis: is used to analyze diode circuit using its
actual characteristics.

1
 Load-Line Analysis (graphical solution)
ID Si From Kirchhoff’s voltage law (KVL) : +E – VD – V0 = 0
Since VO = RID
+ - + +E – VD – V0 = 0 E – VD – RID = 0
VD
R
E +- VO E – VD – RID = 0
Loop
-
: load line equation
P1 ( ID = …….. , VD=0V)

At VD= 0 V:

Since; P1 ( ID = , VD=0V)

P2 ( ID = 0mA. , VD…….)

Since; P2 ( ID = 0mA , VD=E)

2
 Load-Line Analysis (graphical solution)

P1 ( ID = , VD=0V)
P2 ( ID = 0mA , VD=E)

P1

P2

A straight line is defined by the parameters of the network.

It is called the load line because the intersection on the vertical
axes is defined by the applied load R.
3
 Load-Line Analysis (graphical solution)

P1

P2

The maximum ID equals E/R, and the maximum VD equals E.

The point where the load line and the characteristic curve
intersect is the Q-point, which identifies ID and VD for a particular
diode in a given circuit.
4
 Example 2.1

For the given diode configuration and diode characteristics,

determine: VDQ , IDQ and VR.

5
 Example 2.1 - Solution
ID Si From Kirchhoff’s voltage law (KVL): +E – VD – V0 = 0
Since VO = RID
+ - + +E – VD – V0 = 0 E – VD – RID = 0
VD R=0.5K
+ VO
- Loop E – VD – RID = 0
E=10V -
: load line equation
P1 ( ID = …….. , VD=0V)

At VD= 0 V:

Since; P1 ( ID = 20mA , VD=0V)

P2 ( ID = 0mA. , VD…….)

Since; P2 ( ID = 0mA , VD=10V)

6
 Example 2.1 - Solution

The load line is firstly drawn between VD=E=10 V and

ID=E/R=10/0.5K=20mA. The intersection between the load line and
characteristics defines the Q-point as VDQ=0.78V and IDQ=18.5mA.
VR=IDQ R=(18.5mA)(0.5K)=09.25 V.
7
 Diode Configurations

The forward resistance of the diode is usually so small

compared to the other series elements of the network that it
can be ignored.
In general, a diode is in the “on” state if the current
established by the applied sources is such that its direction
matches that of the arrow in the diode symbol, and
VD 0.7V for silicon, VD 0.3V for germanium, and
VD 1.2V for gallium arsenide.
You may assume the diode is “on”, and then find the current
in the diode. If the current flows into the positive terminal of
the diode, then the assumption is right, otherwise, the diode
is “off”.

8
 Series Diode Configurations

Constants Forward Bias

Silicon Diode: VD= 0.7V
Germanium Diode: VD= 0.3V
Analysis (for silicon)
if (E > 0.7 V) Then
The diode is in the “ON”
VD = 0.7 V (or VD = E if E < 0.7 V)
V R = E – VD
ID = IR = IT = VR / R= (E-VD) / R
if (E < 0.7 V) Then
The diode is in the “OFF” (open circuit)
VD = E
ID = IR = IT = 0mA
VR = 0V
9
Equivalent circuit for the “ON” diode
 Series Diode Configurations

Reverse Bias
Diodes ideally behave as open
circuits

Analysis (for silicon)

VD = E
VR = 0V
ID = IR = 0mA

10
Equivalent circuit for the “OFF” diode
 Example 2.4

Determine VD, VR and ID.

11
 Example 2.4-Solutions
Equivalent circuit for the “ON” diode
This diode is Forward Bias, because the
diode is in the “on” state (the current
established by the applied sources is such that
its direction matches that of the arrow in the
diode symbol): Since the applied voltage
establishes a current in the clockwise
direction to match the arrow of the symbol Loop
and the diode is in the “on” state,
Silicon Diode is Forward Bias, E = 8V
R=2.2K
and E =8V 0.7V VD= 0.7V
From Kirchhoff’s voltage law (KVL): +E – VD – VR = 0
VR = E – VD = 8-0.7=7.3V
ID = IR = IT = VR / R= (E-VD) / R
ID = (8-0.7) / 2.2K =3.32mA
12
 Example 2.5

Determine VD, VR and ID.

13
 Example 2.5-Solutions
Equivalent circuit for the “OFF” diode
This diode is Reverse Bias
The current flows into the
positive terminal of the diode,
then the assumption is right,
otherwise, the diode is “off”.
Open circuit. Loop
Removing the diode, we find that E = 8V
the direction of I is opposite to R=2.2K
the arrow in the diode symbol
and the diode equivalent is the Since ID = IR = IT = 0mA
open circuit no matter which VR = RIR = 0V
model is Employed. The diode is From Kirchhoff’s voltage law (KVL):
“off”. Open circuit
+E – VD – VR = 0 +E – VD – 0=0
VD = E = 8V
14
Source Notation

15
 Example 2.6

Determine VD, VR and ID.

Solution

Although the “pressure”

establishes a current with the
same direction as the arrow
symbol, the level of applied
voltage is insufficient to turn
the silicon diode “on.”
(E =0.5V 0.7V )
Diode is OFF Open circuit
From Kirchhoff’s voltage law (KVL): E-VD-VR=0 E-VD-0=0
VD= E=0.5V
Since ID = IR = IT = 0mA
VR = RIR = 0V
16
 Example 2.6

Solution 2

Although the “pressure” establishes a

current with the same direction as the
arrow symbol, the level of applied
voltage is insufficient to turn the
silicon diode “on.” (E =0.5V 0.7V )
Diode is OFF Open circuit

17
Example 2.6

Loop

From Kirchhoff’s voltage law (KVL): E-VD-VR=0 E-VD-0=0

VD= E=0.5V
Since ID = IR = IT = 0mA
VR = RIR = 0V
18
 Example 2.7

Determine Vo and ID. The forward bias voltage for red LED is 1.8 V.

19
 Example 2.7 Solution
Solution
An attack similar to that applied in Example 2.4 will reveal that the
resulting current has the same direction as the arrowheads of the
symbols of both diodes, and the network of the following figure
results because E=12V (0.7V+1.8V)= 2.5 V.

From Kirchhoff’s voltage law

(KVL):
+E – VK1 –VK2 –V0 = 0
V0 = E – VK1 –VK2 = 12-0.7-1.8
Loop V0 = 9.5V
ID = IR = V0 / R= (E-VK1-VK2) / R
Equivalent circuit for the 2 “ON” diodes ID = (12-0.7-1.8) / 680 =13.97mA
20
Example 2.8

Determine ID ,VD2 and Vo.

21
 Example 2.8-Solution

Removing the diodes and determining the direction of the resulting current I
result in the circuit of Fig. 2 . There is a match in current direction for one
silicon diode but not for the other silicon diode. The combination of a short
circuit in series with an open circuit always results in an open circuit and I D
0 A, as shown in Fig. 3

Fig. 1 Fig. 2 Fig. 3

Circuit for Example 2.8 Determining the state Substituting the equivalent
of the diodes of Fig. 1. state for the open diode.

22
 Example 2.8-Solution

The question remains as to what to substitute for the silicon diode. For the
analysis to follow in this and succeeding chapters, simply recall for the actual
practical diode that when ID= 0 A, VD=0V (and vice versa),

Loop
Since ID1 = IR = I = ID2 = 0mA
V0 = RIR = 0V
Equivalent circuit From Kirchhoff’s voltage law (KVL):
E-VD1-VD2-V0=0 VD2= E-VD1=20V
VD2= 20V

23
Example 2.9

Determine I,V1 ,V2 and Vo.

Fig. 2.25

24
 Example 2.9-Solution
The sources are drawn and the current direction indicated as shown
in Fig. 2.26 .
The diode is in the “on” state and the notation appearing in Fig. 2.27
is included to indicate this state. Note that the “on” state is noted
simply by the additional VD= 0.7 V on the figure. This eliminates
the need to redraw the network and avoids any confusion that may

Fig. 2.27
Fig. 2.26
Determining the unknown quantities for the
Determining the state of the diode
network of Fig. 2.25 . KVL, Kirchhoff voltage
for the network of Fig. 2.25 .
loop.
25
 Example 2.9-Solution

From Kirchhoff’s voltage law (KVL) in

Loop1:
E1-V1-VD -V2+ E2=0 V1+V2= E1-VD+E2
Since:
Loop2 V1=R1I ; R1=4.7K
KVL V2=R2I ; R2=2.2K
Loop1 V1+V2= E1-VD+E2 R1I+R2I= E1-VD+E2
(R1+R2)I=E1-VD+E2
E1-VD+E2 10 -0.7+ 5
I= = = 2.07mA
(R1+R2) 4.7K+2.2K

V1=R1I =4.7Kx2.07m=9.73V
V2=R2I =2.2Kx2.07m=4.55V
Applying Kirchhoff’s voltage law to the output section (Loop2) in the clockwise direction
results in
-E2 + V2-V0 =0 V0= V2-E2 = -0.45V

26
 Parallel and Series-Parallel Configurations
Example 2.10

Determine V0,I1 ,I2 and ID2 for the parallel diode

configuration of Fig. 2.28

Fig. 2.28
27
 Example 2.10-Solution

Fig. 2.29
Determining the unknown
quantities for the network
of Fig. 2.28 .

For the applied voltage the “pressure” of the source acts to establish a current through
each diode in the same direction as shown in Fig. 2.29 . Since the resulting current
direction matches that of the arrow in each diode symbol and the applied voltage is
greater than 0.7 V, both diodes are in the “on” state. The voltage across parallel
elements is always the same and
V0= 0.7V

28
 Example 2.10-Solution

From Kirchhoff’s voltage law (KVL) in

A Loop1:
E-VR-VD1 =0 VR= E-VD1=10-0.7=9.3V
Loop1 Since:
VR=RI1 ; R1=0.33K
KVL VR 10 - 0.7
I= = = 28.18mA
R 0.33K

Applying Kirchhoff’s current law to the node A

I1 – ID1-ID2 =0 I1= ID1+ ID2

Assuming diodes of similar characteristics, we have
I1
ID1= ID2 = = 14.09mA
2

29
 Example 2.11

Find the resistor R to ensure a current of 20 mA through the “on”

diode for the given circuit. Both diodes have reverse breakdown
voltage of 3V and average turn-on voltage of 2V.

Fig. 2.30
Network for Example
2.11

30
 Example 2.11-Solution

The application of a positive supply voltage results in a

conventional current that matches the arrow of the green diode and
turns it on.
The polarity of the voltage across the green diode is such that it
reverse biases the red diode by the same amount. The result is the
equivalent network of Fig. 2.31 .

R I =20mA
+ E=8V
Fig. 2.31 -
Operating conditions for the
network of Fig. 2.30 .
From Kirchhoff’s voltage law (KVL) +
- VLED = 2V
E-VR-VLED =0 VR= E-VLED=8-2=6V
Applying Ohm’s law, we obtain
VR=RI

31
 Example 2.13

Determine the currents I1, I2 and ID2 .

Fig. 2.37
Network for Example
2.13

32
 Example 2.13-Solution

A
The applied voltage (pressure)
is such as to turn both diodes
on, as indicated by the resulting
current directions in the network
of Fig. 2.38 . Note the use of the Loop1
abbreviated notation for “on” Loop2
diodes and that the solution is
obtained through an application KVL
of techniques applied to dc KVL
series–parallel networks.

From Kirchhoff’s voltage law (KVL) in Loop2

VD2-VR1=0 VR1= VD2=0.7V Fig. 2.38
Determining the unknown
Applying Ohm’s law, we obtain
quantities for Example 2.13
VR1=R1I1

33
 Example 2.13-Solution

Applying Kirchhoff’s voltage law (KVL) around the indicated loop1 in the
clockwise direction yields

E-VD1-VD2-VR2=0 VR2= E-VD1-VD2=20-0.7-0.7=18.6V

Applying Ohm’s law, we obtain
VR2=R2I2

Applying Kirchhoff’s current law (KCL) in node A

I2-I1-VD2=0 ID2= I2-I1=3.321-0.212=3.11mA

34
AND/OR Gates
Example 2.14
Determine Vo

V1 V2 V0
0V 0V
0V 10V
10V 0V
10V 10V

Fig. 2.39
Positive logic OR gate.

35
 AND/OR Gates/
Example 2.14-Solutions
V1=10V (1 Level) and V2=0 (0 Level)
V1=0 (0 Level) and V2=0 (0 Level) VK
+ -

0.7V
I

V0 = VR V0 = E-VK
+
= RI=0 =VR = RI
(a 0 Level) E 10V =9.3V
I=0
(a 1 Level)

R 1K R 1K
I

36
 AND/OR Gates/
Example 2.14-Solutions
V1=10V (1 Level) and V2=10V (1 Level)
V1=0 (0 Level) and V2=10V (1 Level)
VK
+ -

0.7V
I
VK VK
+ - V0 = E-VK + - V0 = E-VK
+
=VR = RI =VR = RI
0.7V =9.3V E 10V 0.7V =9.3V
+ I (a 1 Level) + (a 1 Level)
E 10V E 10V
R 1K R 1K
I I

37
AND/OR Gates/
Example 2.14-Solutions

V1 V2 V0

0V 0V 0V (0 Level)

0V 10V 9.3V (1 Level)

10V 0V 9.3V (1 Level)

10V 10V 9.3V (1 Level)

38
AND/OR Gates
Example 2.15
Determine Vo

V1 V2 V0
0V 0V
0V 10V
10V 0V
10V 10V

Fig. 2.42
Positive logic AND gate.

39
 AND/OR Gates/
Example 2.15-Solutions
V1=0V (0 Level) and V2=0V (0 Level)
V1=1 (1 Level) and V2=0V (0 Level)
VK
- +

0.7V
I V0 = 0.7V
V0 = 0.7V
VK VK (a 0 Level)
+ - + (a 0 Level) - +
E 10V V0
- V0
0.7V 0.7V
I R R
1K 1K
+ +
I E 10V
- I E 10V
-

40
 AND/OR Gates/
Example 2.15-Solutions
V1=0V (0 Level) and V2=10 (1 Level)
V1=1 (1 Level) and V2=1 (1 Level) VK
- +

0.7V
V0 =E+VR
=E+0=10V
(a 1 Level) I V0 = 0.7V
+
V0
E 10V (a 0 Level)
I=0 +
+ R
R=1K VR =RI=0
E 10V E 10V 1K
+ +
E 10V I E 10V
- -

41
AND/OR Gates/
Example 2.15-Solutions

V1 V2 V0

0V 0V 0.7V (0 Level)

0V 10V 0.7V (0 Level)

10V 0V 0.7V (0 Level)

10V 10V 10V (1 Level)

42
 College of Computer Science
Department of Computer Engineering

CCE121: Semiconductors

Chapter 2b: Diodes applications

This lectures notes are extracted from the text book: "Electronic
Devices and Circuit Theory", 11th edition 2013, by Robert L.
Boylestad and Louis Nashelsky.
2020-2021
Sinusoidal Inputs: Half-Wave Rectification

Half-wave Rectifier

1
Sinusoidal Inputs: Half-Wave Rectification

For t= 0 T/2, the diode is ON.

Diode is substituted with short-circuit equivalence
for ideal diode (reduce complexity).

Conduction region ( 0 T/2)

43
Sinusoidal Inputs: Half-Wave Rectification

For the period, t= T/2 T, the diode is OFF.

Diode is substituted with an open circuit.

Nonconduction region (T/2 T)

44
Sinusoidal Inputs: Half-Wave Rectification

VDC=0.318Vm

The DC output voltage is 0.318Vm,

where Vm = the peak AC voltage.

45
 Sinusoidal Inputs: Half-Wave Rectification

The effect of using a silicon diode with VK=0.7 is shown.

The diode is “ON” when the applied signal is at least 0.7V.
vo= vi –VK
For Vm>>VK : VDC 0.318(Vm-VK)

46
 Sinusoidal Inputs: Half-Wave Rectification

The effect of using a silicon diode with VK=0.7 is shown.

The diode is “ON” when the applied signal is at least 0.7V.
vo= vi –VK
For Vm>>VK : VDC 0.318(Vm-VK)

47
 Example 2.16

a) Sketch dc output vo and determine the dc level of the output.

b) Repeat (a) if the ideal diode is replaced by silicon diode.

Fig. 2.16a

48
 Example 2.16 : Solution

a)
In this situation the diode will conduct during the negative part of the input as shown
in Fig. 2.16b, and vo will appear as shown in the same figure. For the full period, the
dc level is
Vdc = -0.318Vm = -0.318(20 V) = -6.36 V
The negative sign indicates that the polarity of the output is opposite to the defined
polarity of Fig. 2.16a .

Fig. 2.16b

49
 Example 2.16 : Solution

b)
For a silicon diode, the output has the appearance of Fig. 2.16c , and
Vdc -0.318(Vm - 0.7 V) = -0.318(19.3 V)= - 6.14 V
The resulting drop in dc level is 0.22 V, or about 3.5%.

Fig. 2.16c

50
 PIV (PRV)

Because the diode is only forward biased for one-half of the AC

cycle, it is also reverse biased for one-half cycle.

It is important that the reverse breakdown voltage rating of the diode

be high enough to withstand the peak, reverse-biasing AC voltage
and avoid entering the Zener region.
PIV (or PRV) > Vm
PIV = Peak inverse voltage
PRV = Peak reverse voltage
Vm = Peak AC voltage

51
 Full-Wave Rectification

The rectification process can be improved by using a full-wave

rectifier circuit.

Full-wave rectification produces a greater DC output:

Half-wave: Vdc = 0.318Vm

Full-wave: Vdc = 0.636Vm

52
 Full-Wave Rectification – Bridge Network

Network for the period 0 T/2

of the input voltage vi

53
Full-Wave Rectification – Bridge Network

Conduction path for the positive region of vi

Conduction path for the negative region of vi

54
 Full-Wave Rectification – Bridge Network

Since the area above the axis for one full cycle is now twice that
obtained for a half-wave system, the dc level has also been doubled
and
Vdc = 2(0.318Vm)=0.636Vm

VDC=0.636Vm
Full-wave
55
 Full-Wave Rectification – Bridge Network

vi –VK – vo - VK=0 (KVL)

If silicon diode is used,
vo = vi - 2VK

The peak value of the output voltage vo is therefore Vomax = Vm - 2VK

For situations where Vm >> 2VK, the following equation can be applied for
the average value with a relatively high level of accuracy:

VDC=0.636(Vm-VK) Full-wave
56
 Full-Wave Rectification –
Center Tapped Transformer Rectifier

VDC=0.636Vm

57
 Full-Wave Rectification –
Center Tapped Transformer Rectifier

Network conditions for

the positive region of vi

Network conditions for the negative region of vi

58
 Summary of Rectifier Circuits

Vm = peak of the AC voltage.

In the center tapped transformer rectifier circuit, the

peak AC voltage is the transformer secondary voltage
to the tap.
59
 Example 2.17

Determine the output

waveform for the network
and calculate the output dc
level.
Solution

Network for the positive region of vi 60

Example 2.17

61
 Example 2.17-Solution

vo=(1/2) vi
Vomax=(1/2) Vimax =(1/2) 10=5V
VDC=0.636(5V)=3.18 V.

For the negative part the roles of

the diodes are interchanged and
Resulting output vo appears as shown in figure.
62
Full Wave Rectifier with Smoothing Capacitor (AC
to DC Converter)

63
 College of Computer Science
Department of Computer Engineering

CCE121: Semiconductors

Chapter 2c: Zener Diode applications

This lectures notes are extracted from the text book: "Electronic
Devices and Circuit Theory", 11th edition 2013, by Robert L.
Boylestad and Louis Nashelsky.
2020-2021
 Zener Diodes

A Zener diode is a type of diode that permits current

not only in the forward direction like a normal diode,
but also in the reverse direction if the voltage is larger
than the breakdown voltage known as "Zener voltage“
(VZ).
Common Zener voltages are between 1.8 V and 200V.
Zener diode is used as regulator.

64
 Zener Diodes

Approximate equivalent circuits for the Zener diode in the three

possible regions of application.
65
 Example 2.24
Determine the reference voltages provided by the network which uses a white
LED (4V) to indicate power is on. What is the power delivered to the LED and to
the 6 V Zener diode.

66
 Example 2.24- Solution

R R
1.3K 1.3K

White
VDWh +- 4V
+ +
- E=40V VO2 - E=40V VO2
+
VZ1 6V VZ1 - 6V
VO1 VO1
Loop Loop 2
Si VDSi +- 0.7V
Loop1
VZ2 VZ2 +
- 3.3V
3.3V

67
 Example 2.24- Solution

Applying Kirchhoff’s voltage law around the loop 1 in the clockwise direction yields:
Vo1=VZ2+VDSi=3.3+0.7=4V
Applying Kirchhoff’s voltage law around the loop 3 in the clockwise direction yields:
Vo2=Vo1+VZ1=4+6=10V
Applying Kirchhoff’s voltage law around the loop 2 in the clockwise direction yields:
VR=E-Vo2-VDWh=40-10-4=26V
The 4-V across the white LED

The power delivered by the supply

Ps=ExIs=ExIR=(40V)x(20mA)=800mW
The power absorbed by the LED
PLED=VLEDxILED=(4V)x(20mA)=80mW
The power absorbed by the 6-V Zener diode
PZ=VZxIZD=(6V)x(20mA)=120mW

68
 Basic Zener Regulator
The use of the Zener diode as a regulator is so common that three conditions
surrounding the analysis of the basic Zener regulator are considered. The analysis
provides an excellent opportunity to become better acquainted with the response of
the Zener diode to different operating conditions. The basic configuration appears in
Fig. 2.112 . The analysis is first for fixed quantities, followed by a fixed supply voltage
and a variable load, and finally a fixed load and a variable supply.

Fig. 2.112

Vi and R Fixed
The analysis can fundamentally be broken down into two steps.

69
 Basic Zener Regulator

1. Determine the state of the Zener diode by removing it from the network and
calculating the voltage across the resulting open circuit.
Fig. 2.113
Applying step 1 to the network of Fig.
2.112 results in the network of Fig. 2.113
, where an application of the voltage
divider rule results in

Determining the state of the Zener diode.

If Z, Zener diode is on, and the appropriate equivalent

model can be substituted. VL=VZ
If V<Vz, Zener diode is off , and the open-circuit equivalence
is substituted. VL=V

71
 Basic Zener Regulator

2. Substitute the appropriate equivalent circuit and solve for the desired unknowns.
When Zener diode is on (VL=VZ ) When Zener diode is off (VL=V)
the “on” state will result in the equivalent Substituting the open-circuit equivalent
network of the following Figure. results in the same network as in Fig. 2.116,
where we find that

VL=VZ
The Zener diode current must be determined by VL=V
an application of KCL. That is, IR=IL+IZ Fig. 2.116
VR=Vi-VL
where and
I =0
The power dissipated by the Zener diode is and Z
determined by PZ=IZVZ PZ=VZIZ=VZ(0A)=0W 72
 Summary: Basic Zener Regulator

Remove Zener diode from network.

Calculate V across open circuit.

If Z , Zener diode is on, VL=VZ

If V<Vz, Zener diode is off , VL=V

Determining the state of Substituting the Zener equivalent for

the Zener diode. the “on” situation 73
 Example 2.26

(a)For the Zener diode network, determine VL, VR, IZ and PZ.
(b)Repeat part (a) with RL=3 .

74
 Example 2.26 Solutions

(a)determine VL, VR, IZ and PZ. (RL=1.2 )

Since V=8.73V is less than VZ =10V, the diode is in the “off” state, as shown on the
characteristics of Fig. 2.117 . Substituting the open-circuit equivalent results in the
same network as in Fig. 2.116 , where we find that

Fig. 2.116 Fig. 2.117

VL=V= 8.73V VR=Vi-VL=16-8.73=7.27V

Applying Kirchhoff’s voltage law IZ=0
and
around the external loop in the
clockwise direction yields: PZ=VZIZ=VZ(0A)=0W 75
 Example 2.26 Solutions

(b) Repeat part (a) with RL=3 .

Since V=12 V is greater than VZ=10V, the diode is in the “on” state and the network of
Fig. 2.118 results. V =V =10V
L Z
Applying Kirchhoff’s voltage law around
the external loop in the clockwise
direction yields: Vi-VR-VL=0
VR=Vi-VL=16-10=6V
The Zener diode current must be determined
by an application of KCL. That is, IR=IL+IZ
KVL
where
Fig. 2.118
So that, IZ=IR-IL=6mA-3.33mA=2.67mA
The power dissipated is PZ=VZIZ=10V(2.67mA)=26.7mW 76
 Voltage Doubler

This half-wave voltage doubler’s output can be calculated by:

Vout = VC2 = 2Vm
where Vm = peak secondary voltage of the transformer

77
 Voltage Doubler
• Positive Half-Cycle
D1 conducts
D2 is switched off
Capacitor C1 charges to Vm
• Negative Half-Cycle
D1 is switched off
D2 conducts
Capacitor C2 charges to 2Vm
Vout = VC2 = 2Vm

(a) positive half-cycle; (b) negative half-cycle. 78

Voltage Doubler

15
Practical Applications (AC to DC Converter)

80
Practical Applications Battery Charger

81
Practical Applications Battery Charger

82
 Practical Applications - Polarity insurance

(a) Polarity protection for an

expensive, sensitive piece of
equipment;

(b) correctly applied polarity; (c) application of the wrong polarity. 83

Practical Applications - Polarity Detector

Polarity detector using diodes and LEDs.

84
Practical Applications – Exit sign using LEDS

85
Practical Applications – AC Regulator

86
Practical Applications – Square-Wave Generator

87