Chapter 2 – Diode
2
Learning Outcome
At the end of this chapter, students able to:
 Describe a structure of diode, pn junction biasing, I-V
characteristics and diode equivalent circuit
 Explain and analyze the diode series/parallel configuration
with DC supply
 Explain and analyze the operation of clipper
 Explain and analyze the operation of clamper
 Explain and analyze the operation of voltage multiplier
 Explain and analyze the characteristic and application of
zener diode as voltage regulator
Chapter 2 : DIODE
Norsabrina Sihab
Faculty of Electrical Engineering,
Universiti Teknologi MARA
Pulau Pinang
Tel : 04-3823355
Email : norsabrina@ppinang.uitm.edu.my
1
Chapter 2 – Diode
Norsabrina Sihab
ELE232 - Electronics 1
Updated Nov 2013
Chapter 2 – Diode
3
Diode
4
Diode
Diode is created when a pn-junction forms at the boundary
between the two regions that is n-type and p-type material.
P-region has many holes (majority carriers) and only few
thermally generated free electrons (minority carriers).
N-region has many free electrons (majority carriers) and only a
few thermally generated holes (minority carriers).
Depletion Layer -> Area around a p-n junction is called depletion
layer or region which is depleted of free carriers. Free electrons in
the n-region are aimlessly drifting in all directions. At the instant of
p-n junction formation, the free electrons near the junction in the n
region to diffuse across the junction into the p region where they
combine with holes near the junction. When the p-n junction is
formed, the n region loses free electrons as they diffuse across the
junction. This creates a layer of positive charges (pentavalent ions)
near the junction. As the electronics move across the junction, the
p region loses holes as the electrons and holes combine. This
creates a layer of negative charges form the depletion region.
When at equilibrium the depletion region widened, no electrons can
across the p-n junction.
Figure 2.1 - The basic diode structure at the instant of junction formation
showing only the majority and minority carriers.
Norsabrina Sihab
ELE232 - Electronics 1
Updated Nov 2013
Norsabrina Sihab
ELE232 - Electronics 1
Updated Nov 2013
Chapter 2 – Diode
Chapter 2 – Diode
5
Diode
6
Diode
Barrier potential –> Any time there is a positive charge and a
negative charge near each other, there is a forcing acting on the
charges. In the depletion region there are many positive charges
and many negative charges. The forces between the opposite
charges form an electric field. This electric field is a barrier to the
free electrons in n region and external energy must be applied to
get electrons to move across barrier of electric field. The potential
difference of the electric field across the depletion region is the
amount of voltage required to move electrons through the electric
field. The potential difference is called barrier potential.
Figure 2.2 – Formation of depletion region
Norsabrina Sihab
ELE232 - Electronics 1
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Chapter 2 – Diode
Norsabrina Sihab
ELE232 - Electronics 1
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Chapter 2 – Diode
7
Biasing A Diode
8
Biasing A Diode - Forward Bias
Diode is a 2-terminal device that make from p-type and n-type
materials. Ideally conducts current in only one direction.
3 operating conditions:
 No bias - No external voltage is applied: VD = 0V, No current is
flowing: ID = 0A, only a modest depletion layer exists
 Forward bias - External voltage is applied across the p-n
junction in the same polarity as the p- and n-type materials.
 Reverse bias - External voltage is applied across the p-n
junction in the opposite polarity of the p-type and n-type
materials.
Condition that allows current through the p-n junction
External voltage or VBIAS connected to the p region (+VBIAS) and n
region (-VBIAS) where VBIAS > VB (barrier potential)
Positive terminal of VBIAS will push the holes in the p-region towards
the p-n junction. Recombination occurs and number of negative
ions (acceptors) in the p-region near the junction decreases.
Negative terminal of VBIAS will push the free electrons in n-region
towards the junction. Recombination with positive ion and number
of positive ion decreases.
As a result, the number of positive and negative ions decrease so
the width of depletion layer become narrow. e- in n-region easily
move to the p-type. So large number of majority carrier flow across
the junction.
Figure 2.3 – Diode during forward and reverse bias
Norsabrina Sihab
ELE232 - Electronics 1
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Norsabrina Sihab
ELE232 - Electronics 1
Updated Nov 2013
Chapter 2 – Diode
Chapter 2 – Diode
9
Biasing A Diode - Forward Bias
10
Biasing A Diode - Reverse Bias
Condition that prevents current through the p-n junction
External voltage or VBIAS connected to the p region (-VBIAS) and n
region (+ VBIAS) where VBIAS < VB (barrier potential)
Positive terminal of VBIAS will pulls the free electrons away from p-n
junction and positive ions (donors) in n-region increase.
Negative terminal of VBIAS will pulls the free holes from p-region and
number of negative ions (acceptor) in p-region increase.
As a result, the number of positive and negative ions increases so
the width of depletion layer become widen.
Due to widening depletion region, the p-n junction act like a very
poor conductor and allow minority carrier flows (µA). It called
reverse current or leakage current.
Figure 2.4 – A forward biased diode
Norsabrina Sihab
ELE232 - Electronics 1
Chapter 2 – Diode
Updated Nov 2013
Norsabrina Sihab
ELE232 - Electronics 1
Updated Nov 2013
Chapter 2 – Diode
11
Biasing A Diode - Reverse Bias
12
Biasing A Diode
Reverse Saturation Current -> Also called as leakage current
where current in reverse biased condition. The extremely small
current (can be neglected), Is that exist in reverse bias after the
transition current dies out. It caused by the minority carriers in the
n-region and p-regions that produced by thermally generated EHP.
Figure 2.5 – A reverse biased diode
Norsabrina Sihab
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Figure 2.6 – I-V characteristics for diode
Updated Nov 2013
Norsabrina Sihab
ELE232 - Electronics 1
Updated Nov 2013
Chapter 2 – Diode
Biasing A Diode - Breakdown Voltage
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Chapter 2 – Diode
14
Biasing A Diode - Breakdown Voltage
Breakdown Voltage –> If the external reverse bias voltage is
increased to a value called the breakdown, the reverse current
will drastically increase. The reverse-bias potential that results
in this dramatic change in characteristics is called zener voltage,
Vz.
Avalance region (Vz) can be brougt closer to the vertical axis by
increasing the doping levels in the p-type and n-type materials.
As Vz decreases to very low levels, such as -5V, other
mechanism, called Zener Breakdown, will contribute to the
sharp change in the characteristics. It occurs because there is a
strong electric field in the region of the junction that can disrupt
the bonding forces within the atom and generate carriers.
Although the zener breakdown mechanism is a significant
contributor only at lower levels of Vz, this sharp change in
characteristic of a p-n junction called zener diodes.
Norsabrina Sihab
Chapter 2 – Diode
13
The maximum reverse-bias potential can be applied at this
region called Peak Inverse Voltage (PIV) or Peak Reverse
Voltage (PRV) or Breakdown Voltage (VBR).
Diode that operating in this region is called Zener Diode which
normally used as a voltage regulator.
Updated Nov 2013
Norsabrina Sihab
ELE232 - Electronics 1
Updated Nov 2013
Chapter 2 – Diode
15
Diode Model
16
Diode Model
Has 2 terminal – Anode and Cathode
The ideal diode in the non-conduction region (OFF State)
Figure 2.7 – Diode structure and symbol
The ideal diode in the conduction region (ON State)
Figure 2.9 – Diode OFF state
Diode resistance levels - Semiconductors act differently to DC
and AC currents. There are three types of resistances:
1. DC, or static resistance
2. AC, or dynamic resistance
3. Average AC resistance
Figure 2.8 – Diode ON state
Norsabrina Sihab
ELE232 - Electronics 1
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Norsabrina Sihab
ELE232 - Electronics 1
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Chapter 2 – Diode
Chapter 2 – Diode
17
Diode Model
Diode Model
1. DC or Static Resistance, RD
For a specific applied DC voltage VD, the diode has a specific current
ID, and a specific resistance RD.
RD 
rd  
2. AC or Dynamic, Resistance In the reverse bias region:
In the forward bias region:
 The resistance depends on the amount of current (ID) in the
diode.
 The voltage across the diode is fairly constant (26mV for
25C).
 rB ranges from a typical 0.1 for high power devices to 2 for
low power, general purpose diodes. In some cases rB can be
ignored.
The resistance is essentially infinite. The diode acts like an open.
VD
ID
Figure 2.10 – Static Resistance
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rd 
Updated Nov 2013
Chapter 2 – Diode
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Updated Nov 2013
20
Diode Equivalent Circuit
3. Average AC Resistance
AC resistance can be determined by selecting two points on the
characteristic curve developed for a particular circuit.
Norsabrina Sihab
26 mV
 rB
ID
Chapter 2 – Diode
19
Diode Model
rav 
18
1) Piecewise Linear Equivalent Circuit
Total forward voltage, VD across the diode must be greater than VT
before the ideal diode in the equivalent circuit will forward bias.
Vd
(point to point)
I d
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ELE232 - Electronics 1
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Chapter 2 – Diode
Chapter 2 – Diode
21
22
Diode Equivalent Circuit
Diode Equivalent Circuit
2.Simplified Equivalent Circuit (Approximate)
Total forward voltage, VD across the diode must be greater than VT
before the ideal diode in the equivalent circuit will forward bias.
3. Ideal Device
 The barrier potential is negligible, hence once the circuit ON or
short at zero potential current will flow significantly and VD=0V.
Norsabrina Sihab
ELE232 - Electronics 1
Updated Nov 2013
Chapter 2 – Diode
Norsabrina Sihab
Chapter 2 – Diode
23
Diode As A Switch
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24
Diode As A Switch
It can conduct current in only ONE way direction and can act as
switch (ON/OFF).
2 diode conditions – ON & OFF state.
2 basic conditions for diode in ON state.
 Diode must in forward bias condition
 Voltage supply, Vi must be greater than the diode voltage, VD
(Vi>VD)
VSi=0.7V, VGe=0.3V and Videal diode=0V
Norsabrina Sihab
ELE232 - Electronics 1
Updated Nov 2013
Diode in OFF state – act as open circuit. So I=0A.
Norsabrina Sihab
ELE232 - Electronics 1
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Chapter 2 – Diode
Chapter 2 – Diode
25
Diode in Series with DC Supply
26
Diode in Parallel with DC supply
Example 2
Determine ID, Ix and Vo
Check diodes whether ON or OFF
Redraw diode equivalent circuit including others component.
Apply KVL to determine current or voltage
+20V
IX
Example 1
Determine ID, VR and Vo.
Ge
Si
ID
+ Vo
Ge
Si
10V
2.2k
+ Vo
ID
+
VR
-
5.6k
- 5V
-5V
Norsabrina Sihab
ELE232 - Electronics 1
Chapter 2 – Diode
Updated Nov 2013
Norsabrina Sihab
ELE232 - Electronics 1
Updated Nov 2013
Chapter 2 – Diode
27
Exercise
28
Diode Application : Clipper
Basically to clipped-off/eliminate a portion of an AC signal voltage
above or below specific range.
HW rectifier is a basic clipper.
Functions:
1. Altering the shape of the output waveform
2. Circuit transient protection
3. Detection
2 types : 1) series clipper, 2) parallel (shunt) clipper
1) Series Clipper
2 types : a) negative series clipper, b) positive series clipper
The diode in a series clipper circuit “clips” any voltage that does
not forward bias it:

A reverse-biasing polarity

A forward-biasing polarity less than 0.7V for a silicon diode
Norsabrina Sihab
ELE232 - Electronics 1
Updated Nov 2013
Norsabrina Sihab
ELE232 - Electronics 1
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Chapter 2 – Diode
Chapter 2 – Diode
29
1a) Negative Series Clipper
30
1a) Negative Series Clipper (contd)
Clipped off half negative cycle. Diode forward bias during positive
cycle of Vi.
VT is transition voltage. (VT=VDC+Vdiode)
During negative half cycle.
Diode is OFF for all value of Vi.
VO=0V.
Final output
Vi
20
VT=4V
- 20
During positive half cycle
VT=Vdc+VD=4V
if Vi ≤ VT  diode will OFF.
Vo=0V.
If Vi > VT  diode will ON.
KVL : Vi – 4 – Vo =0.
 Vo=Vi-VT=16V
Norsabrina Sihab
Vo
16
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Chapter 2 – Diode
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Chapter 2 – Diode
31
1b) Positive Series Clipper
32
1b) Positive Series Clipper (contd)
Clipped off half positive cycle. Diode forward bias during negative
cycle of Vi.
During positive half cycle
Diode is OFF for all value of Vi.
VO=0V.
During negative half cycle.
VT=- 4-VD=- 4V
if lVil ≤ lVTl  diode OFF. Vo=0V.
If lVil > lVTl diode ON.
KVL : Vi +Vo- 4 =0.
Vo= - Vi+4=-20+4=-16V
Final output
Vi
20
VT= - 4V
- 20
Vo
-16
Norsabrina Sihab
ELE232 - Electronics 1
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Chapter 2 – Diode
Chapter 2 – Diode
33
2a) Negative Parallel Clipper
2a) Negative Parallel Clipper (contd)
During negative half cycle
VT = -0.7-5 = -5.7V
if lVil ≤ lVTl  diode OFF.
 Vo=Vi
If lVil> lVTl  diode ON.

KVL : Vo +0.7+5 =0
Vo = VT = -5.7V
The operation is opposite series clipper.
R
Vi
+
20
+
Si
Vi
Vo
5V
-
-
- 20
During positive half cycle
Diode is OFF for all value of Vi.
VO=Vi=20V
Vi
-
+
0.7V
+
Vi
20
Vo
Si
Vi
5V
Vo
- 20
5V
-
+
-
-
ELE232 - Electronics 1
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Chapter 2 – Diode
Norsabrina Sihab
+
During negative half cycle.
Diode is OFF for all value of Vi.
VO=Vi.
+
Si
Vi
Vo
5V
-
36
Final output
Vi
20
-
- 20
During positive half cycle
VT- Vdc- VD = 0.
 VT=5.7V
If Vi ≤ VT  diode OFF. Vo=Vi.
If Vi > VT  diode ON.
KVL : Vo-0.7-5 =0.
Vo=5.7V
Updated Nov 2013
2b) Positive Parallel Clipper (contd)
R
Vi
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Chapter 2 – Diode
35
2b) Positive Parallel Clipper
20
Final output
R
Vi
R
+
Norsabrina Sihab
34
VT
R
Vi
20
- 20
+
+
Vo
0.7
20
Vo
Vi
5V
-
-
- 20
- 20
Norsabrina Sihab
ELE232 - Electronics 1
Updated Nov 2013
Norsabrina Sihab
ELE232 - Electronics 1
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Chapter 2 – Diode
R
+
+
10
Si
Final Output
Ge
Vi
Vo
5V
Vi
7.7V
-
10
-
VT=5.7V
- 10
During positive half cycle
DGe  OFF for all value of Vi
DSi  ON conditionally
 VT=VDSi+5=5.7V
 If Vi ≤ VT  DSi OFF.
Vo=Vi.
 If Vi > VT  DSi ON.
Vo=5.7V
Norsabrina Sihab
VT= - 8V
During negative half cycle
- 10
DSi  OFF for all value of Vi
DGe  ON conditionally
 VT = -VDGe-7.7 = -8V
Vo
10
If |Vi| ≤ |VT|  DGe OFF.
Vo=Vi.
 If |Vi| > |VT|  DGe ON.
Vo+0.3+7.7=0
Vo=-8V
5.7V

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Chapter 2 – Diode
Updated Nov 2013
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-8V
- 10
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Chapter 2 – Diode
39
Summary of Clipper Circuit
Norsabrina Sihab
38
Combination of Negative and Positive Parallel
Clipper (contd)
Combination of Negative and Positive Parallel Clipper
Vi
Chapter 2 – Diode
37
Updated Nov 2013
40
Exercise
Updated Nov 2013
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ELE232 - Electronics 1
Updated Nov 2013
Chapter 2 – Diode
Chapter 2 – Diode
41
Diode Application : Clamper
42
1) Negative Clamper (contd)
Function -> To clamp or shift a signal to a different DC level
Circuit consist of C,D and R
Final Output
During negative half cycle
Vi
1) Negative Clamper
20
Step 1: Determine Vo using KVL at i/p
Vi+Vo+Vc=0
Vo=–Vi–Vc= – 20 – 24.3= – 44.3V
During positive half cycle
Step 1: Find polarity of VC
Step 2: Determine VO using KVL at o/p
Vo – VD+VDC= 0
Vo=0.7-5= - 4.3V
Step 3: Determine value of VC
Vi-Vc-Vo=0
Vc=24.3V
Updated Nov 2013
Chapter 2 – Diode
+
Vo
5V
-
Final Output
During positive half cycle
Step 1: Find polarity of Vo
Vi+Vc-Vo=0
Vo=20+14.3= 44.3V
Si
R
44
2) Positive Clamper (contd)
+
Vi
Updated Nov 2013
Chapter 2 – Diode
43
C
Vi
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2) Positive Clamper
20
Vo
- 4.3V
- 44.3
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- 20
-
Vi
20
- 20
During negative half cycle (because Diode ON at this cycle)
Step 1: Find polarity of VC
Step 2: Determine VO using KVL at o/p
Vo+ VD- VDC= 0
Vo= 5 - 0.7= 4.3V
Step 3: Determine VC using KVL at i/p
Vi+VDC–VD–VC=0
VC=Vi+VDC–VD=24.3V
- 20
Vo
44.3
4.3
Norsabrina Sihab
ELE232 - Electronics 1
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Norsabrina Sihab
ELE232 - Electronics 1
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Chapter 2 – Diode
Chapter 2 – Diode
45
Example – Design a Clamper
Summary of Clamper Circuit
KVL: +Vo-0.7-5=0
Vo=+5.7V
KVL:
Vi – Vc – Vo=0
Vc=Vi-Vo=15 - 5.7=9.3V
Designed a clamper circuit to
produce output voltage, Vo. Use
silicon diode in your design.
46
During negative cycle
Solution
During positive cycle
Propose design clamper circuit
which D ‘ON’ during positive
cycle. Vo=5.7V
KVL:
+Vi+Vo+Vc=0
Vo= - 15 - 9.3= - 24.3V
Norsabrina Sihab
ELE232 - Electronics 1
Chapter 2 – Diode
Updated Nov 2013
Norsabrina Sihab
Chapter 2 – Diode
47
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Updated Nov 2013
48
Exercise
Exercise
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ELE232 - Electronics 1
Updated Nov 2013
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ELE232 - Electronics 1
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Chapter 2 – Diode
Chapter 2 – Diode
49
Diode Application : Voltage Multiplier
50
Half Wave Voltage Doubler
Half-wave voltage doubler
VP
Function – use clamping action to increase peak rectified
voltage without the necessity of increasing the transformer’s
voltage rating. A voltage doubler is similar to the peak-to-peak
detector but uses rectifier diodes instead of small-signal diodes.
Types – Voltage Doubler (multiply the input peak by factors of
2), Voltage Tripler (multiply the input peak by factors of 3) and
Voltage Quardrupler (multiply the input peak by factors of 4)
Application – in high voltage, low current, high frequencies. Eg
Chathode-ray tubes (CRTs), particle accelerators etc.
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Chapter 2 – Diode
C1
Updated Nov 2013
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2VP
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Chapter 2 – Diode
51
Updated Nov 2013
52
Full-wave Voltage Doubler
1st positive half cycle:
D1 ON while D2 OFF
C1 charging up quickly to the peak
value of input
KVL : -Vp +VD+ VC1 =0
VC1 = VP –VD or approximately Vp
1st positive half cycle:
D2 ON while D1 OFF
C2 charging up quickly to the peak value of input
So KVL : -Vp -VC1 +VD+ VC2 =0
Vo = VC2 = VP+VC1 –VD or approximately 2Vp
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D2 C2
1st negative half cycle
D1 ON while D2 OFF
C1 charging up quickly to
the peak value of input
KVL : -Vi+VD+Vc1=0
So VC1 = Vp –VD or
approximately Vp
Half Wave Voltage Doubler (contd)
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D1
Updated Nov 2013
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ELE232 - Electronics 1
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Chapter 2 – Diode
Chapter 2 – Diode
53
Full-wave Voltage Doubler (contd)
Voltage Tripler
1st negative half cycle
D2 ON while D1 OFF
C2 charging up quickly to the peak
value of input
KVL : +Vi+VC2+VD=0
VC2 = Vp –VD or approximately Vp
Thus the output Vo=VC1+VC2≈2Vp
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Chapter 2 – Diode
54
By connecting another diode-capacitor section to the voltage doubler
creates a voltage tripler.
First two sections act a doubler.
Positive cycle : C1 charge to Vp thru D1.
Negative cycle : C2 charge 2Vp thru D2.
Next positive cycle : C3 charges to 2Vp thru D3.
Tripler output is taken across C1 and C3.
Updated Nov 2013
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Chapter 2 – Diode
55
Updated Nov 2013
56
Exercise
Voltage Quadrupler
By connecting another diode-capacitor section.
First two sections act a doubler.
Positive cycle : C1 charge to Vp thru D1.
Negative cycle : C2 charge 2Vp thru D2.
Next positive cycle : C3 charges to 2Vp thru D3.
Next negative cycle : C4 charges to 2Vp thru D4
Quardrupler output is taken across C2 and C4.
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ELE232 - Electronics 1
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Chapter 2 – Diode
Zener Diode
58
Zener Diode Application : Voltage Regulator
IDZ is opposite from ID which is
designed to work in reverse
bias.
Application : Voltage Regulator
Simplest regulator as shown in figure below.
3 conditions of Vi and load resistance, RL to maintain designed
zener voltage:
1. Vi and RL fixed
2. Vi fixed and RL variable
3. Vi variable and RL fixed
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Chapter 2 – Diode
57
Updated Nov 2013
Chapter 2 – Diode
60
2. Fixed Vi and Variable RL
Step 2: Substitute appropriate
equivalent circuit
Specific range of RL to turn
ON DZ
since RL is minimum, therefore IL is maximum
I L max 
VL
V
 Z
RL RL min
andILmin 
VDR :
V  VL 
VL  Vz
KCL : I R  I Z  I L
IZ  IR  IL
where I L 
VDR :
RV
V  VL  L i
R  RL
 (1)
VL
V
V  VL
and I R  R  i
RL
R
R
power dissipated by zener diode :
if V  VZ ( DZ ON )
Pz  I zVz
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RLVi
R  RL
Updated Nov 2013
VZ
RL max
Once DZ ON, VR remains fixed
Vi  VZ  VR
VZ ( RL  R)  RLVi
since VR is fixed,IR also fixed
VZ
( RL  R)  Vi
RL
IR 
VZ (1 
R
)  Vi
RL
V  VZ
R
 i
RL
VZ
so RL min 
if V  VZ ( DZ OFF )
Norsabrina Sihab
Updated Nov 2013
Chapter 2 – Diode
59
1. Fixed Vi and Fixed RL
Step 1: Determine the state of the
zener diode by removing it from the
network. Calculate voltage across the
resulting open circuit.
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RVZ
Vi  VZ
Norsabrina Sihab
VR
R
KCL : I R  I Z  I L
so I Z  I R  I L
resulting IZmin when ILmax
and IZmax when ILmin because IR is constant
ILmin  IR - IZM
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&
RLmax 
VZ
ILmin
Updated Nov 2013
Chapter 2 – Diode
Chapter 2 – Diode
61
Exercise
3. Fixed RL and Variable Vi
Answer: 10V, 10V,6.3mA, 63mW
I R max  I ZM  I L
3) Determine range of Vi that
will maintain zener diode in
ON state. Answer: 23.67V ~ 36.8V
Vi max  VR max  VZ
VDR :
VL  VZ 
1) Determine VL,VR, IZ and PZ.
The max of Vimax is limited by
the max zener current, IZM
Vi must be sufficiently large
to turn DZ
Min voltage to turn ON is
Vi=Vimin
62
RLVi
R  RL
2)
VZ ( RL  R)  RLVi
Vi min 
RL  R
VZ
RL
a) Determine the range of RL and
IL that will result in VRL being
maintained at 10V Answer: 250 Ω ~
1.2kΩ, 8mA ~ 40mA
b) Determine the max wattage
rating of DZ. Answer: 32mW
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Chapter 2 – Diode
Updated Nov 2013
63
Exercise
Norsabrina Sihab
ELE232 - Electronics 1
Updated Nov 2013
Norsabrina Sihab
ELE232 - Electronics 1
Updated Nov 2013