chapter 3 diode
TRANSCRIPT
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Diodes
1
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Diode• Class of non-linear circuits
– having non-linear v-i Characteristics
• Uses
– Generation of :
• DC voltage from the ac power supply
• Different wave (square wave, pulse) form
generation – Protection Circuits
– Digital logic & memory circuits
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Creating a Diode
• A diode allows current to flow in one
direction but not the other.
• When you put N-type and P-type silicon
together gives a diode its unique
properties.
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Diode
Equivalent circuit in the reverse direction
Equivalent circuit in the forward direction.
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Reverse Bias
• -ve voltage is applied to Anode• Current through diode = 0 (cut off
operation)
• Diode act as open circuit
Forward Bias+ve voltage applied to Anode
• Current flows through diode
• voltage Drop is zero (Turned on)
Operation
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The two modes of operation of ideal diodes
Forward biased
Forward Current 10
mA
Reverse biasedReverse Voltage 10 V
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x .
v D= 1.5v
i D=1. 5
1
= 1.5A
v D= 0
1.5v
1Ω
v D
+
−
i P v
D
+
−
i D= 0
1Ω
1.5v
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Rectifier circuit
Input waveformEquivalent circuit when v i 0
Equivalent circuit when v i ≤ 0
Waveform across diode
Output waveform.
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Exercise 3-3
i D=
10− 0
1k Ω = 10mA
v D=
1
t 2− t
1
∫t 1
t 2
vidt
v D=1
2π [∫0π
10 sinθ + ∫π
2π
0dt ]v D= −
1
2π ∣10cosθ ∣0π
=−
10
2π (− 1− 1)=10
π = 3.18V
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Battery Charger
sin θ =1
2⇒θ = 30
0
Conduction Angle=
π −
2θ=
120
0
one− third of cycle
24sinθ = 12 V
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Figure 3.6 Circuits for Example 3.2. Diodes are ideal , Find the value of I and V
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Example 3.2.
Assumption Both Diodes are conducting
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V = 0, V B= 0
I D2=10− 0
10k = 1mA
I 5k Ω= I D1+ I D2=0− (− 10 )
5k = 2 mA
From
above
equation
I D1
should
be1
mA It is not possible
Node A
Node B
Assumption Both Diodes are conducting
Not Possible
Thus assumption of both diode
conducting is wrong
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Example 3.2(b). Assumption # 2Diodes 1 is not conducting Diodes 2 is conducting
I D2=10− (− 10)
15=
20
15= 1.33mA
V B= V A= 3.3 V , I D1= 0, I D2= 1.33mA
Assumption is correct
V A= 10− (1.33)(5k )= 3.3v
V B= (1.33)(10k )− 10= 3.3v
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Figure E3.4
Diodes are ideal , Find the value of I and V
Fi E3 4
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Figure E3.4
Diodes are ideal , Find the value of I and V
I= 2mA
V= 0V
I= 0A
V= 5VI= 0A
V= -5V
I= 2mA
V= 0V
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Figure E3.4 Diodes are ideal , Find the value of I and V
I= 3mAV= 3V
I= 4mAV= 1V
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Figure P3.2 Diodes are ideal , Find the value of I and V
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Figure P3.2 Diodes are ideal , Find the value of I and V
Diode is conducting
I = 0.6 mA
V = -3V
Diode is cut-off
I = 0 mA
V = 3VDiode is conducting
I = 0.6 mA
V = 3V Diode is cut-off
I = 0 mA
V = -3V
P bl 3 3
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D1 Cut-Off & D2 Conducting
I = 3mA
Problem 3-3
D1 Cut-Off & D2 Conducting
I = 1mA , V=1 V
Diodes are ideal , Find the value of I and V
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Figure P3.4
In ideal diodes circuits, v1 is a 1-kHz, 10V peak sine wave.
Sketch the waveform of vo
i
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In ideal diodes circuits, v1 is a 1-kHz, 10V peak sine wave.
Sketch the waveform of vo
Vp+ = 10V
Vp- = 0V
f = 1 K-Hz
Vp+ = 0V
Vp- = - 10V
f = 1 K-Hz
Vo = 0V
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Figure P3.4 In ideal diodes circuits, v1 s a 1-kHz, 10V peak sine wave.
Sketch the waveform of vo
Vp+ = 10V
Vp- = 0V
f = 1 K-Hz
Vp+ = 10V
Vp- = -10V
f = 1 K-Hz
Vp+ = 10V
Vp- = 0V
f = 1 K-Hz
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Figure P3.4 In ideal diodes circuits, v1 s a 1-kHz, 10V peak sine wave.
Sketch the waveform of vo
Figure P3 4
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Figure P3.4
In ideal diodes circuits, v1 s a 1-kHz, 10V peak sine wave.
Sketch the waveform of vo
Vp+ = 0V
Vp- = -10V
f = 1 K-Hz
V0 = 0V Vp+ = 10V
Vp- = -5V
f = 1 K-Hz
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Figure P3.4 In ideal diodes circuits, v1 s a 1-kHz, 10V peak sine wave.
Sketch the waveform of vo
Vp+ = 10V
Vp- = -5V
f = 1 K-Hz
P bl 3 4(k)
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Problem 3-4(k)
vi
⇒10 V peak @ frequency 1000 H z
v i= 10sin2000 πt
For Vi > 0 V D1 is cutoff D2 is conducting v o =1V For Vi < 0 V is conducting D2 is cutoff v o =v i +1V
- 9 V
P bl 3 4(k)
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Problem 3-4(k)
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Figure P3.6
X = A . BX = A + B
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Problem 3-4 (c)
vi⇒10Vpeak @frequency 1000 H z
vi= 10sin2000 πt
v o =zero
Problem 3 4(f)
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Problem 3-4(f)
+ve Half Cycle with 10 V peak
at 1 KHz
kHz 10-V peak sine wave.
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Problem 3-4(h)
vi⇒10Vpeak @frequency 1000 H z
vi= 10sin2000 πt v o =zero
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vi⇒10Vpeak @frequency 1000 H z
vi= 10sin2000 πt
Problem 3.5
vi is 10 V peak sine wave and I = 100 mA current source. B is battery of
4.5 V . Sketch and label the iB
4.5 v
100 mA
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Solution P3-5
vi> 4.5V ,D1 cutoff all current flows thru battery
Conduction angle
10sinθ = 4.5V⇒θ = sin− 1(0.45)= 26.70 ,153.30
Conduction angle= π − 2θ= 126.60
Fraction of cycle that i B of 100mA flows=126.6
360= 0.35
v i⇒10 Vpeak @ frequency 1000 H zv i= 10sin2000 πt B = 4 .5 V
vi< 4.5V , D
1 conducts D
2 cutoff
All current flows thru D1 , i
B= 0A4.5 v
100 mA
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Problem 3-5
100 mA
4.5 v
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Problem 3-5
4.5
10
100 mA
i Baverage=1
T ∫ i B dt =
1
T [100× 0.35 T ]= 35 mA
REVERSE PO ARITY
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REVERSE POLARITY
PROTECTOR
REVERSE POLARITY
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• The diode in this circuit protects a
radio or a recorder etc... In the event
that the battery or power source is
connected the wrong way round, thediode does not allow current to flow.
REVERSE POLARITY
PROTECTOR
Problem 3-9
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D1& D2 ConductingI 1=1mA
I 3=0.5 mA
I 2 =0.5 mA
V= 0 V
D1=off, D2=OnI 1= I 3=0.66 mA
V = -1.7 V
Problem 3-9
I1
I3
2
I1
I3
2
P bl 3 10
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Problem 3-10
D conductingI=0.225 mA
V=4.5V
D is not conducting
I=0A
V=-2V
P bl 3 16
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Problem 3-16
V RED GREEN
3V On Off D1 conducts0 V Off Off
-3 V Off On D2 conducts
Quiz No 3 DE 28 EE A
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Quiz No 3 DE 28 EE -A
Sketch vO if vi is 8 sin
Find out the conduction angle for the diode &
fraction of the cycle the diode is conducting
Solution Quiz No 3
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Solution Quiz No 3
8V
I1
I2
8= 4I1− 2I2
− 2= − 2I1
+ 3I2
2I2= 2⇒ I 2= 1mA
Vo= 1× 1+ 2= 3V
vi/2
I
I =
8
2
− 2
2 = 1 mA
Vo= 1× 1+ 2= 3V
10-10-07
Conduction angle⇒2θ= 60o
4sinθ = 2⇒θ = 30
Fraction of Cycle the diode conducts= π − 2θ2π
= 13= 33
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22-10-07
Sketch vO if vi is 10 sin
Find out the conduction angle for the diode &
fraction of the cycle the diode is conducts
D1
never conducts
Vi<5V D2 is cut-off, Vo=5V
Vi>5V D2 is conducts
V omax= 5+
10− 5
2= 7.5V
+12 V
5
D1
D2
Conduction angle⇒2θ= 60o
10sin θ = 5⇒θ = 30
Fraction of Cycle the diode conducts=π − 2θ
2π=
1
3= 33
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Quiz No 3 DE 27 CE -B
D1
never conducts
Vi<5V D2 is cut-off, Vo=Vi
Vi>5V D2 is conducts
Sketch vO if vi is 10 sin
Find out the conduction angle for the diode &
fraction of the cycle the diode is conducts
V omax= 5+
10− 5
2
= 7.5V
Conduction angle⇒2θ= 60o
10sin θ = 5⇒θ = 30
Fraction of Cycle the diode conducts=π − 2θ
2π=
1
3= 33
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Problem
• Assume the diodes are ideal,
sketch vo if the input is 10sin
(9)
• Find out the conduction angles for
Diode D1 & D2 (4) and the fractionof the cycle these diodes conduct. (2)
2< < 1
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− 2< vi< 1⇒vo= vi
vi> 1V⇒
v0=
vi− 1
4 × 1+ 1v
ipeak = 10V ⇒v
opeak = 4.25V
vi<− 2V⇒vo=− 2V
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− 2< vi< 1⇒vo= vi
vi
> 1V⇒v0
=v
i− 1
4× 1+ 1
vi= 2V⇒v
o= 1.25V
vi<− 2V⇒vo=− 2V v0=
vi− 1
4× 1+ 1V
-2V
Two-dimensional representation of the silicon crystal
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p y
14 Electrons
Silicon and Germanium
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Silicon Lattice
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Silicon Lattice
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At room temperature, some of the covalent bonds are broken by
thermal ionization.
Each broken bond gives rise to a free electron and a hole, both
of which become available for current conduction.
Intrinsic Semiconductor
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Electrons and holes
Semiconductor Current
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Semiconductor Current
The Doping of Semiconductors
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.
The Doping of Semiconductors
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Valence Electrons
N Type
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N Type
P Type
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P Type
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p-n Junction
• P Junction
– Concentration of holes is high
– Majority charge carrier are hole
• N Junction
– Concentration of electron is high
– Majority charge carrier are electron
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Diffusion Current ID
• Hole diffuse across the junction from
the p side to the n side & similarly
electron
• Two current components add together
to form the diffusion current with
direction from p to n side
Drift Current I
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Drift Current Is
• Diffusion current due to majority carrier
diffusion
• A component due to minority carrier
drift exists across the junction
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(a)The pn junction with no applied voltage (open-circuited
terminals).
(b) The potential distribution along an axis perpendicular to the
F d Bi d C d ti
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Forward Biased Conduction
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• The polarity of applied voltage which can't produce any current is called
Reverse Bias.• The polarity of applied voltage which causes charge to flow through the
diode is called Forward Bias.
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Terminal Characteristics
of
a Junction Diode
The diode i –
v relationship
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The diode i – v relationship with some scales
expanded and others compressed in order to reveal
details.
Terminal Characteristics of a
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Terminal Characteristics of a
Junction Diode
• Forward Biased Region v > 0
• Reversed Biased Region v < 0
• Breakdown Region v < -V ZK
Forward Biased Region
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Forward Biased Region
• Is Saturation current – Scale Current
– Is is constant at a given temperature
– Is is directly proportional to Cross-Sectional region of the diode, Is doubles if cross-sectional area is double
– Is is 10-15 A for small size diode
– Doubles in value for every 10OC rise in temperature
i= I s(e
v
nV T − 1)
Forward Biased Region
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• Thermal Voltage VT
– VT = kT/q• K = Boltzmann‟s constant = 1.38 X 10-23 Joules/Kelvin
• T = Absolute Temperature in Kelvin (273 +Temp in Co)
• q = Magnitude of charge = 1.6 X 10-19 Coulombs
– VT @ 20oC is 25.2mV, ~ 25 mV
• n is 1 or 2 depending on the material andthe physical structure of the diode
– n = 1 for Germanium Diode & n=2 for Silicon
Forward Biased Region
i= I s(ev
nV T −
1)
Forward Biased Region
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b Relationship of the current i to the voltage v
holds good over many decades of current
(seven decades, a factor of 107
i= I s
evnV
T
ln i= ln( I s e
v
nV T
)⇒v
nV T ln I s
v= nV T
lni
I s
Forward Biased Region
i= I s
(ev
nV T − 1)
i >> Is
Forward Biased Region
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I 1= I s e
v1
nV T
I 2= I s e
v2
nV T
I 2
I 1
= e
(v2− v
1)nV
T
(v2− v
1)= nV T
ln I
2
I 1
⇒2. 3 nV T
log I
2
I 1
Forward Biased Region
Forward Biased Region
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• for v drop changes by
for n = 1
for n = 2
0.8v≈ 0.7v
I 2 I
1
= 10
2.3nV T ≈ 60mV
120 mV
¿v< 0.5v⇒cut − in− voltage
v= 0.6v
Forward Biased Region(v2− v1)= 2 .3nV T log
I 2 I 1
Illustrating the temperature dependence of the diode forward characteristi
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At a constant current, the voltage drop decreases by
approximately 2 mV for every 1C increase in
temperature.
Figure E3.9
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If V=1V at 20o C, Find V at
400C and 00C
At 20o C Reverse current Is = 1V/1M Ω= 1μ A
Since the reverse leakage current doubles for every 100 C increase,
At 400 C I = 4*1 = 4 μ A V = 4 μ A * 1MΩ = 4.0 V
At 0 C I = ¼ μ A V = 0.25 V
Is
Forward biased Diode
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Characteristics
Example 3.3• A silicon diode displays a forward
voltage of 0.7 V at a current of 1mA.
Find Is at n=1 & 2
i= I s e
v
nV T ⇒ I s= ie
−vnV
T
η= 1 I s= 10− 3
e
− 0 . 7
25× 10− 3
= 6 . 9× 10− 16
A
η= 2 I s= 10− 3 e
− 0. 7
2× 25× 10− 3
= 8 .3× 10− 10
A
Ex 3.7
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Ex 3.7
Silicon Diode with n=1 has VD=0.7V @
i=1mA. Find voltage drop at i=0.1mA &10mA
i= I s e
vnV
T ⇒ I s= ie
− vnV
T
η= 1 I s= 10− 3 e
− 0 . 7
25× 10− 3
= 6 . 9× 10− 15
A
For i= 0 .1 mA⇒V 1= ηV T ln i I s
= 25× 10− 3 ln 10
− 4
6 .9× 10− 16 ==0.64 V
For i= 10 mA⇒V 1= ηV T lni
I s= 25× 10− 3 ln
10− 2
6 . 9× 10− 16= 0.76 V
Solution P3-18
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(a )
η= 2, Diode current = i= 1000IS
i= I S e
v
nV T ⇒1000 I S = I s e
v
2× 25× 10− 3
v= 0.345 V
(a) At what forward voltage does a diode for which n=2 conduct acurrent equal to 1000Is?
(b) In term if Is what current flows in the same diode when its
forward voltage is 0.7 V
(b)v= 0.7V
i= I S
e
vnV
T = I se0 .7
0 .05 = 1 . 2× 106 I S
P bl 3 23
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Problem 3-23
• The circuit shown utilizes threeidentical diodes having n=1 andIs= 10 -14 A. Find the value of the
current I required to obtain an
output voltage Vo=2 V. Assume n=1
• If a current of 1mA is drawn away
from the output terminal by a
load, what if the change in theoutput voltage. Assume n=1
Solution 3-23
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(b) Load current = 1mA, therefore I
DY = 2.81mA
I DY
I DX
= e
(v DY
− v DX
)
0.025 = e
(v DY
− 2 /3 )
0.025
ΔvoY = vO2− v01= 22.8mV
Info available n= 1 , I S = 10− 14 A,V o= 2V
The voltage across each diode isvo
3= v
DX =
2
3
I DX
= I S e
v DX
ηV T = 10− 14
e
2
3
0.025 = 3.81mA
The circuit shown utilizes three identical diodes
having n=1 and Is= 10 -14 A. Find the value of
the current I required to obtain an output voltage
Vo=2 V.
If a current of 1mA is drawn away from the output terminal
by a load, what if the change in the output voltage.
P bl 3 25
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Problem 3-25
• In the circuit shown,both diode haven=1, but D1 has 10times the junction
area of D2. Whatvalue of V results?
Solution 3-25(a)In the circuit shown, both diode
have n=1, but D1 has 10 times the junction area of D2. What value of
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V 0= V D2− V D1= ηV T ln10 I D2
I D1
. . . . . . . . . . . . . . . . .2
I D1= 10 I S2 e
V D1
ηV T
I D2
I D1=
I S2 e
V D2
ηV T
10 I S2 e
V D1
ηV T
= 0. 1 e
V D2
− V D2
ηV
T . . . . . . . . . . . . . .1
I D1
= I S1
e
V D1
ηV T I D2= I S2 e
V D2
ηV T
I S1= 10 I S2
V 0= V D2− V D1= 0.025× ln80
2
= 92.2 mV
I 1= I D2+ I D1⇒ I D2= I 1− I D1..........3
⇒ I D1= 2mA , I D2= 10− 2= 8mA
V results?
solution 2-25 (b) To obtain a value of 50 mV, what current I2 id needed.
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V o= 50mA , Find ID1 , I D2
I D2= 0.01− I D1
I D2
I D1
= 0 . 1e
V D2
− V D2
ηV T =
I D2
0.01− I D2
= 0 .1e2
I D2
= 4.25mA
I D1=(10− 4.25)= 5.75mA
P bl 3 26
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Problem 3-26
• For the circuit shown,both diodes are
identical, conducting
10mA at 0.7 V and 100
mA at 0.8 V.
• Find „n‟
• Find the value of R for which V = 80 m V.
Solution 3-26 (a)
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Diodes are identical therefore IS ,η ,V
T are same
For Diode 1 ⇒V D1
= 0.7V@I D1
= 10mA
For Diode 2 ⇒V D2= 0.8V@I
D2= 100mA
V D2− V
D1= ηV
T ln
I D2
I D1
0 .8− 0 . 7= η× 0.025× ln100
10
η= 1.739
Find η
V = V D2− V D1= ηV T ln
I D2
I D1
= 0.08= 1.737× 0.025× ln0.01− I D1
I D1
I D1= 1. 4 mA
R=80
1. 4
= 57.1Ω
d R if Vo=80mV
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Problem 3.36
Assuming identical diodes for whichVD =0.7V @ ID=1mA. Find R if V0 = 3 V
V Dx=3
4= 0.75V
I DX = I S e
V DX
ηV T
I D2
I D1
=e
V D2
ηV T
e
V D1
ηV T
= e
(V D2
− V D2
)
ηV T
I D2= I D1¿e
(V D2
− V D2
)
ηV T = 1× e
(0.75− 0 .7 )
25× 10− 3
= 7.389mA
.75= 0 .7+ ηV T lnI D2
10− 3⇒ I D2= 7.389 mA
R=10− 3
7.389× 10− 3
= 947Ω
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Modeling the Diode
Forward
Characteristics
A simple circuit used to illustrate the analysis of circuits in which
the diode is forward conducting.
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g
I D=V DD− V D
R
I D= I
S e
V D
ηV T
Graphical analysis of the circuit using the exponential diode m
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Iterative Analysis using theE ti l M d l
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Exponential Model
Determined the diode current ID andDiode voltage VD with VDD =5V and R
=1000 ohms. Diode has a current of
1mA @ a VD of .7 V, and that its voltagedrop changes by 0.1 V for every decade
change in current.
Solution
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F irst iteration V D= 0.7V
I D= I S e
V D
ηV T
= 4 .3 mA
V 2− V 1= 2.3ηVT log I 2 I 1
ΔV = 2.3VT = 0.1V For Every decade change in current
V 2= V 1+ 0.1log
4 . 3
1. 0 = 0.763 V
S econd iteration V D= 0.763V
I D= I S e
V D
ηV T
= 4.237 mA
V 2− V 1= 2.3ηVT ln I 2 I 1
V 2= 0.763+ 0.1log4.237
4 .3= 0.762 V
Solution
I D
= 4.237mA ,
V D= 0.762V
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The Piecewise-Linear Model
Approximating the diode forward characteristicwith two straight lines: the piecewise-linear model.
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The Piecewise Linear Model
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The Piecewise-Linear Model
• Exponential curve is approx into twostraight lines
• Line No 1 with zero slope & Line 2 witha slope of 1/r d
• The voltage change of less than 50 mV is observed in case the current changefrom 0.1 mA to 10 mA.i D= 0 v D
= 0V
i D
=( v D− V D0 )
r D v
D
≥ V D0
Piecewise-linear model of the diode forward characteristic and its
equivalent circuit representation
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equivalent circuit representation.
Piecewise-l inear model
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The
Constant – VoltageDrop Model
Constant – Voltage DropModel
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Model
• Forward conducting diode exhibitsa constant voltage drop VD
• The voltage change of less than 50mV is observed in case the currentchange from 0.1 mA to 10 mA.
• Model is used when
– Detailed information about diode
characteristics in not available
Constant-voltage-drop model
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The constant-voltage-drop model of the diode forward
characteristics and its equivalent circuit representation
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characteristics and its equivalent-circuit representation.
The Small – Signal Model
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The Small – Signal Model
• A small ac signal is superimposed onthe DC components.
• First determined dc Operating Point
• Then small signal operation around the
operating point
– Small portion of the curve is approximated
as almost linear segment of the diode
characteristics.
The Small – Signal Model
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The Small – Signal Model
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Figure 3.17 Development of the diode small-signal model. Note that the numerical values shown are for a diode with n = 2.
The Small – Signal Model
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In absence of signal
I D= I s e V DηV T
Once signal is applied
v D( t )= V D+ vd (t )
i D( t )= I
sev D
ηV T
i D( t )= I
se[V D+ v
d (t )]
ηV T
i D( t )= I
se
V D
ηV T × e
vd (t )
ηV T
i D(t )= I
De
vd (t )
ηV T
For very small signal vd
ηV T
<<1
i D(t )= I
D(1+
vd
ηV T
)
i D (t )= I D+ id (t )
The Small – Signal Model
The Small – Signal Model
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i D( t )= I D(1+
vd
ηV T )
i D( t )= I D+ id ( t )
id ( t )= I D vd
ηV T
r d =
ηV T
I D
The Small – Signal Model
r d is inversely proportional to I D
Modeling the Diode Forward Characteristic
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Table 3.1 (Continued)
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Exp 3-6
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+
-
VDID
+
-
vd
Exp 3 6V
DD= 10V,v
d = 1V peak amplitude @ 60 Hz
Diode has a current of 1mA @ a V D
of .7 V, n= 2
Find r d ,V D, vd ( t )
Solution
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I D=V
DD
− V D
R = 10−
0.710 = 0.93mA
r d =
ηV T
I D
=2× 250.93
= 53.8Ω
Small signal
v dpeak = v speak
r d
R+ r d
= 5.35mV
Input variation of 10% resulted in output voltage
variation of 0.7+5.4mV(0.8%) Voltage regulation
Exercise 3-16
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Exercise 3 16
• Design a circuit shown so that Vo=3vwhen IL =0 A and Vo changes by 40 mV
per 1mA of diode current.
• (a) Find the value of R• (b) The junction area of each diode
relative to a diode with ).7 V drop at
1mA current. Assume n=1
Excercise 3-16
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r DT
= Δv
o
Δio
=0.04
10− 3
= 40Ω
r DX
= 40/4= 10Ω
I DX
=nV T
r DX
= 2 .5mA
R=15− 3
2.5m= 4.8K Ω
At dc Operating Point VDX= 3 /4= .75V
I D1= 1mA ,V D1 0.7V
I DX
I D1
= I
SX
I S1
e
V DX
− V 1
nV T
⇒ I SX
I S1
= 0.34
The diodes have the junction area 0 .34 times the diode
Why 4 diodes and not 5? Diodes will
not conduct at 0.6 V
Diode Forward Drop inVoltage Regulation
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Voltage Regulation
• Small signal model is used.
• Voltage remains constant in spite of :
– Changes in load current
– Changes in the dc power supply voltages
• One diode provides constant voltage of
0.7 V and for greater voltages diodes
can be connected in series.
Example 3-7
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• A string of three diodes is used toprovide a constant voltage of about 2.1
V. We want to calculate the percentage
change in this regulated voltagecaused by
• (a) a + 10 % change to the power supply
voltage• (b) Connection of a 1 K ohms load
resistance , Assume n=2
Solution Exp 3-7
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Solution Exp 3 7
P 3-53• In a particular cct application, ten “20
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In a particular cct application, ten 20
mA diodes” ( a 20 mA diode is a
diode that provides a 0.7 V drop
when the current thru it is 20 mA)
connected in parallel operate at a
total current of 0.1 A. For the diodes
closely matched, with n=1, what
current flows in each.
What is the corresponding small signal
resistance of each diode and of the
combination?
i Dx=0.1
10==0.01 A
r dx= nV T
I Dx
= 2 . 5Ω
req=2. 5
10= 0.25Ω
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• If each of the 20 mA diode has aseries resistance of 0.2 ohm
associated with the wire bonds to
the junction. What is the equivalent
resistance of the 10 parallel
connected diodes?
What connection resistance would
single diode need in order to be totally
equivalent?
Re q=1
10(2. 5+ 0. 2)= 0.27Ω
The diode i –
v relationship
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Reversed Biased Diode
Leakage current:
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ea age cu e
• In the reverse direction there is asmall leakage cur rent up until the
reverse breakdown vo l tage is reached.
• This leakage is undesirable, obviously
the lower the better.
• Diodes are intended to operate below
their breakdown voltage.
The Reversed Biased Region
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g
i= I s(e
v
nV T − 1
v is negative ∧ >>V T (25mV )
i= − I S
Current in reserved biased diode circuit is due to
leakage current & increases with increase in reverse
voltage
Leakage current is proportional to the junction area
& temperature but doubles for every 10oC rise in
Breakdown Region
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• Once reverse voltage exceeds a threshold value of
diode VZK
, this voltage is called breakdown voltage.
VZK Z – Zener, K – Knee
• At breakdown knee reverse current increases rapidly
with associated small increase in voltage drop
• Diode breakdown is not destructive if power
dissipated by diode is limited by external circuitry.
• Vertical line for current gives property of voltage
regulation
The diode i –
v characteristic with the breakdown region
shown in some detail.
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Zener Diode
Zener Diode
O ti i th R B kd
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• Operation in the Reverse Breakdown
Region
• Very steep i-v cu rve at breakdown w ith
almos t constant vol tage drop region
• Used the design ing vo l tage regu lator
• Diode manu factu red to operate
speci f ical ly in the Breakdown region
called Zener or B reakdown
Zener Diode : Symbol
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I Z
- V Z +
Zener Diode : Symbol
Model: Zener • Manufacturer specify Zener Voltage V at a
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Manufacturer specify Zener Voltage Vz at aspecified Zener test current Iz, the Max.
power that the device can safely dissipate 0.5W @ 6.8 v at max 70mA
• r z Dynamic resistance of the Zener and is the
inverse of the slope of the almost linear i-vcurve at operating point Q
• Lower r z, the more constant Zener Voltage
• The most common range of zener voltage is 3.3 volts to 75 volts,
ΔV z = ΔI z r z
Model for the zener diode.
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Model: Zener
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V z = V zo+ r z I z
I z> I
zk
V z> V zo
es gn ng o e ener s unregulator
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+
-
Vo
Zener regulator
Supply voltage includesa large ripple component
Vo is an output of the zener regulator
that is as constant as possible in spite of
the ripples in the supply voltage VS
and the variations in the load current
Voltage regulator performance can be measured
Line Regulation & Load Regulation
Line Regulation = ΔV o /ΔV s
Load Regulation = ΔVo /ΔIL
Expression of performance : Zener regulator
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I
IL
+
-
Vo
• Only the first term on right hand side is desirable one
Second and third terms depend upon Supply
Voltage Vs and Load current IL
• Line Regulation =
• Load Regulation =
(V s
-V o)
R = (V
o-V
zo)
r z + I L
V o= V zo(R
R+ r z )+ V S (
r z
R+ r z ) - I L(r z ∣∣ R )
ΔV o
ΔV s/ =
r z
(r z + R )
ΔVo
ΔI L= - (r z ∣∣ R )
Expression of performance : Zener regulator
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I
IL
+
-
Vo
• An important consideration for the design
is
• To ensure that current through the zener
diode never becomes too low i.e less
than IZK or Izmin
• Minimum zener current Izmin occurs when
• Supply Voltage Vs is at its minimumVSmin
• Load current IL is at its maximum
ILmax
• Above design can be made be selecting R =
(V s min – V ZO - r z I z min)
(I z min+ I Lmax )
where I Lmax=V Z
R L
The circuit with the zener diode replaced with its equivalent circuit mode
Example 3.8
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Exp 3-8
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Example 3-8
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V =+10 v± 1v
R= 0.5k Ω
V z= 6 .8v I z= 5 mA
r z= 20Ω
I zk = 0 . 2 mA
I RL= 1 mA
a) Find No Load Line RegulationV o∧ ΔV o
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V z= V
zo+ r
z I z
V zo= V z−I zr z=
6.8− 5
×20
×10
−3
=6.7v
Depending upon the manufacturer provide DataFirst calculate Vzo if Vz =6.8 V & Iz=5mA, r Z=20 ohm
Now connecting the Zener diode in the Cct as shownCalculate actual Iz and resulting Vo
Thus establishing operating Point
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Line
Regulation
I z = V − V zo
R+ r z
= 10− 6 .7500+ 20
= 6.35mA
V o= V
zo+ I
z r z = 6 .7+ 6.35× 20× 10− 3= 6.827V ≈ 6.83V
ΔV o ΔV
=38.5
1= 3.85mv /v
Now carry out Small Signal Analysis
Suppress DC source and calculate resultant change in VoUse voltage divider rule
ΔV o= ΔV
+r z
R+ r z
=± 1× 20
520=± 38.5mv
n vO oa res s ance L connected & draws 1mA and load
regulation
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g
1mA drawn by load would decrease
by same amount so
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R L≈6.83 v
1mA= 6.83 k Ω
R L∣∣ R=
20× 6830
6850 = 19.94ΩV Z = V o= V ZO+ I Z r Z = 6 . 7+ 5.35× 20= 6.807 V
I Z =Vs− V Z
R+ R L∣∣r Z
=10− 6.807
500+ 19.94= 6.14 mA
ΔI Z
= 6.35− 6.14= . 21 mA= 210 μA
I z ΔV o= r z ΔI z = 20× −1mA=− 20mV
= ΔV o
ΔI z
=− 20 mV /mA
Checkexact Calculations
by same amount so
Load Regulation
c) for ΔV o R L= 2k Ω
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I R L= V Z
R L= 3 . 4 mA
ΔI Z = − 3 . 4 mA
ΔV o= r Z ΔI Z = − 68 mV
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• 1) Check
Zener at Breakdown region
V o=2000
2500× 10= 8v
2000
500
+10
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10v
V o
10
500
6.63 v
19.8Ω
Δ
+
6.7v−
20
2k Ω
0.5k Ω
A
B
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V oc= 6 .7× 20002020
= 6.63v
Req= 19.8Ω
A
B
d ) RL= 500Ω
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Zener is not
operating
V o=10× 500
1000= 5v
zk
@V o<<Valignl ¿¿¿
¿5<<6.8v
¿¿
L
10 v
500
500
e) Min value of for which the diode still
operates in the breakdown region
R L
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p g
I z
• at BreakdownRegion
+ 10± 1v
500 I z
R L6.7v
0.2 mA
I z= I
zk = 0 .2 mA
V z= V zk = 6.7vV
DD= 9vmin
I =9− 6 . 7
500= 4 . 6mA
I = I zk + I
RL
I RL
= 4 .6− 0. 2= 4 .4 mA
R L=
V zk
I RL
=6 . 7
4 .4m= 1.5k Ω
Problem D3.68Design a 7.5-V zener regulator circuit using a 7.5-V
ifi d t 12 A Th h
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zener specified at 12mA. The zener has anincremental resistance r z = 30 Ω and a knee current
of 0.5mA. The regulator operates from a 10-V supplyand has a 1.2-kΩ load.
(a) What is the value of R you have chosen?
(a) What is the regulator output voltage when thesupply is 10% high? Is 10% low?
(a) What is the output voltage when both the supply is
10% high and the load is removed?
(a) What is the smallest possible load resistor that canbe used while the zener operates at a current nolower than the knee current while the supply is 10%
low?
Solution 3-68
r = 30Ω
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r z
30Ω
I Zk = 0 .5 mAV
Z = 7.5V
I Z = 12 mA7 .5= V ZO+ 12× 30× 10
− 3
⇒V ZO= 7.14 V
I RL=7 .5
1 .2= 6.25 mA
Design a 7.5-V zener regulator circuit usinga 7.5-V zener specified at 12mA. The zener has an incremental resistance r z = 30 Ω anda knee current of 0.5mA. The regulator operates from a 10-V supply and has a 1.2-kΩ l d
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Select I = 10mA
I RL=
7.5
1.2= 6.25 mA
So that I Z = 3.75mA
W hichis> I Zk
R=10− 7 .5
10= 250Ω
kΩ load.
(a) What is the value of R you have chosen?
(a) What is the regulator outputvoltage when the supply is 10%high? Is 10% low?
For ΔV +=± 1V
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ΔV O=± 1×1.2//0.03
0.250+ (1.2//.03 ) =± 0.1V
ThusV O =+7.4V to + 7.6V
With V += 11V and I L= 0
V O= V ZO+11− V O
0.28X0. 03
⇒V O= 7.55V
(a) What is the output voltage whenboth the supply is 10% high andthe load is removed?
(a) What is the smallest possible load resistor that can beused while the zener operates at a current no lower thanthe knee current while the supply is 10% low? IZK=0.5mA,VZO=7.14 V
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11V
V O
R9− 7.155
0.25
= 7.38mA
7.14+ 0.03 X0 .5
7.155V 0.5 mA
250Ω
R Lmin
R Lmin=7.155
7.38− 0 . 5
= 1.04k Ω
1
3
2
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Rectifier Circuit Power Supply
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• Power supply must supply dc voltage to be constantin spite of – variation is ac line voltage
– Variation in current drawn by load, that is variable loadresistance
Rectifier Circuits
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• Filter – Smoothes out pulsating dc but still some time-
dependent components-(ripple) remain in the output
• Voltage Regulation – Reduces ripples
– Stabilizes magnitude of dc output against variation inload current
– Regulation by Zener Diode or Voltage regulator I.C
Half Wave Rectifier
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Transfer characteristic of
the rectifier circuit
Input and output waveforms assuming that rD >
Full Wave Rectifier
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Input and output waveforms.
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Full Wave Rectifier
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• Diode in Reverse biased stateAnode @ - Vs
Cathode @ + Vo
• PIV = 2Vs - VDO
Twice as in case of half wave rectifier
Bridge Rectifier
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The bridge rectifier: (a) circuit; (b) input and output waveforms.
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Bridge Rectifier
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Bridge Rectifier
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Bridge Rectifier
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• Peak Inverse Voltage
– PIV => consider loop D3, R & D2
– VD3(res) = Vo + VD2
– Vo = Vs – 2VD
– PIV = Vs – 2VD + VD = Vs – VD
Half of PIV for Full wave Rectifier
D4
D1
D2 D3
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Figure 3.28 (a) A simple circuit used to illustrate the
effect of a filter capacitor. (b) Input and output
waveforms assuming an ideal diode.
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g
Peak detector with Load
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Figure 3.29 Voltage and current waveforms in thepeak rectifier circuit with CR<<T .
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Charge / Discharge Cycle
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Peak detector with Load
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i L=
V o
Ri D= i
C + i
L
i D= C
dV s
dt
+ i L
Figure 3.30 Waveforms in the full-wave peak
rectifier.
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• When Vr is small
ea ec er : u puVoltage
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When Vr is small
– Vo = Vpeak
– iL is almost constant
– DC components of iL
• Accurate value of output dc voltage
Average Value
i L=V P
R
V o= V P −1
2V r
Charge / Discharge Cyclev o= V P e
−t
CR
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V o= V P − 12 V r
i L=V P
R
e−
T
CR= 1−T
CR
V o= V P − V r ≈ V P e
−T
CR
⇒V r = V P (1− e−
T
CR
)
V r = I L
fC
, provided V r <<V p
V r = V
P (1− 1+T
CR)
I L=V
P
R
V r =V
P
T
CR ⇒
V P
fCR
Peak Rectifier : Ripple Voltage
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• During Discharge cycle
• At the end of discharge cycle
• Since CR >> T
e−
T
CR= 1−T
CR
V o= V P − V r ≈ V P e− T CR
⇒V r = V
P
(1− e−T
CR )
v o= V P e−
t
CR
−T
CR T−T
Peak Rectifier : Ripple Voltage
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Ripple Voltage
V r = I L
fC
, provided V r <<V p
V r = V
P (1− 1+T
CR)
V r =V P
CR
I L= V P
R
e CR= 1−T
CRV r = V P (1− e
−CR
)
V r =
V P T
CR ⇒
V P
fCR
ea ec er : on uc onInterval
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V P
(1−
(ωΔt )2
2 )= V P − V r
⇒ωΔt =√2V
r
V P
V P cos(ωΔt )= V P − V r
Hence ωΔt is small
Cos (wΔt )= 1−(wΔt )2
2 !+ .. .
When Vr<<Vp the conduction angle will be small
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Average Diode Current –During
Conductioni = i + i
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During Discharge Qlost =CV r
i D= i
C + i
L
i Dav
= iCav
+ I L
iCav
= i Dav
− I L
During Charge Qsupplied
= icav Δt
iCav
= i Dav
− I L
Qsupplied= Qsupplied⇒icav Δt = CV r
Qlost= Qsu lied⇒CV r= icav Δt
Average Diode Current –During
Conduction
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i Dav
=2π
√2V r
V P
V P
R+ I
L⇒i
L(1+ π 2V P
V r )
V r
<<V P
⇒ iDav
>> I L
Qlost Qsu lied r cav
V r =V P T CR
=⇒CV r =V P T R
V P T
R= (i Dav
− I L) Δt
Δt = 12πf
2V r
V P
= T 2π
2Vr
V P
V P
T
R= (i Dav
− I L)(T
2π
2Vr
V P )
Deduction
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• As waveform of is almost right angle r
triangleV
r << V
P
i D max= 2i Dav
Observations• Diode current flows for short interval
d t l i h th h l t b
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and must replenish the charge lost by
the capacitor. Discharge interval is long
& discharge is through high resistance
• Maximum diode current
Assuming that i L is almost constant = I L &
CR >>T
r D<< R L
i D=CdV i
dt + i L
i D max
= i L(1+ 2π
2V p
V r )≈ 2i
Dav
Example N0 3-9Consider a peak rectifier fed by a 60 Hz
sinusoidal having a peak value of Vp =
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sinusoidal having a peak value of Vp =
100 V. Let the load resistance R =10 kOhms.
(a) Find the value of the capacitance C
that will result in peak to peak ripple of 2 V
(b) Calculate the fraction of the cycle
during which the diode is conduction
(c) Calculate the average and peak value
of the diode current.
Example 3.9
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• Find value of C for Vr =2V (peak to peak)
• Find fraction of cycles that diode
conducts
• => Diode conducts of cycle
0.2
2π× 100= 3.18
100Sin2Π60 t
10 k Ω
C =V P
V r fR=
100
2× 60× 104= 83 .3μF
ωΔt =2Vr
V P
= 0 .2radian
Solution Exp 3-9
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• Find &
i Dav= I L(1+ π
2V P
V r )
I L=
V P
R=
100
10000= 10 mA
i Dav= 10(1+ π
√2× 100
2 )= 324 mA
imax= 2i Dav= 648 mA
i D max
i Dav
Full wave peak Detector
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In full wave rectifier, the capacitor discharge for almost T/2 time interval.
that mean ripple frequency is twice the
input, soV r =V
P 2 fCR
imax
= 2i Dav
= I L(1+ 2π
V P
2Vr )
i Dav
= I L(1+ π
V P
2Vr )
Applications
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• Peak Rectifier – Peak detector is used
for
– Detecting the peak of the an input signal
for signal processing systems
– Demodulator for amplitude modulated
(AM) signals
.
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Precision Half Wave Rectifier Super Diode
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• Normal Diodes VD= 0.7v are used for
rectifier of input of much larger
amplitude then VD
• For smaller signals detection,
demodulation or rectificationOperational Amplifiers (Op Amp) are
used
Wave form GenerationLimiting Clamping
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• Limiter Circuit – Vo is limited between two levels – upper
(L+) and lower (L-) thresholds
Figure 3.33 Applying a sine wave to a limiter can
result in clipping off its two peaks.
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Figure 3.34 Soft limiting.
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Wave form GenerationLimiting / Clamping
• Double Limiter
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– Clips off both negative & positive peaks• Single Limiter
• Clips off only one side of the input peak
• Application – Limits the inputs to operation Amplifier to a
limit lower than the breakdown voltage of
transistors of input stage of operational
Amplifier
– Half / Full Rectifier for Battery Charger
Figure 3.35 A variety of basic limiting circuits.
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Figure E3.27
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Solution Ex 3-27
(a ) − 5≤ vi≤ 5 ⇒ v = v i
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v R=
[10
10+ 10 ](v i− 5)=
1
2(vi− 5)
v o= 5+ v R=1
2v i+ 2 . 5
(a ) 5≤ vi≤ 5 ⇒ vo v i
(b) V I ≥+ 5 Vo D2 Conduct, D1 cut-off
(c ) v i≤ − 5 V D1 Conducts & D2 is off
v R= [10
10+ 10 ](v i+ 5)=1
2(vi+ 5)
vo=
1
2
(v i+ 5)− 5=
1
2
(vi− 2 . 5)
D C Restorer • The output waveform will have its lower
peak “Clamped” to O V therefore
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peak Clamped to O V therefore
known as “Clamped Capacitor”
• Output waveform will have a finite
average value & is entirely different and
unrelated to the average value of the
input waveform
Application
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T X R
DCRestorers
− 4v
+ 6
− 4
+ 2v
0v
4v
Figure 3.36 The clamped capacitor or dc restorer
with a square-wave input and no load.
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Figure 3.37 The clamped capacitor with a loadresistance R .
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Figure 3.38 Voltage doubler: (a) circuit; (b)
waveform of the voltage across D 1.
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Figure P3.97
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Figure P3.98
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Figure P3.102
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Figure P3.103
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Figure P3.105
+ 4 v i
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Diode Off V0= V i+ V c
Diode On Vo= − 0.7v
D on
− 6
− 0
D off
V C
v o
The Voltage Doubler
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C1
V P sinωt
+ V P -
+
V D1
−
D1
−
2V P
+
+ V P -
V P sinωt
C 1 D1⇒a Clamp circuit
DC Restorer
Special Diode TypeSchottky-Barrier Diode (SBD)
• Shottky-Barrier Diode is formed by
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Shottky Barrier Diode is formed by
bringing metal into contact with amoderately doped „n‟ type
semiconductor material
• Resulting in flow of the conducting
current in one direction from metal
anode to the semiconductor cathodeand acts as an open circuit in the other
direction
Schottky-Barrier Diode (SBD)
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• Gets two important properties
– SBD switches on-off faster due to current
conducts due to majority carrier b
(electrons)
– Forward voltage drop is lower then P-n
junction diode
Varactor
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• Variable Capacitor – Depletion layer acts as junction
capacitance
– Depletion layer varies CapacitanceDepletion Region
Dielectric
← D→
Metallic Plate
Varactor • When a reverse vo ltage is applied to a p-n junction ,
the depletion region, is essentially devoid of carriers
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e dep e o eg o , s esse a y de o d o ca e s
and behaves as the dielectric of a capacitor.
• The depletion region increases as reverse voltage
across it increases; and since capacitance variesinversely as dielectric thickness, the junction
capacitance will decrease as the voltage across the
p-n junction increases.
• By varying the reverse voltage across a p-n
junction the junction capacitance can be varied .
Semiconductor diodes• The tunnel diode, the current through the
device decreases as the voltage is increased
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within a certain range; this property, knownas negative resistance, makes it useful as an
amplifier.
• Gunn diodes are negative-resistance diodes
that are the basis of some microwave
oscillators.
• Light-sensitive or photosensitive diodes can
be used to measure illumination; the voltage
d th d d th t f
SCR (Thyristor)• The Silicon Controlled Rectifier (SCR) is simply a
conventional rectifier controlled by a gate signal.
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• A gate signal controls the rectifier conduction.
• The rectifier circuit (anode-cathode) has a low
forward resistance and a high reverse resistance.
• It is controlled from an off state (high resistance) to
the on state (low resistance) by a signal applied to
the third terminal, the gate.
• Most SCR applications are in power switching,
phase control, chopper, and inverter circuits.
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Photodiode
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If reversed biased PN junction isexposed to incident light – the photons
impacting the junction cause covalent –
bond to break thus give rise to current
known as a photocurrent & is
proportional to the intensity of incident
light.
Converts Light energy into a electrical
signals
Photodiode
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• Photodiode are manufactured usingGallium Arsenide (GaAs)
• Photodiodes are important element of optoelectronics or photonics circuit
(Combination of Electronics & optics)
used for signal processing, storage &transmission
Photodiode : Applications
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• Fiber optics Transmission of telephonic& TV signals
• Opto-storage are CD ROM computer disks
• Wide bandwidth & low signalattenuation.
Light Emitting Diode (LED)• Inverse of Photodiode
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• Converts a forward biased current into light
• GaAs used for manufacturing LEDs
• Used as electronics displays
• Coherent light into a narrow bandwidth laser diodes
Fib O i & CD ROM
LED
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Double heterostructure laser
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Optoisolator
• LED & Photodiode
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Electrical to light Light toelectrical
• Provides complete electrical isolation between
electrical circuits
• Reduces the effects of electrical interference on
signal being fixed within a system
• Reduces risk of shock
• Can be implemented over long distance fiber optics
Laser Pointer
.
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Laser Microphone
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Problem 3-103
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Sketch and label the transfer Characteristics of the circuit shown
over a + 10 V range of the input signal.
All diodes are VD =0.7 V @ 1 mA with
n=1.
What are the slopes of the
characteristic at the extreme + 10 V
levels?
+1 V
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-2 V-2 V
V i
V 0
0<V i< 1 Vo= 0
Problem 3-103
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Ist Sessional• Q No 1 (12 Marks) In the circuit shown, input
voltage is a 1kHz, 10 V peak to peak sine wave. The
diode is an ideal diode.
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d ode s a dea d ode
• (a) Sketch the waveform resulting at output
terminal vO.
• (b) What are its positive and negative peak
values?
• Q No 2 (15 Marks) A circuit utilizes three
Ist Sessional
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Q ( )
identical diodes connected in series having
n=1 and IS= 10-14 A.
• (a) Find the value of current required to
obtain an output voltage of 2 V across thethree diodes combined.
• (b) If a current of 1 mA is drawn away
from the output terminal by a load
• (i) What is the change in output
voltage?
• (ii) What is the value of the load?
• Q No 3 (13 Marks) For the circuit shown,
Ist Sessional
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Q ( ) ,
sketch the output for the sine wave input of
10 volts peak. Label the positive and
negative peak values assuming that CR
>>T.
Ist Sessional
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• Q No 4 (10 Marks) 9.25 V zener diodeexhibits its nominal voltage at a test
current of 28 mA. At this current the
incremental resistance is specified as 7
ohms.
– (a) Find VZO of the zener model.
– (b) Find the zener voltage at a current
of 10 mA.
• Q No 5 (20 Marks) Consider a bridge
rectifier circuit with a filter capacitor C
Ist Sessional
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placed across the load resistor R for the casein which the transformer secondary delivers
a sinusoid of 12 V (rms) having the 60 Hz
frequency and assuming VD = 0.8 V and a
load resistance of 100 ohms.
– Find the value of C that results in a ripple voltage
no larger than 1 V peak to peak.
– Find the diode conduction angle. – Find the load current.
– What is the average load current?
• Q No 6 (10 Marks) In a circuit shown, the
Ist Sessional
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( ) ,
output voltage is 2.4 V. Assuming that the
diodes are identical and are having 0.7 V drop
at 1mA.
– (a) Find the current following through the resistor R.
– (b) What the value of resistor R.
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Figure 3.31 The “superdiode” precision half -wave rectifier and its almost-ideal transfer characteristic. Note that when v I
> 0 and the diode conducts, the op amp supplies the load current, and the source is conveniently buffered, an added
advantage. Not shown are the op-amp power supplies.
Figure P3.82
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Figure P3.91
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Figure P3.92
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Figure P3.93
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Figure P3.105
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Figure P3.105
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Quiz DE28 EE -B
(10 ) 9 2
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(10 Marks) 9.25 V zener diode exhibitsits nominal voltage at a test current of
28 mA. At this current the incremental
resistance is specified as 7 ohms. –(a) Find VZO of the zener model.
–(b) Find the zener voltage at a current
of 10 mA.
Quiz DE 28 EE -A
A di d h i l lt
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A zener diode whose nominal voltageis 10 V at 10 mA has an incremental
resistance of 50 Ω.
(a) What is the value of VZO of the zener
model?