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Chapter 2:
Diode Applications
1
Load--Line Analysis
Load
The load line plots all possible
current (ID) conditions for all
voltages
g applied
pp
to the diode ((VD)
in a given circuit. E / R is the
maximum ID and E is the
maximum VD.
Where the load line and the
characteristic curve intersect is the
Q-point, which specifies a
particular ID and VD for a given
circuit.
E = VD + I D R
When VD = 0
When ID = 0
ID =
VD = E
E
R
2
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Fig. 2.2
Drawing the load line and finding the point of operation.
Ex. 2.1 Employing the diode characteristics, determine:
(a) VDQ and IDQ
(b) VR
E − VD 10 − 1
When VD = 1 V I D =
=
= 18mA
R
0.5k
E = VD + I D R
10 = VD + I D × 0.5k
When VD = 0
When ID = 0
E
10
=
= 20mA
R 0.5k
VD = E = 10
ID =
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Fig. 2.4
Solution to Example 2.1.
E = VD + I D R
10 = VD + I D × 0.5k
When VD = 0
When ID = 0
10
E
=
= 20mA
R 0.5k
VD = E = 10
ID =
Table 2.1
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Series Diode Configurations
Forward Bias
Constants
• Silicon Diode: VD = 0 .7V
• Germanium Diode: VD = 0 .3V
Analysis
• VD = 0.7V (or VD = E if E < 0 .7V)
• VR = E – VD = E – 0.7
• ID = IR = IT = VR / R = (E-0.7)/R
(E-0 7)/R
7
Series Diode Configurations
Reverse Bias
Diodes ideally behave as open circuits
Analysis
• VD = E
• VR = 0 V
• ID = 0 A
8
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Example 2.4. Find VD, VR and ID
VD = 0.7V
VR = E − VD = 8 − 0.7 = 7.3V
ID = IR =
VR
7.3
=
= 3.32mA
R 2.2k
Example 2.5. Find VD, VR and ID with the diode reversed
ID = 0A = IR
VR = I R × R = 0 × 2.2k = 0V
E − VD − VR = 0
E = VD = 8V
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Fig. 2.15
Source notation.
Example 2.6. Find VD, VR and ID
Q 0.5 < VD = 0.7
∴ OFF
ID = 0A
VR = I R × R = 0 × 1.2k = 0V
E − VD − VR = 0
E = VD = 0.5V
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Fig. 2.17
Operating point with E = 0.5 V.
Example 2.7. Determine VO and ID
Si VON = 0.7 V，Red LED VON = 1.8 V
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V O = E − V K 1 − V K 2 = 12 − 0 . 7 − 1 . 8 = 9 . 5V
ID = IR =
VO
9 .5
=
= 13 . 97 mA
R
680
Example 2.8. Determine ID, VD2, and V0
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Fig. 2.23
Substituting the equivalent state for the open diode.
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Fig. 2.24
Determining the unknown quantities for the circuit of Example 2.8.
ID = 0A
V 0 = 0V
V D 2 = E = 20 V
Example 2.9. Determine I, V1, V2 and V0
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Fig. 2.26
Determining the state of the diode for the network of Fig. 2.25.
＋
－
ON
Fig. 2.27
Determining the unknown quantities for the network of Fig. 2.25. KVL, Kirchhoff voltage loop.
KVL
− E1 + V1 + 0 .7 + V 2 − E 2 = 0
− 10 + I × 4 .7 k + 0 .7 + I × 2 .2 k − 5 = 0
I =
10 + 5 − 0 .7 14 . 3
=
= 2 .07 mA
4 .7 k + 2 .2 k 6 .9 k
V1 = I × 4 .7 k = 2 . 07 m × 4 .7 k = 9 .73V
V 2 = 2 . 07 m × 2 .2 k = 4 . 55V
Vo = V 2 − E 2 = 4 .55 − 5 = − 0 . 45V
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Parallel Configurations
23
Parallel Configurations
V = 0.7 V
D
V
=V
= V = 0.7 V
D1
D2
O
V = 10 − 0.7 = 9.3 V
R
I
I
R
=
D1
E−V
D = 10 V − 0.7 V = 28.18 mA
R
0.33kΩ
=I
D2
=
28.18 mA
2
= 14.09 mA
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Example 2.11. Find the R to ensure a current of 20mA through the “on” diode
Reverse breakdown voltage
g =3V
Turn on voltage = 2 V
Fig. 2.31
Operating conditions for the network of Fig. 2.30.
I = 20mA =
R=
E − VLED 8 − 2
=
R
R
6
= 0.3kΩ = 300Ω
20m
For Red LED, VR = 2V < 3V
Red LED is ok
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Example 2.12. Determine the voltage V0
Fig. 2.36
Determining Vo for the network of Fig. 2.35.
Si diode turn on first.
VD = 0.7 V
But, green LED doesn’t
work. 0.7 < 2.2 V
V0 = 12 − 0.7 = 11.3V
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Example 2.13. Determine I1, I2 and ID
ON
20 V
19.3 V
ON
Fig. 2.38
Determining the unknown quantities for Example 2.13.
0.7
= 0.212mA
A
3.3k
Va = E − 0.7 − 0.7 = 18.6V = V2
I1 =
V2
18.6
=
= 3.32mA
5.6k 5.6k
KCL
I D 2 = I 2 − I1 = 3.32 − 0.212 ≈ 3.11mA
I2 =
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Positive logic OR gate. Determine V0.
HI (1)
Fig. 2.40
Redrawn network of Fig. 2.39.
ON
OFF
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Fig. 2.41
Assumed diode states for Fig. 2.40.
HI
V0 = E − VD = 10 − 0.7 = 9.3V
I=
Fig. 2.42
V0 9.3
=
= 9.3mA
R 1k
Positive logic AND gate.
OFF
8V
8V
LO (0)
ON
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Fig. 2.43
Substituting the assumed states for the diodes of Fig. 2.42.
E − V0 10 − 0.7
=
= 9.3mA
1k
1k
Vo = 0.7V
I=
Half--Wave Rectification
Half
The diode only
conducts when it is
in forward bias,
therefore only half
of the AC cycle
passes through the
diode.
The DC output voltage is 0.318Vm, where Vm = the peak AC voltage.
Vdc =
1
π
Vm = 0.318Vm
36
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Fig. 2.44
Half-wave rectifier.
Fig. 2.45
Conduction region (0 → T/2).
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Fig. 2.46
Nonconduction region (T/2 → T).
Fig. 2.47
Half-wave rectified signal.
Vdc =
1
π
Vm = 0.318Vm
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Fig. 2.48
Effect of VK on half-wave rectified signal.
Vdc = 0.318(Vm − Vk )
Ex. 2.16
(a) Sketch the output Vo and determine the dc level of Vo.
(b) Repeat (a) if the ideal diode is replaced by a silicon diode.
(c) Repeat (a) and (b) if Vm is increased to 200V.
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(a) V = −0.318V = −0.318 × (20V ) = −6.36V
dc
m
(b) V = −0.318(V − 0.7) = −0.318 × (20 − 0.7V ) = −6.14V
dc
m
(c)
Vdc = −0.318Vm = −0.318 × (200V ) = −63.6V
Vdc = −0.318(Vm − 0.7) = −0.318 × (200 − 0.7V ) = −63.28V
PIV (PRV)
Because the diode is only forward biased for one-half of the AC cycle, it is
also reverse biased for one-half
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.
PIV (or PRV) > Vm
•
•
•
PIV ≥ Vm
PIV = Peak inverse voltage
PRV = Peak reverse voltage
Vm = Peak AC voltage
44
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Fig. 2.52
Determining the required PIV rating for the half-wave rectifier.
Full--Wave Rectification
Full
Th rectification
The
tifi ti process can b
be iimproved
db
by
using more diodes in a full-wave rectifier
circuit.
Full-wave rectification produces a greater
DC output:
•
•
Half-wave: Vdc = 0.318Vm
Full-wave: Vdc = 0.636Vm
46
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Full--Wave Rectification
Full
Bridge Rectifier
•
•
Four diodes are required
VDC = 0.636 Vm
47
Fig. 2.53
Full-wave bridge rectifier.
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Fig. 2.54
Network of Fig. 2.53 for the period 0 → T/2 of the input voltage vi.
Fig. 2.55
Conduction path for the positive region of vi.
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Fig. 2.56
Conduction path for the negative region of vi.
Fig. 2.57
Input and output waveforms for a full-wave rectifier.
Vdc = 0.636Vm
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Fig. 2.58
Determining VOmax for silicon diodes in the bridge configuration.
Vdc = 0.636(Vm − 2Vk )
Fig. 2.59
Determining the required PIV for the bridge configuration.
PIV ≥ Vm
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Full--Wave Rectification
Full
Center--Tapped Transformer Rectifier
Center
q
Requires
• Two diodes
• Center-tapped transformer
VDC = 0.636(Vm)
55
Fig. 2.60
Center-tapped transformer full-wave rectifier.
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Fig. 2.61
Network conditions for the positive region of vi.
Fig. 2.62
Network conditions for the negative region of vi.
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Fig. 2.63
Determining the PIV level for the diodes of the CT transformer full-wave rectifier.
PIV = Vsec ondary + VR = Vm + Vm = 2Vm
PIV ≥ 2Vm
Example 2.17.
Determine the output waveform and calculate the output dc level and the
required PIV of each diode.
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Fig. 2.66
Redrawn network of Fig. 2.65.
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Fig. 2.67
Resulting output for Example 2.17.
Summary of Rectifier Circuits
Rectifier
Ideal VDC
Realistic VDC
Half Wave Rectifier
VDC = 0.318(Vm
VDC = 0.318Vm – 0.7
Bridge Rectifier
VDC = 0.636(Vm)
VDC = 0.636(Vm) – 2(0.7)
Center-Tapped Transformer
Rectifier
VDC = 0.636(Vm)
VDC = 0.636(Vm) – 0.7
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.
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Diode Clippers
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
65
Fig. 2.68
Series clipper.
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Biased Clippers
Adding a DC source in
series with the clipping
diode changes the effective
forward bias of the diode.
67
Fig. 2.69
Series clipper with a dc supply.
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Fig. 2.70
Determining the transition level for the circuit of Fig. 2.69.
Fig. 2.71
Using the transition voltage to define the “on” and “off” regions.
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Fig. 2.72
Determining vo for the diode in the “on” state.
Fig. 2.73
level.
Sketching the waveform of vo using the results obtained for vo above and below the transition
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Example 2.18
Determine the output waveform for the sinusoidal input.
Fig. 2.75
Determining the transition level for the clipper of Fig. 2.74.
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Fig. 2.76
Sketching vo for Example 2.18.
Example 2.19
Find the output voltage if the applied signal is the square wave.
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Fig. 2.78
vo at vi = +20 V.
Vo = Vi + 5 = 25V
Fig. 2.79
vo at vi = -10 V.
Vo = 0V
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Fig. 2.80
Sketching vo for Example 2.19.
Parallel Clippers
The diode in a parallel clipper
circuit “clips” any voltage that
forward bias it.
DC biasing can be added in
series with the diode to change
the clipping level.
80
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Fig. 2.81
Response to a parallel clipper.
Example 2.20
Determine Vo.
(1) 當0 < Vi < 4V 時，Diode is “on” ，Vo = 4V 。
(2) 當 Vi > 4V 時，Diode is “off” ，Vo = Vi 。
(3) 當 Vi < 0V 時， Diode is “on” ，Vo = 4V。
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Fig. 2.83
Determining the transition level for Example 2.20.
Fig. 2.84
Sketching vo for Example 2.20.
((1)) 當0 < Vi < 4V 時，Diode is “on”
，Vo = 4V 。
(2) 當 Vi > 4V 時，Diode is “off” ，
Vo = Vi 。
(3) 當 Vi < 0V 時， Diode is “on” ，
Vo = 4V。
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Repeat Example 2.20 using Vk = 0.7V
(1) 當0 < Vi < 4V 時，Diode is “on” ，Vo = 4 － 0.7 = 3.3 V 。
Fig. 2.86
Determining vo for the diode of Fig. 2.82 in the “on” state.
(2) 當 Vi > 4V 時，Diode is “off” ，Vo = Vi 。
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Fig. 2.87
Sketching vo for Example 2.21.
Summary of Clipper Circuits
88
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Clampers
A diode and capacitor
p
can be
combined to “clamp” an AC
signal to a specific DC level.
89
Biased Clamper Circuits
The input signal can be any type of
waveform such as sine, square, and
triangle waves.
The DC source lets you adjust the
DC camping level.
90
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Fig. 2.89
Clamper.
Fig. 2.90
Diode “on” and the capacitor charging to V volts.
Vo = 0V
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Fig. 2.91
Determining vo with the diode “off.”
Vo = −2V
Fig. 2.92
Sketching vo for the network of Fig. 2.91.
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Example 2.22 Determine Vo.
Vi = 10V
Vi = −20V
Fig. 2.94
Determining vo and VC with the diode in the “on” state.
Vi = −20V
Diode is “on” ，Vo = 5V 。
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Fig. 2.95
Determining vo with the diode in the “off” state.
Vi = 10V
Fig. 2.96
Diode is “off” ，Vo = 25 + 10 = 35 V 。
vi and vo for the clamper of Fig. 2.93.
Vi = −20V
Diode is “on” ，Vo = 5V 。
Vi = 10V
Diode is “off” ，Vo = 25 + 10 = 35 V 。
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Summary of Clamper Circuits
99
Zener Diodes
The Zener is a diode operated in
reverse bias at the Zener Voltage
((Vz)).
•
•
When Vi ≥ Vz
– The Zener is on
– Voltage across the Zener is Vz
– Zener current: IZ = IR – IRL
– Thee Zener
e e Power:
owe : PZ = VZIZ
When Vi < Vz
– The Zener is off
– The Zener acts as an open circuit
100
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Zener Resistor Values
If R is too large, the Zener diode cannot conduct because the available amount of
current is less than the minimum current rating, IZK. The minimum current is
given by:
ILmin = I R - I ZM
The maximum value of resistance is:
VZ
R Lmax =
I Lmin
If R is too small, the Zener current exceeds the maximum current
rating, IZM. The maximum current for the circuit is given by:
VL
VZ
=
RL
R Lmin
The minimum value of resistance is:
RVZ
R Lmin =
Vi − VZ
I Lmax =
101
Fig. 2.102
Approximate equivalent circuits for the Zener diode in the three possible regions of application.
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Example 2.24.
Determine the reference voltage Vo1 and Vo2.
What is the current through the LED and the power delivered by the supply?
How does the power absorbed by the LED compare to that of the 6-V Zener diode?
Vo1 = VZ 2 + VK = 3.3 + 0.7 = 4V
Vo 2 = Vo1 + VZ 1 = 4 + 6 = 10V
The 4-V across the white LED
I R = I LED =
40 − Vo 2 − VLED 40 − 10 − 4 26
=
=
= 20mA
1.3k
1.3k
1.3k
the power delivered by the supply
PS = E × I S = E × I R = (40V ) × (20mA) = 800mW
the power absorbed by the LED
PLED = VLED × I LED = (4V ) × (20mA) = 80mW
the power absorbed by 6-V Zener diode
PZ = VZ × I Z = (6V ) × (20mA) = 120mW
Example 2.25.
Check its operation and plot the waveform of voltage across the system
for the applied signal.
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Fig. 2.105
Response of the network of Fig. 2.104 to the application of a 60-V sinusoidal signal.
Fig. 2.106
Basic Zener regulator.
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Fig. 2.107
Determining the state of the Zener diode.
V = VL = Vi ×
Fig. 2.108
RL
R + RL
Substituting the Zener equivalent for the “on” situation.
VL = VZ
IR = IZ + IL
IZ = IR − IL
VL
PZ = VZ × I Z
RL
V − VL
V
IR = R = i
R
R
IL =
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Fig. 2.109
Zener diode regulator for Example 2.26.
Determine VL、VR、IZ and PZ
Fig. 2.110
Determining V for the regulator of Fig. 2.109.
V = VL =
1.2
1.2
× Vi =
×16 = 8.73V
1 + 1.2
1 + 1.2
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Fig. 2.111
Resulting operating point for the network of Fig. 2.109.
Zener diode is “off”
VZ = VL = 8.73V
VR = Vi − VL = 16 − 8.73 = 7.27V
Fig. 2.112
Network of Fig. 2.109 in the “on” state.
IZ = 0
PZ = I z × Vz = (0) × 8.73 = 0
I Z = I R − I L = 6 − 3 . 33 = 2 . 67 mA
PZ = I z × Vz = (2.67mA) ×10V = 26.7mW
3
3
× Vi =
×16 = 12V > VZ
1+ 3
1+ 3
VZ = VL = 10V VR = Vi − VL = 16 − 10 = 6V
V = VL =
Zener diode is “on”
IR =
6
10
= 6 mA I Z =
= 3 . 33 mA
1k
3k
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Example 2.27. Determine the range of RL and IL that will result in VRL
being maintained at 10V.
VZ =
RL
Vi
R + RL
R L min =
( R + R L )V Z = R L Vi
1 k × 10
RV Z
=
= 250 Ω
50 − 10
Vi − VZ
RV Z = R L V i − R L V Z = R L (V i − V Z )
Fig. 2.114
VL versus RL and IL for the regulator of Fig. 2.113.
V R = V i − V Z = 50 − 10 = 40 V
V
40
IR = R =
= 40 mA
R
1k
if
I ZM = 32 mA
I L min = I R − I ZM = 40 − 32 = 8 mA
VR
10
R L max =
=
= 1250 = 1 . 25 k Ω
I L min
8m
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Example 2.28. Determine the range of values of Vi that will maintain the
Zener diode in the “on” state.
Fig. 2.116
VZ =
RL
Vi
R + RL
Vi =
220 + 1200
R + RL
VZ =
× 20 = 23 . 67 V
R
220
VL versus Vi for the regulator of Fig. 2.115.
V i max − V Z
R
= I R max × R + V Z = 76 . 67 m × 220 + 20 = 36 . 87 V
I R max =
V i max
IL =
VL
20
=
= 16 . 67 mA
RL
1 .2 k
I ZM = 60 mA
I R max = I L + I ZM = 60 + 16 . 67 = 76 . 67 mA
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Voltage--Multiplier Circuits
Voltage
Voltage multiplier circuits use a combination of diodes and capacitors
to step up the output voltage of rectifier circuits.
•
•
•
Voltage Doubler
Voltage Tripler
Voltage Quadrupler
117
Voltage Doubler
This half-wave voltage doubler’s output can be calculated by:
Vout = VC2 = 2Vm
where Vm = peak secondary voltage of the transformer
118
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Voltage Doubler
•
Positive Half-Cycle
o D1 conducts
o D2 is switched off
o Capacitor C1 charges to Vm
•
Negative Half-Cycle
o D1 is switched off
o D2 conducts
o Capacitor C2 charges to Vm
Vout = VC2 = 2Vm
119
Fig. 2.119
cycle.
•
Double operation, showing each half-cycle of operation: (a) positive half-cycle; (b) negative half-
Positive Half-Cycle
o D1 conducts
o D2 is switched off
o Capacitor C1 charges to Vm
Vout = VC2 = 2Vm
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Fig. 2.119
cycle.
•
Double operation, showing each half-cycle of operation: (a) positive half-cycle; (b) negative half-
Negative Half-Cycle
Half Cycle
o D1 is switched off
o D2 conducts
o Capacitor C2 charges to Vm
Vout = VC2 = 2Vm
Fig. 2.120
Full-wave voltage doubler.
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Fig. 2.121
Alternate half-cycles of operation for full-wave voltage doubler.
Voltage Tripler and Quadrupler
124
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Fig. 2.122
Voltage tripler and quadrupler.
Practical Applications
•
Rectifier Circuits
– Conversions of AC to DC for DC operated circuits
– Battery Charging Circuits
•
Simple Diode Circuits
– Protective Circuits against
– Overcurrent
– Polarity Reversal
– Currents caused by an inductive kick in a relay circuit
•
Zener Circuits
– Overvoltage Protection
– Setting Reference Voltages
126
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Fig. 2.123
Battery charger: (a) external appearance; (b) internal construction.
Fig. 2.124
Electrical schematic for the battery charger of Fig. 2.123.
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Fig. 2.132 (a) Polarity protection for an expensive, sensitive piece of equipment; (b) correctly applied
polarity; (c) application of the wrong polarity.
Fig. 2.132 (continued) (a) Polarity protection for an expensive, sensitive piece of equipment; (b) correctly
applied polarity; (c) application of the wrong polarity.
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Fig. 2.132 (continued) (a) Polarity protection for an expensive, sensitive piece of equipment; (b) correctly
applied polarity; (c) application of the wrong polarity.
Fig. 2.133
Protection for a sensitive meter movement.
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Fig. 2.134 Backup system designed to prevent the loss of memory in a car radio when the radio is removed
from the car.
Fig. 2.135
Polarity dector using diodes and LEDs.
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Fig. 2.136
EXIT sign using LEDs.
Fig. 2.137
Providing different reference levels using diodes.
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Fig. 2.138 (a) How to drive a 6-V load with a 9-V supply (b) using a fixed resistor value. (c) Using a series
combination of diodes.
Fig. 2.138 (continued) (a) How to drive a 6-V load with a 9-V supply (b) using a fixed resistor value. (c)
Using a series combination of diodes.
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Fig. 2.138 (continued) (a) How to drive a 6-V load with a 9-V supply (b) using a fixed resistor value. (c)
Using a series combination of diodes.
70