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7/29/2019 Pertemuan6 Diode Aplc
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Clippers & Clampers
Zener diode application
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Clippers
Clippers or diode limiting is a diode network that have the ability to
clip off a portion on the i/p signal without distorting the remaining part
of the alternating waveform.
Clippers are used to eliminate amplitude noise or to fabricate new
waveforms from an existing signal.
2 general of clippers:
a) Series clippers
b) Parallel clippers
Series Clippers
The series configuration is defined as one where the diode is in series
with the load.
A half-wave rectifier is the simplest form of diode clipper-one resistorand diode.
+
-
ViR
L Vo
-
+
2
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The diode clips any voltage that does not put it in forward
bias. That would be a reverse biasing polarity and a voltage less
than 0.7V for a silicon diode.
Clipper Diode Circuit
3
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VmVi
t0 T/2 T
+
-Vi
V
RL Vo
-
+
The half-wave rectifier with addition of dc supply is shown infollowing figure. The cct known as biased series clipper. The dcsupply have pronounced effect on the o/p of a clipper.
Our initial discussion will be limited to ideal diode.
Biased series clipper
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+ve region turn the diode ON.
-ve region turn the diode OFF.
Vi > V to turn ON the diode
In general diode is open cct (OFF state) and short cct (ON state)
For Vi > V the Vo = ViV
For Vi = V the Vo= 0 V
The complete cct shown above
Vm
Vi
t0 T/2 T
+
-
Vi=Vm
V
RL Vo
-
+
+ -
T/20
Vo
tT
Vm-V
Vi=V (diodes change state)
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Determine the o/p waveform for the network below:
Solution:
Example: Variations of the Clipper Circuit
6
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More Example:
Repeat previous example for the square-wave i/p.
+
-
Vi
V=5 V
RL Vo
-
+
+-Vi
t0 T/2 T
-10
20
7
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Parallel Clippers The diode connection is in parallel configuration
with the o/p. Diode is ideal
+
RL
V
-
+
-
Vi
8
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By taking the output across the diode, the output is now thevoltage when the diode is not conducting.
A DC source can also be added to change the diodes required
forward bias voltage.
Changing Output Perspective
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Example :
Determine the Vo and sketch the o/p waveform for
the below network
+
V=4 V
RL
Vo
-
+
-
Vi
10
t0 T/2 T
Vi
16
-16
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V=VoON4
Vi=VoOFF5
Vi=VoOFF6
Vi=VoOFF7
Vi=VoOFF16
V=VoON3
V=VoON2
V=VoON1
VoDiode stateVi
+
V=4 V
RL
Vo
-
+
-
Vi
+
V=4 V
RL
Vo
-
+
-
Vi
t0 T/2
Vo
16
4
Solution:
+ ve region
t0 T/2 T
Vi
16
-16
11
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Solution (continued):
- ve region
+V=4 V
VRL=0A
Vo
-
+-Vi id=0A
V=VoON-4
V=VoON-5
V=VoON-6
V=VoON-7
V=VoON-16
V=VoON-3
V=VoON-2
V=VoON-1
VoDiode stateVi
t0 T/2
Vo
16
4
T
t0 T/2 T
Vi
16
-16
12
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Example :
Repeat the previous example using a silicon diode with VD=0.7 V
+
V=4 V
RL
Vo
-
+
-Vi
VD=0.7 V
id=0A
V3.3
7.04
VVVi
0VVVi
D
D
Solution:
13
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Solution (continued):
For i/p voltages greater than 3.3 V the diode open cct and Vo=Vi.
For i/p voltages less than 3.3 V the diode short cct and the network result
as/;
+V=4 V
RL
Vo
-
+
-
Vi
VD=0.7 V
id=0AV3.3
7.04Vo
t0 T/2
Vo16
3.3
T
The resulting o/p waveform
14
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15
Clipper Circuits Summary
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Clipper Circuits Summary
16
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Clampers
The clamping network is to clamp a signal to a different dc level. Also
known as dc restorers. The clamping cct is often used in TV receivers as adc restorer.
The network consists of:
a) Capacitor
b) Diode
c) Resistive elementd) Independent dc supply (option)
The magnitude of R and C must be chosen such that the time constant
= RC is large enough to ensure that the voltage across the capacitor does
not discharge significantly during the interval the diode is nonconducting.
Our analysis basis that all capacitor is fully charge and discharge in 5
time constant.
17
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18
+
R Vo
-
+
-
Vi
Vi
t0 T/2 T
V
-V
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19
Operation of clamper+ ve region
+
RVo
-
+
-
Vi
C 0 - T/2: Diode is ON state (short-cct
equivalent)
Assume RC time is small and capacitorcharge to V volts very quickly
Vo=0 V (ideal diode)
- ve region T/2 T: Diode is OFF state (open-cctequivalent)
Both for the stored voltage across
capacitor and applied signal current
through cathode to anode KVL: - V- V- Vo = 0 and
Vo = -2V
+
R Vo
-
+
-
V
C
Vo
-
+V
Vi
t0 T/2 T
V
-V
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20
Tips : Clamping networkTotal swing o/p signal = the total swing i /p signal
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21
Example: Determine Vo for the network below:
+
R=100 kohmVo
-
+
-
Vi
C= 1 uF
V= 5 V
Vi
-20
10
T
0 t1 t2 t3 t4 t
f=1000 Hz
S l ti
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22
Solution:
Step 1:Consider the part of i/p signal that will forward bias the
diode. From network (t1 -t2:-ve region)+
R=100 kohm Vo
-
+-
20 V
Vc
5 V
Step 2:During ON state assume capacitor will charge to a voltage level
determined by the network. Find the store voltage capacitor & obtained
Vo
KVL: -20 +Vc5 = 0
Vc = 25v Vo = 5
Vi
-20
10
T
0 t1 t2 t3 t4t
f=1000 Hz
S l ti ( td)
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23
Solution (cntd):
Step 3:During OFF state assume capacitor will hold on its established
voltage level. From network (t2 -t3:+ve region)
+
R=100 kohm Vo
-
+-
10 V
C
Vc
5 V
KVL
Step 4:Obtained Vo
ms500ms)5(1005istimedischargetotalThe
ms100k)(1u)(100RC:by
determinedisgdischarginofconstantTime
V35Vo0Vo2510
0VoVc10:KVL
Vi
-20
10
T
0 t1 t2 t3 t4t
f=1000 Hz
Solution (cntd):
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Step 5:Checking!!! total swing o/p signal = total swing i/p signal
From network (t2 -t3: +ve region)Vo
35
0 t1 t2 t3 t4t
5
Solution (cntd):
Vi
-20
10
T
0 t1 t2 t3 t4t
f=1000 Hz
24
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Example:Repeat the previous
example using agermanium diode !!!
25
Summary of Clamper Circuits
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Summary of Clamper Circuits
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Zener Diodes
The zener diode is a special type of diodes that is designed to
work in the reverse breakdown region. But it also can operate in the forward bias region.
Zener diode is a main component to design voltage regulator
circuit for DC power supply.
Zener Diodes Characteristic
The I-V characteristics of a diode in Fig 3.19 shows that the
breakdown voltage of a diode is nearly constant over a wide-
range of reverse-bias currents.
27
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Operation region:
Forward bias-operate same as
normal diode
Reverse bias-small current flow
Breakdown-big current flow. This is
the region where the voltage zener is
constant
For normal diode breakdown voltage is
capable to destroy the diode but with
zener diode the current is limited by
connecting series resistor.
0
I
V
Forward
ReverseBreakdown
-Vz0.7
Fig. 3.19
28
Zener Diode
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The Zener is a diode operated in reverse bias at the Zener
Voltage (Vz).
Zener Diode
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Vi and R fixed Fixed DC voltage is applied in network
below, as is the load resistor.
The analysis can be determined with 2
steps.
30
Step 1:
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p
Determine the state of zener diode by removing it from
the network and calculating the voltage across the
resulting open cct.
L
L
L
RR
ViRVV
LR
R
RR
RVi
V:OFFdiodeZenerVzV
Vz-ViV:ONdiodeZenerVzV
Vz
Iz+
- Pzm
RL
Vi
Fig 3.20: Basic zener regulato
V+
-
RL
Vi
Fig 3.21: Determining the state othe zener diode
RV
L
+
-
31
Step 2:
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Subtitute the appropriate equivalent cct and solve for the
desired unknowns.
The ON state will obtained the equivalent cct in
Fig below
LZR III:KCL LRZ III
Vi
R
VIdan
R
VIwhere
R
R
L
L
L
devicefor thespecifiedPPzwhich
VzIzPzzenerbydissipatedPower
ZM
VZ
+
-
RLVi
Fig 3.22
R
VL
+
-
IR
PZM
IZ
IL
32
E l
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33
Example:
a) Determine VL, VR, IZ and PZ in the network below.
b) Repeat part (a) with RL=3 k
VZ=10 V+
-
RL
Vi R=1 k ohmVL
+
-PZM=30 mW
IZ
1.2 kohm
VR+ -
Solution:
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So ut o :
Step 1: Remove zener diode & obtained the zener state
W0VzIzPz
0AIz
7.27V8.73-16V-ViV
8.73VVV
cct)(openstateOFFisdiodeVzV
V73.8k2.1k1
16k2.1
RR
ViRV
LR
L
L
L
V
+
-
RL
Vi=16 V
R=1 k ohm
VL
+
-
IZ
1.2 kohm
IR
0
I
V
Forward
ReverseBreakdown
-Vz0.7
Fig. 3.19
8.73
10
Resulting operating point
34
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Repeat part (a) with RL=3 k
35
Fixed Vi Variable R
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Fixed Vi, Variable RL For an offset Vz a specific range of resistor values need
to be choose to ensure zener diode is ON state.
Too small a load resistance will cause VL < Vz - diode is
OFF state.
Thus the minimum load resistance in previous example
need to be calculate. This can be expressed by the
equation below:
VzVi
RVzRminL
RL > RLmin zener diode is ON state diode canreplaced by Vz source equivalent
36
With R changes the I also changes The table below
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With RL changes the IL also changes. The table belowdescribed relationship between RL ,IL,IR
minL
maxL
R
VzI
maxL
minL
R
VzI
zminLmaxR III
zmaxLminR III
Notes: Izmax = Izm
37
Example:
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p
a) For the network below, determine the range of RL and IL
that will result in VRL being maintained at 10 V.
b) Determine the maximum wattage rating of the diode.
RLVi=50 V
R=1 k ohm IZ
IR
+
-
Vz=10 VIzm = 32 mA
IL
38
Solution:(a)
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k1.25R250:rangeRThe
k1.25m8
10
I
Vz
R
mA8m32-m40I-II
mA40k1
40RVI
25040
k10
1050
10k1
Vz-Vi
RVzR
LL
LminLmax
RLmin
RR
Lmin
zmax
Solution: (b)
320m3210VzIP zmaxzmax
39
Fixed RL Variable Vi
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Fixed RL, Variable Vi For a fixed values of RL in the network below, Vi must
be sufficiently LARGE to turn zener diode ON. The minimum turn-ON voltage Vi=Vimin is expressed
by :
Vz
Iz+
- Pzm
RL
Vi
R
L
L
L
LL
R
VzRRVimin
RR
ViRVzV
40
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The maximum value of Vi is limited by the maximumzener current, Izmax, thus
LIII zmaxRmax
RImaxVi maxR
IL is fixed at Vz/RL and Izmax is the maximum value sothe maximum value of Vi is expressed below:
41
Example:
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Determine the range of values of Vi that will maintain the
zener diode of network below:
RL=1.2k ohmVi
R=220 ohm IZ
IR
+
-
Vz=20 V
Izm = 60 mA
IL
VL
+
-
42
Solution:
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V36.87ViV23.67:ViofrangeThe
V87.3620220m67.76VzRIVimax
mA76.6716.67m60mIIzmaxI
mA16.671200
20
R
Vz
R
VI
V23.671200
202201200
R
VzRRVimin
maxR
LRMax
LL
LL
L
L
RL=1.2k ohmVi
R=220 ohm IZ
IR
+
-Vz=20 V
Izm = 60 mA
IL
VL
+
-
Equivalent cct in the ON state
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Practical Applications of Diode Circuits
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pp
Rectifier CircuitsConversions of AC to DC for DC operated circuits
Battery Charging Circuits
Simple Diode Circuits
Protective Circuits against OvercurrentPolarity Reversal Currents caused by an inductive kick
in a relay circuit
Zener Circuits
Overvoltage Protection
Setting Reference Voltages
45
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References:
1. Thomas L. Floyd, Electronic Devices, Eighth Edition, Prentice Hall, 2002.
2. Robert Boylestad, Electronic Devices and Circuit Theory, Seventh edition,Prentice Hall, 2002.
3. Puspa Inayat Khalid, Rubita Sudirman, Siti Hawa Ruslan,ModulPengajaran Elektronik 1, UTM, 2002.
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