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Chapter 2 : DIODE APPLICATIONSp
Norsabrina SihabNorsabrina SihabFaculty of Electrical Engineering,Universiti Teknologi MARAPulau PinangTel : 04-3823355Email : norsabrina@ppinang.uitm.edu.my
1
Chapter 2 – Diode Applications
Learning Outcome
2
Learning Outcome
At the end of this chapter, students able to:Explain and analyze the operation of half-wave rectifierExplain and analyze the operation of full-wave rectifierp a a d a a y e t e ope at o o u a e ect eExplain and analyze the operation of clipperExplain and analyze the operation of clamperExplain and analyze the operation of diode multiplierExplain and analyze the operation of diode multiplierExplain and analyze the operation of voltage regulatorExplain the basic DC power supply components
Electronics 1Norsabrina Sihab Updated May 2011
Chapter 2 – Diode Applications3
IntroductionIntroductionIt can conduct current in only ONE way direction and can act as switch (ON/OFF).2 di d diti ON & OFF t t2 diode conditions – ON & OFF state.2 basic conditions for diode in ON state.
Diode must in forward bias conditionl l b h h d d lVoltage supply, Vi must be greater than the diode voltage, VD
(Vi>VD)VSi=0.7V, VGe=0.3V and Videal diode=0V
Electronics 1Norsabrina Sihab Updated May 2011
Chapter 2 – Diode Applications4
IntroductionDiode in OFF state – act as open circuit. So I=0A.
Introduction
Diode in Series with DC SupplyCheck diodes whether ON or OFFRedraw diode equivalent circuit including others component.
Diode in Series with DC Supply
Electronics 1Norsabrina Sihab Updated May 2011
q g pApply KVL to determine current or voltage
Chapter 2 – Diode Applications5
Example 1Determine ID, VR and Vo.
Example 2Determine V V2 V and IDetermine ID, VR and Vo.
10VGe Si
+ Vo
5.6kID+ VR
Determine V1, V2, Vo and ID
Solution-5V
-
Solution
Electronics 1Norsabrina Sihab Updated May 2011
Chapter 2 – Diode Applications6
Diode in Parallel with DC supplyDiode in Parallel with DC supplyExample 3
Determine ID, Ix and Vo
+20V
IX
+ Vo
ID Si Ge
+ Vo
2.2k
- 5V
Solution
Electronics 1Norsabrina Sihab Updated May 2011
Chapter 2 – Diode Applications7
Diode As RectifierDiode As RectifierRectifier – convert AC to DC Voltage (not pure DC). Allow current flow in one direction only.2 types – 1) half wave rectifier, 2) full wave rectifier
1) Half Wave Rectifier)Normally used in non-critical low current applicationsMade up of a diode, D and a resistor, RHas abilit to cond ct c ent in one di ection and block c ent inHas ability to conduct current in one direction and block current in the other direction.
+
Vi
+
Vo
-
R
Electronics 1Norsabrina Sihab Updated May 2011
-
Basic rectifier circuit
Chapter 2 – Diode Applications8
Half Wave Rectifier (contd)Half Wave Rectifier (contd)Half-wave average voltage, Vdc
Is determine by calculate the area under the curve and divide it by the period of rectified waveformby the period of rectified waveform.
T
idc dttVT
V
1
).(1
20
= ∫π π
Vi
Vm
[ ]m
m
V
dtdttV
cos2
.0.sin21
0
0
−=
+= ∫ ∫
θπ
ωπ
π
π
[ ]
m
m
V
V )0cos(cos2
2
=
−−−= ππ
πVo
Vm
Vdc=0 318V
mdc VV 318.0=
=π
Where V i i ( k) l f AC lt
Vdc 0.318V
T
Electronics 1Norsabrina Sihab Updated May 2011
Vm is maximum (peak) value of AC voltageVdc is average value of rectified voltage
Chapter 2 – Diode Applications9
Half Wave Rectifier - OperationsHalf Wave Rectifier Operations
Positive half cycle of Vi
Negative half cycle of Vi
Electronics 1Norsabrina Sihab Updated May 2011
Chapter 2 – Diode Applications10
Half Wave Rectifier - Peak Inverse Voltage (PIV)
PIV l k k l ( )Also known as Peak Reverse Voltage (PRV)
Is maximum voltage across the diode in the direction to block current flow.Occurs at the peak of the negative alternation of the input cycle when diode is reverse biased.
KVL: Vm- Vo- PIV = 0Vm = PIV
Electronics 1Norsabrina Sihab Updated May 2011
Chapter 2 – Diode Applications11
2) Full Wave RectifierThe rectification process can be improved by using more diodes in a full-wave rectifier circuit.It can improve 100% of the DC level obtained from a sinusoidal
2) Full Wave Rectifier
It can improve 100% of the DC level obtained from a sinusoidal input. Full-wave rectification produces a greater DC output:
V )(⎟⎞
⎜⎛
mdc
mm
dc
VV
VVV
636.0
)318.0(22
=
=⎟⎠⎞
⎜⎝⎛=π
2 types – a) center tapped transformer, b) bridge network.
Electronics 1Norsabrina Sihab Updated May 2011
Chapter 2 – Diode Applications12
2a) Full Wave Center Tapped Rectifier) ppRequires
Two diodesCenter-tapped transformer
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Chapter 2 – Diode Applications13
Full Wave Center Tapped Rectifier - Operationspp pPositive half cycle of Vi – D1 ‘ON’ & D2 ‘OFF”
Vm
Vdc=0.636Vm
Vo
Negative half cycle of Vi – D1 ‘OFF’ & D2 ‘ON”T/2 T
Negative half cycle of Vi D1 OFF & D2 ON
Electronics 1Norsabrina Sihab Updated May 2011
Chapter 2 – Diode Applications14
Center Tapped Peak Inverse Voltage (PIV)KVL : - Vm + PIV – Vm=0
PIV = 2Vm
Center Tapped Peak Inverse Voltage (PIV)
-- PIV +
Vi=Vm
+R
Vi Vm
- Vm +
Example 4Calculates:Calculates:
i) the DC voltage obtained from a center tapped full wave rectifier for which the peak of rectified voltage is 100Vii) h PIV d l d h di dii) the PIV developed across the diode
Electronics 1Norsabrina Sihab Updated May 2011
Chapter 2 – Diode Applications15
2b) Full Wave Bridge Rectifier
Require four diodes, transformer and resistorVDC = 0.636 Vm
Advantages of FWBRnot used of the center tapped transformer and it requires anot used of the center-tapped transformer and it requires a maximum voltage of Vi across the transformer.PIV required of each diode is half of the center tapped full wave circuit
Electronics 1Norsabrina Sihab Updated May 2011
circuit.
Chapter 2 – Diode Applications16
Full Wave Bridge Rectifier (1) - OperationsFull Wave Bridge Rectifier (1) OperationsPositive half cycle of Vi : D2 & D3 – ‘ON’
D1 D2
D3 D4
Negative half cycle of Vi : D1 & D4 – ‘ON’
D1 D2D1 D2
Electronics 1Norsabrina Sihab Updated May 2011
D3D4
Chapter 2 – Diode Applications17
Full Wave Bridge Rectifier (2) - Operations
Positive half cycle of Vi : D1 & D2 – ‘ON’
Negative half cycle of Vi : D3 & D4 – ‘ON’
Electronics 1Norsabrina Sihab Updated May 2011
Chapter 2 – Diode Applications18
Full Wave Bridge Rectifier - Peak Inverse Voltage (PIV)Full Wave Bridge Rectifier Peak Inverse Voltage (PIV)
KVL : PIV- Vm=0m
PIV = Vm
Summary of Rectifier CircuitsSummary of Rectifier CircuitsSummary of Rectifier CircuitsSummary of Rectifier CircuitsRectifierRectifier VVDCDC PIVPIV
H lf W R tifi V 0 318(V ) VHalf Wave Rectifier VDCDC = 0.318(Vm) Vm
Bridge Rectifier VDCDC = 0.636(Vm) Vm
Electronics 1Norsabrina Sihab Updated May 2011
Center-Tapped Transformer Rectifier VDCDC = 0.636(Vm) 2Vm
Chapter 2 – Diode Applications19
Power Supplies (Voltage Regulators)Power Supplies (Voltage Regulators)Transformer – To step up or step down or isolate the input voltageRectifier – Diodes that convert the AC to DC voltage (half wave or full wave)full wave)Filter – Example capacitor. Use to filter the rectified signal from rectifier to produce a DC voltage with some ripples or AC voltage variationvariation.Regulator – To maintain a constant output voltage with less ripple but remain the same DC value.
Electronics 1Norsabrina Sihab Updated May 2011
Chapter 2 – Diode Applications20
FiltersFilters Is smoothing circuit which to obtain a smoother DC signalTypes – 1) RC filter, 2) L filter, 3) LC filter, 4) π-type filter
Basic Capacitor FilterBasic Capacitor FilterCapacitor is most commonly used of basic filter.Simply capacitor, C connected in parallel with the load resistance.
Electronics 1Norsabrina Sihab Updated May 2011
Chapter 2 – Diode Applications21
Ripples VoltageRipples VoltageRipples voltage occur when capacitor quickly charges at beginning of a cycle and slowly discharges after the positive peak (when reverse bias))DC voltage derive from an AC source signal after rectifying and filtering will have some ripples/variations.Output voltage known as ripples voltage, VrOutput voltage known as ripples voltage, Vr
Vr(p-p) – peak-to-peak ripples voltage
Electronics 1Norsabrina Sihab Updated May 2011
Approximate output voltage of capacitor filter circuit
Chapter 2 – Diode Applications22
Ripples Voltage (contd)DC voltage derive from an AC source signal after rectifying and filtering will have some ripples/variations.
Ripples Voltage (contd)
VVVd −=1
)(
FWfC
IV
VVV
dcm
pprmdc
→−=
= −
4
2 )(
AC DC
HWfC
IV dcm →−=
2 22: mp
rmsVV
Vwhere ==
dcdc VIV ==
Vr – variation in Vo due to charge and discharge
3434
)()(
)(
ppr
Lrmsr
VV
fCRfCV
−=
==
Electronics 1Norsabrina Sihab Updated May 2011
32)(rmsrV
Chapter 2 – Diode Applications23
Ripples Factor, rSmaller ripple respect to dc level meaning better filter circuit.
pp ,
Ripple factor, r
%100% XvoltageDC
rmsinvoltagerippler =32
)()(
pprrmsr
VVwhere −=
%100)( XV
VvoltageDC
dc
rmsr=
32
Ripples factor, r is also depends to the load, RL
Light load Vdc ≅ Vm. Therefore very small ripple factor and capacitor filter provides large DC voltage V
Electronics 1Norsabrina Sihab Updated May 2011
capacitor filter provides large DC voltage, Vdc.Heavy load Vdc < Vm. Therefore bigger ripple factor and capacitor filter provides smaller DC voltage, Vdc.
Chapter 2 – Diode Applications24
ExerciseExercise
1. What is the ripple factor of a sinusoidal signal having peak ripple (Vr(p)) of 2V on an average (Vdc) of 50V? Answer : 2.31%
2. The 35 Vrms ac voltage Vs is derived from an ideal 60Hz main transformer. It is connected to a half-wave rectifier and a 220μF capacitor to form a dc power supply. If the load RL draws an average current 0.15A, determine:
i. the peak-to-peak capacitor ripple voltage Answer : Vr(pp) = 5.6V
ii. the average or dc voltage across the load Answer : Vdc = 43.86V
iii. the percentage ripple factor, %r. Answer : 12.8%
Electronics 1Norsabrina Sihab Updated May 2011
Chapter 2 – Diode Applications25
CLIPPERSCLIPPERSBasically 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 waveform2. Circuit transient protection3. Detection2 types : 1) series clipper 2) parallel (shunt) clipper2 types : 1) series clipper, 2) parallel (shunt) clipper
1) Series Clipper2 types : a) negative series clipper b) positive series clipper2 types : a) negative series clipper, b) positive series clipperThe diode in a series clipperseries clipper circuit “clips” any voltage that does not forward bias it:
A bi i l it
Electronics 1Norsabrina Sihab Updated May 2011
A reverse-biasing polarityA forward-biasing polarity less than 0.7V for a silicon diode
Chapter 2 – Diode Applications26
1a) Negative Series Clipper1a) Negative Series ClipperClipped off half negative cycle. Diode forward bias during positive cycle of Vi.V is transition voltage (V V +V )VT is transition voltage. (VT=VDC+Vdiode)
During positive half cycleVT=Vdc+VD=4Vif Vi ≤ VT diode will OFFif Vi ≤ VT diode will OFF. Vo=0V.If Vi > VT diode will ON. KVL : Vi – 4 – Vo =0.
Electronics 1Norsabrina Sihab Updated May 2011
oVo=Vi-VT=16V
Chapter 2 – Diode Applications27
1a) Negative Series Clipper (contd)
During negative half cycle.d f ll l f
1a) Negative Series Clipper (contd)
Final outputDiode is OFF for all value of Vi.VO=0V.
Vi
20
VT=4V
- 20
VT 4V
Vo
166
Electronics 1Norsabrina Sihab Updated May 2011
Chapter 2 – Diode Applications28
1b) Positive Series Clipper1b) Positive Series ClipperClipped off half positive cycle. Diode forward bias during negative cycle of Vi.
During positive half cycleDiode is OFF for all value of Vi.VO=0V.
Electronics 1Norsabrina Sihab Updated May 2011
Chapter 2 – Diode Applications29
1b) Positive Series Clipper (contd)
During negative half cycle.V 4 V 4V
1b) Positive Series Clipper (contd)
Final outputVT=- 4-VD=- 4Vif lVil ≤ lVTl diode OFF. Vo=0V.If lVil > lVTl diode ON.
Vi
20
KVL : Vi +Vo- 4 =0. Vo= - Vi+4=-20+4=-16V
- 20
VT= - 4V
Vo
-16
Electronics 1Norsabrina Sihab Updated May 2011
Chapter 2 – Diode Applications30
2a) Negative Parallel ClipperThe operation is opposite series clipper.
2a) Negative Parallel Clipper
Vi
20
R
+ +20
Si
5V
+
Vi
-
+
Vo
-
During positive half cycle
- 20
Diode is OFF for all value of Vi.VO=Vi=20V
Vi
R
+ +i
20 Si
5V
+
Vi
-
+
Vo
-
Electronics 1Norsabrina Sihab Updated May 2011
Chapter 2 – Diode Applications31
2a) Negative Parallel Clipper (contd)2a) Negative Parallel Clipper (contd)During negative half cycle
VT = -0.7-5 = -5.7Vf l l l l d d
Final output
if lVil ≤ lVTl diode OFF. Vo=Vi
If lVil> lVTl diode ON. i T
KVL : Vo +0.7+5 =0Vo = VT = -5.7V
Vi
0.7V
R
- +
- 20 5V
Vi
+
Vo
-
Electronics 1Norsabrina Sihab Updated May 2011
Chapter 2 – Diode Applications32
2b) Positive Parallel Clipper2b) Positive Parallel ClipperVi
20
Si
R
+ +
- 20
Si
5V
Vi
-
Vo
-
During positive half cycleVT- Vdc- VD = 0.
VT=5.7VVi
R
+ +T
If Vi ≤ VT diode OFF. Vo=Vi.If Vi > VT diode ON. KVL : V -0 7-5 =0
20
0.7
5V
+
Vi
+
Vo
KVL : Vo-0.7-5 =0. Vo=5.7V - 20
- -
Electronics 1Norsabrina Sihab Updated May 2011
Chapter 2 – Diode Applications33
2b) Positive Parallel Clipper (contd)During negative half cycle.
Diode is OFF for all value of Vi.
2b) Positive Parallel Clipper (contd)Final output
VO=Vi. Vi
20
VT
- 20
Vo
20
- 20
Electronics 1Norsabrina Sihab Updated May 2011
Chapter 2 – Diode Applications34
Test 1 Apr 2010 Q2cTest 1 Apr 2010 Q2c
Electronics 1Norsabrina Sihab Updated May 2011
Chapter 2 – Diode Applications35
Combination of Negative and Positive Parallel ClipperCombination of Negative and Positive Parallel ClipperVi
10
Si
R
+ +
Ge
10
Si
5V
Vi
-
Vo
-
Ge
7.7V
During positive half cycleDG OFF for all value of Vi
- 10
DGe OFF for all value of Vi
DSi ON conditionallyVT=VDSi+5=5.7VIf V ≤ V D OFFIf Vi ≤ VT DSi OFF. Vo=Vi.If Vi > VT DSi ON. Vo=5.7V
Electronics 1Norsabrina Sihab Updated May 2011
o
Chapter 2 – Diode Applications36
Combination of Negative and Positive Parallel Cli ( d)Clipper (contd)
Final Output
Vi
10
V 5 7V
p
During negative half cycle 10
VT=5.7V
VT= - 8V
During negative half cycleDSi OFF for all value of Vi
DGe ON conditionallyVT = -VDGe-7.7 = -8V
- 10
Vo
10T DGe 8
If |Vi| ≤ |VT| DGe OFF. Vo=Vi.If |Vi| > |VT| DGe ON. -8V
5.7V
Electronics 1Norsabrina Sihab Updated May 2011
| i| | T| Ge Vo+0.3+7.7=0Vo=-8V
- 10
Chapter 2 – Diode Applications
Test 1 August 2009 Q3b
37
Test 1 August 2009 Q3b
Electronics 1Norsabrina Sihab Updated May 2011
Chapter 2 – Diode Applications38
Summary of Clipper CircuitSummary of Clipper Circuit
Electronics 1Norsabrina Sihab Updated May 2011
Chapter 2 – Diode Applications39
CLAMPERSCLAMPERSTo clamp or shift a signal to a different DC levelCircuit consist of C,D and R
1) Negative Clamper
During positive half cycleDuring positive half cycleStep 1: Find polarity of VC
Step 2: Determine VO using KVL at o/pV V V 0Vo – VD+VDC= 0Vo=0.7-5= - 4.3V
Step 3: Determine value of VC
Electronics 1Norsabrina Sihab Updated May 2011
Vi-Vc-Vo=0Vc=24.3V
Chapter 2 – Diode Applications40
1) Negative Clamper (contd)1) Negative Clamper (contd)During negative half cycle
Vi
Final Output
20
Step 1: Determine Vo using KVL at i/p Vo
- 20
Vi+Vo+Vc=0Vo=–Vi–Vc= – 20 – 24.3= – 44.3V
- 4.3V
- 44.3
Electronics 1Norsabrina Sihab Updated May 2011
Chapter 2 – Diode Applications41
2) Positive Clamper2) Positive ClamperVi
20
Si
++
C
- 20
5V
RVi
-
Vo
-
During negative half cycle (because Diode ON at this cycle)Step 1: Find polarity of VCStep 2: Determine VO using KVL at o/p
Vo+ VD- VDC= 0Vo= 5 - 0.7= 4.3V
Step 3: Determine VC using KVL at i/pV+V V V 0Vi+VDC–VD–VC=0VC=Vi+VDC–VD=24.3V
Electronics 1Norsabrina Sihab Updated May 2011
Chapter 2 – Diode Applications42
2) Positive Clamper (contd)2) Positive Clamper (contd)During positive half cycle
Step 1: Find polarity of Vo Vi
Final Output
Vi+Vc-Vo=0Vo=20+14.3= 44.3V
i
20
V
- 20
Vo
44.3
4.3
Electronics 1Norsabrina Sihab Updated May 2011
Chapter 2 – Diode Applications43
Example – Design a ClamperDesigned a clamper circuit to produce output voltage, Vo. Use silicon diode in your design.
KVL: +Vo-0.7-5=0Vo=+5.7V
KVL:Vi – Vc – Vo=0Vc=Vi-Vo=15 - 5.7=9.3V
SolutionD i iti l
During negative cycle
During positive cyclePropose design clamper circuit which D ‘ON’ during positive cycle V =5 7Vcycle. Vo=5.7V
KVL:+Vi+Vo+Vc=0
Electronics 1Norsabrina Sihab Updated May 2011
i o c
Vo= - 15 - 9.3= - 24.3V
Chapter 2 – Diode Applications44
Summary of Clamper CircuitSummary of Clamper Circuit
Electronics 1Norsabrina Sihab Updated May 2011
Chapter 2 – Diode Applications
Test 1 August 2010 Q3b
45
Test 1 August 2010 Q3b
Electronics 1Norsabrina Sihab Updated May 2011
Chapter 2 – Diode Applications46
Voltage MultiplierVoltage Multiplier
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.
Electronics 1Norsabrina Sihab Updated May 2011
Chapter 2 – Diode Applications47
Voltage DoublerHalf-wave voltage doubler
Voltage Doubler
D1VP
C2D2 2VP
C1
• During positive cycle : D1 on, D2 Off. C1 charged to the Vp.
• During negative cycle : D2D ff C ’ di hon, D1 off. C1 can’t discharge
so Vp on C1 adds the supply voltage to charge C2 to approximately 2Vp.
Electronics 1Norsabrina Sihab Updated May 2011
approximately 2Vp.
Chapter 2 – Diode Applications48
T1 Feb09 Q3aT1 Feb09 Q3a
Electronics 1Norsabrina Sihab Updated May 2011
Chapter 2 – Diode Applications49
Full-wave voltage doublerD
VDC1
D1
VP
VP
2VP
D21
C2P
• During positive cycle : D1 on, D2 Off. C1 charged to the Vp.
• During negative cycle : D2on, D1 off. C2 charges to approx V Output across theapprox Vp. Output across the two series capacitors approximately 2Vp.
Electronics 1Norsabrina Sihab Updated May 2011
Chapter 2 – Diode Applications50
Voltage Doubler (contd)Voltage Doubler (contd)
Electronics 1Norsabrina Sihab Updated May 2011
Chapter 2 – Diode Applications51
Voltage TriplerVoltage Tripler
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.
Electronics 1Norsabrina Sihab Updated May 2011
Chapter 2 – Diode Applications52
Voltage QuadruplerVoltage 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.p
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.
Electronics 1Norsabrina Sihab Updated May 2011
Chapter 2 – Diode Applications53
ExerciseExercise
Figure below shows another kind of power supply known as voltage lti limultiplier.i. Determine the peak values of output voltages available at the
two terminals indicated as Vo1 and Vo2 if the secondary volatage of the transformer is 120 Vrms.
ii C l l t th i d PIV ti f h di d i thi ltii. Calculate the required PIV rating of each diode in this voltage multiplier circuit.
Show suitable equivalent circuits and equations to justify the calculations. Assume ideal diodes.
(8 marks)
Electronics 1Norsabrina Sihab Updated May 2011
Chapter 2 – Diode Applications54
ZENER REGULATORZENER REGULATORIDZ is opposite from ID which is designed to work in reverse bias.
+
V
Application : Regulator -
VZ > V > 0DZ will OFF
Electronics 1Norsabrina Sihab Updated May 2011
Chapter 2 – Diode Applications55
ZENER REGULATOR (contd)ZENER REGULATOR (contd)Simplest regulator as shown in figure below.3 conditions of Vi and load resistance, RL to maintain designed zener voltage:1. Vi and RL fixed2. Vi fixed and RL variablei L
3. Vi variable and RL fixed
Electronics 1Norsabrina Sihab Updated May 2011
Chapter 2 – Diode Applications56
1. Fixed Vi and Fixed RL1. Fixed Vi and Fixed RLStep 1: Determine the state of the zener diode by removing it from the network Calculate voltage across the
Step 2: Substitute appropriate equivalent circuit
network. Calculate voltage across the resulting open circuit.
zL VV =
LiRL
LRZ
LZR
VVVIdVIh
IIIIIIKCL
−−−−=
+=)1(
:
)(
:
ONDVVifRRVRVV
VDR
L
iLL +==
zzz
LiRR
L
LL
VIPdiodezenerbydissipatedpower
RRIand
RIwhere
=
===
:
Electronics 1Norsabrina Sihab Updated May 2011
)()(
OFFDVVifONDVVif
ZZ
ZZ
<≥ zzz
Chapter 2 – Diode Applications57
2. Fixed Vi and Variable RL2. Fixed Vi and Variable RLSpecific range of RL to turn ON DZ
LL maximumisIthereforeminimum,isRsince
== ZLL
VVI
maxL
ZLmin
minmax
RVIand =
LLL RR
I
VDR :
iLLZ
L
iLL
VRRRVRRVRVV
VDR
=++
==
)(
:Once DZ ON, VR remains fixed
RR fixedalsoIfixed,isVsince=− RZi VVV
iLL
Z
iLLZ
R
VRRRV
VRRRV
=+
+
)(
)(R
:RV
+=
=
LZR
R
IIIKCL
I
Z
Zi
L
iL
Z
VVV
RR
VRRV
−=
=+ )1(
RLminZmax
LmaxZmin
constantisIbecauseIwhenIandIwhenIresulting
−= LRZ IIIso
Electronics 1Norsabrina Sihab Updated May 2011
Zi
ZL
ZL
VVRVRso−
=min Lmin
ZLmaxZMRLmin I
VR&I-II ==
Chapter 2 – Diode Applications58
3. Fixed RL and Variable Vi3. Fixed RL and Variable ViVi must be sufficiently large to turn DZ
Mi lt t t ON i
The max of Vimax is limited by the max zener current, IZM
Min voltage to turn ON is Vi=Vimin
VDR : ZRi
LZMR
VVV
III
+=
+=
maxmax
max
L
iLZL RR
VRVV
VDR
+==
:
ZL
i
iLLZ
VR
RRV
VRRRV+
=
=+
min
)(
LR
Electronics 1Norsabrina Sihab Updated May 2011
Chapter 2 – Diode Applications59
ExerciseExercise1) Determine VL,VR, IZ and PZ.Answer: 10V, 10V,6.3mA, 63mW
3) Determine range of Vi that will maintain zener diode in ON state A 23 67V 36 8VON state. Answer: 23.67V ~ 36.8V
2)
a) Determine the range of RL and IL that will result in VRL being maintained at 10V Answer: 250 Ω ~ 1 2kΩ 8mA ~ 40mA
Electronics 1Norsabrina Sihab Updated May 2011
1.2kΩ, 8mA ~ 40mA
b) Determine the max wattage rating of DZ. Answer: 32mW
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