Get documents about "Chapter 2
Document Sample
Chapter 2 : DIODES
Chapter 2 – Diode Applications
Chapter 2: Diodes 2
Introduction
It can conduct current in only ONE way direction
and can act as switch (ON/OFF).
2 diode conditions – ON & OFF state.
2 basic conditions for diode in ON state.
Diode must in forward bias condition
Voltage supply, Vi must be greater than the
diode voltage, VD (Vi>VD)
Norsabrina Sihab KEE112 Electronics 1 Copyright@JUL2007
Chapter 2 – Diode Applications
Chapter 2: Diodes 3
Introduction
VSi=0.7V, VGe=0.3V and Videal diode=0V
Norsabrina Sihab KEE112 Electronics 1 Copyright@JUL2007
Chapter 2 – Diode Applications
Chapter 2: Diodes 4
Introduction
Diode in OFF state – act as open circuit. So
I=0A.
Norsabrina Sihab KEE112 Electronics 1 Copyright@JUL2007
Chapter 2 – Diode Applications
Chapter 2: Diodes 5
Diode in Series with DC Supply
Check diodes whether ON or OFF
Redraw diode equivalent circuit including others
component.
Apply KVL to determine current or voltage
Norsabrina Sihab KEE112 Electronics 1 Copyright@JUL2007
Chapter 2 – Diode Applications
Chapter 2: Diodes 6
Example 1
Determine ID, VR and Vo.
Ge Si
10V + Vo
+
ID
5.6k VR
-
-5V
Norsabrina Sihab KEE112 Electronics 1 Copyright@JUL2007
Chapter 2 – Diode Applications
Chapter 2: Diodes 7
Example 2
Determine V1, V2, Vo and ID
Ge Si ID
10V + Vo
+ V1 - + V2 -
2.2k
Norsabrina Sihab KEE112 Electronics 1 Copyright@JUL2007
Chapter 2 – Diode Applications
Chapter 2: Diodes 8
Diode in Parallel with DC supply
Example 3 +20V
Determine ID and Vo
Ge Si ID
+ Vo
2.2k
- 5V
Norsabrina Sihab KEE112 Electronics 1 Copyright@JUL2007
Chapter 2 – Diode Applications
Chapter 2: Diodes 9
Diode As Rectifier
Rectifier – 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
Norsabrina Sihab KEE112 Electronics 1 Copyright@JUL2007
Chapter 2 – Diode Applications
Chapter 2: Diodes 10
1) Half Wave Rectifier
Normally used in non-critical low current
applications
Made up of a diode, D and a resistor, R
Has ability to conduct current in one direction
and block current in the other direction.
Norsabrina Sihab KEE112 Electronics 1 Copyright@JUL2007
Chapter 2 – Diode Applications
Chapter 2: Diodes 11
1) Half Wave Rectifier
+ +
R Vo Basic rectifier circuit
Vi
-
-
Norsabrina Sihab KEE112 Electronics 1 Copyright@JUL2007
Chapter 2 – Diode Applications
Chapter 2: Diodes 12
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
waveform.
Norsabrina Sihab KEE112 Electronics 1 Copyright@JUL2007
Chapter 2 – Diode Applications
Chapter 2: Diodes 13
Half Wave Rectifier (contd)
Vi
T
1
Vdc Vi (t ).dt Vm
T 0
2
1
Vm sin t.dt 0.dt
2 0
Vm
cos Vo
2 0
Vm
Vm
cos ( cos0) Vdc=0.318V
2
Vm
Where
T
Vm is maximum (peak) value of AC voltage
Vdc 0.318Vm
Vdc is average value of rectified voltage
Norsabrina Sihab KEE112 Electronics 1 Copyright@JUL2007
Chapter 2 – Diode Applications
Chapter 2: Diodes 14
Half Wave Rectifier - Operations
Positive half cycle of Vi
Norsabrina Sihab KEE112 Electronics 1 Copyright@JUL2007
Chapter 2 – Diode Applications
Chapter 2: Diodes 15
Half Wave Rectifier - Operations
Negative half cycle of Vi
Norsabrina Sihab KEE112 Electronics 1 Copyright@JUL2007
Chapter 2 – Diode Applications
Chapter 2: Diodes 16
Half Wave Rectifier - Peak Inverse Voltage (PIV)
PIV
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.
Norsabrina Sihab KEE112 Electronics 1 Copyright@JUL2007
Chapter 2 – Diode Applications
Chapter 2: Diodes 17
Half Wave Rectifier - Peak Inverse Voltage (PIV)
KVL:
Vm- Vo- PIV = 0
Vm = PIV
- PIV +
- -
I=0A
Vi=Vm R Vo=IR=(0)R=0V
+ +
Norsabrina Sihab KEE112 Electronics 1 Copyright@JUL2007
Chapter 2 – Diode Applications
Chapter 2: Diodes 18
2) Full Wave Rectifier
The 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 input.
Full-wave rectification produces a greater DC
output:
Norsabrina Sihab KEE112 Electronics 1 Copyright@JUL2007
Chapter 2 – Diode Applications
Chapter 2: Diodes 19
2) Full Wave Rectifier
Vm
Vdc 2 2(0.318Vm )
Vdc 0.636Vm
Norsabrina Sihab KEE112 Electronics 1 Copyright@JUL2007
Chapter 2 – Diode Applications
Chapter 2: Diodes 20
2) Full Wave Rectifier
2 types –
a) center tapped transformer,
b) bridge network.
Norsabrina Sihab KEE112 Electronics 1 Copyright@JUL2007
Chapter 2 – Diode Applications
Chapter 2: Diodes 21
2a) Full Wave Center Tapped Rectifier
Requires
Two diodes
Center-tapped transformer
Norsabrina Sihab KEE112 Electronics 1 Copyright@JUL2007
Chapter 2 – Diode Applications
Chapter 2: Diodes 22
Full Wave Center Tapped Rectifier - Operations
Positive half cycle of Vi – D1 ‘ON’ & D2 ‘OFF”
Vo
Vm
Vdc=0.636Vm
Negative half cycle of Vi – D1 ‘OFF’ & D2 ‘ON”
T/2 T
Norsabrina Sihab KEE112 Electronics 1 Copyright@JUL2007
Chapter 2 – Diode Applications
Chapter 2: Diodes 23
Center Tapped Peak Inverse Voltage (PIV)
KVL : - Vm + PIV – Vm=0
PIV = 2Vm - PIV +
-
Vi=Vm
R
+
- Vm +
Norsabrina Sihab KEE112 Electronics 1 Copyright@JUL2007
Chapter 2 – Diode Applications
Chapter 2: Diodes 24
Center Tapped Peak Inverse Voltage (PIV)
Example 4
Calculates:
i) the DC voltage obtained from a center tapped
full wave rectifier for which the peak of
rectified voltage is 100V
ii) the PIV developed across the diode
Norsabrina Sihab KEE112 Electronics 1 Copyright@JUL2007
Chapter 2 – Diode Applications
Chapter 2: Diodes 25
2b) Full Wave Bridge Rectifier
Require four diodes, transformer and resistor
VDC = 0.636 Vm
Norsabrina Sihab KEE112 Electronics 1 Copyright@JUL2007
Chapter 2 – Diode Applications
Chapter 2: Diodes 26
2b) Full Wave Bridge Rectifier
Advantages of FWBR
not 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.
Norsabrina Sihab KEE112 Electronics 1 Copyright@JUL2007
Chapter 2 – Diode Applications
Chapter 2: Diodes 27
Full Wave Bridge Rectifier (1) - Operations
Positive half cycle of Vi : D2 & D3 – ‘ON’
D1 D2
D3
D4
Norsabrina Sihab KEE112 Electronics 1 Copyright@JUL2007
Chapter 2 – Diode Applications
Chapter 2: Diodes 28
Full Wave Bridge Rectifier (1) - Operations
Negative half cycle of Vi : D1 & D4 – ‘ON’
D2
D1
D3
D4
Norsabrina Sihab KEE112 Electronics 1 Copyright@JUL2007
Chapter 2 – Diode Applications
Chapter 2: Diodes 29
Full Wave Bridge Rectifier (2) - Operations
Positive half cycle of Vi : D1 & D2 – ‘ON’
Norsabrina Sihab KEE112 Electronics 1 Copyright@JUL2007
Chapter 2 – Diode Applications
Chapter 2: Diodes 30
Full Wave Bridge Rectifier (2) - Operations
Negative half cycle of Vi : D3 & D4 – ‘ON’
Norsabrina Sihab KEE112 Electronics 1 Copyright@JUL2007
Chapter 2 – Diode Applications
Chapter 2: Diodes 31
Full Wave Bridge Rectifier - Peak Inverse Voltage (PIV)
KVL : PIV- Vm=0
PIV = Vm
Norsabrina Sihab KEE112 Electronics 1 Copyright@JUL2007
Chapter 2 – Diode Applications
Chapter 2: Diodes 32
Summary of Rectifier Circuits
Rectifier VDC PIV
VDC =
Half Wave Rectifier Vm
0.318(Vm)
VDC =
Bridge Rectifier Vm
0.636(Vm)
Center-Tapped VDC =
2Vm
Transformer Rectifier 0.636(Vm)
Norsabrina Sihab KEE112 Electronics 1 Copyright@JUL2007
Chapter 2 – Diode Applications
Chapter 2: Diodes 33
CLIPPERS
Basically to clipped-off/eliminate a portion of an
AC signal voltage above or below specific range.
HW rectifier is a basic clipper.
Functions:
1. Altering the shape of the waveform
2. Circuit transient protection
3. Detection
Norsabrina Sihab KEE112 Electronics 1 Copyright@JUL2007
Chapter 2 – Diode Applications
Chapter 2: Diodes 34
CLIPPERS
2 types :
1) series clipper,
2) parallel (shunt) clipper
Norsabrina Sihab KEE112 Electronics 1 Copyright@JUL2007
Chapter 2 – Diode Applications
Chapter 2: Diodes 35
1) Series Clipper
2 types :
a) negative series clipper,
b) positive series clipper
The diode in a series clipper circuit “clips” any
voltage that does not forward bias it:
A reverse-biasing polarity
A forward-biasing polarity less than 0.7V for
a silicon diode
Norsabrina Sihab KEE112 Electronics 1 Copyright@JUL2007
Chapter 2 – Diode Applications
Chapter 2: Diodes 36
1a) Negative Series Clipper
Clipped off half negative cycle. Diode forward
bias during positive cycle of Vi.
VT is transition voltage. (VT=VDC+Vdiode)
Vi
4V Ideal diode
20
+
+
VO
Vi R
-
-
- 20
Norsabrina Sihab KEE112 Electronics 1 Copyright@JUL2007
Chapter 2 – Diode Applications
Chapter 2: Diodes 37
1a) Negative Series Clipper
During positive half cycle
VT=Vdc+VD=4V
if Vi ≤ VT diode will OFF. Vo=0V.
If Vi > VT diode will ON.
KVL : Vi – VT – Vo =0.
Vo=Vi-VT=16V
Ideal
4V diode
Vi VO
+
20 + + VT - 16
VO
Vi R
-
-
Norsabrina Sihab KEE112 Electronics 1 Copyright@JUL2007
Chapter 2 – Diode Applications
Chapter 2: Diodes 38
1a) Negative Series Clipper (contd)
During negative half cycle.
Diode is OFF for all value of Vi.
VO=0V.
4V Ideal
diode
Vi
+
- + VT -
VO
Vi R
- 20 -
+
Norsabrina Sihab KEE112 Electronics 1 Copyright@JUL2007
Chapter 2 – Diode Applications
Chapter 2: Diodes 39
1a) Negative Series Clipper (contd)
Vi
Final output
20
VT=4V
- 20
Vo
16
Norsabrina Sihab KEE112 Electronics 1 Copyright@JUL2007
Chapter 2 – Diode Applications
Chapter 2: Diodes 40
1b) Positive Series Clipper
Clipped off half positive cycle. Diode forward
bias during negative cycle of Vi.
Ideal
Vi 4V diode
20
+
+
VO
Vi R
-
- 20 -
Norsabrina Sihab KEE112 Electronics 1 Copyright@JUL2007
Chapter 2 – Diode Applications
Chapter 2: Diodes 41
1b) Positive Series Clipper
During positive half cycle
Diode is OFF for all value of Vi.
VO=0V.
Ideal
4V
diode
Vi VO
+
20 + + VT -
VO
Vi R
-
-
Norsabrina Sihab KEE112 Electronics 1 Copyright@JUL2007
Chapter 2 – Diode Applications
Chapter 2: Diodes 42
1b) Positive Series Clipper (contd)
During negative half cycle.
VT+Vdc+VD=0.
VT=- 4V
if lVil ≤ lVTl diode OFF. Vo=0V.
If lVil > lVTl diode ON.
KVL : Vi +Vo+VT =0.
Vo=-16V Ideal
4V
diode
Vi VO
+
- + VT -
VO
Vi R
- 20 -
+ - 16
Norsabrina Sihab KEE112 Electronics 1 Copyright@JUL2007
Chapter 2 – Diode Applications
Chapter 2: Diodes 43
1b) Positive Series Clipper (contd)
Vi
Final output 20
VT= - 4V
- 20
Vo
-16
Norsabrina Sihab KEE112 Electronics 1 Copyright@JUL2007
Chapter 2 – Diode Applications
Chapter 2: Diodes 44
2a) Negative Parallel Clipper
The operation is opposite series clipper.
Vi R
20 + +
Si
Vi Vo
5V
- -
- 20
Norsabrina Sihab KEE112 Electronics 1 Copyright@JUL2007
Chapter 2 – Diode Applications
Chapter 2: Diodes 45
2a) Negative Parallel Clipper
During positive half cycle
Diode is OFF for all value of Vi.
VO=Vi=20V
R
Vi + +
20 Si
Vi Vo
5V
- -
Norsabrina Sihab KEE112 Electronics 1 Copyright@JUL2007
Chapter 2 – Diode Applications
Chapter 2: Diodes 46
2a) Negative Parallel Clipper (contd)
During negative half cycle
VT + VD+VDC =0.
VT= -5-0.7 = -5.7V
if lVil ≤ lVTl diode OFF. If lVil> lVTl diode ON.
Vo=Vi KVL : Vo - VT =0
R
Vo = VT = -5.7V
Vo
Vi - + +
0.7V
Vi VT Vo
5V
- 20 + - -
-5.7
Norsabrina Sihab KEE112 Electronics 1 Copyright@JUL2007
Chapter 2 – Diode Applications
Chapter 2: Diodes 47
2a) Negative Parallel Clipper (contd)
Vi
Final output 20
VT= - 5.7V
- 20
Vo
20
-5.7
Norsabrina Sihab KEE112 Electronics 1 Copyright@JUL2007
Chapter 2 – Diode Applications
Chapter 2: Diodes 48
2b) Positive Parallel Clipper
Vi R
20 + + +
Si
Vi VT Vo
5V
- - -
- 20
Norsabrina Sihab KEE112 Electronics 1 Copyright@JUL2007
Chapter 2 – Diode Applications
Chapter 2: Diodes 49
2b) Positive Parallel Clipper
During positive half cycle
VT- Vdc- VD = 0.
VT=5.7V
If Vi ≤ VT diode OFF. Vo=0V.
If Vi > VT diode ON.
KVL : Vo-VT =0.
Vo=5.7V
Vi R
Vo
+ + +
20
20
Si
Vi VT Vo
+5.7
5V
- - -
- 20
Norsabrina Sihab KEE112 Electronics 1 Copyright@JUL2007
Chapter 2 – Diode Applications
Chapter 2: Diodes 50
2b) Positive Parallel Clipper (contd)
During negative half cycle.
Diode is OFF for all value of Vi.
VO=Vi.
Vi R
Vo
- + +
20
20
Si
Vi VT Vo
5V
+ - -
- 20 - 20
Norsabrina Sihab KEE112 Electronics 1 Copyright@JUL2007
Chapter 2 – Diode Applications
Chapter 2: Diodes 51
2b) Positive Parallel Clipper (contd)
Vi
Final output 20
VT
- 20
Vo
20
- 20
Norsabrina Sihab KEE112 Electronics 1 Copyright@JUL2007
Chapter 2 – Diode Applications
Chapter 2: Diodes 52
Combination of Negative and
Positive Parallel Clipper
Vi R
+ +
10
Si Ge
Vi
Vo
5V 7.7V
-
-
- 10
Norsabrina Sihab KEE112 Electronics 1 Copyright@JUL2007
Chapter 2 – Diode Applications
Chapter 2: Diodes 53
Combination of Negative and
Positive Parallel Clipper
During positive half cycle
DGe OFF for all value of Vi
DSi ON conditionally
VT=VDSi+5=5.7V
If If Vi ≤ VT DSi OFF.
Vo=Vi.
If Vi > VT DSi ON.
Vo=5.7V R
Vi
+ +
10 +
Si Si
Vi VT Vo
5V 7.7V
- - -
- 10
Norsabrina Sihab KEE112 Electronics 1 Copyright@JUL2007
Chapter 2 – Diode Applications
Chapter 2: Diodes 54
Combination of Negative and
Positive Parallel Clipper (contd)
During negative half cycle
DSi OFF for all value of Vi
DGe ON conditionally
VT=VDSi+5=5.7V
If If Vi ≤ VT DGe OFF.
Vo=Vi.
If Vi > VT DGe ON.
Vo=8V
Vi R
- +
10 +
Si Si
Vi VT Vo
5V 7.7V
+ - -
- 10
Norsabrina Sihab KEE112 Electronics 1 Copyright@JUL2007
Chapter 2 – Diode Applications
Chapter 2: Diodes 55
Combination of Negative and
Positive Parallel Clipper (contd)
Vi
Final Output 10
VT=5.7V
VT= -5.7V
- 10
Vo
10
5.7V
-8V
- 10
Norsabrina Sihab KEE112 Electronics 1 Copyright@JUL2007
Chapter 2 – Diode Applications
Chapter 2: Diodes 56
Summary of Clipper Circuit
Norsabrina Sihab KEE112 Electronics 1 Copyright@JUL2007
Chapter 2 – Diode Applications
Chapter 2: Diodes 57
Summary of Clipper Circuit
Norsabrina Sihab KEE112 Electronics 1 Copyright@JUL2007
Chapter 2 – Diode Applications
Chapter 2: Diodes 58
Summary of Clipper Circuit
Norsabrina Sihab KEE112 Electronics 1 Copyright@JUL2007
Chapter 2 – Diode Applications
Chapter 2: Diodes 59
Summary of Clipper Circuit
Norsabrina Sihab KEE112 Electronics 1 Copyright@JUL2007
Chapter 2 – Diode Applications
Chapter 2: Diodes 60
Summary of Clipper Circuit
Norsabrina Sihab KEE112 Electronics 1 Copyright@JUL2007
Chapter 2 – Diode Applications
Chapter 2: Diodes 61
Summary of Clipper Circuit
Norsabrina Sihab KEE112 Electronics 1 Copyright@JUL2007
Chapter 2 – Diode Applications
Chapter 2: Diodes 62
CLAMPERS
To clamp or shift a signal to a different DC
level
Circuit consist of C,D and R
Norsabrina Sihab KEE112 Electronics 1 Copyright@JUL2007
Chapter 2 – Diode Applications
Chapter 2: Diodes 63
1) Negative Clamper
C
Vi
20 +
+
Si
Vi R
Vo
5V
-
-
- 20
Norsabrina Sihab KEE112 Electronics 1 Copyright@JUL2007
Chapter 2 – Diode Applications
Chapter 2: Diodes 64
1) Negative Clamper
During positive half cycle
Step 1: Find polarity of VC
Step 2: Determine VO using KVL at o/p
Vo – VD+VDC= 0
Vo=0.7-5= - 4.3V
Step 3: Find polarity of VC
Vi-Vc-Vo=0 + VC -
Vi Vc=24.3V C
20 +
+
Si
Vi R
Vo
5V
-
-
- 20
Norsabrina Sihab KEE112 Electronics 1 Copyright@JUL2007
Chapter 2 – Diode Applications
Chapter 2: Diodes 65
1) Negative Clamper (contd)
During negative half cycle
+ VC -
C
Vi
20 -
+
Si
Vi R
Vo
5V
+
-
- 20
Step 1: Determine Vo using KVL at i/p
Vi+Vo+Vc=0
Vo=–Vi–Vc= – 20 – 24.3= – 44.3V
Norsabrina Sihab KEE112 Electronics 1 Copyright@JUL2007
Chapter 2 – Diode Applications
Chapter 2: Diodes 66
1) Negative Clamper (contd)
Vi
Final Output 20
- 20
Vo
- 4.3V
Vi(p-p) = 40Vp-p
- 44.3
Norsabrina Sihab KEE112 Electronics 1 Copyright@JUL2007
Chapter 2 – Diode Applications
Chapter 2: Diodes 67
2) Positive Clamper
C
Vi
20 +
+
Si
Vi R
Vo
5V
-
-
- 20
Norsabrina Sihab KEE112 Electronics 1 Copyright@JUL2007
Chapter 2 – Diode Applications
Chapter 2: Diodes 68
2) Positive Clamper
During negative half cycle (because Diode ON at this cycle)
Step 1: Find polarity of VC
Step 2: Determine VO using KVL at o/p
Vo+VD+VDC= 0
Vo= – 0.7 – 5= – 5.7V
Step 3: Determine VC using KVL at i/p
Vi–VDC–VD=0
VC=Vi–VDC–V=14.3V
- VC +
Vi
C
20
-
+
Si=0.7
Vi R
Vo
5V
+
- 20 -
Norsabrina Sihab KEE112 Electronics 1 Copyright@JUL2007
Chapter 2 – Diode Applications
Chapter 2: Diodes 69
2) Positive Clamper (contd)
During positive half cycle
Step 1: Find polarity of Vo
Vi+Vc-Vo=0
Vo=20+14.3=34.3V
- VC +
Vi
C
20
+
+
Si=0.7
Vi R
Vo
5V
-
- 20 -
Norsabrina Sihab KEE112 Electronics 1 Copyright@JUL2007
Chapter 2 – Diode Applications
Chapter 2: Diodes 70
2) Positive Clamper (contd)
Vi
Final Output 20
- 20
Vo
34.3
Vi(p-p) = 40Vp-p
- 5.7
Norsabrina Sihab KEE112 Electronics 1 Copyright@JUL2007
Chapter 2 – Diode Applications
Chapter 2: Diodes 71
Example – Design a Clamper
Designed a clamper circuit to produce output
voltage, Vo. Use silicon diode in your design.
Vi Vo
15 5.7
+ Designed +
Vi clamper Vo
- circuit -
-15 - 24.3
Norsabrina Sihab KEE112 Electronics 1 Copyright@JUL2007
Chapter 2 – Diode Applications
Chapter 2: Diodes 72
Example – Design a Clamper
Solution
During positive cycle During negative cycle
Propose design clamper circuit
+ VC -
which D ‘ON’ during positive
cycle. Vo=5.7V - C +
Si
+ VC -
Vi R
Vo
+ C + 5V
Si +
-
Vi R
Vo
5V
-
-
KVL:
KVL: -Vi-Vo-Vc=0
Vi – Vc – Vo=0 Vo=15 - 9.3= - 24.3V
Vc=Vi-Vo=15 - 5.7=9.3V
Norsabrina Sihab KEE112 Electronics 1 Copyright@JUL2007
Chapter 2 – Diode Applications
Chapter 2: Diodes 73
Summary of Clamper Circuit
Norsabrina Sihab KEE112 Electronics 1 Copyright@JUL2007
Chapter 2 – Diode Applications
Chapter 2: Diodes 74
VOLTAGE MULTIPLIERS
Use clamping action to increase peak rectified
voltage without the necessity of increasing
the transformer’s voltage
Used in high-voltage, low-current application
(example: CRTs)
Common multiplication factors:
2 (voltage doubler)
3 (voltage tripler)
4 (voltage quadrupler)
Norsabrina Sihab KEE112 Electronics 1 Copyright@JUL2007
Chapter 2 – Diode Applications
Chapter 2: Diodes 75
1) Voltage Doubler
Is divided into two type:
1. Half-wave voltage doubler
2. Full-wave voltage doubler
Norsabrina Sihab KEE112 Electronics 1 Copyright@JUL2007
Chapter 2 – Diode Applications
Chapter 2: Diodes 76
1) Voltage Doubler (contd)
Half-Wave Voltage Doubler
During positive half-cycle of secondary voltage:
D1 – forward biased, D2 – reverse biased
C1 is charged to the peak of secondary
voltage (Vp) less the diode drop
Norsabrina Sihab KEE112 Electronics 1 Copyright@JUL2007
Chapter 2 – Diode Applications
Chapter 2: Diodes 77
1) Voltage Doubler (contd)
Half-Wave Voltage Doubler
During negative half-cycle of secondary voltage:
D2 – forward biased, D1 – reverse biased
C1 can’t discharge
Peak voltage on C1 adds to the secondary voltage (Vp)
to charge C2 to approximately 2Vp
If a load resistor is connected across the output, C2 will
discharge during positive half-cycle and again recharge
to 2Vp during negative half-cycle.
Norsabrina Sihab KEE112 Electronics 1 Copyright@JUL2007
Chapter 2 – Diode Applications
Chapter 2: Diodes 78
1) Voltage Doubler (contd)
Full-Wave Voltage Doubler
During positive half-cycle of secondary voltage:
D1 – forward biased, D2 – reverse biased
C1 is approximately charged to the peak
of secondary voltage (Vp)
Norsabrina Sihab KEE112 Electronics 1 Copyright@JUL2007
Chapter 2 – Diode Applications
Chapter 2: Diodes 79
1) Voltage Doubler (contd)
Full-Wave Voltage Doubler
During negative half-cycle of secondary voltage:
D2 – forward biased, D1 – reverse biased
C2 is approximately charged to the peak of secondary
voltage (Vp)
The output voltage is taken across the two capacitors is
series
Norsabrina Sihab KEE112 Electronics 1 Copyright@JUL2007
Chapter 2 – Diode Applications
Chapter 2: Diodes 80
2) Voltage Tripler
During positive half-cycle of secondary During negative half-cycle of secondary
voltage: voltage:
D1 – forward biased, D2 – reverse D2 – forward biased, D1 – reverse
biased biased
C1 charges to Vp C2 charges to 2Vp through D2
During next positive half-cycle of secondary voltage:
C3 charges to 2Vp through D3
Tripler output is taken across C1 and C3
Norsabrina Sihab KEE112 Electronics 1 Copyright@JUL2007
Chapter 2 – Diode Applications
Chapter 2: Diodes 81
3) Voltage Quadrupler
C4 charges to 2Vp during negative half-cycle
The 4Vp output is taken across C2 and C4.
Norsabrina Sihab KEE112 Electronics 1 Copyright@JUL2007
Chapter 2 – Diode Applications
Chapter 2: Diodes 82
ZENER REGULATOR
IDZ is opposite from ID which +
is designed to work in +
V
reverse bias. VZ
-
Application : Regulator -
VZ > V > 0
DZ will OFF
Norsabrina Sihab KEE112 Electronics 1 Copyright@JUL2007
Chapter 2 – Diode Applications
Chapter 2: Diodes 83
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 fixed
2. Vi fixed and RL variable
3. Vi variable and RL fixed
Norsabrina Sihab KEE112 Electronics 1 Copyright@JUL2007
Chapter 2 – Diode Applications
Chapter 2: Diodes 84
1. Fixed Vi and Fixed RL
Step 1: Determine the state of the Step 2: Substitute appropriate
zener diode by removing it from the equivalent circuit
network. Calculate voltage across the
R
resulting open circuit.
+ IZ IL +
Vi
Vz R VL
- -
VL Vz
KCL : I R I Z I L
IZ IR IL (1)
R VL V V VL
VDR : where I L and I R R i
+ + RL R R
RLVi
Vi
V R VL V VL power dissipated by zener diode :
R RL
- Pz I zVz
if V VZ ( DZ ON )
-
if V VZ ( DZ OFF )
Norsabrina Sihab KEE112 Electronics 1 Copyright@JUL2007
Chapter 2 – Diode Applications
Chapter 2: Diodes 85
2. Fixed Vi and Variable RL
Specific range of RL to turn since R L is minimum, therefore I L is maximum
ON DZ R
VL V
I L m ax Z
+ + RL RL m in
Vi VZ
V R VL
and I Lm in
-
R L m ax
VDR : -
RV Once DZ ON, VR remains fixed
V VL L i Vi VZ VR
R RL
VZ ( RL R ) RLVi since VR is fixed, I R also fixed
VR
VZ
( RL R ) Vi IR
RL R
KCL : I R I Z I L
R
VZ (1 ) Vi so I Z I R I L
RL
resulting I Zm in when I Lm ax
R V VZ
i and I Zm ax when I Lm in because I R is constant
RL VZ
VZ
so RL m in
RVZ I Lm in I R - I ZM & R Lm ax
Vi VZ I Lm in
Norsabrina Sihab KEE112 Electronics 1 Copyright@JUL2007
Chapter 2 – Diode Applications
Chapter 2: Diodes 86
3. Fixed RL and Variable Vi
Vi must be sufficiently The max of Vimax is
large to turn DZ limited by the max
Min voltage to turn ON zener current, IZM
is Vi=Vimin
VDR : I R max I ZM I L
VL VZ
RLVi Vi min VR max VZ
R RL
VZ ( RL R ) RLVi
RL R
Vi m in VZ
RL
Norsabrina Sihab KEE112 Electronics 1 Copyright@JUL2007
Chapter 2 – Diode Applications
Chapter 2: Diodes 87
Exercise
1) Determine VL,VR, IZ and PZ. 3) Determine range of Vi that
+ VR - will maintain zener diode in
R=1k IZ IL + ON state
20V
Vz=10V RL=2.7k VL R=220Ω IR
PZM=80mW
- + IZ IL +
Vi Vz=20V RL=1.2kΩ VL
PZM=60mW
-
-
2) R=1k IR
+ IZ IL
Vi=50V Vz=10V RL
PZM=32mW
-
a) Determine the range of RL and
IL that will result in VRL being
maintained at 10V
b) Determine the max wattage
rating of DZ.
Norsabrina Sihab KEE112 Electronics 1 Copyright@JUL2007
