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Dr. N.M. Safri/SEU3003_diode PN
Dr. orlaili Mat Safri
DIODE P
Chapter 2
(ELECTRO ICS)
ELEKTRO IK
SEU 3003
1
Dr. N.M. Safri/SEU3003_diode PN
Diode’s physical structure The I-V characteristics Load line and graphical analysis Internal resistance Diode model Diode with DC power supply – series and parallel connection 7. Basic gates 8. Diodes applications: AC power supply – rectifiers with capacitor filter, clippers and clampers 9. Data sheets 10.Zener diode – simple voltage regulator 11.Other diodes - Photodiodes, LED’s and etc
1. 2. 3. 4. 5. 6.
In this chapter, we will learn:
2
n
Dr. N.M. Safri/SEU3003_diode PN
Fig.1: Basic diode structure
p
pn junction Metal contacts and wire leads
A diode is a single pn junction device with conductive contacts and wire leads connected to each region.
Diode’s Physical Structure
3
Cathode (K)
Dr. N.M. Safri/SEU3003_diode PN
The “arrow” in the symbol points in the direction of conventional current (opposite to electron flow).
The n region is called the cathode and p region is called the anode. anode
Fig.2: Schematic symbol
Anode (A)
There are several types of diodes, but the schematic symbol for a general-purpose diode (rectifier diode) is as shown below.
Diode’s Physical Structure
4
Vbias + －
IF
－
The positive terminal of the source is connected to the anode through the current-limiting resistor. The negative terminal of the source is connected to the cathode. The forward current (IF) is from anode to cathode as indicated.
Dr. N.M. Safri/SEU3003_diode PN
The forward voltage drop (VF) due to the barrier potential is from positive at the anode to negative at the cathode.
Fig.3: Forward bias
－ R +
+
VF
A diode is forward-biased when a voltage source is connected as shown in figure 3.
Diode’s Physical Structure (Forward-bias connection)
5
Fig.4: Reverse bias
Vbias + －
The reverse current is extremely small and can be considered to be zero.
The negative terminal of the source is connected to the anode side of the diode and the positive terminal is connected to the cathode side.
Dr. N.M. Safri/SEU3003_diode PN
Notice that the entire bias voltage (Vbias) appears across the diode.
R
I =0A
Vbias + －
A diode is reverse-biased when a voltage source is connected as shown in figure 4.
Diode’s Physical Structure (Reverse-bias connection)
6
+
－
Vbias
IF
K
Fig.5(a): Forward bias
+
Rlimit
－
A
Ideal diode model
Dr. N.M. Safri/SEU3003_diode PN
Rlimit
K
+ Fig.5(b): Reverse bias
－
Vbias
I =0A
A
Ideal diode model
The ideal model of a diode is a simple switch. When the diode is forward-biased, it acts like a closed (on) switch (Fig. 5 (a)). When the diode is reverse-biased, it acts like an open (off) switch (Fig. 5 (b)). The barrier potential, the forward dynamic resistance, and the reverse current are all neglected.
The I-V Characteristics (The ideal diode model)
7
Dr. N.M. Safri/SEU3003_diode PN
Draw the ideal I-V characteristic curve to depicts the ideal diode operation. At x-axis, indicate VF and VR as positive and negative potential, respectively, and at y-axis, the IF and IR as positive and negative current, respectively.
Activity
8
VR
IR
IF
Dr. N.M. Safri/SEU3003_diode PN
Fig.6: Ideal characteristic curve (blue)
The I-V Characteristics (The ideal diode model)
VF
9
VR = Vbias
IR = 0 A
Vbias IF = R limit
VF = 0 V
Dr. N.M. Safri/SEU3003_diode PN
Complete these equations.
Activity
10
+
－
Vbias
IF
Dr. N.M. Safri/SEU3003_diode PN
11
When the diode is forward-biased, it is equivalent to a closed switch in series with a small equivalent voltage source equal to the barrier potential (0.7 V) with the positive side toward the anode.
The practical model adds the barrier potential to the ideal switch model.
This equivalent voltage source represents the fixed voltage drop (VF) produced across the forward-biased pn junction of the diode and is not an active source of voltage.
Fig.7(a): Forward bias
+
Rlimit
－
Practical diode model VF A + － K
The I-V Characteristics (The practical diode model)
Rlimit
－
Vbias
+
+ I =0A
bias
－V
K
Fig.7(b): Reverse bias
A
Practical diode model
Dr. N.M. Safri/SEU3003_diode PN
12
When the diode is reverse-biased, it is equivalent to an open switch just as in the ideal model. The barrier potential does not affect reverse bias, so it is not a factor.
The I-V Characteristics (The practical diode model)
VR 0.7 V
VF
Dr. N.M. Safri/SEU3003_diode PN
Fig.8: Characteristic curve of silicon diode (blue)
IR
0
IF
The I-V Characteristics (The practical diode model)
13
VR = Vbias
IR = 0 A
Vbias – VF IF = Rlimit
VF = 0.7 V
Dr. N.M. Safri/SEU3003_diode PN
Using Kirchhoff’s voltage law, determine:
Activity
14
Complete diode model 0.7 V r’ A + － d K
Dr. N.M. Safri/SEU3003_diode PN
15
The complete model of a diode consists of the barrier potential, the small forward dynamic resistance (r’d), － IF and the large internal reverse Rlimit Vbias resistance (r’R). + － + The reverse resistance is taken into Fig.9(a): Forward bias account because it provides a path for the reverse current, which is included in this diode model. When the diode is forward-biased, it acts as a closed switch in series with the barrier potential voltage and the small forward dynamic resistance (r’d).
The I-V Characteristics (The complete diode model)
+
-
-
Dr. N.M. Safri/SEU3003_diode PN
Metal contacts and wire leads
－
IF = Imajority - IR
}
Imajority - -- + + -+ - - -+ -+ - + - + -+ p n -+ -+ - +--+ depletion region
IR
The I-V Characteristics (The complete diode model-forward-biased)
}
16
Dr. N.M. Safri/SEU3003_diode PN
17
where IR is the reverse saturation current VF is the applied forward-bias voltage across the diode n is an ideality factor, which is a function of the operating conditions and physical construction; it has a range of 1 and 2 depending a wide variety of factors
T
VF
nV I F = I R e − 1
Using solid-state physics, general characteristics of a semiconductor diode can be defined as follows.
The I-V Characteristics (Shockley’s equation)
Dr. N.M. Safri/SEU3003_diode PN
where k is Boltzmann’s constant = 1.38 Χ 10-23 J/K T is the absolute temperature in kelvins = 273 + the temperature in °C q is the magnitude of electronic charge = 1.6 X 10-19 C
kT VT = q
VT is called the thermal voltage and is determined by
The I-V Characteristics (Thermal voltage (VT))
18
VR
Dr. N.M. Safri/SEU3003_diode PN
Fig.10: Characteristic curve of silicon diode (blue)
*Note the difference between the IF and IR scales.
VF
Slope due to the flow forward resistance
0.7 V (knee voltage)
IR (µA)
0
IF (mA)
The I-V Characteristics (The complete diode model-forward biased)
19
－
Vbias
IR
+
K
Fig.9(b): Reverse bias
－
Rlimit
+
A
r’R
Complete diode model
Dr. N.M. Safri/SEU3003_diode PN
The barrier potential does not affect reverse bias, so it is not a factor.
20
When the diode is reverse-biased, it acts as an open switch in parallel with the large internal reverse resistance (r’R).
The I-V Characteristics (The complete diode model)
VR
IR (µA)
0.7 V (knee voltage)
*Note the difference between the IF and IR scales.
VF
Dr. N.M. Safri/SEU3003_diode PN
Fig.10: Characteristic curve of silicon diode (blue)
Small reverse current due to the high reverse resistance
0
IF (mA)
The I-V Characteristics (The complete diode model)
21
Vbias – 0.7 V IF = Rlimit + r’d
VF = 0.7 V + IFr’d
Dr. N.M. Safri/SEU3003_diode PN
Using Kirchhoff’s voltage law, determine:
Activity
22
Dr. N.M. Safri/SEU3003_diode PN
For a forward-biased diode, as temperature is increased, the forward current increases for a given value of forward voltage. Also, for a given value of forward current, the forward voltage decreases. For a reverse-biased diode, as temperature is increased, the reverse current increases.
The I-V Characteristics (Temperature effects)
23
0.7 V 0.7 V - ∆V
IR (µA)
0
at 25°°C
*Note the difference between the IF and IR scales.
VF
Dr. N.M. Safri/SEU3003_diode PN
Fig.11: Temperature effect on the diode V-I characteristic.
VR
at 25°°C + ∆T
IF (mA)
The I-V Characteristics (Temperature effects)
24
Vbias
－
+
(a)
10 V
5V
(b) Dr. N.M. Safri/SEU3003_diode PN
－
+
1.0 kΩ
1.0 kΩ Vbias
Rlimit
Rlimit
Outline
Determine the forward voltage and forward current for each of the diode model. Also find the voltage across the limiting resistor in each case. Assume r’d = 10 Ω at the determined value of forward current. Fig. (a) Determine the reverse voltage and reverse current for each diode model. Also find the voltage across the limiting resistor in each case. Assume IR = 1 µA. Fig. (b)
Activity
25
－
(a)
10 V
Dr. N.M. Safri/SEU3003_diode PN
Outline
VRlimit = IF Rlimit = (9.21 mA) (1.0 kΩ) = 9.21 V
IF = (Vbias – VF) / (Rlimit + r’d) = (10 V – 0.7 V) / (1.0 kΩ + 10 Ω ) = 9.21 mA
IF = (Vbias – VF) / Rlimit = (10 V – 0.7 V) / 1.0 kΩ = 9.3 mA
VRlimit = IF Rlimit = (9.3 mA) (1.0 kΩ) = 9.3 V
VF = 0.7 V + IFr’d
Complete model
VRlimit = IF Rlimit = (10 mA) (1.0 kΩ) = 10 V
IF = Vbias / Rlimit = 10 V / 1.0 kΩ = 10 mA
VF = 0 V
Ideal model
VF = 0.7 V
Practical model
Vbias
+
1.0 kΩ
Rlimit
Activity (Answer)
26
(b)
5V
VRlimit = IF Rlimit = (1 µ A) (1.0 kΩ) = 1 mV
VRlimit = 0V
Dr. N.M. Safri/SEU3003_diode PN
IR = 1 µ A
IR = 0A
Complete model
VRlimit = 0 V
IR = 0 A
VF = Vbias – VRlimit
Practical model
－
VR = Vbias = 5 V
Ideal model
VR = Vbias = 5 V
Vbias
+
1.0 kΩ
Rlimit
Activity (Answer)
Outline
27
K
A
0.7 V VD (Si) 0.3 V (Ge)
K
K
Dr. N.M. Safri/SEU3003_diode PN
0
+ I =0A
－
0
A
A
VD + －
ID
VD
Cathode (K)
Practical model
Anode (A)
ID
I =0A
K
A
Ideal model
In Summary
A
A
ID (µA)
0
ID (mA)
r’R
VD r’d + －
Outline
VD
28
Reversebiased
Forwardbiased
(knee voltage)
K
K
Complete model
(a)
+
E
ID －
－
(b)
0
Dr. N.M. Safri/SEU3003_diode PN
Fig.12: Series diode configuration: (a) circuit; (b) characteristics
－ R +
+
VD
ID (mA)
Outline
29
VD (V)
It will be used to describe the analysis of a diode circuit using its actual characteristics.
The circuit of Fig. 12 is the simplest of diode configuration.
Load Line and Graphical Analysis
(a)
+
E
ID －
－
(b)
0
IDQ = ?
VDQ = ?
Dr. N.M. Safri/SEU3003_diode PN
Outline
30
VD (V)
Load line
Q-point
Fig.12: Series diode configuration: (a) circuit; (b) characteristics
－ R +
+
VD
ID (mA)
The straight line is called a load line because the intersection on the vertical axis is defined by the applied load R. The analysis to follow is therefore called load-line analysis. analysis
Load Line and Graphical Analysis
－ E = 10 V
+
ID
－
⇒ VD = E - VR ⇒ VD = E - IDR
E – VR – VD = 0
Applying Kirchoff’s voltage law around the close loop, we have
－ R = 0.5 kΩ +
+
VD
Dr. N.M. Safri/SEU3003_diode PN
For vertical-axis, if VD = 0 V, then ID = E/R = 10 V / 0.5 kΩ = 20 mA Outline
31
10 V V (V) D
Load line
For horizontal-axis, if ID = 0 A, then VD = E = 10 V
0
Diode characteristics
Q-point
ID (mA) 20 mA
Example 1 (Complete model)
Load Line and Graphical Analysis
－ E = 10 V
+
ID
－
⇒ VD = E - VR ⇒ VD = E - IDR
E – VR – VD = 0
Dr. N.M. Safri/SEU3003_diode PN
For vertical-axis, if VD = 0 V, then ID = E/R = 10 V / 0.5 kΩ = 20 mA Outline
32
10 V V (V) D
Load line
For horizontal-axis, if ID = 0 A, then VD = E = 10 V
0 V ≅ 0.8 V DQ
Q-point
ID (mA) 20 mA IDQ ≅ 18.3 mA
Applying Kirchoff’s voltage law around the close loop, we have
－ R = 0.5 kΩ +
+
VD
Example 1 (Complete model)
Load Line and Graphical Analysis
－ E = 10 V
+
ID
－
⇒ VD = E - VR ⇒ VD = E - IDR
E – VR – VD = 0
Dr. N.M. Safri/SEU3003_diode PN
For vertical-axis, if VD = 0 V, then ID = E/R = 10 V / 0.5 kΩ = 20 mA Outline
33
10 V V (V) D
Load line
For horizontal-axis, if ID = 0 A, then VD = E = 10 V
0 V = 0.7 V DQ
Q-point
ID (mA) 20 mA IDQ ≅ 18.6 mA
Applying Kirchoff’s voltage law around the close loop, we have
－ R = 0.5 kΩ +
+
VD
Example 1 (Practical model)
Load Line and Graphical Analysis
－ E = 10 V
+
ID
－
⇒ VD = E - VR ⇒ VD = E - IDR
E – VR – VD = 0
Applying Kirchoff’s voltage law around the close loop, we have
－ R = 0.5 kΩ +
+
VD
VDQ = 0 V
Dr. N.M. Safri/SEU3003_diode PN
For vertical-axis, if VD = 0 V, then ID = E/R = 10 V / 0.5 kΩ = 20 mA Outline
34
10 V V (V) D
Load line
For horizontal-axis, if ID = 0 A, then VD = E = 10 V
0
Q-point
ID (mA) IDQ = 20 mA
Example 1 (Ideal model)
Load Line and Graphical Analysis
－ E = 10 V
+
ID
－
0
Diode characteristics
ID (mA)
Dr. N.M. Safri/SEU3003_diode PN
Draw the load line, identify the Q-point and determine VDQ and IDQ.
－ R = 1 kΩ +
+
VD
Example 2 (Practical model)
Activity
Outline
35
VD (V)
－ E = 10 V
+
ID
－
⇒ VD = E - VR ⇒ VD = E - IDR
E – VR – VD = 0
Applying Kirchoff’s voltage law around the close loop, we have
－ R = 1 kΩ +
+
VD
ID (mA)
Q-point
Dr. N.M. Safri/SEU3003_diode PN
For vertical-axis, if VD = 0 V, then ID = E/R = 10 V / 1 kΩ = 10 mA Outline
36
10 V V (V) D
Load line
For horizontal-axis, if ID = 0 A, then VD = E = 10 V
0 V = 0.7 V DQ
10 mA IDQ ≅ 9.3 mA
Example 2 (Practical model)
Activity (Answer)
－ R = 1 kΩ +
－ R = 0.5 kΩ + ID
－
ID
－
+ － E = 10 V
+
VD
+ － E = 10 V
+
VD Q-point
Q-point
Dr. N.M. Safri/SEU3003_diode PN
0 V = 0.7 V DQ
10 mA IDQ ≅ 9.3 mA
ID (mA)
0 V = 0.7 V DQ
20 mA IDQ ≅ 18.6 mA
ID (mA)
Compare Eg. 1 and Eg. 2 (Practical model)
Load Line and Graphical Analysis
Outline
37
10 V VD (V)
Load line
10 V VD (V)
Load line
+
－
－
－
E = 10 V
ID
VD
+
+
E = 10 V
+
ID
－ Q-point
Q-point
0 V ≅ 0.75 V DQ
10 mA IDQ ≅ 9 mA
ID (mA)
0 V ≅ 0.8 V DQ
IDQ ≅ 18.3 mA
20 mA
ID (mA)
10 V V (V) D
Load line
10 V V (V) D
Load line
RD = VD / ID ≅ 83 Ω
At ID = 9 mA, VD = 0.75 V (from the curve),
RD = VD / ID ≅ 44 Ω
At ID = 18.3 mA, VD = 0.8 V (from the curve),
Dr. N.M. Safri/SEU3003_diode PN
Outline
38
In general, The higher the current through a diode, the lower is the dc resistance level.
R = 1 kΩ
+
－
R = 0.5 kΩ
－
+
VD
Determine the dc resistance levels for the diode of Eg. 1 and Eg. 2 (complete model).
Internal resistance
Dr. N.M. Safri/SEU3003_diode PN
Outline
AC or Dynamic Resistance If a sinusoidal rather than a dc input is applied, the varying input will move the instantaneous operating point up and down a region of the characteristics and thus defines a specific change in current and voltage.
DC or Static Resistance The application of a dc voltage to a circuit containing a semiconductor diode will result in an operating point on the characteristic curve that will not change with time. Resistance of a diode, RD = VD / ID
As the operating point (Q-point) of a diode moves from one region to another, the resistance of the diode will also change due to nonlinear shape of the diode characteristic curve.
Internal resistance
39
*Note: Q stand for quiescent, which means “still or unvarying”
∆Id
∆Vd
Q-point (dc operation)
Dr. N.M. Safri/SEU3003_diode PN
AC or Dynamic Resistance
Internal resistance
Outline
40
∆Vd
Q-point (dc operation)
∆Id
Resistance of a diode rd = ∆Vd / ∆Id
∆Vd
Q-point
Outline
In general, the lower the Q-point of operation (smaller current or lower voltage), the higher is the ac resistance. Dr. N.M. Safri/SEU3003_diode PN 41
∆Id
AC or Dynamic Resistance
Internal resistance
Fig.13
+20 V
ID
Si Si
+VD2 －
Dr. N.M. Safri/SEU3003_diode PN
V IR O 5.6 kΩ
Determine ID, VD2 and VO for the series circuit of Fig. 13.
Series Connection
Diode with DC Power Supply
Outline
42
－
10 V
Fig.14
E
+
R D1
I1 0.33 kΩ
Si
ID1 Si
－
VO
Dr. N.M. Safri/SEU3003_diode PN
D2
ID2
+
Determine VO, I1, ID1 and ID2 for the parallel diode configuration of Fig. 14.
Parallel Connection
Diode with DC Power Supply
Outline
43
－
R2
I2 5.6 kΩ
D2 ID2
Si
I1
Dr. N.M. Safri/SEU3003_diode PN
R1
D1
20 V
Fig.15
E
+
3.3 kΩ
Si
Determine I1, I2, and ID2 for the network of Fig. 15.
Series-Parallel Configuration
Diode with DC Power Supply
Outline
44
Fig.16
0V
-5 V D2
D1
Dr. N.M. Safri/SEU3003_diode PN
1 kΩ
VO
Determine VO for the logic gate circuit of Fig. 16. Which diode is in “ON” state? Determine the current that flows through this diode. If the diode is made of silicon, what is the new value of VO?
Basic Gates
Outline
45
Fig.17
0V
-5 V D2
D1
Dr. N.M. Safri/SEU3003_diode PN
-5 V
2.2 kΩ
VO
Determine VO for the logic gate circuit of Fig. 17. Which diode is in “ON” state? Determine the current that flows through this diode. If the diode is made of germanium, what is the new value of VO?
Activity
Outline
46
Fig.18
V2
V1 D2
D1
1 kΩ
VO
5V 5V 5V
Outline
A D logic gate
0V 0V
0V 5V
5V
0V
0V
0V 0V
VO
V2
V1
Dr. N.M. Safri/SEU3003_diode PN
+5 V
Complete the table, according to logic gate circuit of Fig. 18. What is the type of this logic gate circuit?
Activity
47
T t 1 cycle
0 T/2
Vp T t
1. 2. 3. 4.
Dr. N.M. Safri/SEU3003_diode PN
Half-wave rectifier Full-wave rectifier Clippers Clampers
Vp
1 cycle
0 T/2
With time-varying functions, diode can be used as:
1 cycle
0 T/2
Vp
Outline
T t
The diode analysis will now be expanded to include timevarying functions such as the sinusoidal waveform and the square wave. Vi Vi Vi
Diode with AC Power Supply
48
Move
Vp T t
Vi = Vp sin ωt
1 cycle
0 T/2
Vi
－
Vi
+
R
Fig.19
－
Dr. N.M. Safri/SEU3003_diode PN
+ +
－
Vo
RECTIFIER: Half-wave rectification
Diode with AC Power Supply
Outline
49
Move
0 T/2
Vi
Vp
Vp
T/2
0 T/2
Vi
0
Vi
T t
T t
T t
I=0A
R
－
Vo
+
－
Vdc = 0.318 Vp
R
+
Vo
Dr. N.M. Safri/SEU3003_diode PN
Vdc = 0
+
Vi
－
I
Diode is reverse-biased
－
Vi
+
Diode is conducting
RECTIFIER: Half-wave rectification
Diode with AC Power Supply
0
Vo
Vp
T/2
Vp
T/2
0 T/2
Vo
0
Vo
T t
T t
Outline
Vdc 50
Vo = 0 V
T t
Vo = Vi
Dr. N.M. Safri/SEU3003_diode PN
1. Bridge Network 2. Center-Tapped Transformer
Outline
The full-wave rectifier inverts the negative portions of the sine wave so that a unipolar output signal is generated during both halves of the input sinusoid.
RECTIFIER: Full-wave rectification
Diode with AC Power Supply
51
Vp T t
Vi = Vp sin ωt
1 cycle
0 T/2
Vi
Bridge Network
D3
D1 －
Dr. N.M. Safri/SEU3003_diode PN
－
Vi
+
R D4
D2
Fig.20
Vo +
RECTIFIER: Full-wave rectification
Diode with AC Power Supply
Outline
52
0 T/2
Vi
Vp
Vp
T/2
0 T/2
Vi
0
Vi
T t
T t
T t
+ +
D3
D1
D3
I
R
V － o
R
V － o
+
+
I
Diode D1 and D4 are conducting
Diode D2 and D3 are conducting
Vdc = 2 (0.318 Vp) = 0.636 Vp
D4
D2
D4
D2
Dr. N.M. Safri/SEU3003_diode PN
Vdc = 0
Vi
－
－
Vi
+
D1
RECTIFIER: Full-wave rectification
Diode with AC Power Supply
0
Vo
0
Vo
0
Vo
T/2
Vp
T/2
T/2
Vp
53
Vdc
Outline
T t
T t
Vo = Vi
T t
Vo = Vi
Vp T t
Vi = Vp sin ωt
1 cycle
0 T/2
Vi 1:2
+
－ －
Vi
+
－
Vi
Dr. N.M. Safri/SEU3003_diode PN
－
Vi
+
Center-Tapped Transformer
RECTIFIER: Full-wave rectification
Diode with AC Power Supply
Fig.21
D2
R
Vo +
D1
Outline
54
0 T/2
Vi
Vp
Vp
T/2
0 T/2
Vi
0
Vi
T t
T t
T t
－
R
Vo
D1
D2
R
Vo
I
I
Vdc = 2 (0.318 Vp) = 0.636 Vp
+
+
Dr. N.M. Safri/SEU3003_diode PN
Vdc = 0
D2 is conducting D2
+
Vi
+
+
Vi
－ －
1:2
－
Vi
+
o
0
Vo
0
Vo
0
D1 D1 is conductingV － －
Vi
+
Vi
－
－
Vi
+
1:2
RECTIFIER: Full-wave rectification
Diode with AC Power Supply
T/2
Vp
T/2
T/2
Vp
55
Vdc
Outline
T t
T t
Vo = Vi
T t
Vo = Vi
Vp T t
Vi = Vp sin ωt
1 cycle
0 T/2
Vi
－
Vi
+
R
Fig.22
C
－
Dr. N.M. Safri/SEU3003_diode PN
+
－
Vo
+
Outline
If a capacitor is added in parallel with the load resistor of a half-wave rectifier to form a simple filter circuit, we can begin to transform the half-wave sinusoidal output into a dc voltage.
RECTIFIER with FILTER CAPACITA CE
Diode with AC Power Supply
56
0 T/2
Vi
Vp
Vp
T/2
0 T/2
Vi
0
Vi
+
Vi
T t
R －
Vo
Vo = Vi
Dr. N.M. Safri/SEU3003_diode PN
Vdc = Vp – Vr(p-p)/2 = [1 – 1/(2fRC)] Vp
－
C
+
+
Vo
－
R －
C
+
－
Vdc = 0
T t
T t
Vi
+
Diode is conducting
RECTIFIER with FILTER
Diode with AC Power Supply
T/2
0
Vo
T/2
Outline
T t
T t
57
Vdc
Vo = Vc
T t
Vo = Vp = Vc Vo = Vc
0 T/2
Vo
0
Vo
Dr. N.M. Safri/SEU3003_diode PN
The half-wave rectifier of Fig. 19 is an example of the simplest form of diode clipper – one resistor and a diode. Depending on the orientation of the diode, the positive and negative region of the applied signal is “clipped” off.
There are two general categories of clippers: series and parallel.
Clippers are networks that employ diodes to “clip” away a portion of an input signal without distorting the remaining part of the applied waveform.
CLIPPERS
Diode with AC Power Supply
Outline
58
Dr. N.M. Safri/SEU3003_diode PN
Outline
4. It is often helpful to draw the output waveform directly below the applied voltage using the same scales for the horizontal axis and the vertical axis.
3. Determine the applied voltage (transition voltage) that will result in a change of state for the diode from the “off” to the “on” state.
First and mort important: 1.Take careful note of where the output voltage is defined. Next: 2. Try to develop an overall sense of the response by simply noting the “pressure” established by each supply and the effect it will have on the conventional current direction through the diode.
There is no general procedure for analyzing networks, but there are some things one can do to give the analysis some direction.
Tips
59
0 T/2
Vi
T t
Vp = 20 V
－
Vi
+
+
Fig.23
R
Dr. N.M. Safri/SEU3003_diode PN
－
5V
－
Vo
+
Determine the output waveform for the sinusoidal input of Fig. 23?
Activity (1)
Outline
60
0 T/2
Vi
Vp
Vp
T/2
0 T/2
Vi
0
Vi
Answer
T t
T t
T t
+
Vi
－
－
Vi
+
+
R
－
Vo
Dr. N.M. Safri/SEU3003_diode PN
R
+
－
Vo
+
Diode is conducting
0
Vo
T t
Vp + 5
T t
Outline
61
Vo = 0 V
T t
Vo = Vi + 5
Vp + 5
T/2
T/2
0 T/2
5 0
Vo
Vo Diode is not 5V － + conducting at Vi ≤ - 5 V
－
5V
0 T/2
Vi
T t
Vp = 20 V
－
Vi
+
－
Fig.24
R
Dr. N.M. Safri/SEU3003_diode PN
+
5V
－
Vo
+
Determine the output waveform for the sinusoidal input of Fig. 24? Compare the result with the Fig. 24.
Activity (2)
Outline
62
PIV
Dr. N.M. Safri/SEU3003_diode PN
Zener avalanche region
VR
IR (µA)
0 0.7 V (knee voltage)
Small reverse current due to the high reverse resistance
F (mA)
It is the voltage rating that must not be exceeded in the reverse-bias region or the diode will enter the Zener avalanche region. I
Outline
VF
The peak inverse voltage (PIV) rating of the diode is of primary importance in the design of rectification systems.
63
PIV
T t
+
Vi
－
+
I=0A
－
V(PIV)
0
Vi
T/2
Vp
T t
－
Vi
+
－
Vi
+
D2
Vo
+
+
V(PIV)
－
－ －
Vi
+
D1
Dr. N.M. Safri/SEU3003_diode PN
1:2
－
Vo
+
T/2
Vp
T t
Outline
64
Vo = Vi
T t
V(PIV) rating ≥ 2Vp
0
Vo
0 T/2
Vo
V(PIV) rating ≥ Vp
R
PIV rating for full-wave rectifier
0 T/2
Vi
PIV rating for half-wave rectifier
Dr. N.M. Safri/SEU3003_diode PN
Outline
Clamping networks have a capacitor connected directly from input to output with a resistive element in parallel with the output signal. The diode is also in parallel with the output signal but may not have a series dc supply as an added element.
A clamper is a network constructed of a diode, a resistor, and a capacitor that shifts a waveform to a different dc level without changing the appearance of the applied signal.
CLAMPERS
65
Dr. N.M. Safri/SEU3003_diode PN
Outline
5. Check that the total swing of the output matches that of the input.
4. Throughout the analysis, maintain a continual awareness of the location and defined polarity for Vo to ensure that the proper levels are obtained.
3. Assume that during the period when the diode is in the “off” state the capacitor holds on to its established voltage level.
First and mort important: 1. .Start the analysis by examining the response of the portion of the input signal that will forward bias the diode. Next: 2. During the period that the diode is in the “on” state, assume that the capacitor will charge up instantaneously to a voltage level determined by the surrounding network.
Tips for Clamper etworks
66
- 20
0
10
Vi
T
t1
t2 t3
t4
f = 1000 Hz
t
Fig.25
+
5 V－
R 100 kΩ
Dr. N.M. Safri/SEU3003_diode PN
－
Vi
+
C = 1µF
－
Vo
+
Determine Vo for the network of Fig. 25 for the input indicated.
Activity
Outline
67
Dr. orlaili Mat Safri
DIODE P
Chapter 2
(ELECTRO ICS)
ELEKTRO IK
SEU 3003
1
Dr. N.M. Safri/SEU3003_diode PN
Diode’s physical structure The I-V characteristics Load line and graphical analysis Internal resistance Diode model Diode with DC power supply – series and parallel connection 7. Basic gates 8. Diodes applications: AC power supply – rectifiers with capacitor filter, clippers and clampers 9. Data sheets 10.Zener diode – simple voltage regulator 11.Other diodes - Photodiodes, LED’s and etc
1. 2. 3. 4. 5. 6.
In this chapter, we will learn:
2
n
Dr. N.M. Safri/SEU3003_diode PN
Fig.1: Basic diode structure
p
pn junction Metal contacts and wire leads
A diode is a single pn junction device with conductive contacts and wire leads connected to each region.
Diode’s Physical Structure
3
Cathode (K)
Dr. N.M. Safri/SEU3003_diode PN
The “arrow” in the symbol points in the direction of conventional current (opposite to electron flow).
The n region is called the cathode and p region is called the anode. anode
Fig.2: Schematic symbol
Anode (A)
There are several types of diodes, but the schematic symbol for a general-purpose diode (rectifier diode) is as shown below.
Diode’s Physical Structure
4
Vbias + －
IF
－
The positive terminal of the source is connected to the anode through the current-limiting resistor. The negative terminal of the source is connected to the cathode. The forward current (IF) is from anode to cathode as indicated.
Dr. N.M. Safri/SEU3003_diode PN
The forward voltage drop (VF) due to the barrier potential is from positive at the anode to negative at the cathode.
Fig.3: Forward bias
－ R +
+
VF
A diode is forward-biased when a voltage source is connected as shown in figure 3.
Diode’s Physical Structure (Forward-bias connection)
5
Fig.4: Reverse bias
Vbias + －
The reverse current is extremely small and can be considered to be zero.
The negative terminal of the source is connected to the anode side of the diode and the positive terminal is connected to the cathode side.
Dr. N.M. Safri/SEU3003_diode PN
Notice that the entire bias voltage (Vbias) appears across the diode.
R
I =0A
Vbias + －
A diode is reverse-biased when a voltage source is connected as shown in figure 4.
Diode’s Physical Structure (Reverse-bias connection)
6
+
－
Vbias
IF
K
Fig.5(a): Forward bias
+
Rlimit
－
A
Ideal diode model
Dr. N.M. Safri/SEU3003_diode PN
Rlimit
K
+ Fig.5(b): Reverse bias
－
Vbias
I =0A
A
Ideal diode model
The ideal model of a diode is a simple switch. When the diode is forward-biased, it acts like a closed (on) switch (Fig. 5 (a)). When the diode is reverse-biased, it acts like an open (off) switch (Fig. 5 (b)). The barrier potential, the forward dynamic resistance, and the reverse current are all neglected.
The I-V Characteristics (The ideal diode model)
7
Dr. N.M. Safri/SEU3003_diode PN
Draw the ideal I-V characteristic curve to depicts the ideal diode operation. At x-axis, indicate VF and VR as positive and negative potential, respectively, and at y-axis, the IF and IR as positive and negative current, respectively.
Activity
8
VR
IR
IF
Dr. N.M. Safri/SEU3003_diode PN
Fig.6: Ideal characteristic curve (blue)
The I-V Characteristics (The ideal diode model)
VF
9
VR = Vbias
IR = 0 A
Vbias IF = R limit
VF = 0 V
Dr. N.M. Safri/SEU3003_diode PN
Complete these equations.
Activity
10
+
－
Vbias
IF
Dr. N.M. Safri/SEU3003_diode PN
11
When the diode is forward-biased, it is equivalent to a closed switch in series with a small equivalent voltage source equal to the barrier potential (0.7 V) with the positive side toward the anode.
The practical model adds the barrier potential to the ideal switch model.
This equivalent voltage source represents the fixed voltage drop (VF) produced across the forward-biased pn junction of the diode and is not an active source of voltage.
Fig.7(a): Forward bias
+
Rlimit
－
Practical diode model VF A + － K
The I-V Characteristics (The practical diode model)
Rlimit
－
Vbias
+
+ I =0A
bias
－V
K
Fig.7(b): Reverse bias
A
Practical diode model
Dr. N.M. Safri/SEU3003_diode PN
12
When the diode is reverse-biased, it is equivalent to an open switch just as in the ideal model. The barrier potential does not affect reverse bias, so it is not a factor.
The I-V Characteristics (The practical diode model)
VR 0.7 V
VF
Dr. N.M. Safri/SEU3003_diode PN
Fig.8: Characteristic curve of silicon diode (blue)
IR
0
IF
The I-V Characteristics (The practical diode model)
13
VR = Vbias
IR = 0 A
Vbias – VF IF = Rlimit
VF = 0.7 V
Dr. N.M. Safri/SEU3003_diode PN
Using Kirchhoff’s voltage law, determine:
Activity
14
Complete diode model 0.7 V r’ A + － d K
Dr. N.M. Safri/SEU3003_diode PN
15
The complete model of a diode consists of the barrier potential, the small forward dynamic resistance (r’d), － IF and the large internal reverse Rlimit Vbias resistance (r’R). + － + The reverse resistance is taken into Fig.9(a): Forward bias account because it provides a path for the reverse current, which is included in this diode model. When the diode is forward-biased, it acts as a closed switch in series with the barrier potential voltage and the small forward dynamic resistance (r’d).
The I-V Characteristics (The complete diode model)
+
-
-
Dr. N.M. Safri/SEU3003_diode PN
Metal contacts and wire leads
－
IF = Imajority - IR
}
Imajority - -- + + -+ - - -+ -+ - + - + -+ p n -+ -+ - +--+ depletion region
IR
The I-V Characteristics (The complete diode model-forward-biased)
}
16
Dr. N.M. Safri/SEU3003_diode PN
17
where IR is the reverse saturation current VF is the applied forward-bias voltage across the diode n is an ideality factor, which is a function of the operating conditions and physical construction; it has a range of 1 and 2 depending a wide variety of factors
T
VF
nV I F = I R e − 1
Using solid-state physics, general characteristics of a semiconductor diode can be defined as follows.
The I-V Characteristics (Shockley’s equation)
Dr. N.M. Safri/SEU3003_diode PN
where k is Boltzmann’s constant = 1.38 Χ 10-23 J/K T is the absolute temperature in kelvins = 273 + the temperature in °C q is the magnitude of electronic charge = 1.6 X 10-19 C
kT VT = q
VT is called the thermal voltage and is determined by
The I-V Characteristics (Thermal voltage (VT))
18
VR
Dr. N.M. Safri/SEU3003_diode PN
Fig.10: Characteristic curve of silicon diode (blue)
*Note the difference between the IF and IR scales.
VF
Slope due to the flow forward resistance
0.7 V (knee voltage)
IR (µA)
0
IF (mA)
The I-V Characteristics (The complete diode model-forward biased)
19
－
Vbias
IR
+
K
Fig.9(b): Reverse bias
－
Rlimit
+
A
r’R
Complete diode model
Dr. N.M. Safri/SEU3003_diode PN
The barrier potential does not affect reverse bias, so it is not a factor.
20
When the diode is reverse-biased, it acts as an open switch in parallel with the large internal reverse resistance (r’R).
The I-V Characteristics (The complete diode model)
VR
IR (µA)
0.7 V (knee voltage)
*Note the difference between the IF and IR scales.
VF
Dr. N.M. Safri/SEU3003_diode PN
Fig.10: Characteristic curve of silicon diode (blue)
Small reverse current due to the high reverse resistance
0
IF (mA)
The I-V Characteristics (The complete diode model)
21
Vbias – 0.7 V IF = Rlimit + r’d
VF = 0.7 V + IFr’d
Dr. N.M. Safri/SEU3003_diode PN
Using Kirchhoff’s voltage law, determine:
Activity
22
Dr. N.M. Safri/SEU3003_diode PN
For a forward-biased diode, as temperature is increased, the forward current increases for a given value of forward voltage. Also, for a given value of forward current, the forward voltage decreases. For a reverse-biased diode, as temperature is increased, the reverse current increases.
The I-V Characteristics (Temperature effects)
23
0.7 V 0.7 V - ∆V
IR (µA)
0
at 25°°C
*Note the difference between the IF and IR scales.
VF
Dr. N.M. Safri/SEU3003_diode PN
Fig.11: Temperature effect on the diode V-I characteristic.
VR
at 25°°C + ∆T
IF (mA)
The I-V Characteristics (Temperature effects)
24
Vbias
－
+
(a)
10 V
5V
(b) Dr. N.M. Safri/SEU3003_diode PN
－
+
1.0 kΩ
1.0 kΩ Vbias
Rlimit
Rlimit
Outline
Determine the forward voltage and forward current for each of the diode model. Also find the voltage across the limiting resistor in each case. Assume r’d = 10 Ω at the determined value of forward current. Fig. (a) Determine the reverse voltage and reverse current for each diode model. Also find the voltage across the limiting resistor in each case. Assume IR = 1 µA. Fig. (b)
Activity
25
－
(a)
10 V
Dr. N.M. Safri/SEU3003_diode PN
Outline
VRlimit = IF Rlimit = (9.21 mA) (1.0 kΩ) = 9.21 V
IF = (Vbias – VF) / (Rlimit + r’d) = (10 V – 0.7 V) / (1.0 kΩ + 10 Ω ) = 9.21 mA
IF = (Vbias – VF) / Rlimit = (10 V – 0.7 V) / 1.0 kΩ = 9.3 mA
VRlimit = IF Rlimit = (9.3 mA) (1.0 kΩ) = 9.3 V
VF = 0.7 V + IFr’d
Complete model
VRlimit = IF Rlimit = (10 mA) (1.0 kΩ) = 10 V
IF = Vbias / Rlimit = 10 V / 1.0 kΩ = 10 mA
VF = 0 V
Ideal model
VF = 0.7 V
Practical model
Vbias
+
1.0 kΩ
Rlimit
Activity (Answer)
26
(b)
5V
VRlimit = IF Rlimit = (1 µ A) (1.0 kΩ) = 1 mV
VRlimit = 0V
Dr. N.M. Safri/SEU3003_diode PN
IR = 1 µ A
IR = 0A
Complete model
VRlimit = 0 V
IR = 0 A
VF = Vbias – VRlimit
Practical model
－
VR = Vbias = 5 V
Ideal model
VR = Vbias = 5 V
Vbias
+
1.0 kΩ
Rlimit
Activity (Answer)
Outline
27
K
A
0.7 V VD (Si) 0.3 V (Ge)
K
K
Dr. N.M. Safri/SEU3003_diode PN
0
+ I =0A
－
0
A
A
VD + －
ID
VD
Cathode (K)
Practical model
Anode (A)
ID
I =0A
K
A
Ideal model
In Summary
A
A
ID (µA)
0
ID (mA)
r’R
VD r’d + －
Outline
VD
28
Reversebiased
Forwardbiased
(knee voltage)
K
K
Complete model
(a)
+
E
ID －
－
(b)
0
Dr. N.M. Safri/SEU3003_diode PN
Fig.12: Series diode configuration: (a) circuit; (b) characteristics
－ R +
+
VD
ID (mA)
Outline
29
VD (V)
It will be used to describe the analysis of a diode circuit using its actual characteristics.
The circuit of Fig. 12 is the simplest of diode configuration.
Load Line and Graphical Analysis
(a)
+
E
ID －
－
(b)
0
IDQ = ?
VDQ = ?
Dr. N.M. Safri/SEU3003_diode PN
Outline
30
VD (V)
Load line
Q-point
Fig.12: Series diode configuration: (a) circuit; (b) characteristics
－ R +
+
VD
ID (mA)
The straight line is called a load line because the intersection on the vertical axis is defined by the applied load R. The analysis to follow is therefore called load-line analysis. analysis
Load Line and Graphical Analysis
－ E = 10 V
+
ID
－
⇒ VD = E - VR ⇒ VD = E - IDR
E – VR – VD = 0
Applying Kirchoff’s voltage law around the close loop, we have
－ R = 0.5 kΩ +
+
VD
Dr. N.M. Safri/SEU3003_diode PN
For vertical-axis, if VD = 0 V, then ID = E/R = 10 V / 0.5 kΩ = 20 mA Outline
31
10 V V (V) D
Load line
For horizontal-axis, if ID = 0 A, then VD = E = 10 V
0
Diode characteristics
Q-point
ID (mA) 20 mA
Example 1 (Complete model)
Load Line and Graphical Analysis
－ E = 10 V
+
ID
－
⇒ VD = E - VR ⇒ VD = E - IDR
E – VR – VD = 0
Dr. N.M. Safri/SEU3003_diode PN
For vertical-axis, if VD = 0 V, then ID = E/R = 10 V / 0.5 kΩ = 20 mA Outline
32
10 V V (V) D
Load line
For horizontal-axis, if ID = 0 A, then VD = E = 10 V
0 V ≅ 0.8 V DQ
Q-point
ID (mA) 20 mA IDQ ≅ 18.3 mA
Applying Kirchoff’s voltage law around the close loop, we have
－ R = 0.5 kΩ +
+
VD
Example 1 (Complete model)
Load Line and Graphical Analysis
－ E = 10 V
+
ID
－
⇒ VD = E - VR ⇒ VD = E - IDR
E – VR – VD = 0
Dr. N.M. Safri/SEU3003_diode PN
For vertical-axis, if VD = 0 V, then ID = E/R = 10 V / 0.5 kΩ = 20 mA Outline
33
10 V V (V) D
Load line
For horizontal-axis, if ID = 0 A, then VD = E = 10 V
0 V = 0.7 V DQ
Q-point
ID (mA) 20 mA IDQ ≅ 18.6 mA
Applying Kirchoff’s voltage law around the close loop, we have
－ R = 0.5 kΩ +
+
VD
Example 1 (Practical model)
Load Line and Graphical Analysis
－ E = 10 V
+
ID
－
⇒ VD = E - VR ⇒ VD = E - IDR
E – VR – VD = 0
Applying Kirchoff’s voltage law around the close loop, we have
－ R = 0.5 kΩ +
+
VD
VDQ = 0 V
Dr. N.M. Safri/SEU3003_diode PN
For vertical-axis, if VD = 0 V, then ID = E/R = 10 V / 0.5 kΩ = 20 mA Outline
34
10 V V (V) D
Load line
For horizontal-axis, if ID = 0 A, then VD = E = 10 V
0
Q-point
ID (mA) IDQ = 20 mA
Example 1 (Ideal model)
Load Line and Graphical Analysis
－ E = 10 V
+
ID
－
0
Diode characteristics
ID (mA)
Dr. N.M. Safri/SEU3003_diode PN
Draw the load line, identify the Q-point and determine VDQ and IDQ.
－ R = 1 kΩ +
+
VD
Example 2 (Practical model)
Activity
Outline
35
VD (V)
－ E = 10 V
+
ID
－
⇒ VD = E - VR ⇒ VD = E - IDR
E – VR – VD = 0
Applying Kirchoff’s voltage law around the close loop, we have
－ R = 1 kΩ +
+
VD
ID (mA)
Q-point
Dr. N.M. Safri/SEU3003_diode PN
For vertical-axis, if VD = 0 V, then ID = E/R = 10 V / 1 kΩ = 10 mA Outline
36
10 V V (V) D
Load line
For horizontal-axis, if ID = 0 A, then VD = E = 10 V
0 V = 0.7 V DQ
10 mA IDQ ≅ 9.3 mA
Example 2 (Practical model)
Activity (Answer)
－ R = 1 kΩ +
－ R = 0.5 kΩ + ID
－
ID
－
+ － E = 10 V
+
VD
+ － E = 10 V
+
VD Q-point
Q-point
Dr. N.M. Safri/SEU3003_diode PN
0 V = 0.7 V DQ
10 mA IDQ ≅ 9.3 mA
ID (mA)
0 V = 0.7 V DQ
20 mA IDQ ≅ 18.6 mA
ID (mA)
Compare Eg. 1 and Eg. 2 (Practical model)
Load Line and Graphical Analysis
Outline
37
10 V VD (V)
Load line
10 V VD (V)
Load line
+
－
－
－
E = 10 V
ID
VD
+
+
E = 10 V
+
ID
－ Q-point
Q-point
0 V ≅ 0.75 V DQ
10 mA IDQ ≅ 9 mA
ID (mA)
0 V ≅ 0.8 V DQ
IDQ ≅ 18.3 mA
20 mA
ID (mA)
10 V V (V) D
Load line
10 V V (V) D
Load line
RD = VD / ID ≅ 83 Ω
At ID = 9 mA, VD = 0.75 V (from the curve),
RD = VD / ID ≅ 44 Ω
At ID = 18.3 mA, VD = 0.8 V (from the curve),
Dr. N.M. Safri/SEU3003_diode PN
Outline
38
In general, The higher the current through a diode, the lower is the dc resistance level.
R = 1 kΩ
+
－
R = 0.5 kΩ
－
+
VD
Determine the dc resistance levels for the diode of Eg. 1 and Eg. 2 (complete model).
Internal resistance
Dr. N.M. Safri/SEU3003_diode PN
Outline
AC or Dynamic Resistance If a sinusoidal rather than a dc input is applied, the varying input will move the instantaneous operating point up and down a region of the characteristics and thus defines a specific change in current and voltage.
DC or Static Resistance The application of a dc voltage to a circuit containing a semiconductor diode will result in an operating point on the characteristic curve that will not change with time. Resistance of a diode, RD = VD / ID
As the operating point (Q-point) of a diode moves from one region to another, the resistance of the diode will also change due to nonlinear shape of the diode characteristic curve.
Internal resistance
39
*Note: Q stand for quiescent, which means “still or unvarying”
∆Id
∆Vd
Q-point (dc operation)
Dr. N.M. Safri/SEU3003_diode PN
AC or Dynamic Resistance
Internal resistance
Outline
40
∆Vd
Q-point (dc operation)
∆Id
Resistance of a diode rd = ∆Vd / ∆Id
∆Vd
Q-point
Outline
In general, the lower the Q-point of operation (smaller current or lower voltage), the higher is the ac resistance. Dr. N.M. Safri/SEU3003_diode PN 41
∆Id
AC or Dynamic Resistance
Internal resistance
Fig.13
+20 V
ID
Si Si
+VD2 －
Dr. N.M. Safri/SEU3003_diode PN
V IR O 5.6 kΩ
Determine ID, VD2 and VO for the series circuit of Fig. 13.
Series Connection
Diode with DC Power Supply
Outline
42
－
10 V
Fig.14
E
+
R D1
I1 0.33 kΩ
Si
ID1 Si
－
VO
Dr. N.M. Safri/SEU3003_diode PN
D2
ID2
+
Determine VO, I1, ID1 and ID2 for the parallel diode configuration of Fig. 14.
Parallel Connection
Diode with DC Power Supply
Outline
43
－
R2
I2 5.6 kΩ
D2 ID2
Si
I1
Dr. N.M. Safri/SEU3003_diode PN
R1
D1
20 V
Fig.15
E
+
3.3 kΩ
Si
Determine I1, I2, and ID2 for the network of Fig. 15.
Series-Parallel Configuration
Diode with DC Power Supply
Outline
44
Fig.16
0V
-5 V D2
D1
Dr. N.M. Safri/SEU3003_diode PN
1 kΩ
VO
Determine VO for the logic gate circuit of Fig. 16. Which diode is in “ON” state? Determine the current that flows through this diode. If the diode is made of silicon, what is the new value of VO?
Basic Gates
Outline
45
Fig.17
0V
-5 V D2
D1
Dr. N.M. Safri/SEU3003_diode PN
-5 V
2.2 kΩ
VO
Determine VO for the logic gate circuit of Fig. 17. Which diode is in “ON” state? Determine the current that flows through this diode. If the diode is made of germanium, what is the new value of VO?
Activity
Outline
46
Fig.18
V2
V1 D2
D1
1 kΩ
VO
5V 5V 5V
Outline
A D logic gate
0V 0V
0V 5V
5V
0V
0V
0V 0V
VO
V2
V1
Dr. N.M. Safri/SEU3003_diode PN
+5 V
Complete the table, according to logic gate circuit of Fig. 18. What is the type of this logic gate circuit?
Activity
47
T t 1 cycle
0 T/2
Vp T t
1. 2. 3. 4.
Dr. N.M. Safri/SEU3003_diode PN
Half-wave rectifier Full-wave rectifier Clippers Clampers
Vp
1 cycle
0 T/2
With time-varying functions, diode can be used as:
1 cycle
0 T/2
Vp
Outline
T t
The diode analysis will now be expanded to include timevarying functions such as the sinusoidal waveform and the square wave. Vi Vi Vi
Diode with AC Power Supply
48
Move
Vp T t
Vi = Vp sin ωt
1 cycle
0 T/2
Vi
－
Vi
+
R
Fig.19
－
Dr. N.M. Safri/SEU3003_diode PN
+ +
－
Vo
RECTIFIER: Half-wave rectification
Diode with AC Power Supply
Outline
49
Move
0 T/2
Vi
Vp
Vp
T/2
0 T/2
Vi
0
Vi
T t
T t
T t
I=0A
R
－
Vo
+
－
Vdc = 0.318 Vp
R
+
Vo
Dr. N.M. Safri/SEU3003_diode PN
Vdc = 0
+
Vi
－
I
Diode is reverse-biased
－
Vi
+
Diode is conducting
RECTIFIER: Half-wave rectification
Diode with AC Power Supply
0
Vo
Vp
T/2
Vp
T/2
0 T/2
Vo
0
Vo
T t
T t
Outline
Vdc 50
Vo = 0 V
T t
Vo = Vi
Dr. N.M. Safri/SEU3003_diode PN
1. Bridge Network 2. Center-Tapped Transformer
Outline
The full-wave rectifier inverts the negative portions of the sine wave so that a unipolar output signal is generated during both halves of the input sinusoid.
RECTIFIER: Full-wave rectification
Diode with AC Power Supply
51
Vp T t
Vi = Vp sin ωt
1 cycle
0 T/2
Vi
Bridge Network
D3
D1 －
Dr. N.M. Safri/SEU3003_diode PN
－
Vi
+
R D4
D2
Fig.20
Vo +
RECTIFIER: Full-wave rectification
Diode with AC Power Supply
Outline
52
0 T/2
Vi
Vp
Vp
T/2
0 T/2
Vi
0
Vi
T t
T t
T t
+ +
D3
D1
D3
I
R
V － o
R
V － o
+
+
I
Diode D1 and D4 are conducting
Diode D2 and D3 are conducting
Vdc = 2 (0.318 Vp) = 0.636 Vp
D4
D2
D4
D2
Dr. N.M. Safri/SEU3003_diode PN
Vdc = 0
Vi
－
－
Vi
+
D1
RECTIFIER: Full-wave rectification
Diode with AC Power Supply
0
Vo
0
Vo
0
Vo
T/2
Vp
T/2
T/2
Vp
53
Vdc
Outline
T t
T t
Vo = Vi
T t
Vo = Vi
Vp T t
Vi = Vp sin ωt
1 cycle
0 T/2
Vi 1:2
+
－ －
Vi
+
－
Vi
Dr. N.M. Safri/SEU3003_diode PN
－
Vi
+
Center-Tapped Transformer
RECTIFIER: Full-wave rectification
Diode with AC Power Supply
Fig.21
D2
R
Vo +
D1
Outline
54
0 T/2
Vi
Vp
Vp
T/2
0 T/2
Vi
0
Vi
T t
T t
T t
－
R
Vo
D1
D2
R
Vo
I
I
Vdc = 2 (0.318 Vp) = 0.636 Vp
+
+
Dr. N.M. Safri/SEU3003_diode PN
Vdc = 0
D2 is conducting D2
+
Vi
+
+
Vi
－ －
1:2
－
Vi
+
o
0
Vo
0
Vo
0
D1 D1 is conductingV － －
Vi
+
Vi
－
－
Vi
+
1:2
RECTIFIER: Full-wave rectification
Diode with AC Power Supply
T/2
Vp
T/2
T/2
Vp
55
Vdc
Outline
T t
T t
Vo = Vi
T t
Vo = Vi
Vp T t
Vi = Vp sin ωt
1 cycle
0 T/2
Vi
－
Vi
+
R
Fig.22
C
－
Dr. N.M. Safri/SEU3003_diode PN
+
－
Vo
+
Outline
If a capacitor is added in parallel with the load resistor of a half-wave rectifier to form a simple filter circuit, we can begin to transform the half-wave sinusoidal output into a dc voltage.
RECTIFIER with FILTER CAPACITA CE
Diode with AC Power Supply
56
0 T/2
Vi
Vp
Vp
T/2
0 T/2
Vi
0
Vi
+
Vi
T t
R －
Vo
Vo = Vi
Dr. N.M. Safri/SEU3003_diode PN
Vdc = Vp – Vr(p-p)/2 = [1 – 1/(2fRC)] Vp
－
C
+
+
Vo
－
R －
C
+
－
Vdc = 0
T t
T t
Vi
+
Diode is conducting
RECTIFIER with FILTER
Diode with AC Power Supply
T/2
0
Vo
T/2
Outline
T t
T t
57
Vdc
Vo = Vc
T t
Vo = Vp = Vc Vo = Vc
0 T/2
Vo
0
Vo
Dr. N.M. Safri/SEU3003_diode PN
The half-wave rectifier of Fig. 19 is an example of the simplest form of diode clipper – one resistor and a diode. Depending on the orientation of the diode, the positive and negative region of the applied signal is “clipped” off.
There are two general categories of clippers: series and parallel.
Clippers are networks that employ diodes to “clip” away a portion of an input signal without distorting the remaining part of the applied waveform.
CLIPPERS
Diode with AC Power Supply
Outline
58
Dr. N.M. Safri/SEU3003_diode PN
Outline
4. It is often helpful to draw the output waveform directly below the applied voltage using the same scales for the horizontal axis and the vertical axis.
3. Determine the applied voltage (transition voltage) that will result in a change of state for the diode from the “off” to the “on” state.
First and mort important: 1.Take careful note of where the output voltage is defined. Next: 2. Try to develop an overall sense of the response by simply noting the “pressure” established by each supply and the effect it will have on the conventional current direction through the diode.
There is no general procedure for analyzing networks, but there are some things one can do to give the analysis some direction.
Tips
59
0 T/2
Vi
T t
Vp = 20 V
－
Vi
+
+
Fig.23
R
Dr. N.M. Safri/SEU3003_diode PN
－
5V
－
Vo
+
Determine the output waveform for the sinusoidal input of Fig. 23?
Activity (1)
Outline
60
0 T/2
Vi
Vp
Vp
T/2
0 T/2
Vi
0
Vi
Answer
T t
T t
T t
+
Vi
－
－
Vi
+
+
R
－
Vo
Dr. N.M. Safri/SEU3003_diode PN
R
+
－
Vo
+
Diode is conducting
0
Vo
T t
Vp + 5
T t
Outline
61
Vo = 0 V
T t
Vo = Vi + 5
Vp + 5
T/2
T/2
0 T/2
5 0
Vo
Vo Diode is not 5V － + conducting at Vi ≤ - 5 V
－
5V
0 T/2
Vi
T t
Vp = 20 V
－
Vi
+
－
Fig.24
R
Dr. N.M. Safri/SEU3003_diode PN
+
5V
－
Vo
+
Determine the output waveform for the sinusoidal input of Fig. 24? Compare the result with the Fig. 24.
Activity (2)
Outline
62
PIV
Dr. N.M. Safri/SEU3003_diode PN
Zener avalanche region
VR
IR (µA)
0 0.7 V (knee voltage)
Small reverse current due to the high reverse resistance
F (mA)
It is the voltage rating that must not be exceeded in the reverse-bias region or the diode will enter the Zener avalanche region. I
Outline
VF
The peak inverse voltage (PIV) rating of the diode is of primary importance in the design of rectification systems.
63
PIV
T t
+
Vi
－
+
I=0A
－
V(PIV)
0
Vi
T/2
Vp
T t
－
Vi
+
－
Vi
+
D2
Vo
+
+
V(PIV)
－
－ －
Vi
+
D1
Dr. N.M. Safri/SEU3003_diode PN
1:2
－
Vo
+
T/2
Vp
T t
Outline
64
Vo = Vi
T t
V(PIV) rating ≥ 2Vp
0
Vo
0 T/2
Vo
V(PIV) rating ≥ Vp
R
PIV rating for full-wave rectifier
0 T/2
Vi
PIV rating for half-wave rectifier
Dr. N.M. Safri/SEU3003_diode PN
Outline
Clamping networks have a capacitor connected directly from input to output with a resistive element in parallel with the output signal. The diode is also in parallel with the output signal but may not have a series dc supply as an added element.
A clamper is a network constructed of a diode, a resistor, and a capacitor that shifts a waveform to a different dc level without changing the appearance of the applied signal.
CLAMPERS
65
Dr. N.M. Safri/SEU3003_diode PN
Outline
5. Check that the total swing of the output matches that of the input.
4. Throughout the analysis, maintain a continual awareness of the location and defined polarity for Vo to ensure that the proper levels are obtained.
3. Assume that during the period when the diode is in the “off” state the capacitor holds on to its established voltage level.
First and mort important: 1. .Start the analysis by examining the response of the portion of the input signal that will forward bias the diode. Next: 2. During the period that the diode is in the “on” state, assume that the capacitor will charge up instantaneously to a voltage level determined by the surrounding network.
Tips for Clamper etworks
66
- 20
0
10
Vi
T
t1
t2 t3
t4
f = 1000 Hz
t
Fig.25
+
5 V－
R 100 kΩ
Dr. N.M. Safri/SEU3003_diode PN
－
Vi
+
C = 1µF
－
Vo
+
Determine Vo for the network of Fig. 25 for the input indicated.
Activity
Outline
67
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