bnmbnmb b hvh jvjhvjhvhbbnmvv v bvbvbuy v jh b

1. ECE 333 Linear Electronics
Chapter 4 Diodes
Ideal diode  real Si diode  diode circuits  device
modeling  serve as a foundation for modeling
transistors and circuits in future chapters
1

2. §4.1 The Ideal Diode
4.1.1 Current-Voltage Characteristics
Figure 4.1 The ideal diode: (a) diode circuit symbol; (b) i–v
characteristic; (c) equivalent circuit in the reverse direction; (d)
equivalent circuit in the forward direction. 2

3. 4.1.1 Current-Voltage Characteristics
Figure 4.2 The two modes of operation of ideal diodes and
the use of an external circuit to limit (a) the forward current
and (b) the reverse voltage. 3

4. 4.1.2 A Simple Application: The Rectifier
4

5. Ex. 4.1 For the circuit in Fig. 4(a), sketch the transfer characteristics.
Ex. 4.2 For the circuit in Fig4b, sketch the waveform of VD.
4.1.2 A Simple Application: The Rectifier
5

6. Ex. 4.3 In Fig. 4.3a, let VI have a peak value of 10V and R=1 kΩ. Find
the peak value of iD and dc component of v0. (Hint: the average
value of half-sine waves is Vp/π.)
Ans. 10 mA; 3.18 V
T
𝑣𝑎𝑣𝑒 =
0
𝑇
𝑣𝑜(𝑡) 𝑑𝑡/𝑇
𝑣𝑎𝑣𝑒 =
𝑉
𝑝
𝑇 0
𝑇/2
𝑠𝑖𝑛(2𝜋𝑓𝑡) 𝑑𝑡
𝑣𝑎𝑣𝑒 =
𝑉
𝑝
𝑇
1
2𝜋𝑓
−1 −1 − 1 =
𝑉
𝑝
𝜋
𝑖𝐷,𝑝 =
10𝑉
1000Ω
= 10𝑚𝐴
6

7. Example 4.1 vs is a sinusoid with 24-V peak amplitude, find the
fraction of each cycle during which the diode conducts. Also, find
the peak value of the diode current and the max reverse-bias
voltage across the diode.
Ans. The diode conducts when vs exceeds 12 V, conduction angle
is 2θ:
The peak value is:
24𝑐𝑜𝑠𝜃 = 12 2𝜃 = 120𝑜
𝐼𝑑,𝑝 =
24 − 12
100
= 0.12𝐴
7

8. 4.1.3 Another Application: Diode Logic Gates
Figure 4.5 Diode logic gates: (a) OR gate;
(b) AND gate (in a positive-logic system).
8

9. Example 4.2 Assuming the diodes to be ideal, find I and V in the
circuits of Fig. 4.6.
Ans. Two situations: D1 is on or off?
9

10. The characteristic curve consists of three distinct regions:
1. The forward-bias region, determined by v > 0
2. The reverse-bias region, determined by v < 0
3. The breakdown region, determined by v < -VZK
4.2 Terminal Characteristics of Junction Diode
10

11. 4.2.1 The Forward-Bias Region
𝑖 = 𝐼𝑠(𝑒𝑣/𝑉𝑇 − 1)
What is k, T and q?
IS is a constant for a given diode as a given temperature
For high carrier injection:
where n is from 1 to 2.
𝑖 = 𝐼𝑠(𝑒𝑣/𝑛𝑉𝑇 − 1)
𝑉𝑇 =
𝑘𝑇
𝑞
= 0.0862𝑇 (𝑚𝑉)
VT is 25 mV at 20oC
A good approximation: 𝑖 = 𝐼𝑠𝑒𝑣/𝑉𝑇
𝑣 = 𝑉𝑇𝑙𝑛
𝑖
𝐼𝑆
11

12. Because:
4.2.1 The Forward-Bias Region
𝐼1 = 𝐼𝑆𝑒𝑉1/𝑉𝑇
𝐼2 = 𝐼𝑆𝑒𝑉2/𝑉𝑇
𝑉2 − 𝑉1 = 2.3𝑉𝑇𝑙𝑜𝑔
𝐼2
𝐼1
Example 4.3 A Si diode said to be a 1-mA device displays
a forward voltage of 0.7 V at current of 1 mA. What is IS.
How about at a 1-A diode?
𝐼2
𝐼1
= 𝑒(𝑉2−𝑉1)/𝑉𝑇
12

13. Effect of temperature
4.2.1 The Forward-Bias Region
𝑖 = 𝐼𝑠𝑒𝑣/𝑉𝑇
𝑉𝑇 =
𝑘𝑇
𝑞
= 0.0862𝑇 (𝑚𝑉)
13

14. Real diodes exhibit reverse current much larger
than IS. A small-signal diode whose IS is on the
order of 10-14 A to 10-15 A could show a reverse
current on the order of 1 nA.
IS doubles for every 5oC rise in temperature.
Reverse current is mostly due to leakage effects.
It doubles for every 10oC rise in temperature.
4.2.2 The Reverse-Bias Region
𝑖 ≈ −𝐼𝑆
14

15. Ex. 4.9 The diode in Fig. E4.9 is a large high-current
device whose reverse leakage is reasonably
independent of voltage. If V=1 V at 20oC, find the
value V at 400C and 0oC.
4.2.2 The Reverse-Bias Region
Solve:
𝑖 𝑎𝑡 20𝑜
𝐶 =
𝑉
𝑅
= 10−6
𝐴
𝑖 𝑎𝑡 40𝑜𝐶 = 4 × 10−6 𝐴
𝑖 𝑎𝑡 0𝑜
𝐶 = 0.25 × 10−6
𝐴
𝑉 = 𝑖𝑅 = 4.0 𝑉
𝑉 = 𝑖𝑅 = 0.25 𝑉
15

16. • Beyond the breakdown voltage VZR ~ zener
breakdown at the “knee” of i-v curve
4.2.3 The Breakdown Region
Normally not destructive
Used in voltage regulator
16

17. 4.3 Modeling the Diode Forward Characteristic
• Why modeling?
-- We need to analyze the current and voltage in
the circuit.
• Ideal-diode model and exponential model
• This part will serve as a foundation for future
transistor modeling.
17

18. 4.3.1 The Exponential Model
𝐼𝐷 = 𝐼𝑆𝑒𝑉𝐷/𝑉𝑇
𝐼𝐷 =
𝑉𝐷𝐷 − 𝑉𝐷
𝑅
We can solve the two equations
to find VD and ID
18

19. • Graphical Analysis using the Exponential Model
4.3.2 Graphical Analysis
19

20. Ex. 4.4 Determine the current ID and VD with
VDD=5 V and R=1 kΩ. Assume ID=1 mA at VD=0.7V
4.3.3 Iterative Analysis Using the Exponential Model
Solve:
First, we assume VD=V1= 0.7 V
𝐼𝐷 =
𝑉𝐷𝐷 − 𝑉𝐷
𝑅
= 4.3 𝑚𝐴
𝑉2 − 𝑉1 = 2.3𝑉𝑇𝑙𝑜𝑔
𝐼2
𝐼1
𝑉2 = 𝑉1 + 2.3𝑉𝑇𝑙𝑜𝑔
𝐼2
𝐼1
= 0.7 + 0.06𝑙𝑜𝑔4.3 = 0.738
𝐼2 =
𝑉𝐷𝐷 − 0.738
1
= 4.262 𝑚𝐴
Use this ID =4.262 and VD=0.738 𝑉2 = 0.738 + 2.3𝑉𝑇𝑙𝑜𝑔
4.262
4.3
= 0.738 V
20

21. • 4.3.4 The Need for rapid analysis
• 4.3.5 The Constant-Voltage-Drop Model
𝐼𝐷 =
𝑉𝐷𝐷 − 0.7
𝑅
= 4.3 𝑚𝐴
21

22. 4.3.6 The Ideal-Diode Model
𝑉𝐷 = 0 𝑉
𝐼𝐷 =
𝑉𝐷𝐷 − 0
𝑅
= 5 𝑚𝐴
22

23. 4.3.7 The Small-Signal Model
23

24. 4.3.7 The Small-Signal Model
𝑣𝐷 𝑡 = 𝑉𝐷 + 𝑣𝑑(𝑡)
Voltage across the diode:
Induce a current:
𝑖𝐷(𝑡) = 𝐼𝑆𝑒𝑣𝐷(𝑡)/𝑉𝑇
𝑖𝐷(𝑡) = 𝐼𝑆𝑒(𝑉𝐷+𝑣𝑑(𝑡))/𝑉𝑇
𝑖𝐷(𝑡) = 𝐼𝑆𝑒(𝑉𝐷+𝑣𝑑)/𝑉𝑇
Or simplified as:
𝑖𝐷(𝑡) = 𝐼𝑆𝑒𝑉𝐷/𝑉𝑇𝑒𝑣𝑑/𝑉𝑇
𝑖𝐷(𝑡) = 𝐼𝐷𝑒𝑣𝑑/𝑉𝑇
24

25. Because: vd/VT <<1 (that mean vd should be
much less than 26 mV at room temperature)
Use Taylor’s expansion:
or
4.3.7 The Small-Signal Model
𝑖𝐷(𝑡) = 𝐼𝐷𝑒𝑣𝑑/𝑉𝑇
𝑖𝐷(𝑡) = 𝐼𝐷(1 +
𝑣𝑑
𝑉𝑇
)
𝑖𝐷(𝑡) = 𝐼𝐷 +
𝐼𝐷
𝑉𝑇
𝑣𝑑
25

26. 4.3.7 The Small-Signal Model
𝑖𝐷(𝑡) = 𝐼𝐷𝑒𝑣𝑑/𝑉𝑇
𝑖𝐷(𝑡) = 𝐼𝐷 +
𝐼𝐷
𝑉𝑇
𝑣𝑑
𝑖𝐷(𝑡) = 𝐼𝐷 + 𝑖𝑑
𝑖𝑑 =
𝐼𝐷
𝑉𝑇
𝑣𝑑 =
𝑣𝑑
𝑟𝑑
𝑤ℎ𝑒𝑟𝑒 𝑟𝑑 =
𝑉𝑇
𝐼𝐷
rd : diode small-signal resistance
or incremental resistance
26

27. 4.3.7 The Small-Signal Model
𝑟𝑑 = 1/
𝜕𝑖𝐷
𝜕𝑣𝐷 𝑖𝐷=𝐼𝐷
𝑖𝑑 =
𝐼𝐷
𝑉𝑇
𝑣𝑑 =
𝑣𝑑
𝑟𝑑
𝑟𝑑 =
𝑉𝑇
𝐼𝐷
𝑖𝐷(𝑡) = 𝐼𝑆𝑒𝑣𝐷(𝑡)/𝑉𝑇
27

28. Example 4.5: power supply has a dc value of 10
V and 60-Hz sinusoid of 1-V peak amplitude.
What is the dc voltage and amplitude of sine
wave of the diode?
4.3.7 The Small-Signal Model
28

29. Example 4.5: power supply has a dc value of 10 V and 60-Hz
sinusoid of 1-V peak amplitude. What is the dc voltage and
amplitude of sine wave of the diode (VD=0.7 V at ID=1 mA)?
Solve:
Assume: VD ≈ 0.7 V, the diode dc current is
𝐼𝐷 =
10 − 0.7
10
= 0.93 𝑚𝐴
This is very close to 1 mA, so the diode voltage VD is indeed
very close to 0.7 V. So, at this operation point, the diode
incremental resistance rd is:
𝑟𝑑 =
𝑉𝑇
𝐼𝐷
=
25
0.93
= 26.9 Ω
𝑣𝑑 𝑝𝑒𝑎𝑘 = Δ𝑉
𝑠
𝑟𝑑
𝑅 + 𝑟𝑑
= 1
26.9
10000 + 26.9
= 2.68 𝑚𝑉
So, peak amplitude of vd is:
29

30. 4.3.8 Use of the Diode Forward Drop in
Voltage Regulation
• A voltage regulator is a circuit to provide a
constant dc voltage between its output
terminals.
• To avoid: (a) changes in the load current
drawn from the regulator output terminal and
(b) changes in the dc power-supply voltage
30

31. Example 4.6 Consider the circuit in Fig. 4.17. A string of three diodes
is used to provide a constant voltage of about 2.1 V. We want to
calculate the % change in this regulated voltage caused by (a) a ±
10% change in power supply voltage, and (b) connection of a 1-kΩ
load resistance.
Solve:
(a) With no load (RL), the current in the diode
string is:
𝐼𝐷 ≈
10 − 0.7 × 3
1000
= 7.9 𝑚𝐴
𝑟𝑑 =
𝑉𝑇
𝐼𝐷
=
25
7.9
= 3.2 Ω
So, resistance of 3 diodes is: r = 9.6Ω
∆𝑣𝑜 = 2
𝑟
𝑟 + 𝑅
= 2
9.6
9.6 + 1000
= 19 𝑚𝑉 (𝑝𝑒𝑎𝑘 𝑡𝑜 𝑝𝑒𝑎𝑘)
Is this solution precise? 31

32. 𝐼𝐷 ≈
10 − 0.7 × 3
1000
= 7.9 𝑚𝐴
Is the constant voltage model precise?
𝑉𝐷 ≠ 0.7 𝑉
𝑉𝐷 = 𝑉1 + 60𝑚𝑉 l𝑜𝑔
𝐼𝐷
𝐼1
= 0.7+0.06log(7.9)=0.705 V
𝐼𝐷 ≈
10 − 0.705 × 3
1000
= 7.89 𝑚𝐴
𝑟𝑑 =
𝑉𝑇
𝐼𝐷
=
25
7.89
= 3.17 Ω Yes, it is quite precise!
This is very close to 7.9 mA.
32

33. Example 4.6 Consider the circuit in Fig. 4.17. A string of three diodes
is used to provide a constant voltage of about 2.1 V. We want to
calculate the % change in this regulated voltage caused by (a) a ±
10% change in power supply voltage, and (b) connection of a 1-kΩ
load resistance.
Solve:
(b) When a load RL is connected across the
diode string, it draws a current of
approximately 2.1 mA
𝐼𝐿 ≈
0.7 × 3
1000
= 2.1 𝑚𝐴
So, the current through diodes decreases by
2.1 mA, resulting in a decrease in voltage:
∆𝑣𝑜 = −2.1 𝑚𝐴 × 𝑟 = −2.1 × 9.6 = −20 𝑚𝑉
So, this implies that the voltage across each diode
decreases by 6.7 mV: Is this solution precise?
33

34. Let’s use exponential model to solve it.
𝐼𝐿 ≈
0.7 × 3
1000
= 2.1 𝑚𝐴
Let’s use exponential model to solve it.
0.740 0.742 0.744 0.746 0.748 0.750
4.0
4.2
4.4
4.6
4.8
5.0
5.2
5.4
5.6
5.8
6.0
6.2
6.4
6.6
6.8
7.0
7.2
7.4
7.6
7.8
8.0
function
value
VD
[ V ]
left : 10-6VD
right : exp[(VD
-0.7)/0.025]
VD
= 0.743 V
𝐼𝐷 = 𝐼𝑆𝑒𝑉𝐷/𝑉𝑇
𝐼𝑆 = 1𝑒−0.7/𝑉𝑇
𝐼𝐷 + 𝐼𝐿 =
10 − 3𝑉𝐷
1
𝐼𝐿 = 3𝑉𝐷/1
10 − 6𝑉𝐷 = 𝑒 (𝑉𝐷−0.7)/0.025
𝐼𝐿 =
3𝑉𝐷
1
= 2.3𝑚𝐴
So current in diode decreases by 2.29 mA.
∆𝑣𝑜 = −2.3 𝑚𝐴 × 𝑟 = −23 𝑚𝑉 Not much difference
34

35. 4.4 Operation in the Reverse
Breakdown Region-Zener Diodes
Figure 4.18 Circuit symbol for a zener diode.
35

36. 4.4.1 Specifying and Modeling the
Zener Diode
∆V = 𝑟𝑧∆𝐼
rz is the incremental
resistance or dynamic
resistance of Zener
diode
36

37. • VZ = VZ0 + rZ IZ
• For IZ > IZK and VZ > VZ0
4.4.1 Specifying and Modeling the
Zener Diode
37

38. Example 4.7 The 6.8-V zener diode in the circuit of Fig. 4.21(a) is
specified to have VZ = 6.8 V at IZ=5mA, rZ=20Ω, and IZK=0.2mA.
The supply voltage V+ is nominally 10 V but can vary by ±1 V.
(a)Find Vo with no load and with V+ as its nominal value.
Solve: VZ = VZ0 + rZ IZ
VZ0 = VZ - rZ IZ = 6.8 -0.1 = 6.7 V
With no load:
𝐼𝑍 = 𝐼 =
𝑉+ − 𝑉𝑍0
𝑅 + 𝑟𝑍
=
10 − 6.7
0.5 + 0.02
= 6.35 𝑚𝐴
Vo = VZ = VZ0 + rZ IZ =6.7 +0.00635×20=6.83 V
38

39. Example 4.7 The 6.8-V zener diode in the circuit of Fig. 4.21(a) is
specified to have VZ = 6.8 V at IZ=5mA, rZ=20Ω, and IZK=0.2mA.
The supply voltage V+ is nominally 10 V but can vary by ±1 V.
(b) Find the change in Vo resulting from the ±1-V change in V+.
Solve:
So, the line regulation is
∆𝑉
𝑜 = ∆𝑉+
𝑟𝑧
𝑟𝑧 + 𝑅
= ±1
20
20 + 500
= ±38.5 𝑚𝑉
∆𝑉
𝑜 ∆𝑉+ = ±38.5 𝑚𝑉/±1𝑉 = 38.5 𝑚𝑉/𝑉
39

40. Example 4.7 The 6.8-V zener diode in the circuit of Fig. 4.21(a) is
specified to have VZ = 6.8 V at IZ=5mA, rZ=20Ω, and IZK=0.2mA.
The supply voltage V+ is nominally 10 V but can vary by ±1 V.
(c) Find the change in Vo resulting from connectiong a load
resistance RL that draws a current IL=1 mA..
Solve:
When RL draws 1 mA, the zener
diode current will decrease by 1
mA, the corresponding change
in zener diode voltage is:
∆𝑉
𝑜 = 𝑟𝑧 × ∆𝐼𝑧 = 20 × −1
= −20 𝑚𝑉
So, the load regulation is
∆𝑉
𝑜 ∆𝐼𝐿 = −20𝑚𝑉/𝑚𝐴
40

41. (d) Find the change in V0 when RL = 2 kΩ.
(e) Find the value of V0 when RL = 0.5 kΩ.
(f) What is the minimum value of RL for which the diode still
operates in the breakdown region?
Solve: The load current is about 6.8 V/ 2kΩ = 3.4 mA. So the
change in Vo is rz× (- 3.4 mA) = -68 mV
Solve: Assume zener diode is operated in breakdown
region, the load current is 6.8 V/ 0.5 kΩ = 13.6 mA. This is
not possible because it is large than supplied current. So
zener is cut off. Vo = V+ (RL/(RL+R))=5V
Solve: for the zener diode to be operated at the edge of
breakdown region, Iz=Izk=0.2 mA and Vz=Vzk=6.7 V. so the current
through R is (9-6.7)/0.5=4.6. so current in RL is 4.6-0.2=4.4. so
RL=6.7/4.4 = 1.5 kΩ 41

42. 4.5 Rectifier Circuits
Ripple  dc output
𝑣𝑠 = 𝑎𝑐 𝑣𝑜𝑙𝑡𝑎𝑔𝑒 𝑜𝑓 120
𝑁2
𝑁1
Power transformer
dc Power supply 42

43. 4.5.1 The Half-Wave Rectifier
𝑣𝑜 = 0 ; 𝑣𝑠 < 𝑉𝐷
𝑣𝑜 = 𝑣𝑠 − 𝑉𝐷; 𝑣𝑠 < 𝑉𝐷
1. Current-handling capability
required of the diode;
2. Peak inverse voltage (PIV)
that the diode must be able to
withstand without breakdown
PIV = VS (Considering reverse bias)
𝑣𝑠 = 𝑉
𝑠𝑠𝑖𝑛𝜃, 𝜃 = 2𝜋𝑓𝑡
43

44. Ex. 4.19 For Fig. 4.23(a), show and find: (a) the conduction angle
(π-2θ), where conduction begins at an angle θ=sin-1(VD/VS) and
terminates at (π-θ). (b) the average value of vo is 𝑉
𝑜 ≅
1 𝜋 𝑉
𝑠 − 𝑉𝐷 2. (c) the peak diode current is (Vs-VD)/R and PIV.
(given: Vs is 12-V (rms) sinusoidal input, VD=0.7 V, and R=100Ω.
Solve:
π-θ
θ
𝑉
𝑠sinθ
44

45. Ex. 4.19 For Fig. 4.23(a), show and find: (b) the average value of
vo is 𝑉
𝑜 ≅ 1 𝜋 𝑉
𝑠 − 𝑉𝐷 2.
Solve:
π-θ
θ
45

46. Ex. 4.19 For Fig. 4.23(a), show and find: (c) the peak diode
current is (Vs-VD)/R and PIV.
(given: Vs is 12-V (rms) sinusoidal input, VD=0.7 V, and R=100Ω.
Solve:
π-θ
θ
* Rms: root-mean-squared. 12-V root-mean-squared
sinusoid wave is 12 2𝑠𝑖𝑛𝜃. Here, Vs= 12 2
46

47. 4.5.2 The Full-Wave Rectifier
transformer
𝑣𝑠 = 𝑉
𝑠𝑠𝑖𝑛𝜃, 𝜃 = 2𝜋𝑓𝑡
PIV = 2VS - VD
𝑣𝑜 = 𝑣𝑠 − 𝑉𝐷
47

48. 4.5.2 The Full-Wave Rectifier
PIV = VS + VS – VD = 2VS - VD
At this point, D1 feel the PIV
VS
VS - VD
48

49. 4.5.3 The Bridge Rectifier
49

50. 4.5.3 The Bridge Rectifier
What is the PIV?
vD3(reverse) = vo + vD2(forward)
So, PIV = VS - 2VD + VD = VS - VD
𝑣𝑜 = 𝑣𝑠 − 2𝑉𝐷
+
-
𝑣𝑜
D3
D2
50

51. 4.5.4 The Rectifier with a Filter
Capacitor – The peak Rectifier
• The filter capacitor serves to reduce
substantially the variations in output voltage
• Once charge, no way to discharge the capacitor
51

52. Consider real application-- with load resistance: R
52
Assume: CR >> T
𝑖𝐿 = 𝑣𝑜 𝑅
𝑖𝐷 = 𝑖𝐶 + 𝑖𝐿
= 𝐶
𝑑𝑣𝐼
𝑑𝑡
+ 𝑖𝐿

53. Consider real application-- with load resistance: R
53
1. Diode conducts for a brief interval Δt;
2. Assume an ideal diode: from t1 to t2
t1: vI = vo
t2: iD = 0, shortly after the peak of vI
3. During the diode-off interval, C
discharges through R. vo decays
exponentially with time constant CR; at the
end of discharge interval (≈ T), v0 = Vp - Vr
4. When Vr is small, vo is almost constant
and equal to Vp. So, iL is almost constant:
Its dc component IL is
𝐼𝐿 =
𝑉
𝑝
𝑅
* Vr is peak-to-peak ripple voltage

54. 54
Average of vo is
𝑉
𝑜 = 𝑉
𝑝 −
1
2
𝑉
𝑟
Now, we will derive Vr
During the diode-off interval
𝑣𝑜 = 𝑉
𝑝𝑒−𝑡 𝐶𝑅
At the end of discharge interval
𝑉
𝑝 − 𝑉
𝑟 ≈ 𝑉
𝑝𝑒−𝑇 𝐶𝑅
𝑉
𝑟 ≈ 𝑉
𝑝 𝑇/𝐶𝑅
𝑉
𝑟 ≈
𝑉
𝑝
𝑓𝐶𝑅
𝑉
𝑟 ≈
𝐼𝐿
𝑓𝐶
(use Taylor expansion)

55. 55
We can now determine conduction
Interval Δt
𝑉
𝑝 cos 𝜔∆𝑡 = 𝑉
𝑝 − 𝑉
𝑟
(Q: why not using sin(wΔt)?)
use Taylor expansion because Δt is small
cos 𝜔∆𝑡 ≈ 1 −
1
2
(𝜔∆𝑡)2
𝜔∆𝑡 ≈ 2𝑉
𝑟 𝑉
𝑝

56. • From the above results, we can also
determine average diode current during
conduction 𝑖𝐷𝑎𝑣 = 𝐼𝐿(1 + 𝜋 2𝑉
𝑝 𝑉
𝑟) ; this is
much larger than load current during diode
conduction.
• And the peak value of diode current
𝑖𝐷𝑚𝑎𝑥 = 𝐼𝐿 1 + 2𝜋 2𝑉
𝑝 𝑉
𝑟
The above is for half-wave peak rectifier.
56

57. • For a full-wave rectifier circuit with a capacitor,
discharge period T is replaced by T/2 :
57
𝑉
𝑟 ≈
𝑉
𝑝
2𝑓𝐶𝑅
𝑖𝐷𝑎𝑣 = 𝐼𝐿 1 + 𝜋 𝑉
𝑝 2𝑉
𝑟
𝑖𝐷𝑚𝑎𝑥 = 𝐼𝐿 1 + 2𝜋 𝑉
𝑝 2𝑉
𝑟

58. 4.5.5 Precision Half-Wave Rectifier –
The superdiode
58
Negative-feedback path
Of an op amp
Load
𝑣𝑜 = 𝑣𝐼 𝑓𝑜𝑟 ≥ 0

59. 4.6 Limiting and Clamping Circuits
(self-reading)
59
Figure 4.30 General transfer characteristic for a limiter circuit.
Figure 4.31 Applying a sine wave to a limiter can result in clipping off its two peaks.

60. 60
4.6 Limiting and Clamping Circuits
(self-reading)

61. 4.7 Special Diode Types
(self-reading)
• The Schottky-Barrier Diode (SBD)
• Varactors
• Photodiodes
• Light-Emitting Diodes (LEDs)
61

62. Chapter 4 Homework
4.4, 4.9, 4.10, 4.18, 4.23, 4.28, 4.36, 4.54, 4.58,
4.61, 4.76
It is due March 1st (Tuesday).
62