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
§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
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
4.1.2 A Simple Application: The Rectifier
4
4.1.2 A Simple Application: The Rectifier
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.
5
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
𝑖𝐷,𝑝
10𝑉
=
= 10𝑚𝐴
1000Ω
𝑇
𝑣𝑎𝑣𝑒 =
𝑣𝑜 (𝑡) 𝑑𝑡/𝑇
0
𝑣𝑎𝑣𝑒
𝑣𝑎𝑣𝑒
𝑉𝑝
=
𝑇
T
𝑇/2
𝑠𝑖𝑛(2𝜋𝑓𝑡) 𝑑𝑡
0
𝑉𝑝 1
𝑉𝑝
=
−1 −1 − 1 =
𝑇 2𝜋𝑓
𝜋
6
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:
24𝑐𝑜𝑠𝜃 = 12
2𝜃 = 120𝑜
The peak value is:
𝐼𝑑,𝑝 =
24 − 12
= 0.12𝐴
100
7
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
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
4.2 Terminal Characteristics of Junction Diode
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
10
4.2.1 The Forward-Bias Region
𝑖 = 𝐼𝑠 (𝑒 𝑣/𝑉𝑇 − 1)
𝑉𝑇 =
𝑘𝑇
= 0.0862𝑇 (𝑚𝑉)
𝑞
VT is 25 mV at 20oC
A good approximation: 𝑖 = 𝐼𝑠 𝑒 𝑣/𝑉𝑇
𝑣 = 𝑉𝑇 𝑙𝑛
𝑖
𝐼𝑆
What is k, T and q?
IS is a constant for a given diode as a given temperature
For high carrier injection: 𝑖 = 𝐼𝑠 (𝑒 𝑣/𝑛𝑉𝑇 − 1)
where n is from 1 to 2.
11
4.2.1 The Forward-Bias Region
Because:
𝐼1 = 𝐼𝑆 𝑒 𝑉1 /𝑉𝑇
𝐼2 = 𝐼𝑆 𝑒 𝑉2/𝑉𝑇
𝐼2
= 𝑒 (𝑉2−𝑉1)/𝑉𝑇
𝐼1
𝑉2 − 𝑉1 = 2.3𝑉𝑇 𝑙𝑜𝑔
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?
12
4.2.1 The Forward-Bias Region
Effect of temperature
𝑖 = 𝐼𝑠 𝑒 𝑣/𝑉𝑇
𝑘𝑇
𝑉𝑇 =
= 0.0862𝑇 (𝑚𝑉)
𝑞
13
4.2.2 The Reverse-Bias Region
𝑖 ≈ −𝐼𝑆
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.
14
4.2.2 The Reverse-Bias Region
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.
Solve:
𝑖 𝑎𝑡 20𝑜 𝐶 =
𝑉
= 10−6 𝐴
𝑅
𝑖 𝑎𝑡 40𝑜 𝐶 = 4 × 10−6 𝐴
𝑉 = 𝑖𝑅 = 4.0 𝑉
𝑖 𝑎𝑡 0𝑜 𝐶 = 0.25 × 10−6 𝐴
𝑉 = 𝑖𝑅 = 0.25 𝑉
15
4.2.3 The Breakdown Region
• Beyond the breakdown voltage VZR ~ zener
breakdown at the “knee” of i-v curve
Normally not destructive
Used in voltage regulator
16
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
4.3.1 The Exponential Model
𝐼𝐷 = 𝐼𝑆 𝑒 𝑉𝐷 /𝑉𝑇
𝑉𝐷𝐷 − 𝑉𝐷
𝐼𝐷 =
𝑅
We can solve the two equations
to find VD and ID
18
4.3.2 Graphical Analysis
• Graphical Analysis using the Exponential Model
19
4.3.3 Iterative Analysis Using the Exponential Model
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
Solve:
First, we assume VD=V1= 0.7 V
𝑉𝐷𝐷 − 𝑉𝐷
= 4.3 𝑚𝐴
𝑅
𝐼2
𝑉2 − 𝑉1 = 2.3𝑉𝑇 𝑙𝑜𝑔
𝐼1
𝐼2
𝑉2 = 𝑉1 + 2.3𝑉𝑇 𝑙𝑜𝑔 = 0.7 + 0.06𝑙𝑜𝑔4.3 = 0.738
𝐼1
𝑉𝐷𝐷 − 0.738
𝐼2 =
= 4.262 𝑚𝐴
1
𝐼𝐷 =
Use this ID =4.262 and VD=0.738
𝑉2 = 0.738 + 2.3𝑉𝑇 𝑙𝑜𝑔
4.262
= 0.738 V
4.3
20
• 4.3.4 The Need for rapid analysis
• 4.3.5 The Constant-Voltage-Drop Model
𝐼𝐷 =
𝑉𝐷𝐷 − 0.7
= 4.3 𝑚𝐴
𝑅
21
4.3.6 The Ideal-Diode Model
𝑉𝐷 = 0 𝑉
𝐼𝐷 =
𝑉𝐷𝐷 − 0
= 5 𝑚𝐴
𝑅
22
4.3.7 The Small-Signal Model
23
4.3.7 The Small-Signal Model
Voltage across the diode:
𝑣𝐷 𝑡 = 𝑉𝐷 + 𝑣𝑑 (𝑡)
Induce a current:
𝑖𝐷 (𝑡) = 𝐼𝑆 𝑒 𝑣𝐷(𝑡) /𝑉𝑇
𝑖𝐷 (𝑡) = 𝐼𝑆 𝑒 (𝑉𝐷 +𝑣𝑑 (𝑡))/𝑉𝑇
Or simplified as:
𝑖𝐷 (𝑡) = 𝐼𝑆 𝑒 (𝑉𝐷 +𝑣𝑑 )/𝑉𝑇
𝑖𝐷 (𝑡) = 𝐼𝑆 𝑒 𝑉𝐷 /𝑉𝑇 𝑒 𝑣𝑑 /𝑉𝑇
𝑖𝐷 (𝑡) = 𝐼𝐷 𝑒 𝑣𝑑 /𝑉𝑇
24
4.3.7 The Small-Signal Model
Because: vd/VT <<1 (that mean vd should be
much less than 26 mV at room temperature)
𝑖𝐷 (𝑡) = 𝐼𝐷 𝑒 𝑣𝑑 /𝑉𝑇
Use Taylor’s expansion:
𝑣𝑑
𝑖𝐷 (𝑡) = 𝐼𝐷 (1 + )
𝑉𝑇
or
𝐼𝐷
𝑖𝐷 (𝑡) = 𝐼𝐷 + 𝑣𝑑
𝑉𝑇
25
4.3.7 The Small-Signal Model
𝑖𝐷 (𝑡) = 𝐼𝐷 𝑒 𝑣𝑑 /𝑉𝑇
𝐼𝐷
𝑖𝐷 (𝑡) = 𝐼𝐷 + 𝑣𝑑
𝑉𝑇
𝑖𝐷 (𝑡) = 𝐼𝐷 + 𝑖𝑑
𝐼𝐷
𝑣𝑑
𝑖𝑑 = 𝑣𝑑 =
𝑉𝑇
𝑟𝑑
rd : diode small-signal resistance
or incremental resistance
𝑉𝑇
𝑤ℎ𝑒𝑟𝑒 𝑟𝑑 =
𝐼𝐷
26
4.3.7 The Small-Signal Model
𝑖𝐷 (𝑡) = 𝐼𝑆 𝑒 𝑣𝐷(𝑡) /𝑉𝑇
𝐼𝐷
𝑣𝑑
𝑖𝑑 = 𝑣𝑑 =
𝑉𝑇
𝑟𝑑
𝑟𝑑 =
𝑉𝑇
𝐼𝐷
𝜕𝑖𝐷
𝑟𝑑 = 1/
𝜕𝑣𝐷
𝑖𝐷 =𝐼𝐷
27
4.3.7 The Small-Signal Model
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?
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 (VD=0.7 V at ID=1 mA)?
Solve:
Assume: VD ≈ 0.7 V, the diode dc current is
10 − 0.7
𝐼𝐷 =
= 0.93 𝑚𝐴
10
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:
𝑉𝑇
25
𝑟𝑑 =
=
= 26.9 Ω
𝐼𝐷 0.93
So, peak amplitude of vd is:
𝑟𝑑
26.9
𝑣𝑑 𝑝𝑒𝑎𝑘 = Δ𝑉𝑠
=1
= 2.68 𝑚𝑉
𝑅 + 𝑟𝑑
10000 + 26.9
29
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
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:
10 − 0.7 × 3
𝐼𝐷 ≈
= 7.9 𝑚𝐴
1000
𝑉𝑇 25
𝑟𝑑 =
=
= 3.2 Ω
𝐼𝐷 7.9
So, resistance of 3 diodes is: r = 9.6Ω
𝑟
9.6
∆𝑣𝑜 = 2
=2
= 19 𝑚𝑉 (𝑝𝑒𝑎𝑘 𝑡𝑜 𝑝𝑒𝑎𝑘)
𝑟+𝑅
9.6 + 1000
Is this solution precise?
31
Is the constant voltage model precise?
10 − 0.7 × 3
𝐼𝐷 ≈
= 7.9 𝑚𝐴
1000
𝑉𝐷 ≠ 0.7 𝑉
𝐼𝐷
𝑉𝐷 = 𝑉1 + 60𝑚𝑉 l𝑜𝑔
= 0.7+0.06log(7.9)=0.705 V
𝐼1
10 − 0.705 × 3
𝐼𝐷 ≈
= 7.89 𝑚𝐴
1000
This is very close to 7.9 mA.
𝑉𝑇
25
𝑟𝑑 =
=
= 3.17 Ω
𝐼𝐷 7.89
Yes, it is quite precise!
32
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
0.7 × 3
𝐼𝐿 ≈
= 2.1 𝑚𝐴
1000
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
33
decreases by 6.7 mV:
Is this solution precise?
Let’s use exponential model to solve it.
0.7 × 3
𝐼𝐿 ≈
= 2.1 𝑚𝐴
1000
Let’s use exponential model to solve it.
𝐼𝐷 = 𝐼𝑆 𝑒 𝑉𝐷 /𝑉𝑇
10 − 3𝑉𝐷
𝐼𝐷 + 𝐼𝐿 =
1
𝐼𝐿 = 3𝑉𝐷 /1
10 − 6𝑉𝐷 = 𝑒
(𝑉𝐷 −0.7)/0.025
function value
𝐼𝑆 = 1𝑒 −0.7/𝑉𝑇
8.0
7.8
7.6
7.4
7.2
7.0
6.8
6.6
6.4
6.2
6.0
5.8
5.6
5.4
5.2
5.0
4.8
4.6
4.4
4.2
4.0
3𝑉𝐷
𝐼𝐿 =
= 2.3𝑚𝐴
0.740
1
So current in diode decreases by 2.29 mA.
∆𝑣𝑜 = −2.3 𝑚𝐴 × 𝑟 = −23 𝑚𝑉
left : 10-6VD
right : exp[(VD-0.7)/0.025]
VD = 0.743 V
0.742
0.744
0.746
0.748
0.750
VD [ V ]
34
Not much difference
4.4 Operation in the Reverse
Breakdown Region-Zener Diodes
Figure 4.18 Circuit symbol for a zener diode.
35
4.4.1 Specifying and Modeling the
Zener Diode
∆V = 𝑟𝑧 ∆𝐼
rz is the incremental
resistance or dynamic
resistance of Zener
diode
36
4.4.1 Specifying and Modeling the
Zener Diode
• VZ = VZ0 + rZ IZ
• For IZ > IZK and VZ > VZ0
37
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: V = V + r I
Z
Z0
Z Z
VZ0 = VZ - rZ IZ = 6.8 -0.1 = 6.7 V
With no load:
𝑉 + − 𝑉𝑍0
10 − 6.7
𝐼𝑍 = 𝐼 =
=
𝑅 + 𝑟𝑍
0.5 + 0.02
= 6.35 𝑚𝐴
Vo = VZ = VZ0 + rZ IZ =6.7 +0.00635×20=6.83 V
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.
(b) Find the change in Vo resulting from the ±1-V change in V+.
Solve:
𝑟𝑧
+
∆𝑉𝑜 = ∆𝑉
𝑟𝑧 + 𝑅
20
= ±1
20 + 500
= ±38.5 𝑚𝑉
So, the line regulation is
∆𝑉𝑜 ∆𝑉 + = ±38.5 𝑚𝑉/±1𝑉 = 38.5 𝑚𝑉/𝑉
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.
(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:
∆𝑉𝑜 = 𝑟𝑧 × ∆𝐼𝑧 = 20 × −1
= −20 𝑚𝑉
So, the load regulation is
∆𝑉𝑜 ∆𝐼𝐿 = −20𝑚𝑉/𝑚𝐴
40
(d) Find the change in V0 when RL = 2 kΩ.
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
(e) Find the value of V0 when RL = 0.5 kΩ.
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
(f) What is the minimum value of RL for which the diode still
operates in the breakdown region?
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
4.5 Rectifier Circuits
𝑁2
𝑣𝑠 = 𝑎𝑐 𝑣𝑜𝑙𝑡𝑎𝑔𝑒 𝑜𝑓 120
𝑁1
Power transformer
Ripple  dc output
dc Power supply
42
4.5.1 The Half-Wave Rectifier
𝑣𝑠 = 𝑉𝑠 𝑠𝑖𝑛𝜃, 𝜃 = 2𝜋𝑓𝑡
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 43
bias)
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 𝑉𝑜 ≅
1 𝜋 𝑉𝑠 − 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:
θ
π-θ
𝑉𝑠 sinθ
44
Ex. 4.19 For Fig. 4.23(a), show and find: (b) the average value of
vo is 𝑉𝑜 ≅ 1 𝜋 𝑉𝑠 − 𝑉𝐷 2.
Solve:
θ
π-θ
45
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:
θ
π-θ
* Rms: root-mean-squared. 12-V root-mean-squared
sinusoid wave is 12 2𝑠𝑖𝑛𝜃. Here, Vs= 12 2
46
4.5.2 The Full-Wave Rectifier
transformer
𝑣𝑜 = 𝑣𝑠 − 𝑉𝐷
𝑣𝑠 = 𝑉𝑠 𝑠𝑖𝑛𝜃, 𝜃 = 2𝜋𝑓𝑡
PIV = 2VS - VD
47
4.5.2 The Full-Wave Rectifier
VS
VS - VD
At this point, D1 feel the PIV
PIV = VS + VS – VD = 2VS - VD
48
4.5.3 The Bridge Rectifier
49
4.5.3 The Bridge Rectifier
𝑣𝑜 = 𝑣𝑠 − 2𝑉𝐷
What is the PIV?
vD3(reverse) = vo + vD2(forward)
So, PIV = VS - 2VD + VD = VS - VD
D3
+
𝑣𝑜
D2
50
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
Consider real application-- with load resistance: R
Assume: CR >> T
𝑖𝐿 = 𝑣𝑜 𝑅
𝑖𝐷 = 𝑖𝐶 + 𝑖𝐿
𝑑𝑣𝐼
=𝐶
+ 𝑖𝐿
𝑑𝑡
52
Consider real application-- with load resistance: R
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
* Vr is peak-to-peak ripple voltage
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
𝑉𝑝
𝐼𝐿 =
𝑅
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Average of vo is
1
𝑉𝑜 = 𝑉𝑝 − 𝑉𝑟
2
Now, we will derive Vr
During the diode-off interval
𝑣𝑜 = 𝑉𝑝 𝑒 −𝑡
𝐶𝑅
At the end of discharge interval
𝑉𝑝 − 𝑉𝑟 ≈ 𝑉𝑝 𝑒 −𝑇
𝐶𝑅
(use Taylor expansion)
𝑉𝑟 ≈ 𝑉𝑝 𝑇/𝐶𝑅
𝑉𝑝
𝑉𝑟 ≈
𝑓𝐶𝑅
𝐼𝐿
𝑉𝑟 ≈
𝑓𝐶
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We can now determine conduction
Interval Δt
𝑉𝑝 cos 𝜔∆𝑡 = 𝑉𝑝 − 𝑉𝑟
(Q: why not using sin(wΔt)?)
use Taylor expansion because Δt is small
1
cos 𝜔∆𝑡 ≈ 1 − (𝜔∆𝑡)2
2
𝜔∆𝑡 ≈
2𝑉𝑟 𝑉𝑝
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• From the above results, we can also
determine average diode current during
conduction 𝑖𝐷𝑎𝑣 = 𝐼𝐿 (1 + 𝜋 2𝑉𝑝 𝑉𝑟 ) ; this is
much larger than load current during diode
conduction.
• And the peak value of diode current
𝑖𝐷𝑚𝑎𝑥 = 𝐼𝐿 1 + 2𝜋 2𝑉𝑝 𝑉𝑟
The above is for half-wave peak rectifier.
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• For a full-wave rectifier circuit with a capacitor,
discharge period T is replaced by T/2 :
𝑉𝑝
𝑉𝑟 ≈
2𝑓𝐶𝑅
𝑖𝐷𝑎𝑣 = 𝐼𝐿 1 + 𝜋 𝑉𝑝 2𝑉𝑟
𝑖𝐷𝑚𝑎𝑥 = 𝐼𝐿 1 + 2𝜋 𝑉𝑝 2𝑉𝑟
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4.5.5 Precision Half-Wave Rectifier –
The superdiode
Negative-feedback path
Of an op amp
Load
𝑣𝑜 = 𝑣𝐼 𝑓𝑜𝑟
≥0
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4.6 Limiting and Clamping Circuits
(self-reading)
Figure 4.31 Applying a sine wave to a limiter can result in clipping off its two peaks.
Figure 4.30 General transfer characteristic for a limiter circuit.
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4.6 Limiting and Clamping Circuits
(self-reading)
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4.7 Special Diode Types
(self-reading)
•
•
•
•
The Schottky-Barrier Diode (SBD)
Varactors
Photodiodes
Light-Emitting Diodes (LEDs)
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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).
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