Chapter 1:
Diodes and applications
Pham Duy Hung, PhD
Faculty of Electronics and Telecommunications,
VNU-University of Engineering and Technology
Email: hungpd@vnu.edu.vn
7/3/2023
Chapter 1 - Diodes
1
Outline
1. Introduction
2. The ideal diode
3. The real diode
4. Applications
5. Other diode
Textbook: Adel. S. Sedra, Kenneth C. Smith. Microelectronic Circuits. Oxford University
Press. 2011 (Chapter 4).
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Chapter 1 - Diodes
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1. Introduction
• Diode is a semiconductor device conducting the
current in one direction.
• It has two terminals as Anode (A+) and Cathode (K-).
• When positive and negative polarities are at the anode
and cathode, respectively, the diode is forward biased
and is conducting.
• When positive and negative polarities are at the
cathode and anode, respectively, the diode is reverse
biased and is not conducting.
• If the reverse-biasing voltage is sufficiently large, the
diode is in reverse-breakdown region and large current
flows though it.
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2. The Ideal Diode
2.1 Current-Voltage characteristic
• The ideal diode may be considered the most
fundamental nonlinear circuit element
(b) i-v characteristic
(a) Diode circuit symbol
(c) Equivalent circuit in the
reserve direction
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(d) Equivalent circuit in the
forward direction
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2. The Ideal Diode
2.2 Some applications – The rectifier
(a) Rectifier circuit
(c) Equivalent circuit, 𝑣 ≥ 0
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(b) Input waveform
(d) Equivalent circuit, 𝑣 < 0
Chapter 1 - Diodes
(e) Output waveform
5
Exercise 1
Exe4.1-4.3 (P169): For the circuit as follows:
4.1 Sketch the transfer characteristic 𝑣 versus 𝑣
4.2 Sketch the waveform of 𝑣
4.3 Find the peak value of 𝑖 and the DC component 𝑣 if 𝑣 has a peak value of 10V and R=1k
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Exercise 1 (Solution)
4.2 The waveform of 𝒗𝑫
4.1 The transfer characteristic 𝒗𝒐 versus 𝒗𝑰
4.3 Find the peak value of 𝒊𝑫 and the DC component 𝒗𝒐
We have: 𝑣 = 𝑣 + 𝐼 𝑅. 𝐼 max when diode is conducting => 𝑣 = 0 → 𝐼 = = 10𝑚𝐴
DC component 𝑣 :
/
𝑣
𝑣
𝑣
𝑣
𝑇
𝑣
=
𝑠𝑖𝑛𝜔𝑡𝑑𝑡 = −
𝑐𝑜𝑠𝜔 − 𝑐𝑜𝑠0 = −
𝑐𝑜𝑠𝜋 − 𝑐𝑜𝑠0 =
= 3.18𝑉
𝑇
𝜔𝑇
2
2𝜋
𝜋
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2. The Ideal Diode
2.2 Some applications – Diode Logic Gates
The output will be high if one or
more of the inputs are high
Y= A + B + C OR gate
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The output will be high if all of
the inputs are high
Y = A . B . C AND gate
Chapter 1 - Diodes
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Exercise 2
• Exa 4.2 (P171): Assuming
the diodes to be ideal, find
the values of I and V.
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Exercise 3
• Exe 4.4 (P173): Find the values of I and V in
the circuits shown in Fig 1.4.
Figure 1.4
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Exercise 4
• Exe 4.5 (P174): Figure 1.5 shows a circuit for an AC voltmeter. It utilizes a
moving-coil meter that gives a full-scale reading when the average current
flowing through it is 1mA. The meter has a 50 resistance. Find the value of R
that results in the meter indicating a full-scale reading when the input sine-wave
voltage is 20 V peak-to-peak. (Hint: The average value of half-sine waves is Vp /π.)
Figure 1.5
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3. The Real Diode
3.1 The I-V characteristic
• The characteristic curve consists of three
distinct regions:
The forward-bias region, by
The reverse-bias region, by
The breakdown region, by
Knee voltage
Cut-in Voltage
Drop Voltage
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3. The Real Diode
3.1 The I-V characteristic
•
⁄
 is constant (
for a given diode at a given
temperature (Saturation current).
 : thermal voltage
,Note
at room temperature (
)
⁄

or
Where 2.3𝑉 = 60𝑚𝑉
 Voltage drop: 0.7V
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Exercise 5
• Exa 4.3 (P176): A silicon diode said to be a 1mA device displays a forward
voltage of 0.7 V at a current of 1 mA. Evaluate the junction scaling constant .
What scaling constants would apply for a 1A diode of the same manufacture that
conducts 1 A at 0.7 V?
• Solution:
/
/
 We have:
=>
 Diode 1mA: Is=10-3 x e-700/25=10-3 x e-28 A=6.9x 10-16 A
 Diode 1A: Is=1 x e-700/25= e-28 A=6.9x 10-13 A
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Exercise 6-7
• Exe 4.6 (P177): Find the change in diode voltage if the current changes from 0.1
mA to 10 mA.
• Solution: we have V2-V1=2.3 x VT x log(I2/I1)=60 x log (10/0.1)= 120mV
• Exe 4.7 (P177): A silicon junction diode has v = 0.7 V at i = 1 mA. Find the
voltage drop at i = 0.1 mA and i = 10 mA.
• Solution: we have V2-V1=2.3 x VT x log(I2/I1)
o i=0.1mA: V2-700=60xlog(0.1/1) =>V2-700=-60 =>V2=640mV=0.64V.
o i=10mA: V2-700=60xlog(10/1) => V2=700+60=760mV=0.76V.
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3. The Real Diode
3.1 The I-V characteristic
• The Reverse-bias Region: If is
negative and a few time larger than
(~25mV) in magnitude, the diode
current becomes
𝑺
• The Breakdown Region:
The reverse current increase rapidly
with the associated increase in
voltage drop very small
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Chapter 1 - Diodes
Knee voltage
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3. The Real Diode
3.2 Modeling the Diode Forward Characteristics
• The exponential model: The diode operation in the forward region is
provided by the exponential model.
⁄
(Kirchhoff loop)
Two ways for obtaining the solution:
 Graphical Analysis using the exponential model
 Iterative analysis using the exponential model
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3. The Real Diode
3.2 Modeling the Diode Forward Characteristics
• Graphical analysis using the exponential model
⁄
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3. The Real Diode
3.2 Modeling the Diode Forward Characteristics
• Iterative analysis using the exponential model
⁄
Using a simple iterative procedure
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Exercise 8
• Exa4.4 (P180): Determine the current 𝐼 and the diode voltage 𝑉 for the
circuit in beside figure with 𝑉 = 5 V and R = 1k. Assume that the diode has a
current of 1 mA at a voltage of 0.7 V.
• Solution:
 𝐼 =
=
.
= 4.3𝑚𝐴
 We then use the diode equation to obtain a better estimate for 𝑉 using equation 𝑉 − 𝑉 = 2.3𝑉 log .
o 𝑉 = 0.7𝑉, 𝐼 = 1 mA, and 𝐼 = 4.3 mA results in 𝑉 = 0.738 V.
=> Thus the results of the first iteration are 𝐼 = 4.3 mA and 𝑉 = 0.738 V.
 The second iteration proceeds in a similar manner: 𝐼 = 4.262𝑚𝐴; 𝑉 = 0.738𝑉
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3. The Real Diode
3.2 Modeling the Diode Forward Characteristics
• The Constant Voltage Drop Model
It is the simplest and most widely used
diode model.
When forward biased, the diode has a
voltage drop varying in a narrow range,
0.6V to 0.8V
This model assumes
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Exercise 9
Exe 4.10 (P183): For the circuit in beside figure, find 𝐼 and 𝑉 for the case 𝑉 = 5 V and R =
10k. Assume that the diode has a voltage of 0.7 V at 1-mA current. Use (a) iteration and (b) the
constant-voltage-drop model with 𝑉 = 0.7 V.
• Solution for iteration:
 Iteration 1:
.
VD=0.7V, 𝐼 =
=
𝑉 = 𝑉 + 0.06 ∗ log
= 0.7 + 0.06 ∗ log
 Iteration 2: VD=0.678V, 𝐼 =
=> 𝑉 = 𝑉 + 0.06 ∗ log
=
.
• Solution for CVD model
= 0.43𝑚𝐴.
.
= 0.678 𝑉
= 0.432𝑚𝐴.
= 0.678 + 0.06 ∗ log
.
.
−𝑉
5 − 0.7
=
= 0.43𝑚𝐴
𝑅
10𝑘
=> ID=0.43 mA, VD=0.7V
𝐼 =
𝑉
= 0.678 𝑉
Summary: ID=0.43mA, VD=0.678V
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3. The Real Diode
3.2 Modeling the Diode Forward Characteristics
• The small signal model
If
satisfies
,
or
with
The dynamic resistance
→ This is the small signal model of the diode, which applies for
signal that has amplitude smaller than 5mV
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Exercise 10
• Exe4.13(P189): Find the value of the diode small-signal resistance
currents of 0.1 mA, 1 mA, and 10 mA.
at bias
• Solution: The dynamic resistance
o
.
o
o
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Exercise 11
• Exa4.5 (P186): Consider the circuit shown in the below figure for the case in which R = 10k. The
power supply V+ has a DC value of 10 V on which is superimposed a 60Hz sinusoid of 1V peak
amplitude. (This “signal” component of the power-supply voltage is an imperfection in the powersupply design. It is known as the power-supply ripple.) Calculate both the dc voltage of the diode
and the amplitude of the sine-wave signal appearing across it. Assume the diode to have a 0.7-V
drop at 1-mA current.
a) Circuit for the Example
b) Circuit for calculating the dc operating point
c) Small signal equivalent circuit
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Exercise 11 (solution)
• Considering dc quantities only, we assume 𝑉 = 0.7 V and
(
. )
calculate the diode dc current: 𝐼 =
= 0.93𝑚𝐴
• At the operating point, the diode dynamic resistance 𝑟 is:
𝑉
25
𝑟 =
=
= 26.9Ω
𝐼
0.93
• The signal voltage across the diode can be found from the
small-signal equivalent circuit in Fig.(c). Here 𝑣 denotes the
60-Hz 1-V peak sinusoidal component of V+, and 𝑣 is the
corresponding signal across the diode. Using the voltagedivider rule provides the peak amplitude of 𝑣 as follows:
𝑟
26.9
𝑣 𝑝𝑒𝑎𝑘 = 𝑣
=1
= 2.68𝑚𝑉
𝑅+𝑟
10𝑘 + 26.9
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3. The Real Diode
3.3 Zener diode Operation in the Reverse Region
• In breakdown region, a reverse bias (𝑉 ) beyond the
knee voltage (𝑉 ) leads to a large reverse current (𝐼 )
• 𝑟 called the dynamic resistance, is the inverse of the
slope of the almost-linear I-V curve at the Q-point.
Typically, it is in range of a few ohms to a few ten ohms.
(knee current)
• ∆𝐼 and ∆𝑉 on Zener:
• 𝑉 denotes the point at which the straight line of slope
intersects the voltage axis.
• Voltage 𝑉 :
(maximum current)
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Exercise 12
• Exa4.7 (P192): The 6.8-V zener diode in the circuit of Fig. 4.19(a)
is specified to have 𝑉 = 6.8𝑉 at 𝐼 = 5𝑚𝐴, 𝑟 = 20Ω, and 𝐼 =
0.2𝑚𝐴. The supply voltage 𝑉 is nominally 10 V but can vary by
± 1𝑉.
a) Find 𝑉 with no load and with 𝑉 at its nominal value.
∆
b) Find the change in 𝑉 resulting from the ±1𝑉 charge in 𝑉 . Note that
∆
usually expressed in mV/V, is known as line regulation.
,
c) Find the change in 𝑉 resulting from connecting a load resistance 𝑅 that
∆
draws a current 𝐼 =1mA, and hence find the load regulation ( ) in
∆
mV/mA.
d) Find the change in 𝑉 when 𝑅 =2k.
e) Find the value of 𝑉 when 𝑅 =0.5k.
f) What is the minimum value of 𝑅 for which the diode still operates in the
breakdown region?
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Chapter 1 - Diodes
Figure 4.19. (a) Circuit for example; (2) The
circuit with the Zener diode replaced with
its equivalent circuit model.
28
Exercise 12 (Solution)
• 𝑉 = 𝑉 + 𝐼 𝑟 ; 𝑉 =6.8V, 𝐼 = 5 mA, and 𝑟 = 20Ω => 𝑉 = 6.7𝑉
a) Find 𝑉 with no load and with 𝑉 at its nominal value.
𝑉 −𝑉
10 − 6.7
𝐼 =𝐼=
=
= 6.35𝑚𝐴
𝑅+𝑟
500 + 20
→ 𝑉 = 𝑉 + 𝐼 𝑟 = 6.7 + 6.35 ∗ 0.02 = 6.83𝑉
∆
b) Find the change in 𝑉 resulting from the ±1𝑉 charge in 𝑉 (∆𝑉 = ±1𝑉). Note that
, usually expressed
∆
in mV/V, is known as line regulation.
∆𝑉 = ∆𝑉
= ±1
= ±38.5𝑚𝑉, So line regulation: 38.5mV/V
c) Find the change in 𝑉 resulting from connecting a load resistance 𝑅 that draws a current 𝐼 =1mA, and
∆
hence find the load regulation ( ∆ ) in mV/mA.
Because 𝐼 = 1 mA, the zener current 𝐼 will decrease by 1 mA.
∆𝑉 = 𝑟 ∆𝐼 = 20 ∗ −1 = −20𝑚𝑉
∆
The load regulation is: ∆ = −20𝑚𝑉/𝑚𝐴.
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Exercise 12 (Solution)
d) Find the change in 𝑉 when 𝑅 =2k.
.
𝑅 = 2𝑘 → 𝐼 =
= 3.4𝑚𝐴
→ ∆𝐼 = −3.4𝑚𝐴 → ∆𝑉 = 20 ∗ −3.4 = −68𝑚𝑉
e) Find the value of 𝑉 when 𝑅 =0.5k.
.
𝑅 = 0.5𝑘->𝐼 =
= 13.6𝑚𝐴. This is not possible because I=6.35mA (for 𝑉 = 10𝑉). So the Zener must be cut off.
.
If this is indeed the case, then 𝑉 is determined by the voltage divider formed by 𝑅 and R (Fig. a): 𝑉 = 𝑉
= 5𝑉.
Since this voltage is lower than the breakdown voltage of the zener, the diode is indeed no longer operating in
the breakdown region.
f) What is the minimum value of 𝑅 for which the diode still operates in the breakdown region?
For the zener to be at the edge of the breakdown region, 𝐼 = 𝐼 = 0.2𝑚𝐴 and 𝑉 ≅ 𝑉 = 6.7𝑉.
At this point the lowest (worst-case) current supplied through R is I=(9-6.7)/0.5=4.6mA and thus the load current is
𝐼 =4.6 - 0.2 = 4.4 mA. The corresponding value of 𝑅 is
6.7
𝑅 =
≅ 1.5𝑘
4.4
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4. Some application circuits using diodes
4.1 Rectifier circuits
• Half-Wave Rectifier
• Full-Wave Rectifier
• Bridge Rectifier
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Block diagram of a DC power supply
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4. Some application circuits using diodes
4.1 Rectifier circuits
• Half-Wave Rectifier
𝑣 = 0,
𝑣 <𝑉
𝑣 =𝑣 −𝑉 , 𝑣 ≥𝑉
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PIV (Peak Inverse Voltage)=𝑽𝒔
the diode must be able to withstand without breakdown
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Exercise 13
• Exe 4.19 (P197): For the half-wave rectifier circuit in beside figure, show the
following: (a) For the half-cycles during which the diode conducts, conduction
begins at an angle 𝜃 = sin (𝑉 /𝑉 ) and terminates at (𝜋 − 𝜃), for a total
conduction angle of (𝜋 − 2𝜃). (b) The average value (dc component) of 𝑉 is 𝑉 ≅
⁄ 𝑉−
, (c) The peak diode current is (𝑉 − 𝑉 )⁄𝑅. Find numerical
values for these quantities for the case of 12-V (rms) sinusoidal input, 𝑉 =0.7 V,
and R = 100Ω. Also, give the value for PIV.
• Solution: rms (Root mean square) of sinusoidal input is 12V => 𝑉 = 12 ∗ 2
 The conduction angle(𝝅 − 𝟐𝜽)
𝜃 = sin (𝑉 /𝑉 )= sin
.
= 2.4 => The conduction angle: 𝝅 − 𝟐𝜽=175.2
 The average value (DC component) of 𝑽𝒐 : 𝑉 ≅
 The peak current: 𝑰𝑫 = (𝑉 − 𝑉 )⁄𝑅 =
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.
Chapter 1 - Diodes
−
=
−
.
= 5.05𝑉
=0.1267A; 𝑃𝐼𝑉 = 𝑉 = 12 2 = 16.97𝑉
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4. Some application circuits using diodes
4.1 Rectifier circuits
• Half-Wave Rectifier
Ideal Diode
Real Diode?
Filter capacitor
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4. Some application circuits using diodes
4.1 Rectifier circuits
• Half-Wave Rectifier
Ideal diode
is peak-to-peak ripple voltage
𝜔∆𝑡 =
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2𝑉 ⁄𝑉 , where 𝜔 =
Chapter 1 - Diodes
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4. Some application circuits using diodes
4.1 Rectifier circuits
• Full-wave Rectifier
PIV (Peak Inverse Voltage)=𝟐𝑽𝒔 − 𝑽𝑫
the diode must be able to withstand without breakdown
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Exercise 14
• Exe.4.20 (P198): For the full-wave rectifier circuit as the
beside figure, show the following: (a) The output is zero for
an angle of 2
centered around the zerocrossing points of the sin-wave input. (b) The average value
(dc component) of is
. (c)The peak current
through each diode is
.
Find the fraction (percentage) of each cycle during which
, the value of , the peak diode current, and the value of PIV,
all for the case in which is a 12-V (rms) sinusoid,
,
and R = 100 Ω
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Exercise 14 (solution)
rms (Root mean square) of sinusoidal input is 12V =>
a) The
for an angle
.
=
=> The fraction (percentage) of each cycle during which
:
b) The average value (dc component) of
∗
is
c) The peak current through each diode is
d) PIV=2Vs-
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=
=(
∗
10.1V
-0.7)/100=163mA
-0.7=32.2V
Chapter 1 - Diodes
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4. Some application circuits using diodes
4.1 Rectifier circuits
• The Bridge Rectifier
PIV (Peak Inverse Voltage)=𝑽𝒔 − 𝟐𝑽𝑫 + 𝑽𝑫 = 𝑽𝒔 − 𝑽𝑫
the diode must be able to withstand without breakdown
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Exercise 15
• Exe 4.21 (P200): For the bridge rectifier circuit as the beside figure, use
the constant-voltage-drop diode model to show that (a) the average (or
dc component) of the output voltage is
(b) the peak
diode current is
.
Find numerical values for the quantities in (a) and (b) and the PIV for
the case in which is a 12-V (rms) sinusoid,
, and R = 100 Ω.
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Exercise 15 (solution)
• rms (Root mean square) of sinusoidal input is 12V =>
a) The average (or dc component) of the output voltage is
=>
∗
∗
∗ .
b) The peak diode current is
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4. Some application circuits using diodes
4.2 Limiting Circuits
• Limiter circuits

≤𝑣 ≤
𝑣 = 𝐾𝑣
 If 𝑣 >
, 𝑣 is limited to 𝐿
 If 𝑣 <
, 𝑣 is limited to 𝐿
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Transfer characteristic
for a double limiter
circuit
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4. Some application circuits using diodes
4.2 Limiting Circuits
• Limiter circuits
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Exercise 16
• Exe 4.26 (P210): Assuming the diodes to be ideal,
describe the transfer characteristic of the circuit shown
in the beside figure.
𝐷
𝐷
• Solution:
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4. Some application circuits using diodes
4.3 Clamping Circuits
• Clamping Circuits
Clamped capacitor
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4. Some application circuits using diodes
4.3 Clamping Circuits
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4. Some application circuits using diodes
4.3 Clamping Circuits
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5. Some other types of diodes
5.1 Photo diode
Photo Diode
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5. Some other types of diodes
5.2 Light-emitting Diode (LED)
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Exercises:
• Section 4.1 (P216): The ideal diode, from exe. 4.1 to exe. 4.16.
• Section 4.2 (P219): The terminal characteristics of junction diodes, from exe.
4.17 to 4.31.
• Section 4.3 (P221): Modeling the diode forward characteristic, from exe. 4.32 to
4.56.
• Section 4.4 (P224): Operationg in the Reverse breakdown region – Zener diodes,
from exe. 4.57 to exe. 4.64.
• Section 4.5: rectifier circuits, from exe 4.65 to exe. 4.84.
• Section 4.6. limiting and clamping circuits, from exe. 4.85 to exe. 4.95.
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Superdiode
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