Transcript (b).

Chapter1: Diodes
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Figure 2.1 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.
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Electrostatics of PN Junction
NA
ND
Max Electric Field
Depletion width
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Diode Connections
Reverse bias connection
Depletion region expands with reverse bias
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Forward connection
For silicon diodes, the typical forward voltage is 0.7 volts, For germanium
diodes, the forward voltage is only 0.3 volts.
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Typical I-V Characteristics
Thermal Voltage
VT=k*T/q
Figure 2.7 The i–v characteristic of a silicon junction diode.
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The forward-Bias Region: It is entered when the terminal voltage v is positive. In the
forward region the i-v relationship is closely approximated by:
i=is (ev/nVT -1)
(¤)
Where VT=KT/q
k=Boltzmann’s constant= 1.38*10-23 joules/kelvin
T=the absolute temperature in kelvins=273+temperature in °C
q=The magnitude of electronic charge= 1.6*10-19 coulomb
n has a value between 1 and 2
For appreciable current I in the forward direction, specially for i>>Is, (¤) can be
approximated by the relationship
i=is ev/nVT
v=n VT ln(i/iS)
Example: A silicon diode said to be a 1mA device displays a forward voltage of 0.7v
at a current of 1mA. Evaluate the junction scaling constant IS in the event that n is
either 1 or 2.
Solution:
Since i=Is ev/nVT
Then Is=i e-v/nVT
n=1: Is= 10-3e-700/25=6.9*10-16A
N=2: Is= 10-3e-700/50=8.3*10-10A
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Figure 2.10 A simple circuit used to illustrate the analysis of circuits in which the diode is
forward conducting.
Example: Determine the current ID and the diode voltage VD of the circuit with VDD=5V and
R=1kΩ. Assume that the diode has a current of 1mA at a voltage of 0.7V and that its voltage drop
changes by 0.1V for every decade change in current.
Solution: we assume that VD=0.7V Then
I
V2  V1  2.3nVT log 2
I1
ID 
VDD  VD 5  0.7

 4.3mA,
R
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and considering That 2.3nVT =0.1V, we obtain:
by employing the equation
V2  V1  2.3nVT log
I2
I1
V1=0.7V, I1=1mA and I2=4.3mA results in V2=0.763V. Thus results permit to us to get:
ID 
5  0.763
 4.237 mA
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and
V2  0.763  0.1log
4.237
 0.762V
4.3
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Resistance Levels
Figure 2.12 Approximating the diode forward characteristic with two straight lines: the
piecewise-linear model.
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Figure 2.13 Piecewise-linear model of the diode forward characteristic and its equivalent
circuit representation.
The straight-lines (or piecewise linear) model of the fig (a) can be described by:
iD=0
iD=(vD –VDD)/rD
, vD <VDD
, vD >VDD
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Application:
1- In case of DC current, Determine the dc resistance levels for the diode at
(a) ID=2 mA
(b) ID= 20 mA
(c) VD=-10 V
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Figure 2.17 Development of the diode small-signal model. Note that the numerical
values shown are for a diode with n = 2.
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Application 2:
In case of AC current :
(a) Determine the ac resistance at ID= 2 mA.
(b) Determine the ac resistance at ID =25 mA.
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Load Line Analysis
Figure 2.11 Graphical analysis of the circuit in Fig. 3.10 using the exponential diode model.
From the circuit above,
If we consider the 2 equations, we
obtain the Q point in the intersection
with the curve of the diode response
I=f(V)
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Series of Diode configurations
Figure 2.2 The two modes of operation of ideal diodes and the use of an external circuit to limit the
forward current (a) and the reverse voltage (b).
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Figure 2.5 Diode logic gates: (a) OR gate; (b) NAND gate (in a positive-logic system).
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Example: Find the values of I and V in the circuits shown in Fig E2.4
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Application:
1- Determine Vo, I1, ID1, and ID2 for the parallel diode configuration of the figure
shown below
2- Determine I1, I2, and ID2 for the parallel diode configuration of the figure shown
below
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Homework:
Assuming the diodes to be ideal, find the values of I and V in the
circuit
Figure 2.6 Circuits for Example 3.2.
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Zener Diode
Figure 2.20 Circuit symbol for a zener diode and its Model.
Figure 2.21 The diode i–v characteristic with the breakdown region shown in some
detail.
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Application:
1(a) For the Zener diode network, determine VL, VR, IZ, and PZ.
(b) Repeat part (a) with RL = 3 k.
Homework: Ex 42 P130
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Figure 2.23 (a) Circuit for Example 2.8. (b) The circuit with the zener diode replaced with its
equivalent circuit model.
The -6.8v zener diode in the circuit is specified to have Vz =6.8v at Iz=5.mA. rs=20, and Izk=20mA.
The supply voltage V+ is nominally 10v but can vary by ±1v
a) Find V0 with no load and with V+ at its nominal value
b) Find the change in V0 resulting from the ±1 vchange in V+ . Note that (ΔV0/ΔV+), usually
expressed in mV/V, is known line regulation
c) Find the change in V0 resulting from connecting a load resistance RL that draws a current
IL=1mA, and hence find the load regulation (ΔV0/ΔIL), in mV/mA.
d) Find the change inV0 when RL=2kΩ
e) Find the value of V0 when RL=0.5kΩ
f)
What is the minimum value of RL for which the diode still operates in the breakdown region
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Diodes in AC analysis with Ac source
Figure 2.3 (a) Rectifier circuit. (b) Input waveform. (c) Equivalent circuit when vI  0. (d)
Equivalent circuit when vI  0. (e) Output waveform.
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Figure E2.1
Figure E2.2
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Figure 2.4 Circuit and waveforms for Example 3.1.
Example: (a) shows a circuit for charging a -12v battery. If Vs is a sinusoid with 24v peak
amplitude. Find the fraction of each cycle during which the diode conducts. Also, find the peak
value of the diode current and the maximum reverse-bias voltage that appears across the diode
Solution: The diode conducts when Vs exceeds 12v. As shown in Fig2.4 (b). The conduction
angle is 2θ, where θ is given by:
24 cosθ= 12
Thus θ=60° and the conduction angle is 120°, or one-third of a cycle. The peak value of the diode
current is given by
VS max  V0 24  12
Id 
R

100
 0,12 A
The maximum reverse voltage across the diode occurs when Vs is at its negative peak and is equal
to 24+12=36V
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APPLICATIONS OF DIODES:
Diodes are used in so many ways that we will not be able to discuss all of them. The major
applications of the diodes that will be discussed are:
 Rectifiers
 Clippers or Limiters
 Clampers
 Voltage Multipliers
Figure 2.24 Block diagram of a dc power supply.
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Half-wave rectifier
Conduction region (0
T/2).
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No conduction region (T/2
T).
Half-wave rectified signal
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Effect of VK on half-wave rectified signal.
If we consider a network for the example
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The resulting vo for the circuit
If we consider the effect of VK on output of Figure
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Full-wave bridge rectifier
Network of this Figure for the period 0
T/2 of the input voltage vi.
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Conduction path for the positive region of vi.
+
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That’s why if we consider the Input and output waveforms for a full-wave rectifier, we will obtain the following result
Determining VOmax for silicon diodes in the bridge configuration.
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Full-Wave Rectifier using a center-tapped secondary
Network conditions for the positive region of vi.
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Network conditions for the negative region of vi.
Series clipper with a dc supply: if we consider the following figure
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Figure 2.25 (a) Half-wave rectifier. (b) Equivalent circuit of the half-wave rectifier with the diode
replaced with its battery-plus-resistance model. (c) Transfer characteristic of the rectifier circuit. (d)
Input and output waveforms, assuming that rD ! R.
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Figure 2.26 Full-wave rectifier utilizing a transformer with a center-tapped secondary winding: (a)
circuit; (b) transfer characteristic assuming a constant-voltage-drop model for the diodes; (c) input
and output waveforms.
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Figure 2.27 The bridge rectifier: (a) circuit; (b) input and output waveforms.
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Figure 2.28 (a) A simple circuit used to illustrate the effect of a filter capacitor. (b) Input and
output waveforms assuming an ideal diode. Note that the circuit provides a dc voltage equal to the
peak of the input sine wave. The circuit is therefore known as a peak rectifier or a peak detector.
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Figure 2.29 Voltage and current waveforms in the peak rectifier circuit with CR @ T. The diode is
assumed ideal.
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Figure 2.30 Waveforms in the
full-wave peak rectifier.
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Applications: Voltage doubler
Figure 2.38 Voltage doubler: (a) circuit; (b) waveform of the voltage across D1.
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