chapter2 diode see2063
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Cha ter 2
DIODE
Part 1
Resistance Levels
Chapter 2 DIODE, Part 1 Resistance LevelSEE 2063 1
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I-V Characteristics : Silicon vs. Germanium
Peak Inverse Voltage (PIV) isthe maximum reverse-biasPotential that can be a liedbefore entering the Zenerregion.
Si: PIV ~ 1000VGe: PIV ~ 400V
Si: temperature up to 200 C.Ge: temperature below 100 C.
: T~ .Ge: VT~ 0.3V
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Resistance Levels
i) DC or Static Resistance The application of DC voltage to acircuit containing a semiconductordiode will result in an o eratin ointon the characteristic curve that will notchange with time.
The dc resistance levels at the kneeand below will be greater than theresistance levels obtained for the
vertical rise section of thecharacteristics.
Resistance, RDat the operating pointThe lower the current through adiode the higher the dc resistance
level.
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Example : DC or Static Resistance
: e erm ne e c res s ance eve s orthediode at
D=(b) ID= 20mA
(c) VD= -10V
Solution
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ii) AC or Dynamic Resistance
If a sinusoidal voltage(signal) is applied,the varyinginput will move the instantaneous operating point upand down a region of the characteristics and
us e nes a spec c c ange n currenand voltage.
With no applied varying signal, the point of operation-
applied dc levels.Quiescent: still or unvarying
AC Resistance, rd
e ower t e -po nt o operat on(smaller current or lower voltage) thehigher the AC resistance.
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Example : AC or Dynamic Resistance
Solution
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iii) Average AC Resistance
produce a broad swing, the resistanceassociated with the device for this regionis called the average ac resistance.
The average ac resistance is the resistance
determined by a straight line drawn between
maximum and minimum values of inputvoltage.
minimum
The lower the level of currents used todetermine the average resistance the higherthe resistance level.
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Summary of Resistance Levels
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Cha ter 2
DIODE
Part 2
Diode Equivalent Circuits
Chapter 2 DIODE, Part 2 Diode Equivalent Circuits 1
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An equivalent circuit is a combination of elements properly chosen to best represent theActual terminal characteristics of a device, system, or such in a particular operating region.
i) Piecewise-Linear Equivalent CircuitOne technique for obtaining an equivalent circuit for a diode is to approximate the characteristicsof the device by the straight-line segments. For a sloping section,the average ac resistance isthe resistance level appearing in the equivalent circuit of the actual device.
only one direction of conduction through a device, anda reverse-bias condition will result in the open-circuit
state for the device.Since a diode does not reach the conduction stateuntil VDreaches 0.7V with the forward bias, a batteryVTopposing the conduction direction must appear in
.
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ii Sim lified E uivalent Circuit
For most applications, the resistance rav is sufficiently small to be ignored incomparison to the other elements of the network.
It states that a forward-bias silicon diode in an electronic system under dc
conditions has a drop of 0.7V across it in the conduction state at any level ofo e.
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iii Ideal E uivalent Circuit
Let us take it a step further and establish that a 0.7V level can often be ignored incomparison to the applied voltage level.
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Summary of Diode Equivalent Circuits
Chapter 2 DIODE, Part 2 Diode Equivalent Circuits 5
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Cha ter 2
DIODE
Part 3
Diode Notation and Testing
Chapter 2 DIODE, Part 3 Diode Notation and Testing 1
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The notation frequently used for semiconductor diodes.
Various type ofjunction diodes.
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ON state :
The meter has ansource (~2mA) that willdefine the voltage level
~ . .
Digital TesterOL indication:O en
Chapter 2 DIODE, Part 3 Diode Notation and Testing 3
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Chapter 2 DIODE, Part 3 Diode Notation and Testing 4
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Cha ter 2
DIODE
Part 4
Load Line Analysis
Chapter 2 DIODE, Part 4 Load Line Analysis 1
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Load Line AnalysisThe a lied load will have an im act on the oint or re ion of a device. A line can
be drawn on the I-V characteristics of the device that represents the applied load.
Kirchoffs voltage law
Load
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Example 1: Determine (a) VDQ and IDQ (b) VR
Chapter 2 DIODE, Part 4 Load Line Analysis 3
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Example 2: Change R to 2kOhm. Determine (a) VDQ and IDQ (b) VR
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Example 3: Repeat example 1. Determine (a) VDQ and IDQ using approximate
Chapter 2 DIODE, Part 4 Load Line Analysis 5
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Example 4: Repeat example 2. Determine (a) VDQ and IDQ using approximate
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Example 5: Repeat example 1. Determine (a) VDQ and IDQ using ideal diode model
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Approximate and Ideal Diode Models
rav was not employed because rav is typically muchLess than the other series elements of the network.
rav was not employed because rav is typically muchLess than the other series elements of the network.
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Cha ter 2
DIODE
Part 4
Series Diodes Configurations
with DC Inputs
Chapter 2 DIODE, Series Diode Configurations 1
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Series Diode Configuration with DC Inputs
Series diode configuration
(1) Determining the state of the diode (2) Substituting the equivalent
model for the on diode
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(1) Determining the state of the diode
(2) Substituting the equivalent
model for the off diode
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Example 6: Determine VD, VRand ID.
Solution
e app e vo tage esta s es a current nthe clockwise direction to match the arrow
of the symbol and the diode is in the ONs a e .
Series diode configuration
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Example 7: Repeat example 6 with the diode reversed.Determine V V and I .
Solution
in the diode symbol. Diode is in OFF stateresulting to open circuit.
Series diode configuration
1. An open circuit can have any voltage across its terminals, but the current is always 0 A.
2. A short circuit has a 0V drop across its terminals, but the current is limited only by theSurrounding network.
ource notat on
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Example 8: Determine VD, VRand ID.
The level of the applied voltage is
OFF Operating pointSeries diode configuration
insufficient to turn the silicon diodeON. The point of operation on thecharacteristics shown in the figureestablishing the open circuit equivalent.
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Example 9: Determine VD, and ID.
o u on:
The resulting current has the same directionas e arrow ea s o e sym o s o o o es.
Here, E=12V > (0.7V + 0.3V)= 1V.
Determining the state of the diode and
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Example 10: Determine VD, Voand ID.The combination of aShort circuit in seriesWith an open circuit alwaysResults in an open circuit
= .
e erm n ng e s a e o e o e ansubstituting the equivalent model for the diodeSeries diode configuration
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Example 11: Determine I, V1 V2and Vo .
Series diode configuration Determining the state of the diode andsubstituting the equivalent model for the diode
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Cha ter 2
DIODE
Configurations
w t nputs
Chapter 2 DIODE, Parallel and Series Diodes Conf. 1
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Example 12: Determine Vo , I1 , ID1 and ID2 .
Parallel diode configuration
Determining the state of the diode and
substitutin the e uivalent model for the diode
Assuming diodes of similar characteristics,
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Example 13: Determine I.
Parallel diode configurationDetermining the state of the diode and
substituting the equivalent model for the diode
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Example 14: Determine Vo . The applied voltage will turn both diodes on. , , .
the silicon diode would not match 0.3V across the Gediode as required by the fact that the voltage acrossparallel elements must be the same.
Parallel diode configuration
Determining the state of the diode andsubstituting the equivalent model for the diode
Chapter 2 DIODE, Parallel and Series Diodes Conf. 4
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Example 15: Determine I1,I2and ID2 .
Series and Parallel diode configuration
eterm n ng t e state o t e o e an
substituting the equivalent model for the diode
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Cha ter 2
DIODE
Chapter 2 DIODE, AND / OR Gates 1
E ample 16 Determine V
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Example 16: Determine V0 .
Redrawn OR Gates
Positive logic OR GatesFirst, assume that D1 is on and D2 is off.
Vo = E - VD = 10V 0.7V = 9.3V (level 1)With 9.3V at the cathode (-) side of D2 and0V at the anode (+) side, D2 is in the off state.
ur assump on s correc .
Chapter 2 DIODE, AND / OR Gates 2
Example 17: Determine V
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Example 17: Determine V0 .
Positive logic AND Gates
Redrawn OR Gates
With 10V at the cathode side of D1, it is assumed that D1 is in the off state even though
1 .D2 is assumed to be in the on state due to the low voltage at the cathode side and theAvailability of the 10V source through the 1kOhm resistor.
The voltage at Vo is 0.7V due to the forward-biased diode D2. With 0.7V at the anode ofD1 and 10V at the cathode, D1 is definitely in the off state.
Chapter 2 DIODE, AND / OR Gates 3
Ideal Diode
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Ideal Diode
Chapter 2 Ideal Diode 1