© electronics ece 1231 recall-lecture 3 current generated due to two main factors drift –...
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© Electronics© Electronics ECE 1231ECE 1231
Recall-Lecture 3
Current generated due to two main factors Drift – movement of carriers due to the existence of
electric field Diffusion – movement of carriers due to gradient in
concentrations
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Recall-Lecture 3 Introduction of PN junction
Space charge region/depletion region Built-in potential voltage Vbi
Reversed biased pn junction no current flow
Forward biased pn junction current flow due to diffusion of carriers.
Vbi
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CIRCUIT REPRESENTATION OF DIODE
i
vD
The I -V characteristics of the ideal diode.
V = 0V
Conducting state
Reverse bias
Reverse biasedopen circuit
Conducting state short circuit
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CIRCUIT REPRESENTATION OF DIODE – Piecewise Linear Model
i
vDV
Conducting state
Reverse bias
VD = V for diode to turn on.
Hence during conducting state:
=
V
Represented as a battery of voltage = V
Reverse biasedopen circuit
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i
vDV
Conducting state
Reverse bias
CIRCUIT REPRESENTATION OF DIODE – Piecewise Linear Model
VD ≥ V for diode to turn on.
Hence during conducting state:
Reverse biasedopen circuit
=
V
Represented as a battery of voltage = V and forward resistance, rf in series rf
+ VD -
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Diode Circuits: DC Analysis and ModelsDiode Circuits: DC Analysis and Models
Example
Consider a circuit with a dc voltage VPS applied across a resistor and a diode.
Applying KVL, we can write,
or,
The diode voltage VD and current ID are related by the ideal diode equation: (IS is assumed to be known for a particular diode)
Equation contains only one unknown, VD:
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Diode Circuits: Direct ApproachDiode Circuits: Direct Approach
Question
Determine the diode voltage and current for the circuit.
Consider IS = 10-13 A.
and
ITERATION METHOD
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Diode Circuits: Using ModelsDiode Circuits: Using Models
Example
Determine the diode voltage and current using a piecewise linear model.
Assume piecewise linear diode parameters of Vf = 0.6 V and rf = 10 Ω.
Solution:
The diode current is determined by:
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DIODE DC ANALYSIS Find I and VO for the circuit shown below if the diode cut
in voltage is V = 0.7V
I = 0.2325mA
Vo = -0.35V
5 V 5 V
D 1
2 0 k 2 0 k+
VO
-
I
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Find I and VO for the circuit shown below if the diode cut in voltage is V = 0.7V
2 V 8 V
D 1
5 k 2 0 k+
VO
-
I
I = 0.372mA
Vo = 0.14V
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Example 2
a) Determine ID if V = 0.7VR = 4k
b) If VPS = 8V, what must be the value of R to get ID equal to part (a)
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Sinusoidal Analysis
The total input voltage vI = dc VPS + ac vi
iD = IDQ + id
vD = VDQ + vd
IDQ and VDQ are the DC diode current
and voltage respectively.
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Current-voltage Relation
The relation between the diode current and voltage can be written as:
VDQ = dc quiescent voltage
vd = ac component
The -1 term in the equation is neglected.
The equation can be written as:
Diode Circuits: AC Equivalent CircuitDiode Circuits: AC Equivalent Circuit
If vd << VT, the equation can be expanded into linear series as:
The DC diode current Is:
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iD = ID [ 1 + vd/VT]
iD = ID + ID vd / VT = ID + id
where id = ID vd / VT
using Ohm’s law:
I = V/R hence, id = vd / rd compare with id = ID vd / VT
which reveals that rd = VT / ID
CONCLUSION: During AC analysis the diode is equivalent to a resistor, rd
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Example 1Analyze the circuit (by determining VO & vo ).
Assume circuit and diode parameters of
VPS = 5 V, R = 5 kΩ, Vγ = 0.6 V & vi = 0.1 sin ωt
VDQ = V
IDQ
DC ANALYSIS
DIODE = MODEL 1 ,2 OR 3
CALCULATE DC CURRENT, ID
CALCULATE rd
AC ANALYSIS
DIODE = RESISTOR, rd
CALCULATE AC CURRENT, id