Transcript Piecewise Linear Model and AC Analysis

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
Vbi

Forward biased pn junction
 current flow due to diffusion of carriers.
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Analysis of PN Junction Diode in
a Circuit
CIRCUIT REPRESENTATION
OF DIODE
The I -V characteristics of the ideal diode.
i
Reverse
bias
Conducting
state
vD
V = 0V
Reverse biased
open circuit
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Conducting state
short circuit
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CIRCUIT REPRESENTATION OF
DIODE – Piecewise Linear Model
i
Reverse
bias
Conducting
state
vD
V
Reverse biased
open circuit
VD = V for diode to turn on.
Hence during conducting state:
Represented as
a battery of
voltage = V
=
V
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CIRCUIT REPRESENTATION OF
DIODE – Piecewise Linear Model
i
Reverse
bias
Conducting
state
Reverse biased
open circuit
vD
V
VD ≥ V for diode to turn on.
Hence during conducting
state:
+
VD
-
=
V
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rf
Represented as a
battery of voltage =
V and forward
resistance, rf in
series
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Diode 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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Why do you need the Piecewise
Linear Model?
Diode Circuits: Direct Approach
Question
Determine the diode voltage and current
for the circuit.
Consider IS = 10-13 A.
ITERATION
METHOD
and
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Diode 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
D1
20k
5V
+
VO
20k
5V
-
I = 0.2325mA
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Vo = -0.35V
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Find I and VO for the circuit shown below if the diode
cut in voltage is V = 0.7V
I
D1
5k
+
VO
2V
20k
8V
-
I = 0.372mA
Vo = 0.14V
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
a)
Example 2
Determine ID if V = 0.7V
R = 4k
b)
If VPS = 8V, what must be the value of R to get
ID equal to part (a)
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DIODE AC
EQUIVALENT
●
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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Diode Circuits: AC Equivalent Circuit

Current-voltage Relation
The relation between the diode current and
voltage can be written as:
VDQ = dc quiescent voltage
If vd << VT, the equation can be
expanded into linear series as:
vd = ac component
The -1 term in the equation is neglected.
The equation can be written as:
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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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VDQ = V
IDQ
DC equivalent
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rd
id
AC equivalent
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Example 1
Analyze the circuit (by determining VO & vo ).
Assume circuit and diode parameters of
VDQ = V
VPS = 5 V, R = 5 kΩ, Vγ = 0.6 V & vi = 0.1 sin ωt
IDQ
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rd
id
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DC ANALYSIS
DIODE = MODEL 1 ,2
OR 3
CALCULATE DC
CURRENT, ID
AC ANALYSIS
CALCULATE
rd
DIODE = RESISTOR,
rd
CALCULATE AC
CURRENT, id
EXAMPLE 2

Assume the circuit and diode parameters for the circuit
below are
VPS = 10V, R = 20k, V = 0.7V, and vi = 0.2 sin t.
Determine the current, IDQ and the time varying current, id
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