Transcript DIODE AC EQUIVALENT

Recall-Lecture 4

Current generated due to two main factors
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
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 4
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Introduction of PN junction
 Space charge region/depletion region
 Built-in potential voltage Vbi
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Reversed biased pn junction
 no current flow
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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 – Ideal Model
VD = - VS
Reverse-bias
I-V characteristics of ideal model
Forward-bias
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EXAMPLE: Determine the diode voltage and current
in the circuit using ideal model for a silicon diode.
Also determine the power dissipated in the diode.
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CIRCUIT REPRESENTATION OF DIODE
– Piecewise Linear Model
VD = - VS
Reverse-bias
Forward-bias
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I-V characteristics of constant
voltage model
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EXAMPLE: Determine the diode voltage and current
in the circuit (using constant voltage model) for a
silicon diode. Also determine the power dissipated
in the diode. Consider the cut-in voltage V = 0.65 V.
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CIRCUIT REPRESENTATION OF DIODE
– Piecewise Linear Model
VD = - VS
Reverse-bias
I-V characteristics of piecewise model
Forward-bias
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EXAMPLE: Determine the diode voltage and current in
the circuit using piecewise linear model for a silicon
diode. Also determine the power dissipated in the
diode. Consider the cut-in voltage V = 0.65 V and the
diode DC forward resistance, rf = 15 Ω.
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Why do you need to use these
models?
Diode Circuits: Direct Approach
Question
Determine the diode voltage and current
for the circuit.
Consider IS = 10-13 A.
VPS = IDR + VD
5 = (2 x 103) (10-3) [ e ( VD / 0.026) – 1 ] + VD
VD = 0.619 V
ITERATION
METHOD
And ID = 2.19 mA
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IDEAL
MODEL
PIECEWISE
LINEAR
MODEL 1
PIECEWISE
LINEAR
MODEL 2
DIRECT
APPROACH
Diode voltage
VD
0V
0.65 V
0.652 V
0.619 V
Diode current
ID
2.5 mA
2.175 mA
2.159 mA
2.19 mA
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DC Load Line
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A linear line equation
ID versus VD
Obtain the equation using KVL
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The value of
ID at VD = V
Use KVL:
2ID + VD – 5 = 0
ID = -VD + 5 = - VD + 2.5
2
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V
2
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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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Total current
Total voltage
VDQ = DC voltage
vd = ac component
The DC diode current IDQ in term of diode voltage VDQ
If vd << VT , the equation can be expanded into linear series as:
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Therefore, the diode current-voltage relationship can
be represented as
The relationship between the AC components of the diode voltage and diode
current is
Or,
Where,
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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
VPS = 10 V, R = 5 kΩ, Vγ = 0.6 V & vi = 0.2 sin ωt
DC Current
DC Output voltage
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vi
vi
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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 (V).
Determine the current, IDQ and the time varying current, id
ANSWERS
IDQ = 0.465 mA
Id = 9.97 sin t (µA)
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