ENGR-43_Lec-10a_Diode-1_Physics_LoadLine
Download
Report
Transcript ENGR-43_Lec-10a_Diode-1_Physics_LoadLine
Engineering 43
Diodes-1
Bruce Mayer, PE
Registered Electrical & Mechanical Engineer
[email protected]
Engineering-43: Engineering Circuit Analysis
1
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-10a_Diode-1_Physics_LoadLine.pptx
Learning Goals
Understand the Basic Physics of
Semiconductor PN Junctions which
form most Diode Devices
Sketch the VI Characteristics of Typical
PN Junction Diodes
Use the Graphical LOAD-LINE method
to determine the “Operating Point” of
NonLinear (includes Diodes) Circuits
with Resistive Loads
Engineering-43: Engineering Circuit Analysis
2
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-10a_Diode-1_Physics_LoadLine.pptx
Learning Goals
Analyze diode-containing VoltageRegulation Circuits
Use various math models for Diode
operation to solve for Diode-containing
Circuit Voltages and/or Currents
IDEAL and
Learn The difference
PieceWise-Linear
Models
between LARGE-signal
and SMALL-Signal
Circuit Models
Engineering-43: Engineering Circuit Analysis
3
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-10a_Diode-1_Physics_LoadLine.pptx
Diodes are ONE-Way Devices
Diodes exhibit a form of
RECTIFICATION
•
i.e., They allows current to Flow in the
FORWARD direction, But NOT in the
REVERSE direction
– Think of a diode as a“Check-Valve”
for Electrical Current”
– The Voltage Drop Across a
Diode is Called its “Bias” Voltage
FORWARD Bias → Flow ALLOWED
REVERSE Bias → NO Flow Allowed
Engineering-43: Engineering Circuit Analysis
4
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-10a_Diode-1_Physics_LoadLine.pptx
Basic Concepts
Diodes are
extremely Important
Devices as they
perform the
RECTIFICATION
operation
Rectification means
that current can
(usually) Flow in
ONE direction, and
NOT the Other
Engineering-43: Engineering Circuit Analysis
5
The Circuit Symbol
• Notation: vD & iD
represent the
INSTANTANEOUS
diode voltage &
current
Notice
• Anode → + terminal
• Cathode → − terminal
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-10a_Diode-1_Physics_LoadLine.pptx
Diodes are Highly NONlinear
Note
•
•
•
•
Fairly SHARP CORNERS
FLAT Region
Changes in CONCAVITY
Three Distinct
OPERATING REGIONS
Typical RectifyingDiode V-I Curve
Rectification
Engineering-43: Engineering Circuit Analysis
6
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-10a_Diode-1_Physics_LoadLine.pptx
Diode Physics Materials
• Silicon (Si) and Germanium (Ge) are the two most common
single elements that are used to make Diodes. A
compound that is commonly used is Gallium Arsenide
(GaAs), especially in the case of LEDs because of it’s large
bandgap.
• Silicon and Germanium are both group 4 elements,
meaning they have 4 valence electrons. Their structure
allows them to grow in a shape called the diamond lattice.
• Gallium is a group 3 element while Arsenide is a group 5
element. When put together as a compound, GaAs creates
a zincblend lattice structure.
• In both the diamond lattice and zincblend lattice, each atom
shares its valence electrons with its four closest neighbors.
This sharing of electrons is what ultimately allows diodes to
be build. When dopants from groups 3 or 5 (in most cases)
are added to Si, Ge or GaAs it changes the properties of
the material so we are able to make the P- and N-type
materials that become the diode.
Engineering-43: Engineering Circuit Analysis
7
Si
+4
Si
+4
Si
+4
Si
+4
Si
+4
Si
+4
Si
+4
Si
+4
Si
+4
The diagram above shows the 2D
structure of the Si crystal. The
White lines represent the
electronic bonds made when the
valence electrons are shared.
Each Si atom shares one electron
with each of its four closest
neighbors so that its valence band
will have a full 8 electrons.
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-10a_Diode-1_Physics_LoadLine.pptx
N-Type (Negative) Material
+4
+4
+4
+4
+5
+4
+4
+4
+4
When extra valence electrons are
introduced into a material such as silicon
an n-type material is produced. The extra
valence electrons are introduced by
putting impurities or dopants into the
silicon. The dopants used to create an
n-type material are Group V elements.
The most commonly used dopants from
Group V are arsenic, antimony and
phosphorus. The 2D diagram to the left
shows the extra electron that will be
present when a Group V dopant is
introduced to a material such as silicon.
This extra electron is very mobile.
Engineering-43: Engineering Circuit Analysis
8
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-10a_Diode-1_Physics_LoadLine.pptx
P-Type (Positive) Material
+4
+4
+4
+4
+3
+4
+4
+4
+4
p-type material is produced when the
dopant that is introduced is from Group III.
Group III elements have only 3 valence
electrons and therefore there is an electron
missing. This creates a hole (h+), or a
positive charge that can move around in
the material. Commonly used Group III
dopants are aluminum, boron, and gallium.
The 2D diagram to the left shows the hole
that will be present when a Group III dopant
is introduced to a material such as silicon.
This hole is quite mobile in the same way
the extra electron is mobile in a n-type
material.
Engineering-43: Engineering Circuit Analysis
9
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-10a_Diode-1_Physics_LoadLine.pptx
P-N Junction (Diode) Physics
P & N Type Semi
Matls Brought
Together to form
a METALLURICAL
(seamless) Junction
The HUGE
MisMatch in Carrier
Concentrations
Results in e− & h+
Cross DIFFUSION
Engineering-43: Engineering Circuit Analysis
10
Carrier Diffusion
• e− Diffuse in to the
P-Type Material
• h+ Diffuse in to the
N-Type Material
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-10a_Diode-1_Physics_LoadLine.pptx
P-N Junction Physics cont.
In a p-n JCN Carrier
Cross-Diffusion is
SELF-LIMITING
E-Field
• The e−/h+ Diffusion
leaves Behind fixed
IONIZED Atom
Cores of the
OPPOSITE Charge
• The Ion Cores set up
an ELECTRIC FIELD
that COUNTERS the
Diffusion Gradient
Engineering-43: Engineering Circuit Analysis
11
For Si the Field-Filled
Depletion Region
• E-Field 1 MV/m
• Depl Reg Width,
xd = 1-10 µm
• E-fld•dx 0.6-0.7 V
– “built-in” Potential
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-10a_Diode-1_Physics_LoadLine.pptx
P-N Junction Rectifier
A Rectifier is a
“Check Valve” for
Current flow
• Current Allowed in
ONE Direction but
NOT the other
E-Field
Side Issue →
“Bias” Voltage
• A “Bias” Voltage is
just Another name
for EXTERNALLY
APPLIED Voltage
Engineering-43: Engineering Circuit Analysis
12
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-10a_Diode-1_Physics_LoadLine.pptx
P-N Junction Rectifier cont
p-n junction
Rectification
• A small “Forward
Bias” Voltage results
in Large currents
• Any level of
“Reverse” Bias
results in almost NO
current flow
Class Q:
• For Fwd Bias, Which
End is +; P or N???
Engineering-43: Engineering Circuit Analysis
13
E-Field
A: the P end
• The Applied Voltage
REDUCES the
internal E-Field; This
“Biases” The
Junction in Favor of
DIFFUSION
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-10a_Diode-1_Physics_LoadLine.pptx
P-N Junction Rectifier cont.2
• Internal Field
ENHANCED
p-n junction
No Applied Voltage
– Carriers Pulled AWAY
from Jcn; xd grows
Xd
Forward Bias
• Diffusion & E-Field in
Balance, No Current
Flows
Reverse Biased
Engineering-43: Engineering Circuit Analysis
14
• Internal Field
REDUCED
– Carriers PUSHED and
Diffuse to the Jcn
where they are
“injected” into the other
side; xd Contracts
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-10a_Diode-1_Physics_LoadLine.pptx
Properties of Rectifying Junctions
Reverse
Forward
IN914 PN Diode
• IF = 75 000 µA
• IR = 0.025-50 µA
Engineering-43: Engineering Circuit Analysis
15
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-10a_Diode-1_Physics_LoadLine.pptx
The BIASED P-N Junction
“Bias” Simply refers to whether the PN
Junction will allow current to flow or not
• Forward Bias: Vapplied > 0 (Anode Positive)
– In forward bias the depletion region shrinks slightly in
width. With this shrinking the energy required for
charge carriers to cross the depletion region decreases
exponentially. Therefore, as the applied voltage
increases, current starts to flow across the junction.
The barrier potential of the diode is the voltage at
which appreciable current starts to flow through the
diode. The barrier potential varies for different
materials.
Bruce Mayer, PE
The Barrier Potential for Silicon is 0.6-0.7 volts
Engineering-43: Engineering Circuit Analysis
16
[email protected] • ENGR-43_Lec-10a_Diode-1_Physics_LoadLine.pptx
The BIASED P-N Junction
“Bias” Simply refers whether the PN Junction
will allow current to flow or not
• Reverse Bias: Vapplied < 0 (Anode Negative)
– Under reverse bias the depletion region widens. This
causes the electric field produced by the ions to cancel
out the applied reverse bias voltage. A small leakage
current, Is (saturation current) flows under reverse bias
conditions. This saturation current is made up of
electron-hole pairs being PRODUCED in the
depletion region. Saturation current is sometimes
referred to as the scale current because of it’s
relationship to junction temperature
Engineering-43: Engineering Circuit Analysis
17
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-10a_Diode-1_Physics_LoadLine.pptx
Small-Signal Diode
iD
IS
(mA)
• vD ≡ Bias Voltage
• iD ≡ Current through
Diode. iD is
• Negative for
Reverse Bias
• Positive for Forward
Bias
Knee
VBR
~V
vD
• IS ≡ Saturation
Current
• VBR ≡ Breakdown
Voltage
• V ≡ Barrier, or
Knee, Potential
Voltage
(nA)
Engineering-43: Engineering Circuit Analysis
18
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-10a_Diode-1_Physics_LoadLine.pptx
A
Rectifying
Diode
Data
Sheet
Engineering-43: Engineering Circuit Analysis
19
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-10a_Diode-1_Physics_LoadLine.pptx
1N914
Data
Sheet
Page-1
Engineering-43: Engineering Circuit Analysis
20
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-10a_Diode-1_Physics_LoadLine.pptx
1N914
Data
Sheet
Page-2
Engineering-43: Engineering Circuit Analysis
21
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-10a_Diode-1_Physics_LoadLine.pptx
Anode is P-Side
Engineering-43: Engineering Circuit Analysis
22
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-10a_Diode-1_Physics_LoadLine.pptx
Diode Macro Behavior
The Knee Voltage,
𝑉𝜙 , is typically
about 0.6-0.7 V
In the REVERSE
Bias Range, the
current, IS, is on the
order of nA to µA,
depending on the
SIZE of the diode
Temperature
Sensitivity
dV
2 mV
dT
K
Engineering-43: Engineering Circuit Analysis
23
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-10a_Diode-1_Physics_LoadLine.pptx
Diode Model Shockley Eqn
The TransConductance (vD,iD) curve on
the previous slide is
v
characterized by the iD I s e nV 1
SHOCKELY equation:
D
T
• Where
– Is ≡ Reverse Bias SATURATION Current
– n ≡ emission coefficient (sometimes called the
“diode quality factor”)
– VT ≡ Thermal Voltage: VT
kT q
k ≡ Boltzmann’s constant (1.38x10−23 J/K)
q ≡ Charge on an electron (1.6x10−19 Coul)
Engineering-43: Engineering Circuit Analysis
24
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-10a_Diode-1_Physics_LoadLine.pptx
Diode Model Shockley Eqn
Note the Shockely Eqn does NOT
predict Reverse-Bias BreakDown
Also when vD is Large and Negative
v D
nV
iD I s e T 1 iD I s e 10 1 I s 0 1 I s
• But we saw from the vi curves that Is is
NOT constant
• Also reverse Bias Currents are
often much larger than Is
Engineering-43: Engineering Circuit Analysis
25
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-10a_Diode-1_Physics_LoadLine.pptx
Diode Model Shockley Eqn
when vD is Large and Positive
e
vD
nVT
1 e
vD
nVT
1 e
vD
nVT
iD I s e
vD
nVT
• At 300k VT ≈ 26 mV
• if vD = 0.2V (200 mV), 200mV
1.526mv
e
168.7
and n ≈ 1.5 Then
Thus the simplified Eqn is quite
accurate in the FORWARD-Bias Region
Engineering-43: Engineering Circuit Analysis
26
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-10a_Diode-1_Physics_LoadLine.pptx
Diode Model Shockley Eqn
Finally If we know iD & IS then we can
iD
solve the Shockely
vD nVT ln 1
Equation for vD
Is
• Note that this eqn is NOT defined for
Negative Diode currents and thus
applies only to the
Forward Bias Region
vD by Schockely Eqn
450
400
350
vD (mV)
For n = 1.5,
Is = 25 µA
300
250
>> iD = linspace(0,1000,500); % in mA
>> vD = n*VT*log(iD/Is +1);
>> plot(iD,vD)
>> plot(iD,vD, 'LineWidth', 3), grid,
xlabel('iD (mA)'), ylabel('vD (mV)'),
title('vD by Schockely Eqn')
200
150
100
50
0
Engineering-43: Engineering Circuit Analysis
27
0
100
200
300
400
500
600
700
800
900
iD (mA)Bruce Mayer, PE
[email protected] • ENGR-43_Lec-10a_Diode-1_Physics_LoadLine.pptx
1000
Example: Exercise 10.2
Consider a diode
under Fwd-Bias so
vD
that this Eqn
nVT
applies: iD I s e
In this case
• VT = 26 mV
• n=1
Find ∆vD so that the
current iD 2
Doubles: i 2
By Fwd Bias eqn
v D v D
nVT
iD 2 I s e
2
vD
iD1
nV
I s e T
Cancelling Is and
dividing the
exponential
e
v D v D v D
nVT
nVT
e
v D
nVT
D1
Engineering-43: Engineering Circuit Analysis
28
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-10a_Diode-1_Physics_LoadLine.pptx
2
Example: Exercise 10.2
Taking the ln of both
sides of the last eqn
v D
nVT
ln e
ln 2
Thus
v D
ln 2
nVT
Then
v D ln 2nVT
Engineering-43: Engineering Circuit Analysis
29
Recalling Values for
n & VT
v D ln 2 1 26mV
18.02 mV
Next, Find ∆vD so
that the current
increases 10X
In the above eqn
replace 2 with 10
v D ln 10 1 26mV
59.9 mV
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-10a_Diode-1_Physics_LoadLine.pptx
ZENER Diodes
“Zeners” exploit the
reverse-bias BreakDown to effect
Voltage Regulation
A typical vi curve
Vz is specified
explicitly
The Rev-Bias BrkDwn portion of the vi
curve is designed to
be as VERTICAL as
possible
Ckt Designers use
Zeners to Regulate
(or pin) a voltage
point at certain level
Engineering-43: Engineering Circuit Analysis
30
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-10a_Diode-1_Physics_LoadLine.pptx
Zener Diode Voltage Regulation
A Typical Zener Ckt
V “Pinned at Vref
Engineering-43: Engineering Circuit Analysis
31
A Zener diode is used in this
circuit to regulate the voltage of
the current supply provided by
V+ to the desired Vref voltage
required by this IC circuit. The
small size and low cost of the
Zener (compared to other
techniques, such as a linear
regulator or reference chip)
make it ideal in this type of
application.
Note that This solution would
NOT work in an application with
a very high V+ (forcing the
Zener device to dissipate more
than a few watts).
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-10a_Diode-1_Physics_LoadLine.pptx
Source-Resistor vs Diode LoadLine
When We analyzed
The NONlinear
LINEAR DC Circuits,
problem can be
we arrived at Linearsolved
Algebra Equations
NUMERICALLY
(e.g. MATLAB fzero)
• i.e.; n-Eqns in
n-Unknowns
But the Diode vi
curve is highly nonLinear, so the
previous methods
do NOT apply
Engineering-43: Engineering Circuit Analysis
32
OR it can be solved
GRAPHICALLY
using so-called
Load-Line Analysis
• IF the powering ckt is
LINEAR
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-10a_Diode-1_Physics_LoadLine.pptx
Digression: LINEAR Devices
V-Source:VS 0 i VS
I-Source: I S 0 v I S
0 x b
0 x b
Resistor: iR GvR or vR RiR
m x 0
Capacitor: iC C dvC dt 0 mx 0
Inductor: vL L diL dt 0 mx 0
Engineering-43: Engineering Circuit Analysis
33
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-10a_Diode-1_Physics_LoadLine.pptx
Source-Resistor-Diode LoadLine
The simplest
LoadLine analysis is
a SERIES Circuit
with
vR
• A Power Source
(usually Voltage)
• A controlling element
(in this case a diode)
• A Resistive Load
A pictorial
Representation
Engineering-43: Engineering Circuit Analysis
34
By ohm VSS vR vD
& KVL
RiD vD
We need to find iD
and vD
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-10a_Diode-1_Physics_LoadLine.pptx
Source-Resistor-Diode LoadLine
From the KCL eqn
Notice for the
Resistor
vR
VSS Ri D vD
• If iD=0, then vD = VSS
• If vD=0, then
iD = VSS/R
Engineering-43: Engineering Circuit Analysis
35
Plotting these two
(vD,iD)points on a
vD-iD curve produces
a STRAIGHT-line
that describes the
Resistor Behavior
The Diode Curve
can plotted on top
of the Resistor curve
• The Crossing Pt is
the solution, or
OperatingBrucePoint
Mayer, PE
[email protected] • ENGR-43_Lec-10a_Diode-1_Physics_LoadLine.pptx
Constructing the LoadLine
Recall the the KCL Eqn: VSS Ri D vD
Taking iD as the dependent Variable
VSS
1
iD v D
R
R
of form
y mx b
The above Eqn is a straight line
Need only Two Points to construct the
circuit operating line
vD
iD
Pt
(load line) so make
B
0 Vss/R
a “T-Table”
A
Vss
0
Engineering-43: Engineering Circuit Analysis
36
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-10a_Diode-1_Physics_LoadLine.pptx
LoadLine graphic Conceptually
(Also Called Quiescent, or Q, Point)
Engineering-43: Engineering Circuit Analysis
37
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-10a_Diode-1_Physics_LoadLine.pptx
Vsrc + Resistor + Diode Math
VSS
1
Recall
iD v D
R
R
Eqn for iD:
v
Also the
nV
iD I s e 1
Shockley Eqn:
D
T
vD
Equating
vD V SS
nVT
Is e
1
the iD’s:
R
R
The Equation is TRANSCENDTAL in vD
Engineering-43: Engineering Circuit Analysis
38
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-10a_Diode-1_Physics_LoadLine.pptx
Vsrc + Resistor + Diode Math
For use in MATLAB fzero Collect all
terms on one side of Eqn:
vD
vD V SS
nVT
Is e
1 0
R
R
Note that the Eqn has been “ZeroED”
If we know R, n, VT, and VSS (vD is the
unknown) then we can create an
anonymous function for input into fzero
Engineering-43: Engineering Circuit Analysis
39
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-10a_Diode-1_Physics_LoadLine.pptx
Load Line Example
Consider Ckt
8.3 Ω
1.5 V
• IS = 2x10−14 amps
• n = 1.5
Plot Load-Line and
Diode Curve on
Same Graph yields
the operating point
LoadLine Analysis
200
180
140
120
iD (mA)
For the Diode Use
the Shockely Eqn in
Fwd Bias with
160
100
80
60
40
20
0
0
0.5
1
vD (volts)
Engineering-43: Engineering Circuit Analysis
40
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-10a_Diode-1_Physics_LoadLine.pptx
1.5
LoadLine Example Q-pt
LoadLine Analysis
200
180
160
140
iD (mA)
120
100
80
60
≈4340
20
0
0
Engineering-43: Engineering Circuit Analysis
41
0.5
1
vD (volts)
≈1.15
1.5
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-10a_Diode-1_Physics_LoadLine.pptx
Check Answer by MATLAB
MATLAB Code
MATLAB Results
% Bruce Mayer, PE
% ENGR43 * 07Jan12
% LoadLine_Example_120107.m
%
Is = 2E-14 % Amps
n = 1.5
vD = linspace(0,1.2,500); % 0-1.2V
iD = Is*(exp(vD/(n*VT)-1)); % in Amps
R = 8.3; % ohms
Vss = 1.5 % Volts
vR = [0, 1.5] % Resistor Voltage Pts
iR = [Vss/R, 0] % Resistor Current Pts
%
% plot in mA
plot(vD, 1000*iD, vR, 1000*iR, 'LineWidth',3),
grid,...
xlabel('vD (volts)'), ylabel('iD (mA)'),
title('LoadLine Analysis')
%
% Solve Numerically as check
% fcn to Zero
Zfcn = @(x) -x/R+Vss/R - Is*(exp(x/(n*VT)-1))
vDQ = fzero(Zfcn,1.1)
iDQma = 1000*(-vDQ/R+Vss/R)
vDQ =
1.1461
iDQma =
42.6349
Thus the Graphical
Estimate is quite
good for the Q
Point.
Engineering-43: Engineering Circuit Analysis
42
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-10a_Diode-1_Physics_LoadLine.pptx
Zener V-Reg Load Line Analysis
A regulating Circuit
Solving for iD again
yields the LoadLine
VSS
1
iD v D
R
R
• Note that the Zener
LoadLine has a
NEGATIVE y-intercept
The LoadLine T-table
By KVL
vD Ri D VSS 0
Engineering-43: Engineering Circuit Analysis
43
Pt
vD
iD
B
0
−Vss/R
A
−Vss
0
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-10a_Diode-1_Physics_LoadLine.pptx
Example 10.3 Zener Reg
• R = 1 kΩ
• Could find Zener
Diode vi curve from
the Data Sheet
Analyze this Ckt for
OutPut is the
Diode Voltage
• VSS,lo = 15V
• VSS,hi = 20V
Engineering-43: Engineering Circuit Analysis
44
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-10a_Diode-1_Physics_LoadLine.pptx
Example 10.3 Zener Reg
Pt
vD
iD
Blo
0
−15 mA
Alo −15 V
Bhi
0
Ahi −20V
0
−20 mA
0
Engineering-43: Engineering Circuit Analysis
45
VSS
1
iD v D
R
R
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-10a_Diode-1_Physics_LoadLine.pptx
Check Sensitivity of Zener LL
Find
dvD
vD 10 10.5V
15 20V
dVSS VSS
dvD 0.5 V
10%
dVSS
5 V
Thus, for example, a
4V change in the
Input, VSS, results in
only a 0.4V change
in the OutPut, vo
Engineering-43: Engineering Circuit Analysis
46
Most REAL Zeners have
much STEEPER DownSlope than that shown in
the Book, so they have
much BETTER output
Sensitivity
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-10a_Diode-1_Physics_LoadLine.pptx
Complicated Circuit LL Analysis
Any LINEAR Circuit, no matter how
complicated allows Diode LoadLine
Analysis to determine the Q-Point
Recall that ANY linear ckt can be
“Thévenized” into a V-src and Series
resistance or impedance
Thus to use the LL on this type of Ckt,
simply disconnect the Diode, Thévenize
the LINEAR ckt-fragment and then
reconnect the Diode
Engineering-43: Engineering Circuit Analysis
47
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-10a_Diode-1_Physics_LoadLine.pptx
Complicated Circuit LL Analysis
The Thévenization
process concept
Put into y = mx+b
form to produce the
LoadLine (iD is
dependent variable)
1
iD
RT
Then by Ohm & KVL
(ClockWise current)
as before
RT iD vD VSS 0
Engineering-43: Engineering Circuit Analysis
48
VSS
vD
RT
Now can apply the
normal LoadLine
Methods
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-10a_Diode-1_Physics_LoadLine.pptx
Example: Thévenization
Given Ckt
Now Detach the
NONlinear element
(the Diode) and
Thévenize
Find Q-point Given
Zener vi Curve
←VT
First Redraw
←RT
Engineering-43: Engineering Circuit Analysis
49
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-10a_Diode-1_Physics_LoadLine.pptx
Example: Thévenization
For VT have simple
Voltage-Divider
6
VT 24V
20V
6 1.2
Since have INDEP
Src find RT by
Source DeActivation
←VT
←RT
Now ReAttach Diode
for LoadLine Ckt
• Setting 24V Src = 0
6kΩ 1.2kΩ
RT
1kΩ
6 1.2kΩ
Engineering-43: Engineering Circuit Analysis
50
vD
iD
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-10a_Diode-1_Physics_LoadLine.pptx
Example: Thévenization
Then KVL on the
Thévenized Ckt
VSS
1
iD v D
R
R
1
20V
iD
vD
1kΩ
1kΩ
Then the T-Table
Pt
vD
iD
B
0
−20 mA
A
− 20 V
The Q-Point is then
the Intersection of
the Line and Curve
• See next slide
0
Engineering-43: Engineering Circuit Analysis
51
Next Draw this
LoadLine on the
Same Graph as the
Zener vi Curve
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-10a_Diode-1_Physics_LoadLine.pptx
Example: Thévenization
Thus the Operating, or Q,
Point for the Thevenized
Ckt
• vD = −10 V
• iD = −10 mA
Engineering-43: Engineering Circuit Analysis
52
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-10a_Diode-1_Physics_LoadLine.pptx
Example: Thévenization
Also Find the
ORIGINAL Source
Current
Note that vL =
−vD = +10V
IS
10V
Engineering-43: Engineering Circuit Analysis
53
Using KCL and Ohm
Find
24 10V
IS
11.67 mA
1.2 kΩ
Then the Source
Power
PS VS I S 24V 11.67 mA
PS 280 mW
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-10a_Diode-1_Physics_LoadLine.pptx
All Done for Today
William
Bradford
Shockley
Born February 13, 1910 • London, England
Died August 12, 1989 (aged 79) • Stanford, CA
Engineering-43: Engineering Circuit Analysis
54
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-10a_Diode-1_Physics_LoadLine.pptx
Engineering 43
Appendix
Diode vi Curves
Bruce Mayer, PE
Registered Electrical & Mechanical Engineer
[email protected]
Engineering-43: Engineering Circuit Analysis
55
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-10a_Diode-1_Physics_LoadLine.pptx
Fig 10.8 Diode characteristic
for Exercise 10.3
Engineering-43: Engineering Circuit Analysis
56
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-10a_Diode-1_Physics_LoadLine.pptx
Figure P10.20
for HW
Reference
Engineering-43: Engineering Circuit Analysis
57
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-10a_Diode-1_Physics_LoadLine.pptx
Vsrc + Resistor + Diode Math
For use in MATLAB fzero Multiply
Both sides by R/vD:
vD
R
vD V SS
nVT
I s e
1
R
vD
R
Yields
V SS
1
vD
Engineering-43: Engineering Circuit Analysis
58
vD
RI s nVT
e
1
vD
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-10a_Diode-1_Physics_LoadLine.pptx