Lecture 6Solid-State Diodes and Diode Circuits
1
Goals
• Understand diode structure and basic layout• Explore various diode models including the mathematical model, the
ideal diode model, and the constant voltage drop model• Define regions of operation of the diode (forward, reverse bias, and
reverse breakdown)• Apply the various types of models in circuit analysis
2
Diode Introduction
• A diode is formed by interfacing an n-type semiconductor with a p-type semiconductor.
• A pn junction is the interface between n and p regions.
Diode symbol
3
Diode Layout
4
Space Charge Region Formation at the pn Junction
5
Diode i-v Characteristics
Turn-on voltage marks point of significant current flow. Is is called the reverse saturation current.
6
where IS = reverse saturation current (A) vD = voltage applied to diode (V)q = electronic charge (1.60 x 10-19 C)k = Boltzmann’s constant (1.38 x 10-23 J/K)T = absolute temperaturen = nonideality factor (dimensionless)VT = kT/q = thermal voltage (V) (25 mV at room temp.)
IS is typically between 10-18 and 10-9 A, and is strongly temperature dependent due to its dependence on ni
2. The nonideality factor is typically close to 1, but approaches 2 for devices with high current densities. It is assumed to be 1 if the value of “n”
Diode Equation
iD IS expqvD
nkT
1
IS exp
vD
nVT
1
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Diode Voltage and Current Calculations (Example)
Problem: Find diode voltage for diode with given specifications
Given data: IS=0.1 fA ID= 300 A
Assumptions: Room-temperature dc operation with VT=0.025 V
Analysis:
With IS=0.1 fA
With IS=10 fA
With ID= 1 mA, IS=0.1 fA
V718.0)A16-10A4-103V)ln(10025.0(11ln
SIDI
TnVDV
V603.0DV
V748.0DV
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Diode Current for Reverse, Zero, and Forward Bias
• Reverse bias:
• Zero bias:
• Forward bias:
iD IS expvD
nVT
1
IS 0 1 IS
iD IS expvD
nVT
1
IS 1 1 0
iD IS expvD
nVT
1
IS exp
vD
nVT
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Ideal Diode
A semiconductor diode consists of a PN junction and has (2) terminals, an anode(+) and a cathode(-).The diode is a unidirectional device; Current flows only from anode to cathodeForward bias : v>0, any current will flow (ideal diode acts as a short circuit)Reverse Bias: v < 0, i=0 (open circuit)
+-Anode
CathodeForward bias
Vanode > vcathode
Reverse biasVanode < vcathode
Break point
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Diode Circuit Analysis: Basics
V and R may represent Thevenin equivalent of a more complex 2-terminal network.Objective of diode circuit analysis is to find quiescent operating point for diode, consisting of dc current and voltage that define diode’s i-v characteristic.
Loop equation for given circuit is:
This is also called the load line for the diode. Solution to this equation can be found by:
1. Simplified analysis with ideal diode model.
2. Simplified analysis using constant voltage drop model.
3. Graphical analysis using load-line method.
4. Analysis with diode’s mathematical model.
DD VRIV
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1.Analysis using Ideal Model for Diode
If diode is forward-biased, voltage across diode is zero. If diode is reverse-biased, current through diode is zero.
vD =0 for iD >0 and vD =0 for vD < 0 Thus diode is assumed to be either on or off.
Analysis is conducted in following steps:
1. Select diode model.2. Identify anode and cathode of diode and label
vD and iD.3. Guess diode’s region of operation from circuit.4. Analyze circuit using diode model appropriate
for assumed operation region.5. Check results to check consistency with
assumptions.
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Analysis using Ideal Model for Diode: Example
Since source is forcing positive current through diode assume diode is on.
Q-point is(1 mA, 0V)
0
m1k10
V)010(
D
D
I
AI
our assumption is right. our assumption is right.
Since source is forcing current backward through diode assume diode is off. Hence ID =0 . Loop equation is:
Q-point is (0, -10 V)
V10
01010 4
D
DD
V
IV
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Two-Diode Circuit Analysis
Analysis: Ideal diode model is chosen. Since 15V source is forcing positive current through D1 and D2 and -10V source is forcing positive current through D2, assume both diodes are on.
Since voltage at node D is zero due to short circuit of ideal diode D1,
Q-points are (-0.5 mA, 0 V) and (2.0 mA, 0 V)
But, ID1 <0 is not allowed by diode, so ?.
mA50.1k10
V)015(1
I mA00.2k5
V)10(02
DI
211 DI
DII mA50.025.1
1
DI
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Two-Diode Circuit Analysis (contd.)
Since current in D1 is zero, ID2 = I1,
Q-points are D1 : (0 mA, -1.67 V):off
D2 : (1.67 mA, 0 V) :on
Analysis: Since current in D2 but that in D1 is invalid, the second guess is D1 off and D2 on.
V67.17.16151
000,10151
mA67.115,000
V251
0)10(2
000,51
000,1015
ID
V
I
DII
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Break point analysis
VV
VKK
KV
VVx
in
in
3
).12
2(2
2
The break point is wherei=0 and v=0 transition from ‘ON” to “OFF”OR
It is the point after which the diode will start conducting
Range of V after which the diode will be “ON”
Vin
1K
2K
2V
Vx1K
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2.Analysis using Constant Voltage Drop Model for Diode
Analysis:
Since 10V source is forcing positive current through diode assume diode is on.
vD = Von for iD >0 and vD = 0 for vD < Von.
mA94.0k10
V)6.010(k10
V)10(
onD
VI
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3. Graphical analysis using load-line method.
Linear Circuit
Nonlinear element
+
-
iI and V must satisfy both the linear
And non-linear circuit equations
For the linear circuit
Vth(VOC)
R
Ino(ISC) R
i
+
i
Finding Thevenin or Norton equivalent of the linear circuit
Load-Line Analysis (Example)
Problem: Find Q-pointGiven data: V=10 V, R=10k.Analysis:
To define the load line we use,VD= 0VD= 5 V, ID =0.5 mA
These points and the resulting load line are plotted.Q-point is given by intersection of load line and diode characteristic:
Q-point = (0.95 mA, 0.6 V)
1010 4DD VI
mA1)k10/V10( DI
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4. Analysis using Mathematical Model for Diode
Ideal Diode
)1(
9.25 voltageThermalq
KT
current saturation ReverseI
)21(factor ideality n
eTemperaturT
)/.38x10Constant(1 sBoltzmann'K
)(1.6x10 chargeelectron q
)1(
s
23-
19-
T
D
D
nV
v
sD
T
o
o
nKT
qv
sD
eIi
mVV
n
K
KJ
C
eIi
Iterative Method based on Mathematical Model
)1( T
D
nV
v
sD eIi
Linear Circuit
Nonlinear element
+
-
iI and V must satisfy both the linear
And non-linear circuit equations
VOC=Vth
i
ISC =VOC/R
i
vva
ia
ib
Repeat e.
for solve circuit,linear theinto Plug d.
ln
find andequation diode theinto Put c.
b.
for valuea Assume a.
R
vViv
I
inVv
vi
iR
vVi
v
aocba
S
aTa
aa
aDoc
D
D
Linear circuit equation
Exponential diode equation
Example: Analysis using Mathematical Model for Diode
DD
DT
DSD
VV
VnV
VII
140exp101010
140exp101exp
134
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Problem: Find Q-point for given diode characteristic.Given data: IS =10-13 A, n=1, VT
=0.0025 VAnalysis:
is load line, given by a transcendental equation. A numerical answer can be found by using iterative method.
Use the technique in the previous slide
Q-point = ( 0.9426 mA, 0.5742 V)
Since, usually we don’t have accurate saturation current and significant tolerances for sources and passive components, we need answers precise up to only 2or 3 significant digits.
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Summary: Diode circuit linear models
a. Ideal diode
v
i
0 ,0
0 ,0
iv
iv
b. Constant voltage model
0 ,
0 ,
iVv
iVv
D
D
vVD
For Silicon diodes
0.65 < vD < 0.8
+
VD
i
ideal
ideal
ideal
VDO
rD
c. Piecewise linear model
vVDO
i1/rD
Summary: Diode circuit linear models