Diodes
Sections 3.3.1 – 3.3.8
1
Modeling Diode Characteristics
• Exponential model nonlinearity makes circuit analysis
difficult.
• Two common approaches are graphical analysis and iterative
analysis
• For simple battery, resistor, diode circuit (see below) graphical
analysis can be performed by simply plotting relationship of
diode and resistor equations on i-v plane
• The plot of the straight line resistor equation is commonly
referred to as the load line
• Intersection of diode equation and load line represents the
operating point of the circuit
• Graphical analysis is good for only simple circuits
2
ID = ISe
VD / nVT
VDD − VD
ID =
R
Figure 3.10 A simple circuit used to illustrate the analysis of circuits in which the diode is forward conducting.
3
Figure 3.11 Graphical analysis of the circuit in Fig. 3.10 using the exponential diode model.
4
Iterative Analysis
• Iterative analysis can be performed using a simple
procedure
• For example consider VDD=5V and R = 1Kohm.
Assume diode current of 1mA at a voltage of 0.7V
and assume voltage drop changes 0.1V for every
decade drop in current (Ans. VD = 0.762V)
• Gives precise answer but can be time consuming
especially if done by hand calculations (note: we can
use a circuit simulator like SPICE to help out)
5
Piecewise Linear Model
• Analysis can be greatly simplified if we can
find a linear approximation of diode equation
• We can approximate the diode forward
characteristics with two straight lines (see
below)
• The piecewise linear model can be represented
by an equivalent circuit that includes a battery
and a resistance
6
iD = 0, vD ≤ VD 0
iD = (vD − VD 0 ) / rD , vD ≥ VD 0
Note: VD0=0.65V and rD=20 ohms for above example
Figure 3.12 Approximating the diode forward characteristic with two straight lines: the piecewise-linear model.
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Figure 3.13 Piecewise-linear model of the diode forward characteristic and its equivalent circuit representation.
8
• Repeating analysis for previous circuit with equivalent circuit
substituted for diode
VDD − VD 0
ID =
R + rD
5 − 0.65
ID =
= 0.00426 A = 4.26mA
1000 + 20
VD = VD 0 + I D rD = 0.65 + .00426 * 20 = 0.735V
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Constant Voltage Drop Model
• Even simpler model uses a vertical straight line to
approximate fast rising part of exponential curve (i.e.
forward biased diode exhibits constant voltage drop
typically assumed to be 0.7V) (see below)
• Because of its simplicity, this model is commonly used
in initial stages of analysis and design
• For previous circuit VD = 0.7V and ID = 4.3mA
5 − 0.7
ID =
= 0.0043 A = 4.3mA
1000
10
Figure 3.15 Development of the constant-voltage-drop model of the diode forward characteristics. A vertical straight line
(B) is used to approximate the fast-rising exponential. Observe that this simple model predicts VD to within ±0.1 V over the
current range of 0.1 mA to 10 mA.
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Figure 3.16 The constant-voltage-drop model of the diode forward characteristics and its equivalent-circuit representation.
12
• Example 1
• Design the circuit for figure below to provide an output
voltage of 2.4V. Assume diodes have 0.7V drop at 1mA
and that ∆V=0.1V/decade change in current
• Ans: R = 760 ohms
Figure E3.12
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Small Signal Model
• In many circuit applications a diode is biased
at an operating point with some small ac signal
superimposed on the dc bias
• For this situation a good solution is to find dc
bias using one of previous methods and then
model small signal variations by resistance
equal to 1/slope of diode curve at the operating
point
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Figure 3.17 Development of the diode small-signal model. Note that the numerical values shown are for a diode with n = 2.
15
• The value of rd can be determined as follows:
ID = ISe
VD / nVT
vD (t ) = VD + vd (t )
iD (t ) = I S e
v D / nVT
iD (t ) = I S e (VD + vd )/ nVT = I s eVD / nVT e vd / nVT = I D e vd / nVT
vd
⟨⟨1
nVT
Assume v is small
⎛
vd
iD (t ) ≅ I D ⎜⎜1 +
⎝ nVT
⎞
⎟⎟
⎠
d
Small Signal appox.
16
⎛
vd
iD (t ) ≅ I D ⎜⎜1 +
⎝ nVT
⎞
⎟⎟ = I D + id
⎠
vd
ID
id (t ) =
vd =
nVT
rd
nVT
rd =
ID
Small-signal resistance
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For circuit in (a) R=10Kohms, V+=10V DC + 60Hz sine wave
with 1V peak amplitude (i.e power supply ripple) Calculate
DC voltage of diode and amplitude of sinusoidal signal
appearing across it. Assume diode to have 0.7V drop at 1mA
and n=2.
Figure 3.18 (a) Circuit for Example 3.6. (b) Circuit for calculating the dc operating point. (c) Small-signal equivalent
circuit.
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10 − 0.7
ID =
= 0.93mA
10
Note: close to 1mA so diode voltage will be close to assumed value of 0.7V
nVT
2 * .025
=
= 53.8Ω
rd =
ID
0.00093
Note: signal voltage can be found from small-signal model in figure (c) above
rd
0.0538
vd ( peak ) = VS
= 1*
= 5.35mV
R + rd
10 + 0.0538
Note: since signal is small, use of small-signal model is valid
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Example 2
• Find the value of the diode small-signal
resistance rd at bias currents of 0.1mA, 1mA,
and 10mA. Assume n=1
• Ans: 250ohms, 25ohms, 2.5ohms
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Example 3
• Consider a diode with n=2 biased at 1mA.
Find the current as a result of changing the
voltage by -20mV. Calculate using smallsignal model and exponential model
• Ans: -0.40ma, -0.33mA
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