4.4.1 I M P R O V E D P I E C E W I S E L I N E A R M O D E L S
FOR NONLINEAR ELEMENTS *
The accuracy of the results of a piecewise liner analysis depends on the accuracy
of the model used. In this section, we will discuss the process of creating more
precise models of nonlinear elements when increased accuracy is desired.
To illustrate the process, let us use the diode as an example of a nonlinear
element. Thus far, we used the simple, ideal diode model. It is obvious from the
preceding example that the major effect in the model is when the voltage vD
across the diode is positive, and above about 0.6 V. Substantial improvement
can be made by adding a 0.6-V source in series with the ideal diode, as shown in
Figure 4.33a. The corresponding i v characteristic for the improved piecewise
linear model is shown in Figure 4.33b. Let us work a simple example using this
piecewise linear model.
+
iD
iD
vD
vD
+
0.6 V
0.6 V
(b)
(a)
F I G U R E 4.33 Improved
piecewise linear diode models.
+
iD
iD
Slope
+
vD
0.6 V
-
vD
0.6 V
Rd
(c)
214c
(d)
1
= ----Rd
e x a m p l e 4. 14 a n o t h e r e x a m p l e u s i n g p i e c e w i s e
l i n e a r m o d e l i n g Let us rework the example containing a voltage source,
resistor, and diode in Figure 4.16 using the piecewise linear model for the diode from
Figure 4.33b. The behavior of this model, comprising an ideal diode in series with a
voltage source, can also be summarized in two statements:
Diode ON (vertical segment): vD = 0.6 V for iD > 0
Diode OFF (horizontal segment): iD = 0
for vD < 0.6 V
(4.47)
Let us determine iD for E = 3 V and E = −5 V, given that R = 500 . According to
the piecewise linear method, we will focus on one straight-line segment at a time, using
linear analysis within each segment.
Vertical segment When iD and vD are in the vertical segment of their characteristic,
the circuit shown in Figure 4.34b results, and we can write
iD =
E − 0.6 V
R
.
(4.48)
Horizontal segment Figure 4.34c shows the corresponding circuit when the diode is
operating as an open circuit. In this segment,
iD = 0.
(4.49)
Combining the results Intuition tells us that the vertical segment applies when
E > 0.6 V (the diode turns on) and the horizontal segment applies otherwise (diode
is off). Thus, when E = 3 V, Equation 4.48 applies, and
iD =
E − 0.6 V
R
=
3 − 0.6
500
= 4.8 mA.
Comparing to Equation 4.40, notice that the value of iD predicted by this improved
model is slightly lower than that predicted by the ideal diode model. The ideal diode
model did not account for the 0.6-V drop across the diode, and so overestimated
the current.
Equation 4.49 applies when E = −5 V, so
iD = 0.
214d
+
vR
iD
R
+
+
E
-
vD
0.6 V
-
(a) Complete model
vR
+
iD
R
F I G U R E 4.34 Piecewise linear
analysis in the vertical and horizontal straight-line segments using the
diode model containing a voltage
source.
+
E
-
+
vD
0.6 V
(b) Vertical segment
+
vR
iD
R
+
E
-
+
0.6 V
vD
-
(c) Horizontal segment
Further improvement in accuracy can be realized by adding a series resistor Rd of
suitable value to the ideal diode and voltage source, as shown in Figure 4.33c.
The specific choice of resistor value depends on the application; one should
strive to make the characteristic match over the range of diode current expected
in the specific circuit (see Figure 4.35). We will illustrate the use of this model in
an example. More examples using these and other more complicated piecewise
models will appear throughout the book, and specifically in Chapters 7 and 16.
214e
t h e d i o d e r e s i s t a n c e Choose values for Rd
for the piecewise linear diode model in Figure 4.33c assuming that the resistance must
provide a reasonable match for currents up to 0.4 A and 1 A. Assume VTH = 0.025 V
and Is = 10−12 A.
e x a m p l e 4. 15
iD (A)
Figure 4.35 plots the v i characteristics for the diode. The figure shows that the resistance
value Rd1 = 0.1 provides a good match for the diode v i characteristics up to 1 A,
while the resistance value Rd2 = 0.2 provides a better match in the smaller current
range from zero to 0.4 A.
1.0
0.9
0.8
Slope = 10, Rd1 = 0.1 Ω
0.7
0.6
0.5
0.4
Slope = 5, Rd2 = 0.2 Ω
0.3
F I G U R E 4.35 Choosing a value
of the resistance in the piecewise
linear diode model.
0.2
0.1
0.0
0.0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
1.1
vD (V)
214f
e x a m p l e 4.16
a
more
complicated
piecewise
l i n e a r m o d e l Let us further rework the previous example using the piecewise linear model for the diode from Figure 4.33c. The behavior of this model,
comprising an ideal diode in series with a voltage source and a resistor, can be
summarized in two statements:
Diode ON (vertical segment): vD = 0.6 V + iD Rd for iD > 0.
Diode OFF (horizontal segment): iD = 0
for vD < 0.6 V.
(4.50)
Again, let us determine iD for E = 3 V and E = −5 V, given that R = 500 and
Rd = 10 .
Vertical segment When iD and vD are in the vertical segment of their characteristic,
the circuit shown in Figure 4.36b results, and we can write
iD =
E − 0.6 V
R + Rd
.
(4.51)
Horizontal segment Figure 4.24c shows the corresponding circuit when the diode is
operating as an open circuit. In this segment,
iD = 0.
(4.52)
Combining the results For E = 3 V, Equation 4.51 applies, and so
iD =
E − 0.6 V
R + Rd
=
3 − 0.6 V
500 + 10
= 4.7 mA.
Equation 4.52 applies when E = −5 V, so
iD = 0.
As illustrated using the diode, increasingly better fits to an actual nonlinear
device characteristic can be obtained by introducing more and more ideal
elements. For example, for the diode, increasingly better fits to the actual
diode characteristic can be obtained by introducing more and more ideal diodes,
batteries, and resistors. But again a price is paid; increased accuracy of the
model brings increased complexity. The proper compromise between simplicity and accuracy is not always obvious. Start with the simplest model, then add
complexity to see if the solution changes in major ways.
214g
+
vR
iD
R
+
E
-
+
vD
0.6 V
Rd
-
(a) Complete model
vR
+
iD
R
+
E
-
+
vD
0.6 V
Rd
-
F I G U R E 4.36 Piecewise linear
analysis in the vertical and horizontal straight-line segments using the
diode model containing a voltage
source and a resistor.
(b) Vertical segment
+
vR
iD
R
+
E
-
+
vD
0.6 V
Rd
(c) Horizontal segment
214h