082-Engineering-Mathematics-Anthony-Croft-Robert-Davison-Martin-Hargreaves-James-Flint-Edisi-5-2017

422 Chapter 12 Applications of differentiation
I
I
+
–
R
V
V s —R
V s
V s V
Point at which both
the diode and the
resistor equations
are satisfied
(a)
(b)
Figure12.15
Asimple non-linear circuit: (a)seriesdiode--resistorcircuit;(b)resistorload line
superimposed on diode characteristic.
Solution
There are several ways to solve this problem. A difficulty exists because the diode
I--V relationship is non-linear. One possibility is to draw a load line for the resistor
superimposedonthediodeI--V characteristic,asshowninFigure12.15(b).Theload
lineisanequationfortheresistorcharacteristicwrittenintermsofthevoltageacross
the diode,V, and the current through the diode,I. Itisgiven by
V s
−V=IR
I = − 1 R V +V s
R
This is a straight line with slope − 1 R and vertical intercept V s
.WhenV =0,I=
V
R
s
. This corresponds to all of the supply voltage being dropped across the resistor.
R
WhenV = V s
,I = 0. This corresponds to all of the supply voltage being dropped
across the diode. Therefore, these two limits correspond to the points within which
the circuit must operate. The solution to the circuit can be obtained by determining
the intercept of the diode characteristic and the load line. This is possible because
both the resistor characteristic and diode characteristic are formulated in terms ofV
andI,andsoanysolutionmusthavethesamevaluesofI andV forbothcomponents.
If an accurate graph is used, it is possible to obtain a reasonably good solution. An
alternative approach is to use the Newton--Raphson technique. Combining the two
component equations gives
−V +V s
=RI s
(e 40V −1)
NowR =2.2×10 4 ,I s
=10 −14 ,V s
=2andso
−V +2 = 2.2 ×10 4 ×10 −14 (e 40V −1)
Now, define f (V) by
f(V)=2.2×10 −10 (e 40V −1)+V −2
We wishtosolve f (V) = 0.We have
f ′ (V ) = 2.2 ×10 −10 ×40e 40V +1 = 8.8 ×10 −9 e 40V +1
Choose aninitial guess ofV 1
= 0.5:
V 2
=V 1
− f(V 1 )
f ′ (V 1
) =0.5−2.2 ×10−10 (e 20 −1) +0.5 −2
= 0.7644
8.8 ×10 −9 e 20 +1