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CHAPTER 10
Exercises
E10.1
Solving Equation 10.1 for the saturation current and substituting values,
we have
4
15
exp(
/
) 1
10
exp(0.600/0.026) 1
9.502 10
A
D
s
T
i
I
vn
V
−
−
=
−
=
−
=×
Then for
0.650
v
=
V, we have
15
exp(
/
) 1
9.502 10
exp(0.650/0.026) 1
0.6841 mA
DD
iI
n
−
=−
=
×
×
−
=
Similarly for
0.700
=
V,
4.681
=
mA.
E10.2
The approximate form of the Shockley Equation is
exp(
/
)
=
.
Taking the ratio of currents for two different voltages, we have
11
12
22
exp(
/
)
exp (
)/
exp(
/
)
iv
vv
==
−
Solving for the difference in the voltages, we have:
ln(
/
)
Vii
∆=
Thus to double the diode current we must increase the voltage by
0.026ln(2)
18.02 mV
=
and to increase the current by an order of
magnitude we need
0.026ln(10)
59.87 mV
=
E10.3
The load line equation is
.
SS
Ri
=+
The loadline plots are shown on
the next page.
From the plots we find the following operating points:
(a)
1.1 V
9 mA
DQ
(b)
1.2 V
13.8 mA
(c)
0.91 V
4.5 mA
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E10.4
Following the methods of Example 10.4 in the book, we determine that:
(a) For
1200
,
600
, and
12 V.
TT
L
RR
V
=Ω
=
Ω
=
(b) For
400
,
300
, and
6 V.
=
The corresponding load lines are:
3
At the intersections of the load lines with the diode characteristic we
find (a)
9.4 V
LD
vv
=−
≅
; (b)
6.0 V
≅
.
E10.5
Writing a KVL equation for the loop consisting of the source, the
resistor, and the load, we obtain:
15
100(
)
D
ii
v
−
The corresponding load lines for the three specified values of
i
L
are
shown:
At the intersections of the load lines with the diode characteristic, we
find (a)
10 V;
o
=
(b)
10 V;
=
(c)
5 V.
=
Notice that
the regulator is effective only for values of load current up to 50 mA.
E10.6
Assuming that
1
and
2
are both off results in this equivalent circuit:
Because the diodes are assumed off, no current flows in any part of the
circuit, and the voltages across the resistors are zero.
Writing a KVL
equation around the lefthand loop we obtain
1
10
=
V, which is not
consistent with the assumption that
1
is off.
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E10.7
Assuming that
D
1
and
2
are both on results in this equivalent circuit:
Writing a KVL equation around the outside loop, we find that the voltage
across the 4k
Ω
resistor is 7 V and then we use Ohm’s law to find that
i
1
equals 1.75 mA.
The voltage across the 6k
Ω
resistance is 3 V so
x
is
0.5 mA.
Then we have
21
1.25
DD
ii
=−
=
−
mA, which is not consistent
with the assumption that
2
is on.
E10.8
(a) If we assume that
1
is off, no current flows, the voltage across the
resistor is zero, and the voltage across the diode is 2 V, which is not
consistent with the assumption.
If we assume that the diode is on, 2 V
appears across the resistor, and a current of 0.5 mA circulates clockwise
which is consistent with the assumption that the diode is on.
Thus the
diode is on.
(b) If we assume that
2
is on, a current of 1.5 mA circulates
counterclockwise in the circuit, which is not consistent with the
assumption.
On the other hand, if we assume that
2
is off we find that
2
3
v
where as usual we have referenced
2
positive at the anode.
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This note was uploaded on 11/02/2010 for the course ECE 3710 taught by Professor Haris during the Spring '07 term at Georgia Institute of Technology.
 Spring '07
 Haris

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