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CHAPTER 10
Exercises
E10.1 Solving Equation 10.1 for the saturation current and substituting values,
we have
iD
Is 
exp(vD / nVT )  1
10 4

exp(0.600 / 0.026)  1
 9.502  10 15 A
Then for vD  0.650 V, we have

iD  I s exp(vD / nVT )  1  9.502  1015  exp(0.650 / 0.026)  1
 0.6841 mA
Similarly for vD  0.700 V, iD  4.681 mA.
E10.2 The approximate form of the Shockley Equation is iD  Is exp(vD / nVT ) .

Taking the ratio of currents for two different voltages, we have
iD 1 exp(vD 1 / nVT )
  exp (vD 1  vD 2 ) / nVT 
Solving for the difference in the voltages, we have:
vD  nVT ln(iD 1 / iD 2 )
Thus to double the diode current we must increase the voltage by
vD  0.026ln(2)  18.02 mV and to increase the current by an order of
magnitude we need vD  0.026ln(10)  59.87 mV
E10.3 The load line equation is VSS  RiD  vD . The load-line plots are shown on
the next page. From the plots we find the following operating points:
(a) VDQ  1.1 V IDQ  9 mA

(b) VDQ  1.2 V IDQ  13.8 mA
(c) VDQ  0.91 V IDQ  4.5 mA
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication
is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
1
or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
E10.4 Following the methods of Example 10.4 in the book, we determine that:
(a) For RL  1200 , RT  600 , and VT  12 V.
(b) For RL  400 , RT  300 , and VT  6 V.
The corresponding load lines are:
2
At the intersections of the load lines with the diode characteristic we
find (a) vL  vD  9.4 V ; (b) vL  vD  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(iL  iD )  vD
The corresponding load lines for the three specified values of iL are
shown:
At the intersections of the load lines with the diode characteristic, we

find (a) vo  vD  10 V; (b) vo  vD  10 V; (c) vo  vD  5 V. Notice that
the regulator is effective only for values of load current up to 50 mA.
E10.6 Assuming that D1 and D2 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 left-hand loop we obtain vD 1  10 V, which is not
consistent with the assumption that D1 is off.
3
E10.7 Assuming that D1 and D2 are both on results in this equivalent circuit:
Writing a KVL equation around the outside loop, we find that the voltage
across the 4-kΩ resistor is 7 V and then we use Ohm’s law to find that
iD1 equals 1.75 mA. The voltage across the 6-kΩ resistance is 3 V so ix is
0.5 mA. Then we have iD 2  ix  iD 1  1.25 mA, which is not consistent
with the assumption that D2 is on.
E10.8 (a) If we assume that D1 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 D2 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 D2 is off we find that
vD 2  3 where as usual we have referenced vD 2 positive at the anode.
This is consistent with the assumption, so D2 is off.
(c) It turns out that the correct assumption is that D3 is off and D4 is
on. The equivalent circuit for this condition is:
4
For this circuit we find that iD 4  5 mA and vD 3  5 V. These results
are consistent with the assumptions.
E10.9 (a) With RL = 10 kΩ, it turns out that the diode is operating on line
segment C of Figure 10.19 in the book. Then the equivalent circuit is:
We can solve this circuit by using the node-voltage technique, treating vo

as the node voltage-variable. Notice that vo  vD . Writing a KCL
equation, we obtain
vo  10 vo  6 vo
  0
2000 12 10000
Solving, we find vD  vo  6.017 V. Furthermore, we find that
iD  1.39 mA. Since we have vD  6 V and iD  0, the diode is in fact
operating on line segment C.
(b) With RL = 1 kΩ, it turns out that the diode is operating on line
segment B of Figure 10.19 in the book, for which the diode equivalent is
an open circuit. Then the equivalent circuit is:
Using the voltage division principle, we determine that vD  3.333 V.

Because we have 6  vD  0, the result is consistent with the assumption
that the diode operates on segment B.
5
E10.10 The piecewise linear model consists of a voltage source and resistance in
series for each segment. Refer to Figure 10.18 in the book and notice
that the x-axis intercept of the line segment is the value of the voltage
source, and the reciprocal of the slope is the resistance. Now look at
Figure 10.22a and notice that the intercept for segment A is zero and
the reciprocal of the slope is (2 V)/(5 mA) = 400 Ω. Thus as shown in
Figure 10.22b, the equivalent circuit for segment A consists of a 400-Ω
resistance.
Similarly for segment B, the x-axis intercept is +1.5 V and the reciprocal
slope is (0.5 mA)/(5 V) = 10 kΩ.
For segment C, the intercept is -5.5 V and the reciprocal slope is 800 Ω.
Notice that the polarity of the voltage source is reversed in the
equivalent circuit because the intercept is negative.
E10.11 Refer to Figure 10.25 in the book.
(a) The peak current occurs when the sine wave source attains its peak
amplitude, then the voltage across the resistor is Vm VB  20  14  6 V
and the peak current is 0.6 A.
(b) Refer to Figure 10.25 in the book. The diode changes state at the
instants for which Vm sin(t ) VB . Thus we need the roots of
20 sin(t )  14. These turn out to be t1  0.7754 radians and
t2    0.7754 radians.
1.591 1.591T
The interval that the diode is on is t2  t1    0.2532T .
 2
Thus the diode is on for 25.32% of the period.
E10.12 As suggested in the Exercise statement, we design for a peak load

voltage of 15.2 V. Then allowing for a forward drop of 0.7 V we require
Vm  15.9 V. Then we use Equation 10.10 to determine the capacitance
required. C  (ILT ) /Vr  (0.1 /60) / 0.4  4167 F.
E10.13 For the circuit of Figure 10.28, we need to allow for two diode drops.
Thus the peak input voltage required is Vm  15 Vr /2  2  0.7  16.6 V.
6
Because this is a full-wave rectifier, the capacitance is given by Equation
10.12. C  (ILT ) /(2Vr )  (0.1 /60) / 0.8  2083 F.
E10.14 Refer to Figure 10.31 in the book.
(a) For this circuit all of the diodes are off if 1.8  vo  10 . With the
diodes off, no current flows and vo  vin . When vin exceeds 10 V, D1 turns
on and D2 is in reverse breakdown. Then vo  9.4  0.6  10 V. When vin
becomes less than -1.8 V diodes D3, D4, and D5 turn on and
vo  3  0.6  1.8 V. The transfer characteristic is shown in Figure
10.31c.
(b) ) For this circuit both diodes are off if 5  vo  5 . With the diodes
off, no current flows and vo  vin .
When vin exceeds 5 V, D6 turns on and D7 is in reverse breakdown. Then

v 5
a current given by i  in (i is referenced clockwise) flows in the
2000
circuit, and the output voltage is vo  5  1000i  0.5vin  2.5 V
When vin is less than -5 V, D7 turns on and D6 is in reverse breakdown.

v 5
Then a current given by i  in (still referenced clockwise) flows in
2000
the circuit, and the output voltage is vo  5  1000i  0.5vin  2.5 V
E10.15 Answers are shown in Figure 10.32c and d. Other correct answers exist.
E10.16 Refer to Figure 10.34a in the book.
(a) If vin (t )  0, we have only a dc source in the circuit. In steady state,

the capacitor acts as an open circuit. Then we see that D2 is forward
conducting and D1 is in reverse breakdown. Allowing 0.6 V for the
forward diode voltage the output voltage is -5 V.
(b) If the output voltage begins to fall below -5 V, the diodes conduct
large amounts of current and change the voltage vC across the capacitor.
Once the capacitor voltage is changed so that the output cannot fall
7
below -5 V, the capacitor voltage remains constant. Thus the output
voltage is vo  vin  vC  2sin(t )  3 V.
(c) If the 15-V source is replaced by a short circuit, the diodes do not
conduct, vC = 0, and vo = vin.
E10.17 One answer is shown in Figure 10.35. Other correct answers exist.
E10.18 One design is shown in Figure 10.36. Other correct answers are possible.
E10.19 Equation 10.22 gives the dynamic resistance of a semiconductor diode as

rd  nVT / IDQ .
IDQ (mA) rd (Ω)
0.1 26,000
1.0 2600
10 26
E10.20 For the Q-point analysis, refer to Figure 10.42 in the book. Allowing for
a forward diode drop of 0.6 V, the diode current is
V  0.6
IDQ  C
RC
The dynamic resistance of the diode is
nVT
rd 
IDQ
the resistance Rp is given by Equation 10.23 which is
1
Rp 
1 / RC  1 / RL  1 / rd
and the voltage gain of the circuit is given by Equation 10.24.
Rp
Av 
R  Rp
Evaluating we have
VC (V) 1.6 10.6
IDQ (mA) 0.5 5.0
rd (Ω) 52 5.2
Rp (Ω) 49.43 5.173
Av 0.3308 0.04919
8
Problems
P10.1 A one-way valve that allows fluid to flow in one direction but not in the
other is an analogy for a diode.
P10.2
P10.3
P10.4 The Shockley equation gives the diode current iD in terms of the applied
voltage vD :
 v  
iD  I s exp  D   1
  nVT  
where I s is the saturation current, and n is the emission coefficient
which takes values between 1 and 2. The voltage VT is the thermal
voltage given by
kT
VT 
q
where T is the temperature of the junction in kelvins, k  1.38  1023
joules/kelvin is Boltzmann's constant, and q  1.60  1019 coulombs is the
magnitude of the electrical charge of an electron.
P10.5 VT  kT / q  (1.38  10 23T ) /(1.60  10 19 )

Temperature C Absolute temperature VT (mV)
20 293 25.3
175 448 38.6
9
P10.6 For vD  0.6 V, we have
iD  0.5  10 3  I s [exp(v D / nVT )  1]
 Is exp vD nVT 
Thus, we determine that:
iD
Is   4.874  10 9 A
exp(v D / nVT )
Then, for vD  0.65 V, we have
iD  I s [exp(v D / nVT )  1]  1.308 mA
Similarly, for vD  0.700 V, we find iD  3.421 mA.
P10.7* The approximate form of the Shockley Equation is iD  Is exp(vD / nVT ) .

Taking the ratio of currents for two different voltages, we have
  exp (vD 1  vD 2 ) / nVT 
Solving for n we obtain:
vD1  vD 2 0.600  0.680
n   1.336
VT ln(iD 1 / iD 2 ) 0.026 ln(1 / 10)
Then, we have
iD 1
Is   3.150  10 11 A
exp(v D 1 / nVT )
P10.8*
or transmission in any form or by any means, electronic, mechanical, photocopying, 10
recording, or likewise. For information regarding permission(s), write to:
P10.9
P10.10 At T  175 K , we have:

v D  0.65  0.002(150  25)
 0.40 V
P10.11 VT  kT / q  (1.38  10 23  373) /(1.60  10 19 )  32.17 mV

iD 1
Is   421 .9  0 9 A
[exp(v D 1 / nVT )  1]
iD 2  I s [exp(v D 2 / nVT )  1]  4.733 mA
P10.12 Using the approximate form of the Shockley Equation, we have

103  Is exp  0.600 nVT  (1)
102  Is exp  0.700 nVT  (2)
Dividing the respective sides of Equation (2) by those of Equation (1), we
have
I exp  0.700 nVT 
10  s  exp  0.100 nVT 
I s exp  0.600 nVT 
ln 10   0.100 nVT
n  0.100 VT ln 10   1.670
P10.13 For part (a), Equation 10.3 gives the diode voltage in terms of the current
as
vD  nVT ln iD / Is   1
For part (b) with a 100-Ω resistance in series, the terminal voltage is
v  v D  100iD
A MATLAB program to obtain the desired plots is:
log10_of_id = -5:0.01:-2;
id = 10.^log10_of_id;
vd = 2*0.026*log((id/20e-9) + 1);
v = vd + 100*id;
semilogy(v,id)
hold
semilogy(vd,id)
(Note in MATLAB log is the natural logarithm.) The resulting plots are
shown:
The plot for part (a) is a straight line. The series resistance is relatively
insignificant for currents less than 0.1 mA and is certainly significant for
currents greater than 1 mA.
P10.14*
With the switch open, we have:

iD 1  103  Is exp v nVT   1
 Is exp v nVT 
Thus, we determine that:
103 103
Is    9.5  1014 A
exp v VT  exp  0.6 0.026
With the switch closed, by symmetry, we have:
iD 1  iD 2  0.5 mA
0.5  103  Is exp v VT 
0.5  103
v  nVT ln  582 mV
Is
Repeating the calculations with n  2 , we obtain:
Is  9.75  109 A
v  564 mV
P10.15* (a) By symmetry, the current divides equally and we have

IA  IB  100 mA
(b) We have
iD  Is exp v nVT   1
 Is exp v nVT 
Solving for I s , we obtain
iD
Is 
exp v nVT 
For diode A, the temperature is TA  300 K , and we have
kTA 1.38  1023  300
VTA    25.88 mV
q 1.6  1019
0.100
I sA   1.792  1013 A
exp  0.700 0.02588
For diode B, we have T  305 K , and

IsB  2IsA  3.583  1013 A
VTB  26.31 mV
Applying Kirchhoff's current law, we have
0.2  IA  IB
0.2  1.792  1013  exp v 0.02588  3.583  1013  exp v 0.02631 
Solving for v by trial and err, we obtain v  697.1 mV , IA  87 mA , and

IB  113 mA .
P10.16* The load-line equation is

Vs  Rs ix  v x
Substituting values, this becomes
3  ix  v x
Next, we plot the nonlinear device characteristic equation
ix  exp(v x )  1 /10
and the load line on the same set of axes.
The MATLAB commands are:
clear
%plot load line
vx=0:0.01:3;
ix= 3-vx;
plot(vx,ix)
hold
%plot nonlinear device characteristic
ix=(exp(vx)-1)/10;
plot(vx,ix)
grid minor
ylabel('ix (A)')
xlabel('vx (V)')
Finally, the solution is at the intersection of the load line and the
characteristic as shown:
P10.17 The load-line equation is

Substituting values this becomes
4  ix  v x
Next we plot the nonlinear device characteristic equation
i x  v x  2v x2
and the load line on the same set of axes. The MATLAB commands are:
clear
%plot load line
vx=0:0.01:4;
ix= 4-vx;
plot(vx,ix)
hold
ix=vx+2*(vx.^2);
plot(vx,ix)
grid minor
axis([0 4 0 5])
ylabel('ix (A)')
xlabel('vx (V)')
P10.18 The load-line equation is

Substituting values, this becomes
15  3i x  v x
in which ix is in milliamperes and vx is in volts. Next, we plot the nonlinear
device characteristic equation and the load line on the same set of axes
using the commands:
clear
%plot load line
vx=0:0.01:15;
ix= 5-vx/3;
plot(vx,ix)
hold
ix=0.01./(1-vx/4).^3;
plot(vx,ix)
axis([0 5 0 5])
grid minor
ylabel('ix (mA)')
xlabel('vx (V)')
P10.19 The load line equation is

Substituting values’ this becomes

8  2i x  v x
Next we plot the nonlinear device characteristic equation i x  v x3 / 4 and
the load line on the same set of axes.
The MATLAB commands are:
clear
%plot load line
vx=0:0.01:3;
ix= 4-vx/2;
plot(vx,ix)
hold
ix=(vx.^3)./4;
plot(vx,ix)
grid minor
ylabel('ix (A)')
xlabel('vx (V)')
P10.20 (a) In a series circuit, the total voltage is the sum of the voltages across
the individual devices. Thus, we add the characteristics horizontally. The
overall volt-ampere characteristic is
(b) In a parallel circuit, the overall current is the sum of the currents
through the individual devices. Thus, we add the characteristics
vertically. The overall volt-ampere characteristic is
P10.21 The load-line construction is:
At the intersection of the characteristic and the load line, we have

v  1.8 V and i  2.64 mA.
P10.22 A Zener diode is a diode intended for operation in the reverse breakdown
region. It is typically used to provide a source of constant voltage. The
volt-ampere characteristic of an ideal 5.8-V Zener diode is:
P10.23* The circuit diagram of a simple voltage regulator is:
P10.24 Refer to Figure 10.14 in the book. As the load resistance becomes
smaller, the reverse current through the Zener diode becomes smaller in
magnitude. The smallest load resistance for which the load voltage
remains at 10 V corresponds to zero diode current. Then, the load
current is equal to the current through the 100- resistor given by
iL  (15  10) / 100  50 mA. Thus, the smallest load resistance is
RL  vo / iL  200 .
P10.25 We need to choose Rs so the minimum reverse current through the Zener
diode is zero. Minimum current through the Zener occurs with minimum
Vs and maximum iL. Also, we can write:
V v L
iZ  s  iL
Rs
Substituting values, we have
85
iZ  0   0.15
Rs
Solving for the resistance we find Rs  20 .
Maximum power dissipation in the resistance occurs for maximum Vs.
(V  v ) 2 (12  5) 2
Pmax  s max L   2.45 W
Rs Rs
P10.26 The diagram of a suitable regulator circuit is
or transmission in any form or by any means, electronic, mechanical, photocopying,20
We must be careful to choose the value of R small enough so I z remains
positive for all values of source voltage and load current. (Keep in mind
that the Zener diode cannot supply power.) From the circuit, we can
write
V V
Iz  s L  IL
R
Minimum I z occurs for IL  100 mA and Vs  8 V . Solving for the
maximum value allowed for R, we have
V V 85
Rmax  s L   30 
Iz  I L 0  0.1
Thus, we must choose the value of R to be less than 30  . We need to
allow some margin for component tolerances and some design margin.
However, we do not want to choose R too small because the current and
power dissipation in the diode becomes larger as R becomes smaller.
Thus, a value of about 24  would be suitable. (This is a standard value.)
With this value of R, we have
V 5
IR max  s max  208 mA
R
IZ max  IR max  208 mA
PR max  IR max  R  1.04 W
2
PZ max  5IZ max  1.04 W
P10.27 Refer to the solution to Problem P10.26. In the present case, we have
Rmax  10  , and we could choose R  8.2  because this is a standard
value and we need to provide some design margin. With this value, we
have
IR max  IZ max  610 mA
PR max  PZ max  3.05 W
P10.28 Refer to the solution to Problem P10.26. In the present case, we have
Rmax  3.0  , and we would choose R  2.4  because this is a standard
value and we need to provide some design margin. With this value, we
have
IR max  IZ max  2.08 A
PR max  PZ max  10.4 W

P10.29 First, find the Thévenin equivalent for the circuit as seen looking back
from the terminals of the nonlinear device:
Then, write the KVL equation for the equivalent circuit:
Vt  Rt iD  vD
Next, plot this equation on the device characteristics and find the value
of vD and iD at the intersection of the load line and the device
characteristic. After the device current and voltage have been found,
the original circuit can be solved by standard methods.
P10.30* The Thévenin resistance is Rt Voc / Isc  10 /2  5 . Also the Thévenin

voltage is Vt Voc  10 V. The load line equation is 10  5iab  vab . We plot
the load line and nonlinear device characteristic and find the solution at
the intersection as shown:
P10.31 If we remove the diode, the Thévenin equivalent for the remaining circuit
consists of a 5-V source in series with a 1-k resistance. The load line is
At the intersection of the characteristic and the load line, we have the
device current i1  3.0 mA. Then, applying KCL to the original circuit, we
have i2  10  i1  7.0 mA.
consists of a 10-V source in series with a 5-k resistance. The load line
is

v  1.65 V and i  1.67 mA.
consists of a 4-V source in series with an 800- resistance. The load line
is

v  1.80 V and i  2.75 mA.
P10.34 An ideal diode acts as a short circuit as long as current flows in the
forward direction. It acts as an open circuit provided that there is
reverse voltage across it. The volt-ampere characteristic is shown in
Figure 10.15 in the text. After solving a circuit with ideal diodes, we
must check to see that forward current flows in diodes assumed to be on,
and we must check to see that reverse voltage appears across all diodes
assumed to be off.
P10.35 The equivalent circuit for two ideal diodes in series pointing in opposite
directions is an open circuit because current cannot flow in the reverse
direction for either diode.
The equivalent circuit for two ideal diodes in parallel pointing in opposite
directions is a short circuit because one of the diodes is forward
conducting for either direction of current flow.
P10.36* (a) D1 is on and D2 is off. V  10 volts and I  0.

(b) D1 is on and D2 is off. V  6 volts and I  6 mA.
(c) Both D1 and D2 are on. V  30 volts and I  33.6 mA.
10
P10.37 (a) The diode is on, V  0 and I   2 mA.
5000
(b) The diode is off, I  0 and V  5 V.
6
(c) The diode is on, V  0 and I   2 mA.
3000
(d) The diode is on, I  3 mA and V  6 V.
P10.38 (a) D1 is on, D2 is on, and D3 is off. V  7.5 volts and I  0.
(b) Vin D1 D2 D3 D4 V I
0 on on on on 0 0
2 on on on on 2V 2 mA
6 off on on off 5V 5 mA
10 off on on off 5V 5 mA
The plot of V versus Vin is:
P10.39 (a) The output is high if either or both of the inputs are high. If both
inputs are low the output is low. This is an OR gate.
(b) The output is high only if both inputs are high. This is an AND gate.
P10.40
P10.41 When the sinusoidal source is positive, D2 is on and D1 is off. Then, we

have vo (t )  2.5 sin(2t ).
When the source is negative, D1 is on and D2 is off. Then, we have
vo (t )  2.5 sin(2t ).
P10.42 If a nonlinear two-terminal device is modeled by the piecewise-linear

approach, the equivalent circuit of the device for each linear segment
consists of a voltage source in series with a resistance.
P10.43
v  Rai Va
P10.44 We know that the line passes through the points (2 V, 5 mA) and (3 V, 15
mA). Thus, the slope of the line is -1/R = (-10 mA)/(1 V), and we have R =
100 . Furthermore, the intercept on the voltage axis is at v = 1.5 V.
Thus, the equivalent circuit is
10.45* The equivalent circuits for each segment are shown below:
For the circuit of Figure P10.45a, we can determine by trial and error (or
by a load-line analysis) that the device operates on the middle line
segment. Thus, the equivalent circuit is:
15  0.756
i   4.68 mA
3000  44.4
v  0.756  44.4i  0.964 V
For the circuit of Figure P10.45b, we can determine by trail and error
that the device operates on the upper right-hand line segment. Thus, the
equivalent circuit is:
2  6.8
i   10.4 mA
50  800
v  800i  6.8  1.48 V
P10.46 In the forward bias region (iD > 0), the equivalent circuit is a short
circuit. Thus in the equivalent circuit, the voltage source is zero and the
resistance is zero.
In the reverse bias region (0 > vD > -10 V) the equivalent circuit is an open
circuit. Thus in the equivalent circuit, the voltage is indeterminate and
the resistance is infinite.
In the reverse breakdown region (0 > iD), the equivalent circuit consists
of a 10-V voltage source in series with zero resistance.
P10.47* For small values of iL, the Zener diode is operating on line segment C of
Figure 10.19, and the equivalent circuit is
Writing a KCL equation at node A, we obtain:

v L  13 v L  6
  iL  0
100 12
Solving we obtain
vL  6.75  10.71iL
This equation is valid for vL  6 V. When 0  vL  6 V, the Zener diode
operates on line segment B, for which the Zener is modeled as an open
circuit and we have
vL  13  100iL
Plotting these equations results in
P10.48 (a) Assuming that the diode is an open circuit, we can compute the node
voltages using the voltage-division principle.
200 300
v 1  16  8V v 2  16  12 V
200  200 300  100
Then, the voltage across the diode is v D  v 1  v 2  4 V. Because vD is
less than Vf  0.7 V, the diode is in fact operating as an open circuit.
(b) Assuming that the diode operates as a voltage source, we can use KVL
to write:
v1  v 2  0.7
Placing a closed surface around the diode to form a super node and
writing a KCL equation gives
v 1  16 v 1 v  16 v 2
  2  0
200 200 100 300
Solving these equations, we find v 1  10.686 V and v 2  9.986 V. Then,
writing a KCL equation at node 1 gives the diode current.
16  v 1 v
iD   1  26.86 mA
200 200
Because the diode current is negative, the diode operation is not
consistent with the model.
P10.49 Half-wave rectifier with a capacitance to smooth the output voltage:
Full-wave circuits:
P10.50 The peak value of the ac source is Vm  15 2  21.21 V. Thus the PIV is
21.21 V and the peak current is 21.21/50 = 424.3 mA.
P10.51 The dc output voltage is equal to the peak value of the ac source, which is
v L  15 2  21.21 V. The load current is i L  v L / RL  424 .3 mA. The
charge that passes through the load must also pass through the diode.
The charge is Q  i LT  0.4243 / 60  7.071 mC. The peak inverse voltage
is 42.42 V. The charge passes through the diode in a very short interval,
thus the peak diode current is much larger than the load current.
P10.52 The diode is on for VB Vm sin(t ) . Substituting values and solving, we
find that during the first cycle after t = 0 the diode is on for
arcsin(12 / 24)   / 6  t    arcsin(12 / 24)  5 / 6
The current is given by
24 sin(t )  12
i (t )   48 sin(t )  24 A
0.5
The charge passing through the circuit during the first cycle is
5 / 6 5 / 6
 48  48 3  16
Q1   [ 48 sin(t )  24]dt   cos(t )  24t  
 / 6     / 6 
The average current is the charge passing through the circuit in 1 second
(or 60 cycles). Also, we have   120 Thus
48 3  16 48 3  16 24 3
I avg  60  60   8  5.232 A
 120 
Then, the time required to fully charge the battery is
100
T   19.1 hours
5.232
P10.53 (a) The integral of Vm sin(t ) over one cycle is zero, so the dc voltmeter
reads zero.
t T / 2
 Vm Vm
T /2 T
1   1 
(b) Vavg  V
  m sin(t )dt   0dt    cos(t ) 
T  0 T /2  T  t 0 
1 
T /2 T
 2V
(c) Vavg   Vm sin(t )dt   Vm sin(t )dt   m
T  0 T /2  
P10.54* For a half-wave rectifier, the capacitance required is given by Equation

10.10 in the text.
I T 0.25 1 60 
C  L   20833 F
Vr 0.2
For a full-wave rectifier, the capacitance is given by Equation 10.12 in the
text:
I T 0.25 1 60 
C  L   10416 F
2Vr 2  0.2
P10.55* The output voltage waveform is:
The peak voltage is approximately 10 V. Assuming an ideal diode, the ac

source must have a peak voltage of 10 V. The circuit is:
The capacitance required is given by Equation 10.10 in the text.

I T 0.1 1 60 
C  L   833 F
Vr 2
P10.56 As in Problem P10.55, the peak voltage must be 10 V. For a full-wave

rectifier, the capacitance is given by Equation 10.12 in the text:
I T 0.1 1 60 
C  L   417 F
2Vr 2 2 
The circuit diagram is:
P10.57 As in Problem P10.55, the peak voltage must be 10 V. For a full-wave

rectifier, the capacitance is given by Equation 10.12 in the text:
I LT 0.1 1 60 
C    417 F
2Vr 2 2 
The circuit diagram is:
P10.58 If we allow for a forward diode drop of 0.8 V, the peak ac voltage must
be 10.8 V. Otherwise, the circuit is the same as in the solution to
Problem P10.55.
P10.59 (a) The current pulse starts and ends at the times for which
vs (t ) VB
20 sin(200t )  12
Solving we find that
sin1 (0.6) T
tstart   1.024 ms and tend   tstart  3.976 ms
200 2
Between these two times the current is
20 sin(200t )  12
i (t ) 
80
A sketch of the current to scale versus time is
(b) The charge flowing through the battery in one period is
tend tend
20 sin(200t )  12
Q  i (t )dt   dt
tstart t
start
80
t
 1 12t  end
  cos(200t ) 
 800 80 tstart
Q  194 C
Finally, the average current is the charge divided by the period.
Q 194  106
Iavg    19.4 mA
T 10  103
P10.60 (a) With ideal diodes and a large smoothing capacitance, the load voltage
equals the peak source voltage which is Vm = 12 V. Then the PIV is 2Vm =
24 V.
(b) Here again with ideal diodes and a large smoothing capacitance, the
load voltage equals the peak source voltage which is Vm = 12 V. However,
the PIV is only Vm = 12 V.
P10.61 (a) The circuit operates as three full-wave rectifiers with a common load
and shared diodes. Thus, the load voltage at any instant is equal to the
largest source-voltage magnitude. The plot of the load voltage is
T is the period of the sinusoidal sources.
(b) The minimum voltage occurs at t T / 12 and is given by
Vmin Vm cos T / 12 Vm cos( / 6)  0.866Vm . Thus the peak-to-peak ripple
is 0.134Vm.
The average load voltage is given by
1 T
Vavg   v L t dt
T 0
However, since v L (t ) has 12 intervals with the same area, we can write:
1 T / 12 12 T /12
Vavg   v L t dt   Vm cos(t )dt
T / 12 0 T 0
6 sin( / 6)
Vavg Vm  0.955Vm

(c) To produce an average charging current of 30 A, we require
Vavg  12  0.1  30  15 V
(d) In practice, we would need to allow for forward drops of the diodes,
drops across the slip rings, and resistances of the stator windings and
wiring.
P10.62 A clipper circuit removes or clips part of the input waveform. An

example circuit with waveforms is:
We have assumed a forward drop of 0.6V for the diode.
P10.63
P10.64 Refer to Figure P10.64 in the book. When the source voltage is positive,
diode D3 is on and the output vo(t) is zero. For source voltages between 0
and 5 V, none of the diodes conducts and vo(t) = vs(t). Finally, when the
source voltage falls below -5 V, D1 is on and D2 is in the breakdown region
so the output voltage is 5 V. The waveforms are:
P10.65
P10.66
P10.67
P10.68 A clamp circuit adds or subtracts a dc component to the input waveform

such that either the positive peak or the negative peak is forced to
assume a predetermined value. An example circuit that clamps the
positive peak to +5 V is shown below:
We have allowed a forward drop of 0.6 V for the diode.

P10.69* Refer to the circuit shown in Figure P10.69 in the book. If the output
voltage attempts to become less than -5 V, the Zener diode breaks down
and current flows, charging the capacitance. Thus, the negative peak is
clamped to -5 V. The input and output waveforms are:
P10.70* A suitable circuit is:
P10.71 A suitable circuit is:
P10.72 This is a clamp circuit that clamps the positive peaks to zero.
P10.73
The capacitor C1 and diode D1 act as a clamp circuit that clamps the
negative peak of vA t  to zero. Thus, the waveform at point A is:
Diode D2 and capacitor C2 act as a half-wave peak rectifier. Thus, the
voltage across RL is the peak value of vA t  . Thus, vL t   2Vm . This is
called a voltage-doubler circuit because the load voltage is twice the peak
value of the ac input. The peak inverse voltage is 2Vm for both diodes.
P10.74* A suitable circuit is:
We must choose the time constant RC >> T, where T is the period of the
input waveform.
P10.75 A suitable circuit is:
We must choose R1 to ensure that the 5.6-V Zener is in the breakdown

region at all times and choose the time constant R2C T , where T is the
period of the input waveform.
P10.76 (a) A suitable circuit is:
We choose the resistors R1 and R2 to achieve the desired slope.
1 R2
Slope  
3 R1  R2
Thus, choose R1  2R2 . For example, R1  2 k and R2  1 k.
(b) A suitable circuit is:
Other resistor values will work, but we must make sure that D2
remains forward biased for all values of vin , including vin  10 V .
To achieve the desired slope (i.e., the slope is 0.5) for the transfer
characteristic, we must have R1  R2 .
P10.77 IDQ represents the dc component of the diode current with no signal
applied to the circuit, and id(t) represents the changes from the Q-point
current when the signal is applied. Furthermore, iD(t) is the total diode
current. Thus, we have
iD (t )  IDQ  id (t )  4  0.5cos(200t ) mA
P10.78 The small signal equivalent circuit of a diode is a resistance known as the
dynamic resistance. The dynamic resistance is the reciprocal of the slope
of the iD versus vD characteristic at the operating point.
P10.79 Dc sources voltage sources are replaced by short circuits in a small-signal

equivalent circuit. By definition the voltage across a dc voltage source is
constant. Thus, even if there is ac current flowing through the dc source
the ac voltage across it is zero as is the case for a short circuit.
P10.80 We should replace dc current sources by open circuits in a small-signal

equivalent circuit. The current through a dc current source is constant.
Thus, the ac current must be zero even if we apply an ac voltage. Zero
current for a non-zero applied voltage implies that we have an open
circuit.
P10.81* A plot of the device characteristic is:
Clearly this device is not a diode because it conducts current in both
directions. The dynamic resistance is given by:
1
 di  8
rD   D   2
 dvD  3vD
A plot of the dynamic resistance versus vD is:
P10.82 We are given
 10 6
iD  for  7 V  v D  0
1  vD 7 2
A plot of this is:

The dynamic resistance is:
rD  (dv D / di D ) 1  3.5  10 6 (1  v D / 7) 3
To find the dynamic resistance at a given Q-point, we evaluate this

expression for vD VDQ .
For I DQ  0.5 mA, we have VDQ  6.68695 V and rD  22.7 .
For I DQ  10 mA, we have VDQ  6.930 V and rD  3.5 .
P10.83 We are given vD(t) = 4 + 0 .02cos(t) V and iD(t) = 7 + 0.2cos(t) mA.

The dynamic resistance is the ratio of the ac voltage amplitude to the ac
current amplitude.
vd 0.02
rD    100 
id 0.2  10 3
The Q-point results if we set the ac signals to zero. Thus, we have
VDQ  4 V and I DQ  7 mA
P10.84 Dynamic resistance is given by
1
 di  dv
rD   D   D
 dvD  diD
Because voltage is constant for changes in current, the dynamic
resistance is zero for an ideal Zener diode in the breakdown region.
P10.85* To find the Q-point, we ignore the ac ripple voltage and the circuit
becomes:
Thus, we have:
85
I sQ   150 mA
20
ILQ  5 100  50 mA
IDQ  IsQ  ILQ  100 mA
The small-signal or ac equivalent circuit is:
where rD is the dynamic resistance of the Zener diode. Using the

voltage-division principle, the ripple voltage across the load is
Rp
v Lac Vripple 
R  Rp
1
where Rp  is the parallel combination of the load resistance
1 RL  1 rD
and the dynamic resistance of the diode. Substituting values, we find
Rp
VLac  10  103  1 
20  Rp
Solving, we find Rp  0.202  . Then, we have:
1
Rp  0.202  which yields rD  0.202  .
1 100  1 rD
Practice Test
T10.1 (a) First, we redraw the circuit, grouping the linear elements to the left
of the diode.
Then, we determine the Thévenin equivalent for the circuit looking back
from the diode terminals.
Next, we write the KVL equation for the network, which yields
VT  RT iD  v D . Substituting the values for the Thévenin voltage and
resistance, we have the load-line equation, 3  200iD  v D . For iD  0 , we
have vD  3 V which are the coordinates for Point A on the load line, as
shown below. For v D  0 , the load-line equation gives iD  15 mA which
are the coordinates for Point B on the load line. Using these two points
to plot the load line on Figure 10.8, we have
The intersection of the load line and the diode characteristic gives the
current at the operating point as iD  9.6 mA.
(b) First, we write the KCL equation at the top node of the network,
which yields iD  v D / 25  40 mA. For iD  0 , we have v D  1 V which are
the coordinates for Point C on the load line shown above. For v D  0 , the
load-line equation gives iD  40 mA which plots off the vertical scale.
Therefore, we substitute iD  20 mA , and the KCL equation then yields
v D  0.5 V . These values are shown as point D. Using Points C and D we
plot the load line on Figure 10.8 as shown above. The intersection of the
load line and the diode characteristic gives the current at the operating
point as iD  4.2 mA.
T10.2 If we assume that the diode is off (i.e., an open circuit), the circuit
becomes
Writing a KCL equation with resistances in k, currents in mA, and
v  12 v x  (16)
voltages in V, we have x   0 . Solving, we find that
1 2
v x  2.667 V. However, the voltage across the diode is v D  v x , which
must be negative for the diode to be off. Therefore, the diode must be
on.
With the diode assumed to be on (i.e. a short circuit) the circuit

becomes
Writing a KCL equation with resistances in k, currents in mA and

v  12 v x  (16) v x
voltages in V, we have x    0 . Solving, we find that
1 2 4
v x  2.286 V. Then, the current through the diode is
v
iD  i x  x  0.571 mA. Of course, a positive value for iD is consistent
4
with the assumption that the diode is on.
T10.3 We know that the line passes through the points (5 V, 2 mA) and (10 V, 7
mA). The slope of the line is -1/R = -i/v  (-5 mA)/(5 V), and we have
R  1 k. Furthermore, the intercept on the voltage axis is at v  3 V.
Thus, the equivalent circuit for the device is
T10.4 The circuit diagram is:
Your diagram may be correct even if it is laid out differently. Check to

see that you have four diodes and that current flows from the source
through a diode in the forward direction then through the load and
finally through a second diode in the forward direction back to the
opposite end of the source. On the opposite half cycle, the path should
be through the other two diodes and through the load in the same
direction as before. Notice in the diagram that current flows downward
through the load on both half cycles.
T10.5 An acceptable circuit diagram is:
Your diagram may be somewhat different in appearance. For example, the

4-V source and diode B can be interchanged as long as the source polarity
and direction of the diode don't change; similarly for the 5-V source and
diode A. The parallel branches can be interchanged in position. The
problem does not give enough information to properly select the value of
the resistance, however, any value from about 1 k to 1 M is acceptable.
T10.6 An acceptable circuit diagram is:
The time constant RC should be much longer than the period of the
source voltage. Thus, we should select component values so that
RC  0.1 s.
T10.7 We have
kT 1.38  10 23  300
VT    25.88 mV
q 1.60  10 19
nVT 2  25.88  10 3
rd    10.35 
I DQ 5  10 3
The small-signal equivalent circuit for the diode is a 10.35  resistance.
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tormented by them that they can hardly eat at all.”
Thomas Wilson, who was Bishop of the Isle of Man about a

century since, was a particularly benevolent man. To supply the poor
with clothing, he kept in constant employment at his own house
several tailors and shoemakers. On one occasion, in giving orders to
one of his tailors to make him a cloak, he directed that it should be
very plain, having simply a button and loop to keep it together. “But,
my lord,” said the tailor, “what would become of the poor button-
makers, if every one thought in that way? they would be starved
outright.” “Do you say so, John?” replied the bishop; “why then
button it all over, John.”
Temperance.—Temperance puts wood on the fire, flour in the

barrel, meat in the tub, vigor in the body, intelligence in the brain,
and spirit in the whole composition of man.
The following anecdote was told by Lord Mansfield, a celebrated

English judge. He had turned away his coachman for certain small
thefts, and the man begged his lordship to give him a character that
he might obtain another place.
“What kind of a character can I give you?” said his lordship.
“Oh, my lord, any character your lordship pleases to give me, I
shall most thankfully receive.”
His lordship accordingly sat down and wrote as follows:
“The bearer, John ——, has served me three years in the capacity
of coachman. He is an able driver and a sober man. I discharged
him because he cheated me. Mansfield.”
John thanked his lordship and went off. A few mornings
afterwards, when his lordship was stepping into his coach, a man in
a handsome livery made him a low bow. To his surprise, he
recognised his late coachman.
“Why, John,” said his lordship, “you seem to have got an excellent
place; how could you manage this with the character I gave you?”
“Oh! my lord,” said John, “it was an exceedingly good character;
my new master, on reading it, said he observed your lordship
recommended me for a good driver and a sober man.” “These,” said
he, “are just the qualities I want in a coachman. I observe his
lordship adds that he discharged you for cheating him. Hark you,
sirrah, I’m a Yorkshireman; I defy you to cheat me.”
When Capt. Clapperton, the African traveller, breakfasted with the

Sultan Bautsa, he was treated with a large broiled water rat, and
alligators’ eggs both fried and stewed.
Good Measure.—“I don’t know how it is,” said a person who was
fond of writing poetry for the public journals, but whose productions
had always met with a rejection—“I have written a great deal, but my
pieces have never been published.”
“Perhaps,” replied his friend, “there were faults in your effusions
that you were not aware of, but which were easily detected by the
hawk-eyed editors. The measure might not have been correct.”
“There it is now,” rejoined the disappointed poet; “I can always
write the first line well enough; but I am often perplexed about the
second. Now, this is poetry, but it don’t seem to jingle to my
satisfaction.
‘Tread lightly, stranger, o’er this hallowed dust,
For if you don’t mend your ways—lay like me you must.’”
“Pshaw!” exclaimed the critic, “that’s bad measure.”

“Bad measure! why, man, you’re mistaken, it’s very good measure
—it’s more than enough!”
“Boy,” said a gentleman to a lad in the West, “boy, is there any

game where you live?” “Yes,” said the lad, “there’s a power of
turkies, a heap of squirrels, and a right smart sprinkle of deer.”
A retort.—An old miser, owning a farm, found it impossible one

day to do his work without assistance and accordingly offered any
man food for performing the requisite labor. A half-starved pauper
hearing of the terms, accepted them. Before going into the fields in
the morning, the farmer invited his help to breakfast; after finishing
the meal, the old skin-flint thought it would be saving time if they
should place the dinner upon the breakfast-table. This was readily
agreed to by the unsatisfied stranger, and dinner was soon
despatched. ‘Suppose now,’ said the frugal farmer, ‘we take supper;
it will save time and trouble, you know.’ ‘Just as you like,’ said the
eager eater, and at it they went. ‘Now we will go to work,’ said the
satisfied and delighted employer. ‘Thank you,’ replied the delighted
laborer, ‘I never work after supper!’
An Illustration.—There was once a converted Indian, who,

being asked if he believed in the Trinity, said he did. He was then
asked his reason. He said he would answer in his Indian way. ‘We go
down to the river in winter, and we see it covered with snow; we dig
through the snow and we come to ice; we chop through the ice and
we come to water;—snow is water, ice is water, and water is water,’
said he; ‘therefore the three are one.’
The Scottish Thistle.—The origin of this national badge is thus

handed down by tradition:—When the Danes invaded Scotland, it
was deemed unwar-like to attack an enemy in the pitch darkness of
night, instead of a pitched battle by day; but on one occasion the
invaders resolved to avail themselves of this stratagem; and, in order
to prevent their tramp from being heard, they marched bare-footed.
They had thus neared the Scottish force unobserved, when a Dane
unluckily stepped with his naked foot upon a superbly prickled thistle,
and instinctively uttered a cry of pain, which discovered the
assailants to the Scots, who ran to their arms, and defeated the foe
with a terrible slaughter. The thistle was immediately adopted as the
insignia of Scotland.
Osceola.—It is stated that the name of Osceola was given to that
famous chief by an old lady in a frontier village, who had newly
arrived in the country, and had never seen an Indian. On his
approach, she broke forth in utter astonishment—“Oh see! oh la!
what a funny looking man!”
To Correspondents.
College Hill Poughkeepsie, July 30th, 1842.
Mr. Merry:—
I have made out the following answers to some of your
puzzles in the August No. of the Museum, which it will be
gratifying to me to know are correct.
Yours respectfully.
William ——.
To the first, of 6 letters,—Stable.

To the second, of 16 letters,—Washington Irving.
To the third, of 13 letters,—North Carolina.
To the fourth, of 15 letters,—Marie Antoinette.
☞ Our friend William is a good Yankee, and has, therefore,

guessed right.
Mr. Merry:—
Sir,—If you think the following puzzle worthy a place in
your excellent magazine, by inserting it you will confer a great
favor on
A Subscriber.
I am a word of 16 letters.
My 4, 5, 13, 15, 8, 13, is a city in Spain.

My 13, 14, 12, is a river in Russia.
My 1, 2, 16, 3, 5, 1, is a part of the body.
My 4, 11, 1, 2, is a very troublesome insect.
My 7, 8, 9, 9, 8, 5, 4, is a boy’s name.
My 10, 8, 13, is what all are guilty of.
My 6, 12, 3, 7, is something very common in cold
weather.
My whole is a person who has created some excitement
of late.
MERRY’S MUSEUM.
V O L U M E I V . — N o . 5 .
SKETCHES OF BIBLE SCENES.
Ruins of Jericho as they now appear.
Jericho was situated twenty miles northeast of Jerusalem. It was

taken by Joshua, who received orders from God to besiege it soon
after his passage over Jordan. There was a most remarkable
fulfilment of Joshua’s denunciation against any who should rebuild it:
“Cursed be the man before the Lord that riseth up and buildeth this
city Jericho; he shall lay the foundations thereof in his first-born, and
in his youngest son shall he set up the gate of it.” This warning
prevented the Jews from building on the spot where the ancient city
had stood; but about five hundred years after, Hiel of Bethel
undertook to rebuild it, and lost his eldest son in laying the
foundations, and his youngest when he hung up the gates.
The modern village, called Eicha, is situated in the midst of a
plain, and is very miserable and filthy, being composed of hovels
made of four stone walls, covered with cornstalks and gravel. The
few gardens around seem to contain nothing but tobacco and
cucumbers. About two miles from the village may be seen
foundations of hewn stones and portions of walls, which render it
probable that it was the site of the ancient city.
The Scriptures speak of Jericho as the city of palm trees, and
Josephus everywhere describes them as being very abundant and
large. The region also produced honey, the cypress tree, and the
common fruits of the earth in great abundance. The sycamore tree
likewise flourished there.
Of all these productions, which so distinguished the plains of
Jericho, few now remain. The groves of palms have all disappeared,
and only one solitary palm tree lingers in all the plain. The sycamore
too is nowhere seen, and honey, if found at all, is very rare.
In the time of the crusades the sugarcane was cultivated at
Jericho, but is now unknown there.
Askelon.
This is a city in the land of the Philistines, on the coast of the

Mediterranean Sea. It was once a place of great importance and
note among the Philistines, and was one of their seats of
government. It is also famous for a temple dedicated to Apollo, at
which Herod, the grand-father of Herod the Great, officiated as
priest. After the death of Joshua, the tribe of Judah took the city of
Askelon.
The wine that is made in this city was very much esteemed, and
the cypress tree was also common. This was very much admired by
the ancients for its grace and beauty. The modern town is called
Scalona, and is a small and uninteresting place.
Bethlehem.
This town is situated about five miles and a half southeast of

Jerusalem. There is no doubt that the village called by the Arabs Beit
Lahen, which means “House of Flesh,” is the same as the ancient
Bethlehem, which the Jews called “House of Bread.”
The present inhabitants of Bethlehem are all Christians, and they
amount to three thousand souls. The town has gates at some of the
principal streets; the houses are solidly built, but are not large. There
are many olive gardens, fig orchards and vineyards round about, and
the adjacent fields, though stony and rough, produce, nevertheless,
good crops of grain. Here was the scene of the beautiful narrative of
Ruth gleaning in the field of Boaz, after his reapers. The inhabitants,
besides their agriculture, employ themselves in carving beads,
crucifixes, and models of the holy sepulchre and other similar articles
in olive wood and mother-of-pearls. Indeed, the neatest and most
skilfully wrought specimens of these articles come from Bethlehem.
About thirty rods from the village stands a large convent, occupied
by Greeks, Latins, and Armenians. It encloses the church built by the
empress Helena, over the spot where, according to tradition, our
Savior was born. Vast numbers of pilgrims come to view the place,
especially at Easter, when such multitudes assemble, that the church
is often crowded to suffocation, and contests frequently ensue
between the different sects. On one occasion, the privilege of saying
mass at the altar on Easter day was fought for at the door of the
sanctuary itself, with drawn swords.
The pretended place of the nativity is a grotto or cave beneath the
church, very splendidly ornamented with a marble pavement,
recesses decorated with sculpture and painting, and massy silver
lamps of exquisite workmanship. Just beneath the marble altar, upon
the pavement, is a star, formed of inlaid stones, which marks the
spot of the Savior’s birth, and is said to be placed immediately
underneath the point where the star of the East became fixed, to
direct the wise men in the object of their search.
This cave is not the only celebrated spot within the precincts of
the church. One grotto is pointed out as the tomb of the Innocents;
another possesses some interest as having been the abode of St.
Jerome for many years. Another is shown as the spot where Joseph
sat during the birth of Christ, and another is said to be the place
where the Virgin Mary hid herself and her son from the fury of Herod.
The most remarkable spot in the neighborhood of Bethlehem is
the reservoir called Solomon’s pools. There are three of them, of
great magnitude, the waters of which are discharged from one to
another, and conveyed by an aqueduct to Jerusalem.
Bethlehem is celebrated in the Old Testament as the birthplace
and city of David; and in the New as that of David’s greater Son,
Christ, the Savior of the world. For 1800 years the earth has
renewed her carpet of verdure, and seen it again decay; yet the
skies and the fields, the rocks and the hills, and the valleys around
remain unchanged, and are still the same as when the glory of the
Lord shone about the shepherds, and the song of a multitude of the
heavenly host resounded among the hills, proclaiming, “Glory to God
in the highest, and on earth, peace, good-will toward men!”
To preserve a friend, three things are required—to honor him,
present; to praise him, absent; and assist him in his necessity.
Merry’s Adventures.
chapter xxiv.
Although I did not know what was before me, and had no
scheme even for providing myself with bread for a single day, I felt
an indescribable degree of delight at my release from prison. To be
shut up by our fellow-men, as if unworthy of enjoying light and liberty,
is very hard to bear: to know that one is innocent of crime—and yet
to be cast into a dungeon, and made the companion of the wicked
and the degraded—is calculated to beget a deep sense of injustice.
Such, indeed, was my feeling while in prison; and even when I was
free, it still mingled with my joy, impressing me with a sad
consciousness that even in society, and surrounded by laws
designed to protect us from wrong, we are not wholly secure, and
may be called upon, through the weakness or wickedness of our
fellow-men, to suffer the most bitter pangs.
I, however, resisted these feelings and poured out my gratitude to
Bill Keeler—my deliverer. On inquiry, I learned of him, that while at
Salem, he had accidentally heard of my imprisonment; and though
he supposed me guilty of some misdemeanor, he still gathered all
the money he could, and pushed off on foot to New York, to obtain
my release. The success of his endeavors has already been
detailed.
Having talked over the events already laid before the reader, Bill
asked me what I intended to do. I told him that I had formed no plan.
He then urged me to go back with him to Salem; but as I seemed
very reluctant to do so, his mind appeared to be turned to some
other project. We walked along the street for a considerable distance
in silence, and with an uncertain and sauntering gait—my companion
evidently in great perplexity. At last his countenance brightened, and
turning round on his heel, he led me on, with a decided step, in a
direction opposite to that which we had pursued.
“Well, well,” said Bill, cheerfully, “when one door shuts, another
opens: if the mountain doesn’t come to you, you must go to the
mountain. How would you like to become a traveller, Bob?”
“I should like it of all things.”
“So I thought—and I’ll get it all fixed.”
“But how am I to pay the expenses?”
“I brought a couple of friends with me, who’ll do that for you: they’r
queer chaps, but you’ll learn to like ’em. You remember old Sarah’s
cave? well, as I was climbing among the rocks just below it, a few
days ago, in search of a woodchuk that had just duv into his burrow,
a large stone gave way under my feet, and down the ledge I went,
for more than three rod. A great mass of rubbish came down with
me, and it’s a kind of miracle I wan’t smashed. I was a little stunned,
but by-and-by I came to myself. There I lay, half covered with stones,
leaves and gravel. Thinks I, what’s this all about? Just then I put out
my hand to get up, and I felt something mighty cold. Well, what do
you think it was? Why, ’twas a rattle-snake, and just by his side lay
seven others! It was cold weather, and they were as straight and stiff
as bean poles. Well, says I, there’s nothin made in vain—so I took
tew on ’em, and doubled ’em up and put ’em into one of my stockins,
and carried ’em home.
“When I got there, I took ’em out and laid ’em on the harth, and
when they got warm they began to squirm. Well—my wife—Hepsey
—(you remember Hepsey?—by the way—she sent her love to you,
Bob—though I’d forgot that)—she made a dreadful screechin about
it, and little Bob, he set up his pipes, and the cat stuck up her back,
and Jehu barked as if there’d been an attack of the Indians!
“Well, pretty soon the two critters began to stick out their tongues
and their eyes grew as bright as a couple of lightnin-bugs in a foggy
night. They then put their tails this way and that, and finally rolled
themselves into a heap, and set up such a rattlein as I never heard
afore. It was as much as to say—let every man look out for his own
shins! Everybody cleared—wife, baby, cat and dog—except myself.
Takin’ the varmin in the tongs, one by one, I threw ’em out the
winder, into a snow-bank, just to keep ’em cool and civil. I then made
a box, and put ’em in, and fitted a pane of glass in the top, so you
could look in and see ’em. Well, I brought the box and the two
sarpints along with me, thinkin that when you got out of prison, they
might be of sarvice.”
“What do you mean?” said I, in the greatest wonder.
“Mean? why, that you should take this box under your arm, and
travel over the world, as independent as a lord. The sarpints will be
meat and drink and clothin and lodgin, and a welcome to boot. I
thought it likely, when I set out, from what I heerd, that you’d got into
some scrape, and that it might be necessary for you to be scarce in
these parts; so I thought the snakes would suit your case exactly.
You needn’t look so sour, fir I don’t expect you to eat ’em. But hear
my story. I was three days in going from Salem to York, and when I
got there, I had tew dollars more in my pocket than when I set out,
and I lived like a prince all the time! And how do you think ’twas
done? Why, by the sarpints, to be sure! When I put up at the tavern
at night, I set the box down by my side in the bar-room, and took my
fife, and began to play Yankee Doodle.
“Pretty soon everybody got round me, and then I teld ’em about
the sarpints, and how they might see ’em fer sixpence apiece. Well, I
got sixpences as thick as nuts in November. Now, Bob, you’ve had a
good eddication, and can tell all about sarpints, and make up a good
story, and you can travel all over the world, and come home as rich
as a Jew. So you may have ’em, and I shall be happy to think that
you’re travelling like a gentleman, while I go home to pound my
lapstone and take care of my family.”
“I thank you a thousand times, my dear Bill,” said I; “but I fear this
will not do for me. You can turn your hand to anything, but I am a
helpless creature, compared with yourself!”
“No, no,” said my friend earnestly. “You’ll do well enough when
you get your hand in. You must try, at least. Here, take my penknife,
if you haint got one. A penknife’s a mighty good thing—no man need
to feel low-sperited with a penknife in his pocket. When I’m away and
feel kind o’ humsick, I take out my penknife, and get a stick and go to
cuttin on’t, and it turns out a whistle, or a walkin-stick, or somethin
else, and all the time I am as contented as a cow a stealin corn-
stalks. A penknife’s a friend in need, and no man should ever be
without one. You must take my fife, too, Bob, for you can play it well.
It will make you welcome everywhere—as we catch flies with
molasses, you can catch customers with music.”
To all this, I still replied that I doubted my success, and feared to
undertake the scheme. “Faint heart never won fair lady,” said Bill.
“Nothing venture, nothing have. You won’t succeed if you don’t try: a
man never fails, when success is matter o’ life and death. If you set
out, you won’t starve. You’ll be like Seth Follet’s eel—you must go
ahead.”
“Well, tell me the story of the eel.”
“Why, didn’t you never bear of Seth Follet’s eel? Seth had a long
aqueduct, made of logs, with an auger-hole bored thro’ ’em, to carry
the water from a spring on a hill, to his house. After a while the water
wouldn’t run, because the hole in the logs had got filled up with mud.
Well, Seth was a queer genius; so he got an eel and put into the hole
in the logs at one end. The critter went along pretty well for a time,
but by-and-by he came to the mud. He then thought he’d turn about,
but he couldn’t do that, for he just fitted the hole, you know! Then he
thought he’d back out, but he couldn’t do that nother, for an eel’s a
thing that can’t work both ways. Well now, what should he do? Why,
there was only one thing to be done—to go ahead; and ahead he
went—and cleared out the aqueduct!”
I could not help laughing heartily at this anecdote, and I confess
that the reasoning of Bill seemed to be fraught with good sense. We
spent the night together at the little tavern where he had left his box,
and in the morning I concluded to adopt his scheme. Bill departed,
the tears standing in his eyes—and taking the serpents, strapped
across my shoulders, I set out on my adventures.
I am not going to give a detail of my travels, at present. I am
afraid my readers are weary of my long story; and beside, I have
promised to bring my narrative to a close in my next number. I must,
therefore, pass lightly over my adventures as a showman; I must say
little of my experiences as a travelling merchant, and come down to
a period several years subsequent to my parting with Bill Keeler, as
just related. The war with England, declared by the United States in
1812, was then raging, and circumstances led me to take a part in it.
The events to which I allude, will be given in the next chapter.
Rivers.
Rivers have their rise in little rills, which gush from the sides of
mountains. Several of these unite, and form a stream; and these
again meeting, form a rivulet; and several rivulets form a river, which
sometimes runs for many thousand miles, and makes all the country
fertile through which it passes.
When a river descends from high land to that which is lower, it
often falls over rocks and precipices,—it is then called a cascade; or,
if very large, a cataract. Some of these are so large, that the water
breaks into spray before it reaches the ground, and the sound of it
may be heard for several miles.
Some rivers overflow their banks at certain seasons, owing to the
melting of the snow on the mountain tops, or the fall of heavy rains.
The river Nile overflows its banks; and, when the waters subside,
very great crops of rice and corn immediately spring up, as food for
man. There are very few parts of the earth in which rivers are not
found; and great, indeed, is their use to mankind.
Shall I tell you what a river is like? It is like the life of man—small
at first; the little stream is like a little child, and plays among the
flowers of a meadow; it waters a garden, or turns a puny mill.
As it flows on it gathers strength; and, like a child in youth, it
becomes turbulent and impatient as it swells along. Now, like a
roaring cataract, it shoots headlong down many a rock; then it
becomes a sullen and gloomy pool, buried in the bottom of a glen.
Recovering breath by repose, it again dashes on, till, tired of
uproar and mischief, it quits all that it has swept along, and leaves
the valley, through which it has passed, strewed with its rejected
waste.
Now, again, it travels more slowly. It passes through the busy
haunts of men, lending its service on every side, and, advancing in
its course, becomes stately and grand. Now, instead of breaking
over obstacles, it twines round them, and it thus passes along a
more quiet course.[13]
At last it leaves the busier world, and slowly and silently travels
on; till, at the end, it enters the vast abyss of ocean, which seems
spread out, like eternity, to receive it.
[13] Pliny.

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Electrical Engineering Principlesand And Applications

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Digital Information Age An Introduction to Electrical

Electrical Engineering in Context Smart Devices Robots

Then for vD  0.650 V, we have

Similarly for vD  0.700 V, iD  4.681 mA.

E10.2 The approximate form of the Shockley Equation is iD  Is exp(vD / nVT ) .

(a) VDQ  1.1 V IDQ  9 mA

At the intersections of the load lines with the diode characteristic, we

(b) If we assume that D2 is on, a current of 1.5 mA circulates

We can solve this circuit by using the node-voltage technique, treating vo

Using the voltage division principle, we determine that vD  3.333 V.

E10.11 Refer to Figure 10.25 in the book.

E10.12 As suggested in the Exercise statement, we design for a peak load

E10.14 Refer to Figure 10.31 in the book.

When vin exceeds 5 V, D6 turns on and D7 is in reverse breakdown. Then

When vin is less than -5 V, D7 turns on and D6 is in reverse breakdown.

E10.16 Refer to Figure 10.34a in the book.

(a) If vin (t )  0, we have only a dc source in the circuit. In steady state,

E10.19 Equation 10.22 gives the dynamic resistance of a semiconductor diode as

P10.5 VT  kT / q  (1.38  10 23T ) /(1.60  10 19 )

Similarly, for vD  0.700 V, we find iD  3.421 mA.

P10.7* The approximate form of the Shockley Equation is iD  Is exp(vD / nVT ) .

P10.10 At T  175 K , we have:

P10.11 VT  kT / q  (1.38  10 23  373) /(1.60  10 19 )  32.17 mV

P10.12 Using the approximate form of the Shockley Equation, we have

With the switch open, we have:

P10.15* (a) By symmetry, the current divides equally and we have

For diode B, we have T  305 K , and

Applying Kirchhoff's current law, we have

Solving for v by trial and err, we obtain v  697.1 mV , IA  87 mA , and

P10.16* The load-line equation is

P10.17 The load-line equation is

P10.18 The load-line equation is

P10.19 The load line equation is

Substituting values’ this becomes

At the intersection of the characteristic and the load line, we have

P10.26 The diagram of a suitable regulator circuit is

PZ max  5IZ max  1.04 W

PR max  PZ max  3.05 W

PR max  PZ max  10.4 W

Then, write the KVL equation for the equivalent circuit:

P10.30* The Thévenin resistance is Rt Voc / Isc  10 /2  5 . Also the Thévenin

At the intersection of the characteristic and the load line, we have

At the intersection of the characteristic and the load line, we have

P10.36* (a) D1 is on and D2 is off. V  10 volts and I  0.

(b) The diode is off, I  0 and V  5 V.

(d) The diode is on, I  3 mA and V  6 V.

P10.38 (a) D1 is on, D2 is on, and D3 is off. V  7.5 volts and I  0.

The plot of V versus Vin is:

P10.41 When the sinusoidal source is positive, D2 is on and D1 is off. Then, we

P10.42 If a nonlinear two-terminal device is modeled by the piecewise-linear

Writing a KCL equation at node A, we obtain: