Analysis and design of linear circuits

LIBROS UNIVERISTARIOS
Y SOLUCIONARIOS DE
MUCHOS DE ESTOS LIBROS
LOS SOLUCIONARIOS
CONTIENEN TODOS LOS
EJERCICIOS DEL LIBRO
RESUELTOS Y EXPLICADOS
DE FORMA CLARA
VISITANOS PARA
DESARGALOS GRATIS.


The Analysis and Design of Linear Circuits, Sixth Edition

Problem 4-1
Find the voltage gain v O 'v S and current gain i O 'i x in Figure P4-1 for r = 10 kΩ.
i
500 Ω iO
100 Ω X

vS

400 Ω

iX 2 kΩ
r@

vO

Solution:
The solution is presented in the following MATLAB code.
clear all
format short eng
syms vs ix io vo
r = 10e3;
% Find the current ix
ix = vs/(100+400);
% Find the current io and the voltage vo
io = -r*ix/(500+2e3);
vo = io*2e3;
% Compute the gains
Kv = simplify(vo/vs)
Ki = simplify(io/ix)
Kv =
-16
Ki =
-4

Answer:
K V = −16 and K I = −4.

Solutions Manual


The Analysis and Design of Linear Circuits, Sixth Edition

Solutions Manual

Problem 4-2

Find the voltage gain v O 'v 1 and the current gain i O 'i S in Figure P4-2. For i S = 2 mA, find the
power supplied by the input current source and the power delivered to the 2-kΩ load resistor.
i
100 Ω 1
O

iS

v1

100 Ω

100 i1
2 kΩ

2 kΩ

vO

Solution:
The solution is presented in the following MATLAB code.
clear all
syms is v1 i1 io vo
% Find i1 and v1
i1 = is/100/(1/100+1/100);
v1 = i1*100;
% Find io and vo
io = -100*i1/2e3/(1/2e3+1/2e3);
vo = io*2e3;
% Find the gains

Kv = simplify(vo/v1)
Ki = simplify(io/is)
% Find the source and load powers
v1_num = subs(v1,is,2e-3);
is = 2e-3;
ps_num = is*v1_num
vo_num = Kv*v1_num;
io_num = Ki*is;
po_num = vo_num*io_num
Kv =
-1000
Ki =
-25
ps_num =
200.0000e-006
po_num =
5

Answer:
K V = −1000 and K I = −25.
For i S = 2 mA, p S = 200 µW and p L = 5 W.


The Analysis and Design of Linear Circuits, Sixth Edition

Solutions Manual

Problem 4-3
Find the voltage gain v O 'v S and current gain i O 'i x in Figure P4-3 for g = 10!3 S.
1 kΩ

vS

iX

3 kΩ

iO

500 Ω

vX

g@
vX

10 kΩ

2 kΩ

Solution:
The solution is presented in the following MATLAB code.
clear all
syms vs ix vx io vo
g = 1e-3;
% Find ix and vx
ix = vs/(1e3+3e3);
vx = 3e3*ix;
% Find io and vo
io = g*vx/(500+2e3)/(1/10e3+1/(500+2e3));

vo = io*2e3;
% Find the gains
Kv = simplify(vo/vs)
Ki = simplify(io/ix)
Kv =
6/5
Ki =
12/5

Answer:
K V = 1.2 and K I = 2.4.

vO


The Analysis and Design of Linear Circuits, Sixth Edition

Solutions Manual

Problem 4-4
(a) Find the voltage gain v O 'v S and current gain i O 'i x in Figure P4-4.
(b) Validate your answers by simulating the circuit in OrCAD.
6.3 kΩ

1.5 kΩ

Solution:
(a) The solution is presented in the
following MATLAB code.

vS

iX

clear all
syms vs ix io vo
% Find ix
ix = vs/1.5e3;
% Find io and vo
io = -50*ix;
vo = io*2.2e3;
% Find the gains
Kv = simplify(vo/vs)
Ki = simplify(io/ix)
Kv =
-220/3
Ki =
-50

(b) The following OrCAD simulation verifies the answer in Part (a).

Answer:
(a) K V = −73.33 and K I = −50.
(b) The OrCAD simulation verifies the answer in Part (a).

50iX

iO

2.2 kΩ

vO


The Analysis and Design of Linear Circuits, Sixth Edition

Problem 4-5
Find the voltage gain v O 'v S in Figure P4-5.
10 kΩ

iO

vX
vS

50 vX
1 kΩ

vO

Solution:
The solution is presented in the following MATLAB code.
clear all
syms vs vx vo
% Solve for vx in terms of vs
Eqn1 = 'vx-(vs-(-50*vx))';
vx = solve(Eqn1,'vx');
% Find vo
vo = -50*vx;
% Find the gain

Kv = simplify(vo/vs)
Kv_num = double(Kv)
Kv =
50/49
Kv_num =
1.0204e+000

Answer:
K V = 50/49 = 1.02

Solutions Manual


The Analysis and Design of Linear Circuits, Sixth Edition

Solutions Manual

Problem 4-6
Find an expression for the current gain i O 'i S in Figure P4-6. Hint: Apply KCL at node A.
iS

iO

A

RS

vS

βiE

RE

RC
iE

Solution:
The solution is presented in the following MATLAB code.
clear all
syms vs is ie io B Rs Re Rc vA
% Write a KCL expression at node A
Eqn1 = '(vA-vs)/Rs + vA/Re - B*ie';
% Write additional equations relating the variables in the circuit
Eqn2 = 'ie -vA/Re';
Eqn3 = 'io + B*ie';
Eqn4 = 'is -(vs-vA)/Rs';
% Solve the equations
Soln = solve(Eqn1,Eqn2,Eqn3,Eqn4,'io','is','ie','vA');
io = simplify(Soln.io);
is = simplify(Soln.is);
% Calculate the gain
Ki = io/is
Ki =
B/(-1+B)

Answer:
β
KI =
β −1

The Analysis and Design of Linear Circuits, Sixth Edition

Solutions Manual

Problem 4-7
(a) Find the voltage v O in Figure P4-7.
(b) Validate your answer by simulating the circuit in
OrCAD.

0.001 vX
A

1 kΩ

B

2.2 kΩ

Solution:
(a) The solution is presented in the following MATLAB
code.

10 V

clear all
syms vx vo
% Use node-voltage analysis
Eqn1 = '(vx-10)/2.2e3 + vx/1.5e3 + (vx-vo)/1e3 + 1e-3*vx';
Eqn2 = '-1e-3*vx + (vo-vx)/1e3 + vo/3.3e3';
% Solve the equations for vo

Soln = solve(Eqn1,Eqn2,'vx','vo')
vo = double(Soln.vo)
Soln =
vo: [1x1 sym]
vx: [1x1 sym]
vo =
4.3980e+000

(b) The following OrCAD circuit verifies the answer in Part (a).

Answer:
(a) v O = 4.398 V.
(b) The OrCAD simulation verified the results in Part (a).

1.5 kΩ

vX 3.3 kΩ

vO


The Analysis and Design of Linear Circuits, Sixth Edition

Problem 4-8
(a) Find an expression for the current gain i O 'i S in Figure P4-8.
(b) Find an expression for the voltage gain v O 'v S in Figure P4-8.
iO
R1
R2

vS

vX

μvX

RL

Solution:
The solution is presented in the following MATLAB code.
clear all
syms vs is vx io vo mu R1 R2 RL
% Write a node-voltage equation
Eqn1 = '(vx-vs)/R1 + (vx-(-mu*vx))/R2';
% Write additional equations relating the circuit parameters
Eqn2 = 'vo + mu*vx';
Eqn3 = 'is - (vs-vx)/R1';
Eqn4 = 'vo - io*RL';
% Solve the equations
Soln = solve(Eqn1,Eqn2,Eqn3,Eqn4,'vx','vo','is','io');
vo = simplify(Soln.vo);
is = simplify(Soln.is);
io = simplify(Soln.io);
% Compute the gains
Ki = simplify(io/is)
Kv = simplify(vo/vs)
Ki =
-mu*R2/RL/(1+mu)
Kv =
-mu*R2/(R2+R1+R1*mu)

Answer:

− μR2
(1 + μ ) RL
− μR2
(b) K V =
(1 + μ ) R1 + R2

(a) K I =

Solutions Manual


The Analysis and Design of Linear Circuits, Sixth Edition

Solutions Manual

Problem 4-9
Find an expression for the voltage gain v O 'v S in Figure P4-9.

RS

A

vX

gvX

vS

RO

vO

Solution:
The solution is presented in the following MATLAB code.
clear all
syms vs vx Rs g Ro vo
% Write a node-voltage equation
Eqn1 = '(vo-vs)/Rs - g*vx + vo/Ro';
% Write additional equations relating the parameters in the circuit
Eqn2 = 'vx - (vs-vo)';
% Solve the equations
Soln = solve(Eqn1,Eqn2,'vx','vo');
vo = simplify(Soln.vo);
% Calculate the gain
Kv = simplify(vo/vs)
Kv =
Ro*(1+g*Rs)/(Ro+g*Rs*Ro+Rs)

Answer:

KV =

gRO RS + RO
gRO RS + RO + RS


The Analysis and Design of Linear Circuits, Sixth Edition

Solutions Manual

Problem 4-10
(a) Find an expression for the voltage gain v O 'v S in Figure P4-10.
(b) Let R S = 10 kΩ, R L = 10 kΩ and μ = 100. Find the voltage gain v O 'v S as a function of R F .
What is the voltage gain when R F is an open circuit, a short circuit, and for R F = 100 Ω?
(c) Simulate the circuit in OrCAD by varying R F from 1 Ω to 10 MΩ. Read your output for R F =
100 Ω. How does your answer compare with part (b)?
RF
RS
vX
vS

μvX

RL

vO

Solution:
The solutions to Parts (a) and (b) are presented in the following MATLAB code.
clear all
disp('Part (a)')
syms vs vx vo vA Rs Rf mu RL
% Write a node-voltage equation
Eqn1 = '(vA-vs)/Rs + (vA-mu*vx)/Rf';
% Write other equations relating the circuit parameters
Eqn2 = 'vA - (vx + mu*vx)';
Eqn3 = 'vo - mu*vx';

% Solve the equations
Soln = solve(Eqn1,Eqn2,Eqn3,'vA','vx','vo');
vo = simplify(Soln.vo);
% Calculate the gain
Kv = simplify(vo/vs)
disp('Part (b)')
Kv = simplify(subs(Kv,{Rs,RL,mu},{10e3,10e3,100}))
Rf_num = [0 100];
Kv_num = subs(Kv,Rf,Rf_num)
Kv_inf = limit(Kv,Rf,inf)
Part (a)
Kv =
mu/(Rf+Rf*mu+Rs)*Rf
Part (b)
Kv =
100/(101*Rf+10000)*Rf
Kv_num =
0.0000e-003
497.5124e-003
Kv_inf =
100/101


The Analysis and Design of Linear Circuits, Sixth Edition

Solutions Manual

(c) The OrCAD circuit and simulation results are shown below.
PARAMETERS:
R1 resistor = 1

10k
Vs
1V

Rf

{resistor}
RL
10k

E1

+
-

V

+
-

E

0

Answer:

μRF
(1 + μ ) RF + RS
100 RF

(b) K V =
. For R F as an open circuit, a short circuit, and for R F = 100 Ω, we have
101RF + 10000
K V equal to 100/101, 0, and 0.4975, respectively.
(c) The OrCAD results are presented above. With R F as an open circuit or short circuit, the
voltage gain values approach the correct values. For R F = 100 Ω, we have K V = 0.4975, which
agrees with the calculations in Part (b).
(a) K V =


The Analysis and Design of Linear Circuits, Sixth Edition

Solutions Manual

Problem 4-11
Find the Thévenin equivalent circuit that the load R L sees in Figure P4-11.
iS

RS

vS

r@
iS

RL

vO

Thévenin circuit

Solution:
Since the circuit has a dependent source, we cannot reliably use the look-back technique to
compute the Thévenin resistance. We need to find the open-circuit voltage and the short-circuit
current. The solution is presented in the following MATLAB code.
clear all
syms vs Rs Rp r is vT RT isc
Eqn1 = 'is - (vs-r*is)/Rs';
Eqn2 = 'vT - r*is';
Eqn3 = 'isc - r*is/Rp';
% Solve the equations
Soln = solve(Eqn1,Eqn2,Eqn3,'is','vT','isc');
vT = simplify(Soln.vT)
isc = simplify(Soln.isc);
RT = simplify(vT/isc)
vT =
r*vs/(Rs+r)
RT =
Rp

Answer:
R T = R P and vT =

rVS
.
RS + r


The Analysis and Design of Linear Circuits, Sixth Edition

Problem 4-12
Find R IN in Figure P4-12.
R
iS

r@
iS

RIN
Solution:
Find the ratio of the input voltage to the input current using MATLAB.
clear all
syms is R r iin vin Rin
iin = is;
vin = is*R+r*is;
Rin = simplify(vin/iin)
Rin =
R+r

Answer:
R IN = R + r.

Solutions Manual


The Analysis and Design of Linear Circuits, Sixth Edition

Problem 4-13
Find R IN in Figure P4-13.

iS

R

βi S

RIN

Solution:
Find the ratio of the input voltage to the input current using MATLAB.
clear all
syms is ix vin R Rin B
Eqn1 = 'ix - (is + B*ix)';
ix = solve(Eqn1,'ix');
vin = R*ix;
Rin = vin/is
Rin =
-R/(-1+B)

Answer:
R
RIN =
.
1− β

Solutions Manual


The Analysis and Design of Linear Circuits, Sixth Edition

Problem 4-14
Find the Thévenin Equivalent circuit seen by the load in Figure P4-14.
i

RS
iX
RO
vS

βiX

v

Load

Solution:
The solution is presented in the following MATLAB code.
clear all
syms vs Rs Rx ix B Ro vT isc RT
% Find the open-circuit voltage and the short-circuit current
ix = vs/(Rs+Rx);
vT = -B*ix*Ro
isc = -B*ix;
RT = simplify(vT/isc)
vT =
-B*vs/(Rs+Rx)*Ro
RT =
Ro

Answer:

R T = R O and vT =

− βRO vS
.
RS + RX

Solutions Manual


The Analysis and Design of Linear Circuits, Sixth Edition

Problem 4-15
Find the Norton Equivalent circuit seen by the load in Figure P4-15.
i

iS

vX

gvX

RO

v

Load

Solution:
The solution is presented in the following MATLAB code.
clear all

syms is vx g Ro iN vT RN
% Find the short-circuit current
vx = 0;
iN = is
% Find the open-circuit voltage
Eqn1 = 'vx - (is-g*vx)*Ro';
vx = solve(Eqn1,'vx');
vT = simplify(vx);
% Find the Norton resistance
RN = simplify(vT/iN)
iN =
is
RN =
Ro/(1+g*Ro)

Answer:
RO
and i N = i S.
RN =
1 + gRO

Solutions Manual


The Analysis and Design of Linear Circuits, Sixth Edition

Solutions Manual

Problem 4-16
The circuit parameters in Figure P4-16 are R B = 50 kΩ, R C = 4 kΩ, β = 120, V γ = 0.7 V, and V CC

= 15 V. Find i C and v CE for v S = 2 V. Repeat for v S = 5 V.

RC
iC
RB i B
vCE

VCC

vS
Solution:
The solution is presented in the following MATLAB code.
clear all
RB = 50e3;
RC = 4e3;
B = 120;
Vg = 0.7;
Vcc = 15;
% Determine the operating mode and find iC and vCE
disp('vs = 2 V')
vs = 2;
if vsdisp('Cutoff Mode')
iC = 0;
vCE = Vcc;
else
iB = (vs-Vg)/RB;
iC = B*iB;
vCE = Vcc-iC*RC;
if vCE<0

disp('Saturation Mode')
vCE = 0;
iC = Vcc/RC;
else
disp('Active Mode')
end
end
iC
vCE
% Determine the operating mode and find iC and vCE
disp('vs = 5 V')
vs = 5;
if vsdisp('Cutoff Mode')
iC = 0;
vCE = Vcc;
else


The Analysis and Design of Linear Circuits, Sixth Edition

Solutions Manual

iB = (vs-Vg)/RB;
iC = B*iB;
vCE = Vcc-iC*RC;
if vCE<0
disp('Saturation Mode')
vCE = 0;
iC = Vcc/RC;

else
disp('Active Mode')
end
end
iC
vCE
vs = 2 V
Active Mode
iC =
3.1200e-003
vCE =
2.5200e+000
vs = 5 V
Saturation Mode
iC =
3.7500e-003
vCE =
0.0000e-003

Answer:
For v S = 2 V, the transistor is in the active mode and we have i C = 3.12 mA and v CE = 2.52 V.
For v S = 5 V, the transistor is in saturation mode and we have i C = 3.75 mA and v CE = 0 V.


The Analysis and Design of Linear Circuits, Sixth Edition

Solutions Manual

Problem 4-17
The circuit parameters in Figure P4-16 are R C = 3 kΩ, β = 100, V γ = 0.7 V, and V CC = 5 V.

Select a value of R B such that the transistor is in the saturation mode when v S $ 2 V.

RC
iC
RB i B
vCE

VCC

vS
Solution:
The solution is presented in the following MATLAB code.
clear all
RC = 3e3;
B = 100;
Vg = 0.7;
Vcc = 5;
vs = 2;
% In saturation mode vCE = 0 and iC = isc = Vcc/RC
vCE = 0;
iC = Vcc/RC;
% Find iB at the edge of the active mode
iB = iC/B;
% Find RB at the edge of the active mode
RB = (vs-Vg)/iB
RB =
78.0000e+003

If R B decreases, then i B increases, which pushes the transistor further into the saturation mode.
Answer:

R B < 78 kΩ.


The Analysis and Design of Linear Circuits, Sixth Edition

Solutions Manual

Problem 4-18
The parameters of the transistor in Figure P4-18 are β = 50 and V γ = 0.7 V. Find i C and v CE for
v S = 0.8 V. Repeat for v S = 2 V.

10 kΩ
iC
10 kΩ
vS

iB

vCE

20 kΩ

15 V

Solution:
The solution is presented in the following MATLAB code.
clear all
RB = 10e3;
B = 50;
Vg = 0.7;

% Find a Thevenin equivalent for the output loop
Vcc = 20e3*15/(20e3+10e3)
RC = 1/(1/20e3+1/10e3)
% Determine the operating mode and find iC and vCE
disp('vs = 0.8 V')
vs = 0.8;
if vsdisp('Cutoff Mode')
iC = 0;
vCE = Vcc;
else
iB = (vs-Vg)/RB;
iC = B*iB;
vCE = Vcc-iC*RC;
if vCE<0
disp('Saturation Mode')
vCE = 0;
iC = Vcc/RC;
else
disp('Active Mode')
end
end
iC
vCE
% Determine the operating mode and find iC and vCE
disp('vs = 2 V')
vs = 2;
if vsdisp('Cutoff Mode')
iC = 0;

The Analysis and Design of Linear Circuits, Sixth Edition

Solutions Manual

vCE = Vcc;
else
iB = (vs-Vg)/RB;
iC = B*iB;
vCE = Vcc-iC*RC;
if vCE<0
disp('Saturation Mode')
vCE = 0;
iC = Vcc/RC;
else
disp('Active Mode')
end
end
iC
vCE
Vcc =
10.0000e+000
RC =
6.6667e+003
vs = 0.8 V
Active Mode
iC =
500.0000e-006
vCE =

6.6667e+000
vs = 2 V
Saturation Mode
iC =
1.5000e-003
vCE =
0.0000e-003

Answer:
For v S = 0.8 V, the transistor is in the active mode and we have i C = 500 µA and v CE = 6.667 V.
For v S = 2 V, the transistor is in saturation mode and we have i C = 1.5 mA and v CE = 0 V.


The Analysis and Design of Linear Circuits, Sixth Edition

Solutions Manual

Problem 4-19
Two OP AMP circuits are shown in Figure P4-19. Both claim to produce a gain of either ± 10.
(a) Show that the claim is true.
10 kΩ
vIN
1 kΩ
(b) A practical source with a series
vO
resistor of 1 kΩ is connected to the input
vIN 1 kΩ
vS
of each circuit. Does the original claim
9 kΩ

still hold? If it does not, explain why?
Solution:
(a) The solution is presented in the
following MATLAB code.

1 kΩ

Source

Circuit 1

Circuit 2

clear all
format short eng
disp('Part (a)')
disp('Circuit 1')
% Circuit 1 is a non-inverting amplifier
R1 = 9e3;
R2 = 1e3;
K1 = (R1+R2)/R2
% Circuit 2 is an inverting amplifier
R1 = 1e3;
R2 = 10e3;
K2 = -R2/R1
Part (a)
Circuit 1
K1 =
10.0000e+000
K2 =

-10.0000e+000

(b) The practical source does not change the gain of Circuit 1, but it does change the gain of
Circuit 2.
% Circuit 2 is an inverting amplifier
R1 = 1e3+1e3;
R2 = 10e3;
K2 = -R2/R1
K2 =
-5.0000e+000

The gain is reduced for Circuit 2 because the source resistor is in series with the OP AMP's input
resistor, which effectively increases the value of R 1 used to compute the gain for the inverting
amplifier.
Answer:
(a) K 1 = 10 and K 2 = −10.
(b) K 1 = 10 and K 2 = −5. The source resistor loads only Circuit 2.

vO


The Analysis and Design of Linear Circuits, Sixth Edition

Solutions Manual

Problem 4-20
Suppose the output of the practical source shown in Figure P4-19 needs to be amplified by – 100
and you can use only the two circuits shown. How would you connect the circuits to achieve
this? Explain why.
1 kΩ

10 kΩ

vIN

vO

vS

vIN

1 kΩ
vO

9 kΩ
1 kΩ

Source

Circuit 1

Circuit 2

Solution:
Connect the practical source to the input of Circuit 1 and then connect the output of Circuit 1 to
the input of Circuit 2. The practical source does not change the gain of Circuit 1, so it still
provides a gain of 10. The output of Circuit 1 will not change the gain of Circuit 2, since the
output resistance of an OP AMP circuit is very small. The overall gain will be (10)(−10) = −100,
as requested.

Answer:
Presented above in the Solution.


The Analysis and Design of Linear Circuits, Sixth Edition

Solutions Manual

Problem 4-21
(a) Find v O in terms of v S in Figure P4-21.
(b) Validate your answer by simulating the circuit in OrCAD.

22 kΩ
vS

33 kΩ

330 kΩ

56 kΩ

vO

Solution:
(a) Find the Thévenin equivalent of the input circuit and then analyze the circuit as an inverting
OP AMP. The solution is presented in the following MATLAB code.
clear all
% Find the Thevenin voltage
syms vs vo
vT = 56e3*vs/(22e3+56e3);

RT = 33e3+1/(1/22e3+1/56e3);
% Find the output voltage in terms of vs
Rf = 330e3;
Kv = -Rf/RT;
vo = Kv*vT
vo_num = vpa(vo,4)
vo =
-840/173*vs
vo_num =
-4.855*vs

(b) The following OrCAD simulation verifies the results presented in Part (a).

Answer:
(a) v O = −4.86 v S.
(b) The OrCAD simulation verifies the results of Part (a).