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Unformatted text preview: ,fx'g Chapter 1
Review of Linear Circuit Theory 1.2 1.4 Plot the vi characteristic of the following circuit when Va = 5 V. ' R = 1 k0 i
>—o a 4. Va 5 V V a) A 2 kn resistor is connected across the terminals a—a’ in the
circuit shown above. Find the operating point of the resistor using
the graphical method. Verify the result using Uhm’ Law. Find the Thevenin equivalent of the circuit shown below, as seen
from the terminals a—a’. R2 = 20 kn 11 5mA R1=10 Id: In the following circuit, use superposition to find the Thevenin
equivalent seen at the terminals bb’ The circuit shown below contains a dependent source. Find the
Thevenin equivalent of the circuit as seen at the terminals a—a’ The circuit shown below contains a dependent source. Find the
Thevenin equivalent of the circuit as seen at the terminals bb’.
Use the "test source" method to find RTh The circuit shown below contains a current—dependent current source.
I Find the equivalent resistance seen looking into the terminals aa’.
Use the "test source" method. a
1.8 Find the Norton equivalent of the circuit shown below.
R = 22 k0
a
9 V
Va
""""" x5 3, 1.9 Find the Norton equivalent of the following circuit: R2 = 5 k0
a
11 = 8 mA R1 = 12 kn
a)
1.10 The circuit shown below contains a dependent source. Find the Norton equivalent of the circuit as seen at terminals aa’. This
circuit is the same one analyzed in Prob. 1.5. Measurements with a voltmeter are made 'at the terminals of a circuit
that contains only resistors, dc current sources, and fixed voltage
sources. The meter has an internal resistance of 1 MD. With no
other loads applied, the measured voltage is 12 V. With an
additional external 1 Mn resistance connected in parallel with the
meter, the measured voltage is 10 V. Find the Thevenin equivalent
of the resistive circuit as seen at the measured terminals. A circuit is powered from a 10V voltage supply, as shown below. When the circuit drives a 10 kﬂ resistive load at 5 V, it draws 2 mA
from this supply. Find the power dissipated in the load; find the power dissipated in the circuit.
2 mA Voltage
supply SOLUTIONS T0 PROBLEMS Chapter 1  Since the circuit is resistive (i.e., contains only a resistor and a
constant voltage source), it’s vi
characteristic is a straight line.
The vi characteristic can be plotted
by finding two points. For i = 0 (open circuit), v = Va =
5 V (vaxis intercept of the v—i
characteristic). For v = 0 (short
circuit), i = Va/R = 5 mA (i—axis in—
tercept). The resulting vi charac
teristic is shown below. A i (M) 0123456v(V) The v—i characteristic of a 2 kn resistor consists of a straight
line with slope '(1 mA)/(2 V) = 0.5
mA/V passing through the origin: 73 "““i‘tiaA; , Thevenin circuit operating
point O 1 2 3 4 5 6 V (V) The point of intersection, i.e., the
operating point, is seen to be 3.3 V; 1.6 mA. An analytical solution con
firms this result. From Uhm’s Law:
5 V
i = ——————~ = 1.67 mA
g 1 k0 + 2 kn
v = (1.66 mA)(2 kﬂ) = 3.33 V
Alternatively, from the voltage
divider relation:
2 kn
v = 5 V ——~——~— = 3.33 V
1 k0 + 2 k0 The Thevenin voltage is equal to the voltage at the terminals aa’ ' under opencircuit conditions (i.e., with no other elements connected to the terminals): R2 =.20 kﬂ' i = 0 I1 I1
0 50v
5 mA — Under open circuit conditions, all of , ,
the current 11 flows through R1, To find vTh' find the separate causing a voltage drop contributions of V1 and V2 to vac.
IlRl = (5mA)(10kﬂ) = 50 v Here is the circuit with V1 set to
to develop across R1. No voltage zero and V2 ”°""i
drop is developed across R2, since R
the current through it is zero. R1 2 . b
Thus, vac = 50 V. The Thevenin +
resistance is equal to the resistance v
seen at the terminals aa’ with 11 9 R4 0C1
set to zero (open circuit): .b:
R2 = 20 k0 From the voltage divider relation
v R4
v = ——
0C1 2 R2 + R4 Here is the circuit with V2 set to
zero and V1 ”on": By inspection, RTh = R1 + R2 = 30 k0.
0 Alternative method for finding RThi Find the current flowing through a
shortcircuit applied to the termi
nals aa’: R = 20 k0 . .
2 a Because a short curcuut appears directly across the series combina—
iSC tion of R2 and R4 (voltage drop equal
to zero), vgcz = 0. Thus VTh is 1.7mA
equal to v0c1 alone:
v R4
VTh ’ V0C1 — 2 R2 + R4 The current flowing through the ‘
shortcircuit is also equal to the T0 flnd RTh: set b°th v1 and V2 t° current flowing through R2. From the zero (short circuits) and. f'"d th?
current divider relation: resistance seen at the terminals bb R1
‘ = I ————— R R2
'SC 1 R1 + R2 1 b
10 kﬂ
= 5 mA —— = 1.67 mA ., (= RTh
10 kﬂ + 20 kﬂ ‘
.b’
' 50 V
RTh=ﬂ=——=30kll ' .
'SC 167 mA By Inspection, RTh = R4R2. Note that the R3R1 combination is
bypassed by the V2=0 short circuit. 74 Alternatively, RTh could be found
by applying a short circuit to the
terminals bb’ and using superposi—
tion to find the contributions of V1
and V2 to isc. Find the open—circuit voltage at the terminals a—a’. From KVL around the input loop, (R1 + R2)l2 + R403 4 1)i2 = 0
Solving for i2 yields i2 = 0.
Consequently, the dependent source is
set to zero as well (open circuit),
and RTh in the remaining circuit
becomes just R3. ° Alternative method:
With V1 on, the current flowing
through.a short circuit applied to i
a
l
i
l
l
l
l
i
i
i The dependent source pulls a current the aa’ terminals becomes ~ﬁi2, 5°
/,m1 p32 up from ground through R3, so that RTh = 19$ = —ﬁ32R3 = 3
1_W} that vac = —pi2R3. Find i2 by apply— isc pi2 ing KVL to the llinput loop" of the
circuit. Note that a total current
of (p + 1)i2 flows through R4, with 1.6 Ste 1: Re resent v R and
i2 entering R4 via R2 and ﬁiz enter— p p S, 1’ ing R4 via the dependent source. R2 by a Simpler Thevenln curcu1t consisting of one resistor and one From KVL: .
V1 = (R1 + R2)i2 + R4(ﬁ + 1)i2 voltage source.
Solving for i2 yields R R
. v1 1l 2
=——————
2 R1 + R2 + (ﬂ + 1)R4
vs R2
so that ﬁR _______
vac = pi2R3 = v 3 R1 + R2 1 R1 + R2 + (p + 1)R4
where VTh E vgc. T
To find RTh: set V1 to zero (short circuit) and find the resistance seen
looking into the terminals aa’. Next find VOC (i.e., VTh at .the
/’NE terminals bb’) by applying KVL to
i :7 the llinput loop”. Note that no
current flows through the open 5 75 circuit at the terminals where v1 is
defined, hence no current flows
through R1R2, and the voltage drop
across the resistance R1R2 is zero.
Applying KVL yields VS __Eg___ = V1 + 9mV1R3
R1+R2
or R 1
v1,= Vs 2 R1 + R2 (1 + nga)
The open—circuit voltage thus becomes
R2 9mR3
R1 + R2 (1 + 9mR3)
To find RTh: set V3 to zero and
apply a ”test" voltage source to b— b’. The value of RTh is determined by Vtest/ltest Note that a test
current source could also be used, but a test voltage source simplifies
the computation because it directly
fixes v1 to a known value. voc = 9mV1R3 = vs Since i1 = O, the voltage drop across
R1lR2 is zero, so that v1 = ‘Vtest The current ltest has two compo
nents, one through R3 and one through
the dependent source: . Vtest _ vtest 'test= R ’ 9mV1 ' + gmvtest
3 R3 The ratio of Vtest to itest can be computed from this equation, i.e., —1
Vt t 1 _ 1
RTh= es = ['R—+gm] =R3H—
'test 3 76 Find RTh by applying a test source to the terminals aa’. In
this case, a test current source
works best because it directly fixes
i1 to the value itest' R1 R2 i2 a Applying KCL to node "a" yields
ltest = i2  ﬂi1 = i2  pitest
Solving for i2 results in
i2 = (P + 1)ltest
Adding up the drops across R1 and R2 yields
Vtest = 31R1 + i2R2 .
= 'testRl + (ﬂ + 1)‘testR2
so that
v
Rn. = .test = R1 + (p + 1m
ltest The value of R2 as seen from a—a’ has
essentially been multiplied by the
factor (p + 1) via the action of the dependent source. The Norton equivalent consists of a current source and parallel
resistance. The value of the source
is equal to the shortcircuit current
measured at terminals aa’: R = 22 kﬂ a where isc = Va/R = (9 V)/(22 k0) =
0.41 mA/V. The Norton resistance is
found by. setting the Va source to
zero (short circuit), yielding RN = R
= 22 kﬂ. The complete Norton equivalent circuit is shown below. The Norton current is found by applying a shortcircuit to the aa’
terminals: R2 = 5 kn a From current division: ' R1 I 12 kne A 5 65 A
I = """—— = In = . m
5C R1 + R2 1 17 kn The Norton resistance is found by
setting I1 to zero (open circuit) and
observing the net resistance at the
terminals a—a’: R2 = 5 kn a II = 0 R1 = 12 kﬂ ¢== RN J
a RN = R1 + R2 = 5 kn + 12 kn = 17 kn.
The complete Norton equivalent of the
circuit is shown below. 77 The opencircuit voltage of this circuit was found in Prob. 1.5:
. ﬁR3
R1 + R2 + (ﬁ+ 1)R4
Similarly, the Thevenin resistance,
_computed with V1 set to zero, was
found to be RTh = R3. These elements
of the Thevenin equivalent circuit
can be used to find the element
values of the Norton equivalent
circuit. Specifically, RN = R3, and
voc ﬁV1
R3 . R1 + R2 + (p + 1)R4
The Norton equivalent of the original Voc = ﬁ32R3 = V1 circuit can be represented in the
following form:
ﬁV1 3
R1 + R2 + (p + 1)R4 R3
3, Here is a summary of the measurements. With the meter alone,
the current out of the resistive
circuit is (12 V)/( 1 M0) = 12 pA.: 12 pA Resistive 12 V
circuit  With an additional 1 Mn resistor
connected, the current out of the resistive circuit becomes
10 V (1 M“)! l(1 M“)
20 pA = 20 pA Resistive {O V circuit 'For the general Thevenin equivalent
shown below, v = VTh  iRTh. RTh i+ + VTh v Applying the known data results in:
12 V = VTh  (12 FA)RTh: and
10 V = VTh  (20 pA)RTh.
Simultaneous solution of these equa
tions yields VTh = 15 V; RTh = 250 kﬂ The power extracted from the supply is equal to (10 V)(2 mA) = 20
mW. The power dissipated in the load
is (5 V)2/(1o kn) = 2.5 mW. (Alterna
tively, (5 V)(Op5 mA) = 2.5 mW). The
power dissipated in the circuit is
equal to the difference: 20 mW  2.5
mW = 17.5 mW If the opamp is ideal, the current into the v+ terminal will be
zero, and no voltage drop will occur
across R2. The voltage at v+ will
thus equal VIN. The remaining 78 circuit will multiply v+ by the non
inverting gain, so that the output
becomes
R3 + R1 19 kn
———————v = R1 1” 3 kn VOUT = VIN = 6.3VIN If the op—amp is ideal, the current into the v+ terminal will be
zero. This current is identical to
the iIN flowing through R2, hence lIN
= O and Rin = N. [:::::] An ideal opamp saturates when VOUT reaches VCC or VEE As shown in
Prob. 2.1, the circuit has voltage
gain Av = 6.3, hence VDUT will equal
VCC for V C 10 V
VIN 2 — = * 5 1.6 V
Av 6.3
Similarly, the opamp will saturate
at VEE for
VE 15 V
VIN S ——— = — 2 2.4 V
Av 6.3 Since the op—amp is ideal, the current into the v+ terminal will be
zero. As a result, no current flows
through R2, and the voltage drop
across R2 becomes zero. With v+ held
at ground potential by R2, the cir
cuit functions as a simple inverting
amplifier with ' R3 30 k0 VOUT avmnkn VIN = 7.5 VIN AVD 0.68 V  0.70 V = —1.5 mV/°C
AT _ 10 0c ...
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 Spring '09
 Horenstein

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