Model the circuit as:
a)
Weneedtokeepv2 acrossR2 as4.8≤v2 ≤5.4
a)
Weneedtokeepv2 acrossR2 as4.8≤v2 ≤5.4
For I = 0.3 A
0.1 A
display is active
display is not active
KCLata:v2−15+v2 +I=0
RR
12
Assumed that maximum I results in minimum v2 and visa-versa.
Assumed that maximum I results in minimum v2 and visa-versa.
Then
v = 4.8V whenI=0.3A
v = 4.8V whenI=0.3A
2 5.4V whenI =0.1A
Substitute these corresponding values of v2 and I into the KCL equation and solve for the resistances
4.8 −15 + 4.8 + 0.3 = 0
RR
12
5.4 −15 + 5.4 + 0.1 = 0
5.4 −15 + 5.4 + 0.1 = 0
RR
b)
⇒R = 7.89Ω, R = 4.83Ω
12
I = 15−4.8 = 1.292 A ⇒ P
I = 15−4.8 = 1.292 A ⇒ P
= (1.292)2(7.89) = 13.17 W
= (5.4)2 =6.03W
4.83
4.83
12
1max 1max
R 7.89 R
I = 5.4 =1.118A
R2 max 4.83
⇒ P
R2 max
R2 max
maximum supply current =
c) No; if the supply voltage (15V) were to rise or drop, the voltage at the display would drop
below 4.8V or rise above 5.4V.
The power dissipated in the resistors is excessive. Most of the power from the supply is dissipated in the resistors, not the display.
c) No; if the supply voltage (15V) were to rise or drop, the voltage at the display would drop
below 4.8V or rise above 5.4V.
The power dissipated in the resistors is excessive. Most of the power from the supply is dissipated in the resistors, not the display.
I R = 1.292 A
1max
4-25
DP 4-2
Express the voltage of the 8 V source in terms of its node voltages to get vb − va = 8 . Apply KCL to the supernode corresponding to the 8 V source:
va−v1+va+vb+vb−(−v2)=0 ⇒ 2va−v1+2vb+v2=0 RRRR
Express the voltage of the 8 V source in terms of its node voltages to get vb − va = 8 . Apply KCL to the supernode corresponding to the 8 V source:
va−v1+va+vb+vb−(−v2)=0 ⇒ 2va−v1+2vb+v2=0 RRRR
Nextsetva =0toget
⇒ 2va−v1+2(va+8)+v2=0
⇒ 4va−v1+v2+16=0
⇒ va=v1−v2−4 4
0=v1−v2−4 ⇒ v1−v2=16V 4
⇒ va=v1−v2−4 4
0=v1−v2−4 ⇒ v1−v2=16V 4
Forexample,v1 =18Vandv2 =2V.
4-26
DP 4-3
a)
pplyKCLtoleftmesh: ApplyKCLtorightmesh:
light.
b) From the equation for I, we see that decreasing R increases I:
tryR =50Ω ⇒ I =45mA(won'tlight) tryR =25Ω ⇒ I =61mA ⇒ willlight
Now check R±10% to see if the lamp will light and not burn out:
−10% → 22.5Ω → I = 63.63 mAlamp will
+10% → 27.5Ω → I = 59.23 mAstay on
DP 4-4
a)
pplyKCLtoleftmesh: ApplyKCLtorightmesh:
light.
b) From the equation for I, we see that decreasing R increases I:
tryR =50Ω ⇒ I =45mA(won'tlight) tryR =25Ω ⇒ I =61mA ⇒ willlight
Now check R±10% to see if the lamp will light and not burn out:
−10% → 22.5Ω → I = 63.63 mAlamp will
+10% → 27.5Ω → I = 59.23 mAstay on
DP 4-4
−5+50i1 +300 (i1 −I)=0
(R+2)I+300(I−i1)= 0
I = 150
1570 + 35 R
We desire 50 mA ≤ I ≤ 75 mA so if R = 100 Ω, then I = 29.59 mA fi l amp so the lamp will not
I = 150
1570 + 35 R
We desire 50 mA ≤ I ≤ 75 mA so if R = 100 Ω, then I = 29.59 mA fi l amp so the lamp will not
Solving for I:
Equivalent resistance: R = R1 || R2 || (R3 + R4 )
Voltage division in the equivalent circuit: v1 = R (25)
10+R
We require vab = 10 V. Apply the voltage division principle in the left circuit to get:
We require vab = 10 V. Apply the voltage division principle in the left circuit to get:
4-27
(RR (R+R))
10=R4 v1=R4 × 1 2 3 4 ×25
R3 + R4 R3 + R4
This equation does not have a unique solution. Here’s one solution:
This equation does not have a unique solution. Here’s one solution:
10+RR (R+R)
()
1234
DP 4-5
chooseR =R = 25ΩandR +R =20Ω
12 34
then10=R4× (12.520) ×25⇒R=18.4Ω 20 10+(12.520) 4
andR3 +R4 =20⇒R3 =1.6Ω
Apply KCL to the left mesh:
(R +R )i − R i −v =0 131321
Apply KCL to the left mesh:
− R3 i1 + (R2 + R3 ) i2 + v2 = 0
then10=R4× (12.520) ×25⇒R=18.4Ω 20 10+(12.520) 4
andR3 +R4 =20⇒R3 =1.6Ω
Apply KCL to the left mesh:
(R +R )i − R i −v =0 131321
Apply KCL to the left mesh:
− R3 i1 + (R2 + R3 ) i2 + v2 = 0
Solving for the mesh currents using Cramer’s rule:
v1 −R3 (R1+R3) v1
v1 −R3 (R1+R3) v1
−v2 (R2 +R3) −R3 −v2
i1 = and i2 =
i =[2v1−v2]1000andi=[−2v2+v1]1000 ⇒i=i−i=v1+v2
1 2 3000
okay okay
i =[2v1−v2]1000andi=[−2v2+v1]1000 ⇒i=i−i=v1+v2
1 2 3000
okay okay
∆∆
where ∆ = (R + R ) (R + R ) − R 2
13233
TryR1 = R2 =R3 =1kΩ=1000Ω.Then∆=3MΩ.Themeshcurrentswillbegivenby
TryR1 = R2 =R3 =1kΩ=1000Ω.Then∆=3MΩ.Themeshcurrentswillbegivenby
3×106 2 3×106
1
Now check the extreme values of the source voltages:
Now check the extreme values of the source voltages:
ifv1 =v2 = 1V ⇒ i = 23 mA
ifv1 =v2 = 2V ⇒ i = 43 mA
4-28
Exercises
Ex 5.3-1
R = 10 Ω and is = 1.2 A.
Ex 5.3-2
R=10Ωandis =−1.2A. Ex 5.3-3
R = 8 Ω and vs = 24 V. Ex 5.3-4
R=8Ωandvs =−24V. Ex 5.4-1
Ex 5.4-2
im = 25 − 3 (5)=5−3=2A 3+2 2+3
Ex 5.4-3
R=10Ωandis =−1.2A. Ex 5.3-3
R = 8 Ω and vs = 24 V. Ex 5.3-4
R=8Ωandvs =−24V. Ex 5.4-1
Ex 5.4-2
im = 25 − 3 (5)=5−3=2A 3+2 2+3
Ex 5.4-3
Chapter 5 Circuit Theorems
v = 20 (15)+20− 10
m 10+20+20 10+(20+20) 5
m 10+20+20 10+(20+20) 5
(2)=6+20(−2)=−2V
v =3 3 (5)− 3 (18)=5−6=−1A
m 3+(3+3) 3+(3+3)
5-1
Ex 5.5-1
5-2
Ex 5.5-2
Ex 5.6-1
ia=2ia−12 ⇒ ia=−3A
6
voc =2ia =−6V
12+6ia =2ia ⇒ ia =−3A
voc =2ia =−6V
12+6ia =2ia ⇒ ia =−3A
3isc =2ia
Rt =−6=3Ω −2
Rt =−6=3Ω −2
⇒ isc =2(−3)=−2A
3
5-3
Ex 5.6-2
ia=2ia−12 ⇒ ia=−3A
6
voc =2ia =−6V
12+6ia =2ia ⇒ ia =−3A
3isc =2ia ⇒ isc =2(−3)=−2A 3
Rt =−6=3Ω −2
voc =2ia =−6V
12+6ia =2ia ⇒ ia =−3A
3isc =2ia ⇒ isc =2(−3)=−2A 3
Rt =−6=3Ω −2
5-4
Ex 5.6-3
Ex 5.7-1
Rt =12×24=12×24= 8Ω
12+24 36
voc= 24 (30)=20V 12+24
i= 20 8+R
voc = 6 (18)=12V 6+3
Rt =2+(3)(6)=4Ω 3+6
For maximum power, we require RL =Rt =4Ω
voc= 24 (30)=20V 12+24
i= 20 8+R
voc = 6 (18)=12V 6+3
Rt =2+(3)(6)=4Ω 3+6
For maximum power, we require RL =Rt =4Ω
Then
p=oc= =9W
p=oc= =9W
v2 122
max
4R 4(4)
t
5-5
Ex 5.7-2
and
Rt = 20 Ω
⇒ voc = pmax4Rt = 5(4)20=20V
⇒ voc = pmax4Rt = 5(4)20=20V
1
isc = 3 (5.6)=
isc = 3 (5.6)=
v2
50 (5.6)=5A
50+1+5
Rt =3+150(30)=3+25=28Ω 150+30
Rt =3+150(30)=3+25=28Ω 150+30
1+ 1 + 1
3 150 30
R i 2 (28)52
pmax= t sc =
=175W
Ex 5.7-3
() The power increases as Rt decreases so choose Rt = 1 Ω. Then
p = i v = 100 (5) = 13.9 W max (1+5)2
Ex 5.7-4
From the plot, the maximum power is 5 W when R = 20 Ω. Therefore:
() The power increases as Rt decreases so choose Rt = 1 Ω. Then
p = i v = 100 (5) = 13.9 W max (1+5)2
Ex 5.7-4
From the plot, the maximum power is 5 W when R = 20 Ω. Therefore:
44
10R 100R
p = i v = L
(10) = L
R+RR+R 2
t Lt L Rt+RL
pmax = oc
4 Rt
5-6
Problems
Section 5-3: Source Transformations
P5.3-1
(a)
Section 5-3: Source Transformations
P5.3-1
(a)
5-7
(b) −9−4i−2i+(−0.5)=0
i = −9+(−0.5)=−1.58 A
i = −9+(−0.5)=−1.58 A
∴Rt =2Ω
vt =−0.5V
vt =−0.5V
4+2
v=9+4i=9+4(−1.58)=2.67 V
(c) ia =i=−1.58A
P5.3-2
(checked using LNAP 8/15/02)
Finally,applyKVL:
−10+3ia +4ia −16=0
3
∴ ia = 2.19A
(checked using LNAP 8/15/02)
(checked using LNAP 8/15/02)
5-8
P5.3-3
Source transformation at left; equivalent resistor for parallel 6 and 3 Ω resistors:
Equivalents for series resistors, series voltage source at left; series resistors, then source transformation at top:
Source transformation at left; series resistors at right:
Parallel resistors, then source transformation at left:
Source transformation at left; equivalent resistor for parallel 6 and 3 Ω resistors:
Equivalents for series resistors, series voltage source at left; series resistors, then source transformation at top:
Source transformation at left; series resistors at right:
Parallel resistors, then source transformation at left:
5-9
Finally, apply KVL to loop
P5.3-4
−6+ i (9+19)−36−vo = 0
i=5/2 ⇒vo =−42+28 (5/2)=28V
(checked using LNAP 8/15/02)
(checked using LNAP 8/15/02)
− 4 − 2000ia − 4000ia +10 −2000ia −3=0
∴ia =375μA
(checked using LNAP 8/15/02)
(checked using LNAP 8/15/02)
5-10
P5.3-5
P5.3-6
−12−6ia +24−3ia−3=0⇒ia=1A
(checked using LNAP 8/15/02)
(checked using LNAP 8/15/02)
A source transformation on the right side of the circuit, followed by replacing series resistors
with an equivalent resistor:
Source transformations on both the right side and the left side of the circuit:
Source transformations on both the right side and the left side of the circuit:
5-11
Replacing parallel resistors with an equivalent resistor and also replacing parallel current sources
with an equivalent current source:
Finally, va =50(100)(0.21)=100(0.21)=7V 50+100 3
(checked using LNAP 8/15/02)
Finally, va =50(100)(0.21)=100(0.21)=7V 50+100 3
(checked using LNAP 8/15/02)
5-12
Section 5-4 Superposition
P5.4–1
Consider 6 A source only (open 9 A source)
Consider 9 A source only (open 6 A source)
Consider 6 A source only (open 9 A source)
Consider 9 A source only (open 6 A source)
Use current division:
v1 =6 15 ⇒v=40V 20 15 + 30 1
v1 =6 15 ⇒v=40V 20 15 + 30 1
Use current division:
v2 =9 10 ⇒v=40V
v2 =9 10 ⇒v=40V
20
∴v=v1 +v2 =40+40=80V
∴v=v1 +v2 =40+40=80V
10 + 35 2
P5.4-2
(checked using LNAP 8/15/02)
20i1 + 12+4i1 +12i1 =0 ⇒i1 =−1/3mA
20i1 + 12+4i1 +12i1 =0 ⇒i1 =−1/3mA
Consider 12 V source only (open both current sources)
KVL:
Consider 3 mA source only (short 12 V and open 9 mA sources)
KVL:
Consider 3 mA source only (short 12 V and open 9 mA sources)
Current Division:
i =3 16 = 4 mA
i =3 16 = 4 mA
2 16+20 3
5-13
Consider 9 mA source only (short 12 V and open 3
mA sources)
Current Division:
i=−9 12 =−3mA
3 24+12
∴i=i1+i2+i3 =−1/3+4/3−3=−2mA
P5.4–3
Consider 30 mA source only (open 15 mA and short 15 V sources). Let i1 be the part of i due to
the 30 mA current source.
(checked using LNAP 8/15/02)
i = 30 2 = 6 mA ⇒ i = i 6 = 2 mA
a 2+8 1
a 6+12
Consider 15 mA source only (open 30 mA source and short 15 V source) Let i2 be the part of i
Consider 15 mA source only (open 30 mA source and short 15 V source) Let i2 be the part of i
due to the 15 mA current source.
i = 15 4 = 6 mA ⇒ i = i 6 = 2 mA
b 4+6 2
b 6+12
5-14
Consider 15 V source only (open both current sources). Let i3 be the part of i due to the 15 V
voltage source.
i=−2.5 6||6 =−103=−0.5mA
3
(6||6)+12 3+12
Finally,
P5.4–4
P5.4–4
i=i1+i2+i3 =2+2−0.5=3.5mA
Consider 10 V source only (open 30 mA source and
short the 8 V source)
(checked using LNAP 8/15/02)
Letv1 bethepartofva duetothe 10 V voltage source.
v = 100||100 (10) 1 (100||100)+100
= 50 (10)=10 V 150 3
Letv2 bethepartofva duetothe 8 V voltage source.
v = 100||100 (8) 1 (100||100)+100
= 50 (8)= 8 V 150 3
Letv1 bethepartofva duetothe 10 V voltage source.
v = 100||100 (10) 1 (100||100)+100
= 50 (10)=10 V 150 3
Letv2 bethepartofva duetothe 8 V voltage source.
v = 100||100 (8) 1 (100||100)+100
= 50 (8)= 8 V 150 3
Consider 8 V source only (open 30 mA source and
short the 10 V source)
5-15
Consider 30 mA source only (short both the 10 V
source and the 8 V source)
Let v2 be the part of va due to the
30 mA current source.
v3 =(100||100||100)(0.03) = 100 (0.03) =1 V
3
(checked using LNAP 8/15/02)
Let i1 be the part of ix due to the 8 V voltage source.
Apply KVL to the supermesh: 6(i1)+3(i1)+3(i1)−8=0
i1 = 8 = 2 A 12 3
Let i2 be the part of ix due to the 2 A current source.
Apply KVL to the supermesh: 6(i2)+3(i2 +2)+3i2 =0
i2 =−6=−1 A 12 2
v3 =(100||100||100)(0.03) = 100 (0.03) =1 V
3
(checked using LNAP 8/15/02)
Let i1 be the part of ix due to the 8 V voltage source.
Apply KVL to the supermesh: 6(i1)+3(i1)+3(i1)−8=0
i1 = 8 = 2 A 12 3
Let i2 be the part of ix due to the 2 A current source.
Apply KVL to the supermesh: 6(i2)+3(i2 +2)+3i2 =0
i2 =−6=−1 A 12 2
Finally,
P5.4-5
Consider 8 V source only (open the 2 A source)
P5.4-5
Consider 8 V source only (open the 2 A source)
va =v1+v2+v3 =10+8+1=7V
33
Consider 2 A source only (short the 8 V source)
Finally,
ix =i1+i2 =2−1=1 A
326
5-16
Section 5-5: Thèvenin’s Theorem
P5.5-1
(checked using LNAP 8/15/02)
5-17
P5.5-2
The circuit from Figure P5.5-2a can be reduced to its Thevenin equivalent circuit in four steps:
(a) (b)
(c) (d)
Comparing (d) to Figure P5.5-2b shows that the Thevenin resistance is Rt = 16 Ω and the open circuit voltage, voc = −12 V.
The circuit from Figure P5.5-2a can be reduced to its Thevenin equivalent circuit in four steps:
(a) (b)
(c) (d)
Comparing (d) to Figure P5.5-2b shows that the Thevenin resistance is Rt = 16 Ω and the open circuit voltage, voc = −12 V.
5-18
P5.5-3
The circuit from Figure P5.5-3a can be reduced to its Thevenin equivalent circuit in five steps:
The circuit from Figure P5.5-3a can be reduced to its Thevenin equivalent circuit in five steps:
(b)
(c)
(a)
(d)
(e)
Comparing (e) to Figure P5.5-3b shows that the Thevenin resistance is Rt = 4 Ω and the open
circuit voltage, voc = 2 V.
(checked using LNAP 8/15/02)
(checked using LNAP 8/15/02)
5-19
P5.5-4
Find Rt:
Find Rt:
Write mesh equations to find voc:
R = 12(10+2) =6Ω
t 12+(10+2)
Mesh equations:
12i1 +10i1 −6(i2 −i1)=0
6(i2 −i1)+3i2 −18=0 28i1 =6i2
Mesh equations:
12i1 +10i1 −6(i2 −i1)=0
6(i2 −i1)+3i2 −18=0 28i1 =6i2
9i2 −6i1 =18
36i1=18 ⇒ i1=1A
2
i =141=7 A
2323
Finally,
v =3i +10i =37+101=12 V
oc 2 1 3 2
(checked using LNAP 8/15/02)
5-20
P5.5-5
Find voc:
Notice that voc is the node voltage at node a. Express the controlling voltage of the dependent source as a function of the node voltage:
va = −voc Apply KCL at node a:
−6−voc +voc +−3v =0 8 4 4 oc
−6+voc +2voc −6voc =0 ⇒ voc =−2 V Find Rt:
We’ll find isc and use it to calculate Rt. Notice that the short circuit forces
va = 0 Apply KCL at node a:
Find voc:
Notice that voc is the node voltage at node a. Express the controlling voltage of the dependent source as a function of the node voltage:
va = −voc Apply KCL at node a:
−6−voc +voc +−3v =0 8 4 4 oc
−6+voc +2voc −6voc =0 ⇒ voc =−2 V Find Rt:
We’ll find isc and use it to calculate Rt. Notice that the short circuit forces
va = 0 Apply KCL at node a:
−6−0+0+−30+i =0
8 4 4 sc
isc = 6 = 3 A
84
Rt =voc =−2=−8Ω
isc 34 3
(checked using LNAP 8/15/02)
5-21
P5.5-6
Find voc:
Apply KCL at the top, middle node: 2va −va = va +3+0 ⇒ va =18 V 36
The voltage across the right-hand 3 Ω resistor is zero so: va = voc = 18 V Find isc:
ApplyKCLatthetop,middlenode: 2va −va =va +3+va ⇒ va =−18V 363
Find voc:
Apply KCL at the top, middle node: 2va −va = va +3+0 ⇒ va =18 V 36
The voltage across the right-hand 3 Ω resistor is zero so: va = voc = 18 V Find isc:
ApplyKCLatthetop,middlenode: 2va −va =va +3+va ⇒ va =−18V 363
Apply Ohm’s law to the right-hand 3 Ω resistor :
Finally: Rt =voc =18=−3Ω isc −6
Finally: Rt =voc =18=−3Ω isc −6
isc = va = −18 = −6 V
33
(checked using LNAP 8/15/02)
(checked using LNAP 8/15/02)
5-22
P5.5-7
(b)LetR1 =R2 =1kΩ.Then
625Ω=Rt =1000 ⇒ d =1000−2=−0.4A/A
(b)LetR1 =R2 =1kΩ.Then
625Ω=Rt =1000 ⇒ d =1000−2=−0.4A/A
(a)
−vs +R1ia +(d+1)R2ia =0
i=
a
vs
R +(d+1)R
R +(d+1)R
12
v = (d+1)R2vs
oc R +(d+1)R
12
ia = vs R1
i =(d+1)i =(d+1)vs
ia = vs R1
i =(d+1)i =(d+1)vs
sc a
R1
−ia−dia+vT −iT=0
R2
R1ia =−vT
iT =(d+1)vT +vT =R2(d+1)+R1×vT
R1 R2 R1 R2
R =vT =
t
R1 R2
i R +(d+1)R
i R +(d+1)R
and
5=v =(d+1)vs ⇒ v =−0.4+25=13.33V
oc d +2 s −0.4+1
(checked using LNAP 8/15/02)
(checked using LNAP 8/15/02)
d+2 625
T12
5-23
P5.5-8
From the given data:
6= 2000 v
R+2000 oc v =1.2V
t ⇒ oc
R =−1600Ω
2=vt
4000
R+4000 oc
P5.5-9
v =
R voc
Rt +R
v = 8000 (1.2)=1.5 V
−1600 + 8000
t
When R = 8000 Ω,
From the given data:
0.004= voc
R +2000
t
v =24V
⇒ oc
i= voc
Rt +R
⇒
24 =0.006=6 mA ⇒ i≤6 mA
24 =0.006=6 mA ⇒ i≤6 mA
0.003= oc
t
(a) When i = 0.002 A:
0.002 = 24
v
R =4000Ω
t
R +4000
R = 8000 Ω
(b) Maximum i occurs when R = 0:
P5.5-10
The current at the point on the plot where v = 0 is the short circuit current, so isc = 20 mA.
The voltage at the point on the plot where i = 0 is the open circuit voltage, so voc = −3 V.
The slope of the plot is equal to the negative reciprocal of the Thevenin resistance, so
−1=0−0.002 ⇒ Rt=−150Ω Rt −3−0
The slope of the plot is equal to the negative reciprocal of the Thevenin resistance, so
−1=0−0.002 ⇒ Rt=−150Ω Rt −3−0
4000+R
4000
5-24
P5.5-11
−12+6000ia +2000ia +1000ia =0
ia = 4 3000 A
voc =1000 ia = 4 V 3
ia = 0 due to the short circuit −12+6000isc =0 ⇒ isc =2 mA
voc =1000 ia = 4 V 3
ia = 0 due to the short circuit −12+6000isc =0 ⇒ isc =2 mA
v4
Rt= oc=3=667Ω
Rt= oc=3=667Ω
isc .002
4
4
ib= 3
667+R
ib = 0.002 A requires 4
R= 3 −667=0 0.002
(checked using LNAP 8/15/02)
ib = 0.002 A requires 4
R= 3 −667=0 0.002
(checked using LNAP 8/15/02)
5-25
P5.5-12
10=i+0 ⇒ i=10A
voc +4i−2i=0
⇒ voc=−2i=−20V i+isc =10 ⇒ i=10−isc
4i+0−2i=0 ⇒ i=0 ⇒ isc =10A Rt =voc =−20=−2Ω
−2=iL=−20 ⇒RL=12Ω RL −2
(checked using LNAP 8/15/02)
⇒ voc=−2i=−20V i+isc =10 ⇒ i=10−isc
4i+0−2i=0 ⇒ i=0 ⇒ isc =10A Rt =voc =−20=−2Ω
−2=iL=−20 ⇒RL=12Ω RL −2
(checked using LNAP 8/15/02)
isc 10
5-26
Section 5-6: Norton’s Theorem
P5.6-1
When the terminals of the boxes are open-circuited, no current flows in Box A, but the resistor in Box B dissipates 1 watt. Box B is therefore warmer than Box A. If you short the terminals of each box, the resistor in Box A will draw 1 amp and dissipate 1 watt. The resistor in Box B will be shorted, draw no current, and dissipate no power. Then Box A will warm up and Box B will cool off.
When the terminals of the boxes are open-circuited, no current flows in Box A, but the resistor in Box B dissipates 1 watt. Box B is therefore warmer than Box A. If you short the terminals of each box, the resistor in Box A will draw 1 amp and dissipate 1 watt. The resistor in Box B will be shorted, draw no current, and dissipate no power. Then Box A will warm up and Box B will cool off.
P5.6-2
(checked using LNAP 8/16/02)
5-27
P5.6-3
P5.6-4
To determine the value of the short circuit current, isc, we connect a short circuit across the terminals of the circuit and then calculate the value of the current in that short circuit. Figure (a) shows the circuit from Figure 5.6-4a after adding the short circuit and labeling the short circuit current. Also, the meshes have been identified and labeled in anticipation of writing mesh equations. Let i1 and i2 denote the mesh currents in meshes 1 and 2, respectively.
In Figure (a), mesh current i2 is equal to the current in the short circuit. Consequently, i2 = isc . The controlling current of the CCVS is expressed in terms of the mesh currents as
P5.6-4
To determine the value of the short circuit current, isc, we connect a short circuit across the terminals of the circuit and then calculate the value of the current in that short circuit. Figure (a) shows the circuit from Figure 5.6-4a after adding the short circuit and labeling the short circuit current. Also, the meshes have been identified and labeled in anticipation of writing mesh equations. Let i1 and i2 denote the mesh currents in meshes 1 and 2, respectively.
In Figure (a), mesh current i2 is equal to the current in the short circuit. Consequently, i2 = isc . The controlling current of the CCVS is expressed in terms of the mesh currents as
ia =i1−i2 =i1−isc
3i1−2(i1−i2)+6(i1−i2)−10=0 ⇒ 7i1−4i2 =10 (1)
Apply KVL to mesh 2 to get
5i2−6(i1−i2)=0 ⇒ −6i1+11i2=0 ⇒ i1=11i2 6
Substituting into equation 1 gives
711i −4i =10 ⇒ i =1.13A ⇒ i =1.13A
3i1−2(i1−i2)+6(i1−i2)−10=0 ⇒ 7i1−4i2 =10 (1)
Apply KVL to mesh 2 to get
5i2−6(i1−i2)=0 ⇒ −6i1+11i2=0 ⇒ i1=11i2 6
Substituting into equation 1 gives
711i −4i =10 ⇒ i =1.13A ⇒ i =1.13A
Apply KVL to mesh 1 to get
622 2 sc
5-28
Figure (a) Calculating the short circuit current, isc, using mesh equations.
To determine the value of the Thevenin resistance, Rt, first replace the 10 V voltage source by a 0 V voltage source, i.e. a short circuit. Next, connect a current source across the terminals of the circuit and then label the voltage across that current source as shown in Figure (b). The Thevenin resistance will be calculated from the current and voltage of the current source as
Rt =vT iT
In Figure (b), the meshes have been identified and labeled in anticipation of writing mesh equations. Let i1 and i2 denote the mesh currents in meshes 1 and 2, respectively.
In Figure (b), mesh current i2 is equal to the negative of the current source current. Consequently, i2 = iT . The controlling current of the CCVS is expressed in terms of the mesh
Apply KVL to mesh 2 to get
5i2+vT−6(i1−i2)=0 ⇒ −6i1+11i2=−vT
To determine the value of the Thevenin resistance, Rt, first replace the 10 V voltage source by a 0 V voltage source, i.e. a short circuit. Next, connect a current source across the terminals of the circuit and then label the voltage across that current source as shown in Figure (b). The Thevenin resistance will be calculated from the current and voltage of the current source as
Rt =vT iT
In Figure (b), the meshes have been identified and labeled in anticipation of writing mesh equations. Let i1 and i2 denote the mesh currents in meshes 1 and 2, respectively.
In Figure (b), mesh current i2 is equal to the negative of the current source current. Consequently, i2 = iT . The controlling current of the CCVS is expressed in terms of the mesh
Apply KVL to mesh 2 to get
5i2+vT−6(i1−i2)=0 ⇒ −6i1+11i2=−vT
currents as
Apply KVL to mesh 1 to get
Apply KVL to mesh 1 to get
ia =i1−i2 =i1+iT
3i1−2(i1−i2)+6(i1−i2)=0 ⇒ 7i1−4i2=0 ⇒ i1=4i2 (2)
7
3i1−2(i1−i2)+6(i1−i2)=0 ⇒ 7i1−4i2=0 ⇒ i1=4i2 (2)
7
Substituting for i1 using equation 2 gives
−64i +11i =−v ⇒ 7.57i =−v
−64i +11i =−v ⇒ 7.57i =−v
Finally,
722T 2T
Rt =vT =−vT =−vT =7.57Ω iT −iT i2
Rt =vT =−vT =−vT =7.57Ω iT −iT i2
5-29
Figure (b) Calculating the Thevenin resistance, Rt = vT , using mesh equations.
iT
To determine the value of the open circuit voltage, voc, we connect an open circuit across the terminals of the circuit and then calculate the value of the voltage across that open circuit. Figure (c) shows the circuit from Figure 4.6-4a after adding the open circuit and labeling the open circuit voltage. Also, the meshes have been identified and labeled in anticipation of writing mesh equations. Let i1 and i2 denote the mesh currents in meshes 1 and 2, respectively.
In Figure (c), mesh current i2 is equal to the current in the open circuit. Consequently, i2 = 0 A . The controlling current of the CCVS is expressed in terms of the mesh currents as
To determine the value of the open circuit voltage, voc, we connect an open circuit across the terminals of the circuit and then calculate the value of the voltage across that open circuit. Figure (c) shows the circuit from Figure 4.6-4a after adding the open circuit and labeling the open circuit voltage. Also, the meshes have been identified and labeled in anticipation of writing mesh equations. Let i1 and i2 denote the mesh currents in meshes 1 and 2, respectively.
In Figure (c), mesh current i2 is equal to the current in the open circuit. Consequently, i2 = 0 A . The controlling current of the CCVS is expressed in terms of the mesh currents as
ia =i1 −i2 =i1 −0=i1
3i1 −2(i1 −i2 )+6(i1 −i2 )−10=0 ⇒ 3i1 −2(i1 −0)+6(i1 −0)−10=0
⇒ i1=10=1.43A 7
Apply KVL to mesh 2 to get
5i2+voc−6(i1−i2)=0 ⇒ voc=6(i1)=6(1.43)=8.58V
Figure (c) Calculating the open circuit voltage, voc, using mesh equations. Asacheck,noticethatRt isc =(7.57)(1.13)=8.55≈voc
(checked using LNAP 8/16/02) 5-30
3i1 −2(i1 −i2 )+6(i1 −i2 )−10=0 ⇒ 3i1 −2(i1 −0)+6(i1 −0)−10=0
⇒ i1=10=1.43A 7
Apply KVL to mesh 2 to get
5i2+voc−6(i1−i2)=0 ⇒ voc=6(i1)=6(1.43)=8.58V
Figure (c) Calculating the open circuit voltage, voc, using mesh equations. Asacheck,noticethatRt isc =(7.57)(1.13)=8.55≈voc
(checked using LNAP 8/16/02) 5-30
Apply KVL to mesh 1 to get
P5.6-5
To determine the value of the short circuit current, Isc, we connect a short circuit across the terminals of the circuit and then calculate the value of the current in that short circuit. Figure (a) shows the circuit from Figure 4.6-5a after adding the short circuit and labeling the short circuit current. Also, the nodes have been identified and labeled in anticipation of writing node equations. Let v1, v2 and v3 denote the node voltages at nodes 1, 2 and 3, respectively.
In Figure (a), node voltage v1 is equal to the negative of the voltage source voltage. Consequently, v1 = −24 V . The voltage at node 3 is equal to the voltage across a short, v3 = 0 .
The controlling voltage of the VCCS, va, is equal to the node voltage at node 2, i.e. va = v2 . The voltage at node 3 is equal to the voltage across a short, i.e. v3 = 0 .
To determine the value of the short circuit current, Isc, we connect a short circuit across the terminals of the circuit and then calculate the value of the current in that short circuit. Figure (a) shows the circuit from Figure 4.6-5a after adding the short circuit and labeling the short circuit current. Also, the nodes have been identified and labeled in anticipation of writing node equations. Let v1, v2 and v3 denote the node voltages at nodes 1, 2 and 3, respectively.
In Figure (a), node voltage v1 is equal to the negative of the voltage source voltage. Consequently, v1 = −24 V . The voltage at node 3 is equal to the voltage across a short, v3 = 0 .
The controlling voltage of the VCCS, va, is equal to the node voltage at node 2, i.e. va = v2 . The voltage at node 3 is equal to the voltage across a short, i.e. v3 = 0 .
Apply KCL at node 2 to get
v1−v2=v2−v3 ⇒ 2v1+v3=3v2 36
Apply KCL at node 3 to get
v2 −v3 +4v =i ⇒ 9v =i
Figure (a) Calculating the short circuit current, Isc, using mesh equations.
To determine the value of the Thevenin resistance, Rth, first replace the 24 V voltage source by a 0 V voltage source, i.e. a short circuit. Next, connect a current source circuit across the terminals of the circuit and then label the voltage across that current source as shown in Figure (b). The Thevenin resistance will be calculated from the current and voltage of the current source as
Rth =vT iT
Also, the nodes have been identified and labeled in anticipation of writing node equations. Let v1, v2 and v3 denote the node voltages at nodes 1, 2 and 3, respectively.
v1−v2=v2−v3 ⇒ 2v1+v3=3v2 36
Apply KCL at node 3 to get
v2 −v3 +4v =i ⇒ 9v =i
Figure (a) Calculating the short circuit current, Isc, using mesh equations.
To determine the value of the Thevenin resistance, Rth, first replace the 24 V voltage source by a 0 V voltage source, i.e. a short circuit. Next, connect a current source circuit across the terminals of the circuit and then label the voltage across that current source as shown in Figure (b). The Thevenin resistance will be calculated from the current and voltage of the current source as
Rth =vT iT
Also, the nodes have been identified and labeled in anticipation of writing node equations. Let v1, v2 and v3 denote the node voltages at nodes 1, 2 and 3, respectively.
6 32 sc 6a sc
⇒ −48=3va ⇒ va=−16V
⇒ i =9(−16)=−24A sc 6
⇒ i =9(−16)=−24A sc 6
5-31
In Figure (b), node voltage v1 is equal to the across a short circuit, i.e. v1 = 0 . The
controlling voltage of the VCCS, va, is equal to the node voltage at node 2, i.e. va = v2 . The
voltage at node 3 is equal to the voltage across the current source, i.e. v3 = vT .
Apply KCL at node 2 to get
v1−v2=v2−v3 ⇒ 2v1+v3=3v2 ⇒ vT=3va 36
Apply KCL at node 3 to get
Apply KCL at node 2 to get
v1−v2=v2−v3 ⇒ 2v1+v3=3v2 ⇒ vT=3va 36
Apply KCL at node 3 to get
v2−v3+4v2+iT=0 ⇒ 9v2−v3+6iT=0
63
⇒ 9va −vT +6iT =0
⇒ 3vT −vT +6iT =0 ⇒ 2vT =−6iT
Rt =vT =−3Ω iT
⇒ 9va −vT +6iT =0
⇒ 3vT −vT +6iT =0 ⇒ 2vT =−6iT
Rt =vT =−3Ω iT
Finally,
To determine the value of the open circuit voltage, voc, we connect an open circuit across the terminals of the circuit and then calculate the value of the voltage across that open circuit. Figure (c) shows the circuit from Figure P 4.6-5a after adding the open circuit and labeling the open circuit voltage. Also, the nodes have been identified and labeled in anticipation of writing node equations. Let v1, v2 and v3 denote the node voltages at nodes 1, 2 and 3, respectively.
In Figure (c), node voltage v1 is equal to the negative of the voltage source voltage. Consequently, v1 = −24 V . The controlling voltage of the VCCS, va, is equal to the node voltage
at node 2, i.e. va = v2 . The voltage at node 3 is equal to the open circuit voltage, i.e. v3 = voc . Apply KCL at node 2 to get
To determine the value of the open circuit voltage, voc, we connect an open circuit across the terminals of the circuit and then calculate the value of the voltage across that open circuit. Figure (c) shows the circuit from Figure P 4.6-5a after adding the open circuit and labeling the open circuit voltage. Also, the nodes have been identified and labeled in anticipation of writing node equations. Let v1, v2 and v3 denote the node voltages at nodes 1, 2 and 3, respectively.
In Figure (c), node voltage v1 is equal to the negative of the voltage source voltage. Consequently, v1 = −24 V . The controlling voltage of the VCCS, va, is equal to the node voltage
at node 2, i.e. va = v2 . The voltage at node 3 is equal to the open circuit voltage, i.e. v3 = voc . Apply KCL at node 2 to get
Figure (b) Calculating the Thevenin resistance, Rth = vT , using mesh equations.
iT
5-32
v1−v2=v2−v3 ⇒ 2v1+v3=3v2 ⇒ −48+voc=3va
36
Apply KCL at node 3 to get
v2−v3+4v2=0 ⇒ 9v2−v3=0 ⇒ 9va=voc 63
Combining these equations gives
3(−48+voc)=9va =voc ⇒ voc =72V
Apply KCL at node 3 to get
v2−v3+4v2=0 ⇒ 9v2−v3=0 ⇒ 9va=voc 63
Combining these equations gives
3(−48+voc)=9va =voc ⇒ voc =72V
Figure (c) Calculating the open circuit voltage, voc, using node equations.
As a check, notice that
Rth Isc =(−3)(−24)=72=Voc
(checked using LNAP 8/16/02)
Section 5-7: Maximum Power Transfer P5.7-1
a) For maximum power transfer, set RL equal to the Thevenin resistance:
RL =Rt =100+1=101Ω
b) To calculate the maximum power, first replace the circuit connected to RL be its Thevenin equivalent circuit:
Rth Isc =(−3)(−24)=72=Voc
(checked using LNAP 8/16/02)
Section 5-7: Maximum Power Transfer P5.7-1
a) For maximum power transfer, set RL equal to the Thevenin resistance:
RL =Rt =100+1=101Ω
b) To calculate the maximum power, first replace the circuit connected to RL be its Thevenin equivalent circuit:
5-33
vL = 101 (100) = 50 V
101 + 101
v2 502
pmax = L = =24.75W
Reduce the circuit using source transformations:
Then (a) maximum power will be dissipated in resistor R when: R = Rt = 60 Ω and (b) the value of that maximum power is
P =i2(R) = (0.03)2(60) =54mW
v2 502
pmax = L = =24.75W
Reduce the circuit using source transformations:
Then (a) maximum power will be dissipated in resistor R when: R = Rt = 60 Ω and (b) the value of that maximum power is
P =i2(R) = (0.03)2(60) =54mW
The voltage across RL is
Then
P5.7-2
P5.7-2
RL 101
max
R
5-34
P5.7-3
v=vRL
L SR+R
SL
v2 v2R
∴p=L=SL
L RL (RS +RL)2
By inspection, pL is max when you reduce RS to get the
smallest denominator.
∴setRS =0
∴setRS =0
P5.7-4
Find Rt by finding isc and voc:
Find Rt by finding isc and voc:
The current in the 3 Ω resistor is zero because of the short circuit. Consequently, isc = 10 ix.
Apply KCL at the top-left node to get
ix +0.9=10ix ⇒ ix =0.9=0.1A 9
isc =10ix =1A
ix +0.9=10ix ⇒ ix =0.9=0.1A 9
isc =10ix =1A
so
Next
Apply KCL at the top-left node to get
5-35
ix +0.9=10ix ⇒ ix =0.9=0.1A
9
Apply Ohm’s law to the 3 Ω resistor to get
voc =3(10ix)=30(0.1)=3V
For maximum power transfer to RL:
RL =Rt =voc =3=3Ω
Apply Ohm’s law to the 3 Ω resistor to get
voc =3(10ix)=30(0.1)=3V
For maximum power transfer to RL:
RL =Rt =voc =3=3Ω
isc 1
The maximum power delivered to RL is given by
The required value of R is
R=R =8+ (20+120)(10+50) =50Ω t (20+120)+(10+50)
R=R =8+ (20+120)(10+50) =50Ω t (20+120)+(10+50)
oc
170+30
200 200
v2 32 3
p = oc = = W
p = oc = = W
P5.7-5
max
4R 4(3) 4
t
v = 170 (20)10− 30 (20)50
170+30
=170(20)(10)−30(20)(50)=4000=20V
The maximum power is given by
v2 202
p=oc= =2W
max
4 R
4 (50)
t
5-36
PSpice Problems
SP5-1
(a)
(b)
(b)
vo = 0.3333 v1 + 0.3333 v2 + 33.33 i3
7−18
7=0.3333(10)+0.3333(8)+33.33i3 ⇒ i3 = 3 = 3 =30mA
100 100
a = 0.3333
b = 0.3333
c =33.33 V/A
3
5-37
SP5-2
Before the source transformation:
Before the source transformation:
NAME
V_V1 V_V2
After the source transformation:
V_V1 V_V2
After the source transformation:
CURRENT
-3.000E-02 -4.000E-02
-3.000E-02 -4.000E-02
VOLTAGE SOURCE CURRENTS
VOLTAGE SOURCE CURRENTS
NAME CURRENT
V_V2 -4.000E-02
V_V2 -4.000E-02
5-38
SP5-3
voc = −2 V
VOLTAGE SOURCE CURRENTS
NAME
V_V3 V_V4
V_V3 V_V4
CURRENT
-7.500E-01 7.500E-01
isc = 0.75 A
Rt = −2.66 Ω
-7.500E-01 7.500E-01
isc = 0.75 A
Rt = −2.66 Ω
5-39
SP5-4
voc = 8.571 V
VOLTAGE SOURCE CURRENTS
VOLTAGE SOURCE CURRENTS
NAME
CURRENT
V_V5
V_V6
X_H1.VH_H1 9.434E-01
V_V6
X_H1.VH_H1 9.434E-01
-2.075E+00
1.132E+00
isc = 1.132 A
Rt = 7.571 Ω
5-40
Verification Problems
VP5-1
Use the data in the first two lines of the table to determine voc
and Rt:
0.0972= voc R+0
0.0972= voc R+0
0.0438= oc
t
v =39.9V
⇒ oc
v
R=410Ω
t
R +500
Now check the third line of the table. When R= 5000 Ω:
i = voc = 39.9 = 7.37 mA Rt +R 410+5000
which disagree with the data in the table.
The data is not consistent.
Use the data in the table to determine voc and isc: voc =12V (line1ofthetable)
Now check the third line of the table. When R= 5000 Ω:
i = voc = 39.9 = 7.37 mA Rt +R 410+5000
which disagree with the data in the table.
The data is not consistent.
Use the data in the table to determine voc and isc: voc =12V (line1ofthetable)
t
VP5-2
isc =3mA
so Rt =voc =4kΩ
isc
Next, check line 2 of the table. When R = 10 kΩ:
i = voc = 12 = 0.857 mA
so Rt =voc =4kΩ
isc
Next, check line 2 of the table. When R = 10 kΩ:
i = voc = 12 = 0.857 mA
(line3ofthetable)
Rt+R 10103 +5103
()()
which agrees with the data in the table.
0.001 = i = voc = 12 ⇒ R = 8000 Ω
0.001 = i = voc = 12 ⇒ R = 8000 Ω
To cause i = 1 mA requires
I agree with my lab partner’s claim that R = 8000 causes i = 1 mA.
I agree with my lab partner’s claim that R = 8000 causes i = 1 mA.
Rt+R 10103 +R
()
5-41
VP5-3
Design Problems
DP5-1
The equation of representing the straight line in Figure DP 5-1b is v = −Rt i + voc . That is, the slope of the line is equal to -1 times the Thevenin resistance and the "v - intercept" is equal to the
opencircuitvoltage.Therefore:Rt =− 0−5 =625Ω andvoc =5V. 0.008 − 0
Try R1 =R2 =1kΩ.(R1 ||R2 mustbesmallerthanRt =625Ω.)Then 5= R2 vs =1vs ⇒vs =10V
Design Problems
DP5-1
The equation of representing the straight line in Figure DP 5-1b is v = −Rt i + voc . That is, the slope of the line is equal to -1 times the Thevenin resistance and the "v - intercept" is equal to the
opencircuitvoltage.Therefore:Rt =− 0−5 =625Ω andvoc =5V. 0.008 − 0
Try R1 =R2 =1kΩ.(R1 ||R2 mustbesmallerthanRt =625Ω.)Then 5= R2 vs =1vs ⇒vs =10V
60 60
i=voc =611 =11=54.55mA
11
i=voc =611 =11=54.55mA
11
Rt +R
The measurement supports the prelab calculation.
The measurement supports the prelab calculation.
(110)+40 60+40
R1 +R2 2
12
and
Now vs, R1, R2 and R3 have all been specified so the design is complete.
DP5-2
The equation of representing the straight line in Figure DP 5-2b is v = −Rt i + voc . That is, the slope of the line is equal to -1 times the Thevenin resistance and the "v - intercept" is equal to the
open circuit voltage. Therefore: Rt = − 0 − (−3) = 500 Ω and voc = −3 V. −0.006 − 0
Now vs, R1, R2 and R3 have all been specified so the design is complete.
DP5-2
The equation of representing the straight line in Figure DP 5-2b is v = −Rt i + voc . That is, the slope of the line is equal to -1 times the Thevenin resistance and the "v - intercept" is equal to the
open circuit voltage. Therefore: Rt = − 0 − (−3) = 500 Ω and voc = −3 V. −0.006 − 0
625=R + RR =R +500 ⇒ R =125Ω
3R+R3 3
12
From the circuit we calculate
Rt =R3(R1+R2)andvoc =− R1R3
R1 +R2 +R3 R1 +R2 +R3
Rt =R3(R1+R2)andvoc =− R1R3
R1 +R2 +R3 R1 +R2 +R3
is
so
500Ω=R3(R1 +R2)and−3V=− R1R3
R1 +R2 +R3 R1 +R2 +R3
R1 +R2 +R3 R1 +R2 +R3
is
5-42
Try R3 =1kΩ and R1 +R2 =1kΩ. Then Rt =500Ωand
−3=−1000R1 is =−R1 is ⇒6=R1is
2000 2
ThisequationcanbesatisfiedbytakingR1 =600Ωandis =10mA.Finally,R2 =1kΩ-400Ω=
ThisequationcanbesatisfiedbytakingR1 =600Ωandis =10mA.Finally,R2 =1kΩ-400Ω=
400 Ω. Now is, R1, R2 and R3 have all been specified so the design is complete.
DP5-3
The slope of the graph is positive so the Thevenin resistance is negative. This would require
R3 + R1R2 < 0 , which is not possible since R1, R2 and R3 will all be non-negative.
R1 +R2
Is it not possible to specify values of vs, R1, R2 and R3 that cause the current i and the
Is it not possible to specify values of vs, R1, R2 and R3 that cause the current i and the
voltage v in Figure DP 5-3a to satisfy the relationship described by the graph in Figure DP 5-3b.
DP5-4
The equation of representing the straight line in Figure DP 5-4b is v = −Rt i + voc . That is, the slope of the line is equal to the Thevenin impedance and the "v - intercept" is equal to the open
circuit voltage. Therefore: Rt = − −5 − 0 = −625 Ω and voc = −5 V. 0 − 0.008
The open circuit voltage, voc, the short circuit current, isc, and the Thevenin resistance, Rt, of this circuit are given by
DP5-4
The equation of representing the straight line in Figure DP 5-4b is v = −Rt i + voc . That is, the slope of the line is equal to the Thevenin impedance and the "v - intercept" is equal to the open
circuit voltage. Therefore: Rt = − −5 − 0 = −625 Ω and voc = −5 V. 0 − 0.008
The open circuit voltage, voc, the short circuit current, isc, and the Thevenin resistance, Rt, of this circuit are given by
v = R2(d+1) v
oc
oc
R +(d+1)R s
12
and
LetR1 =R2 =1kΩ.Then
LetR1 =R2 =1kΩ.Then
R1
R= R1R2
,
isc =(d+1)vs
t R +(d+1)R
12
and
−625Ω=Rt =1000 ⇒ d=1000−2=−3.6A/A
d+2 −625
−5=(d+1)vs ⇒ v =−3.6+2(−5)=−3.077V d + 2 s −3.6 +1
−5=(d+1)vs ⇒ v =−3.6+2(−5)=−3.077V d + 2 s −3.6 +1
Now vs, R1, R2 and d have all been specified so the design is complete.
5-43
Exercises
Ex. 6.4-1
Chapter 6: The Operational Amplifier
vs +vs −vo +0 = 0
RR
12
vo = 1+ R2
vo = 1+ R2
Ex. 6.4-2
vR
s1
a)
a)
va=R2 vs
R+R
12
va +va −v0 +0=0
va +va −v0 +0=0
1+R4
v R vR+RR
vo=1+R4 ⇒ vo= R2
a 3 s123
a 3 s123
R3 R4
b) WhenR>>R then R2 −R2 =1andvo −1+R4
21R+R Rv R
122s3
6-1
Ex. 6.5-1
vo +vo −vs +0=0
RR
21
vo= R2
vR+R s12
vR+R s12
Ex. 6.6–1
Rf
R R out Rin
R R out Rin
v−v v
in out+in+0=0⇒v=1+ v
in out+in+0=0⇒v=1+ v
f11
when Rf =100kΩ and R1 =25kΩ then
vout = 1+100⋅103 = 5
v 25⋅103
in
6-2
Ex. 6.7-1
10×103 R2 10×103 R1
v3 =− 3 3 v2 +1+ 3 1+
10×10 R2 +10×10 10×10
3 v1
10×10
R2 R1
=− 3 v2 +21+
3 v1
10×10
R2 +10×10
Werequire v =(4)v −1v ,so
3152
4=21+ R1
10×103 1
10×103 1
⇒ R =10×103 =10kΩ
and
R2 ⇒ R +10×103 =5R ⇒ R =2.5kΩ
1=
5 R2+10×103 2 2 2
5 R2+10×103 2 2 2
6-3
Ex. 6.7-2
As in Ex 6.7-1
As in Ex 6.7-1
R2 R1
v3 =− 3 v2 +21+
R2 +10×10
3 v1
10×10
Werequire v =(6)v −4v ,so
3152
6=21+ R1 ⇒ R =20×103 =20kΩ
10×103 1
and
Ex. 6.8-1
Ex. 6.8-1
R2 ⇒ 4R +40×103 =5R ⇒ R =40kΩ
4=
5 R2+10×103 2 2 2
5 R2+10×103 2 2 2
Analysis of the circuit in Section 6.7 showed that output offset voltage = 6 v
os
ForaμA741opamp, vos ≤ 1mVand ib1 ≤ 80nAso
ForaμA741opamp, vos ≤ 1mVand ib1 ≤ 80nAso
+ (50×103 ) i
b1
−3 3−9
output offset voltage = 6 v + (50×103 )i ≤ 6 (10 )+(50.10 )(80×10 ) = 10 mV
output offset voltage = 6 v + (50×103 )i ≤ 6 (10 )+(50.10 )(80×10 ) = 10 mV
Ex. 6.8-2
os b1
v = −R2 v +1+R2v +Ri
o RinRos2b1
11
WhenR =10kΩ, R = 2kΩ, v ≤ 5mVand i ≤ 500nAthen
WhenR =10kΩ, R = 2kΩ, v ≤ 5mVand i ≤ 500nAthen
2 1 os b1
output offset voltage ≤ 6 5×10 + 10×10 500.10 ≤ 35×10−3 = 35 mV
( −3)( 3)( −9)
6-4
Ex. 6.8-3
Analysis of this circuit in Section 6.7 showed that output offset voltage = 6 v os
For a typical OPA1O1AM, vos = 0.1 mV and ib = 0.012 nA so
Analysis of this circuit in Section 6.7 showed that output offset voltage = 6 v os
For a typical OPA1O1AM, vos = 0.1 mV and ib = 0.012 nA so
+ (50 ×103 ) i
b1
Ex. 6.8-4
output offset voltage
≤ 6 0.1×10−3 + (50×103 )0.012×10−9
≤ 0.6×10−3 + 0.6×10−6 − 0.6×10−3 = 0.6 mV
Writing node equations
After some algebra
After some algebra
v−−vs+v−−vo+ v− =0
Ra Rb Ri +Rs
Ri
vo −−A
v−
R+Rv−v
is+o−=0
R0 Rb
R0(Ri +Rs)+ARiRf
A =vo =
vs Rf +R0 R+R +R Rf +R0+Ri+Rs −ARiRa
vs Rf +R0 R+R +R Rf +R0+Ri+Rs −ARiRa
v ()(is)a( )
Forthegivenvalues,Av =−2.00006V/V.
6-5
Problems
Section 6-4: The Ideal Operational Amplifier P6.4-1
Section 6-4: The Ideal Operational Amplifier P6.4-1
P6.4-2
(checked using LNAP 8/16/02)
Apply KVL to loop 1:
−12+3000i +0+2000i = 0 11
amp are zero so
i = i = 2.4mA o1
i = −i = −2.4mA 21
va =i2(1000)+0=−2.4V Apply Ohm’s law to the 4 kΩ resistor
vo =va −io (4000)
= −2.4−(2.4×10−3 )(4000)= −12 V
(checked using LNAP 8/16/02)
−12+3000i +0+2000i = 0 11
amp are zero so
i = i = 2.4mA o1
i = −i = −2.4mA 21
va =i2(1000)+0=−2.4V Apply Ohm’s law to the 4 kΩ resistor
vo =va −io (4000)
= −2.4−(2.4×10−3 )(4000)= −12 V
(checked using LNAP 8/16/02)
⇒i= 12 =2.4mA
1
5000
The currents into the inputs of an ideal op
The currents into the inputs of an ideal op
6-6
P6.4-3
The voltages at the input nodes of an ideal op amp
areequalso va =−2V.
Apply KCL at node a:
vo −(−2)+12−(−2)=0 ⇒v =−30 V
o
io =−2−vo =3.5mA 8000
(checked using LNAP 8/16/02) The voltages at the input nodes of an ideal
opampareequalso v=5V.
Apply KCL at the inverting input node of
the op amp:
Apply KCL at node a:
vo −(−2)+12−(−2)=0 ⇒v =−30 V
o
io =−2−vo =3.5mA 8000
(checked using LNAP 8/16/02) The voltages at the input nodes of an ideal
opampareequalso v=5V.
Apply KCL at the inverting input node of
the op amp:
8000 4000
Apply Ohm’s law to the 8 kΩ resistor
Apply Ohm’s law to the 8 kΩ resistor
P6.4-4
−va−5−0.1×10−3−0=0 ⇒ v=4V
10000 a
3000 4000
Apply Ohm’s law to the 20 kΩ resistor
Apply Ohm’s law to the 20 kΩ resistor
i=va =1mA
20000 5
(checked using LNAP 8/16/02)
(checked using LNAP 8/16/02)
P6.4-5
The voltages at the input nodes of an ideal op amp are equal so
va =0V.ApplyKCLatnodea: −vo −0−12−0−2⋅10−3 =0
The voltages at the input nodes of an ideal op amp are equal so
va =0V.ApplyKCLatnodea: −vo −0−12−0−2⋅10−3 =0
⇒vo =−15V
Apply KCL at the output node of
the op amp:
io+ vo + vo =0⇒io=7.5mA 6000 3000
Apply KCL at the output node of
the op amp:
io+ vo + vo =0⇒io=7.5mA 6000 3000
(checked using LNAP 8/16/02)
6-7
P6.4-6
The currents into the inputs of an ideal op amp are zero and the voltages at the input nodes of anidealopampareequalso va =2.5V.
The currents into the inputs of an ideal op amp are zero and the voltages at the input nodes of anidealopampareequalso va =2.5V.
Apply Ohm’s law to the 4 kΩ resistor:
ia = va = 2.5 =0.625mA
4000 4000
Apply KCL at node a:
ib =ia =0.625mA
Apply KVL:
vo =8000ib +4000ia
= (12×103 )(0.625×10−3 )= 7.5 V
P6.4-7
vo =8000ib +4000ia
= (12×103 )(0.625×10−3 )= 7.5 V
P6.4-7
(checked using LNAP 8/16/02)
v −0 v −0
− s − a +0=0 ⇒ va =−
− s − a +0=0 ⇒ va =−
R2
RRR
vs
12 1
0−v 0−v R2 +R3 R2 +R3
i= a+ a=− v= v
oRRRRaRRs
232313
v −0 v −0 R4
− o − a +0=0 ⇒ v =−
v =
R2R4
v
RRoRaRRs
43 3 13
6-8
P6.4-8
The node voltages have been labeled using:
1. The currents into the inputs of an ideal op amp are zero and the voltages at the input nodes of an ideal op amp are equal.
2. KCL
3. Ohm’s law Then
v0 =11.8−1.8=10V and
io = 10 =2.5mA 4000
P6.4-9
KCL at node a:
The node voltages have been labeled using:
1. The currents into the inputs of an ideal op amp are zero and the voltages at the input nodes of an ideal op amp are equal.
2. KCL
3. Ohm’s law Then
v0 =11.8−1.8=10V and
io = 10 =2.5mA 4000
P6.4-9
KCL at node a:
(checked using LNAP 8/16/02)
va −(−18)+ va
4000 8000
+0=0 ⇒ va =−12V
The node voltages at the input nodes of
ideal op amps are equal so vb = va .
Voltage division:
vo = 8000 vb =−8V 4000 + 8000
vo = 8000 vb =−8V 4000 + 8000
(check using LNAP 8/16/02)
6-9
Section 6-5: Nodal Analysis of Circuits Containing Ideal Operational Amplifiers
P6.5-1
KCLatnodeb: vb−2+ vb +vb+5=0 ⇒ vb=−1V 20000 40000 40000 4
Thenodevoltagesattheinputnodesofanidealopampareequalsove =vb =−1 V. 4
KCLatnodeb: vb−2+ vb +vb+5=0 ⇒ vb=−1V 20000 40000 40000 4
Thenodevoltagesattheinputnodesofanidealopampareequalsove =vb =−1 V. 4
KCLatnodee:
ve +ve −vd =0 ⇒ vd =10ve =−10 V
1000 9000 4
P6.5-2
Apply KCL at node a:
0=va−12+ va +va−0 ⇒va=4V
6000 6000 6000
Apply KCL at the inverting input of the op amp:
Apply KCL at node a:
0=va−12+ va +va−0 ⇒va=4V
6000 6000 6000
Apply KCL at the inverting input of the op amp:
(checked using LNAP 8/16/02)
−va −0+0+0−vo = 0
6000 6000
⇒ vo =−va =−4V
Apply KCL at the output of the op amp:
i−0−vo+ vo =0 o 6000 6000
i−0−vo+ vo =0 o 6000 6000
(checked using LNAP 8/16/02)
⇒i=− vo
o
o
=1.33mA
3000
6-10
P6.5-3
Apply KCL at node a:
va−v0+va +va−0 =0⇒v =R1+1+1v= R2R3+R2R4+R3R4v
va−v0+va +va−0 =0⇒v =R1+1+1v= R2R3+R2R4+R3R4v
RRR04RRRaRRa
432 43223
Apply KCL at the inverting input of
the op amp:
−va −0−vs −0 = 0 RR 21
⇒va =−R2 vs R
−va −0−vs −0 = 0 RR 21
⇒va =−R2 vs R
1
= −R2R3+R2R4+R3R4 vs
Plug in values ⇒ yields vo = − 30+900+30 = −200 V/V
vs 4.8
P6.5-4
Ohm’s law:
i = v1 −v2
R2
KVL:
v = (R +R +R )i = R +R +R (v −v ) 0123R12
2
P6.5-4
Ohm’s law:
i = v1 −v2
R2
KVL:
v = (R +R +R )i = R +R +R (v −v ) 0123R12
2
RR
13
123
6-11
P6.5-5
1a12 a1112
v−v+v−v+0=0 ⇒v =1+Rv−Rv
RR RR
1777
v2−vb −v1−v2+0=0 ⇒v=1+R2v−R2v
RRbR2R1
2777
−vb−vc +vc−0+0 = 0 ⇒v = R6 v
R R c R+R b
46 46
−va−vc+vc−v0+0=0 ⇒v =−R5 v +(1+R5)v
RR 0 Ra Rc
35 3 3
v =RR+R(R+R)(1+R)v−R(1+R)+R(R+R)Rv 051635225163521
R3R7 R3(R4+R6) R7 R3 R7 R3 (R4+R6)R7
i0 = vc−v0 = R5
v =RR+R(R+R)(1+R)v−R(1+R)+R(R+R)Rv 051635225163521
R3R7 R3(R4+R6) R7 R3 R7 R3 (R4+R6)R7
i0 = vc−v0 = R5
6-12
P6.5-6
KCLatnodeb:
KCLatnodea:
So
KCLatnodea:
So
va + vc =0 ⇒ v =−5v
20×103 25×103 c 4a
v −−5v v−(−12) v v+0 a a
v −−5v v−(−12) v v+0 a a
a
40 ×103
40 ×103
+
a
10 ×103
10 ×103
+ a + 4 =0 ⇒ v=−12V
20 ×103 10 ×103 a 13
vc =−5va =−15.
4 13
6-13
P6.5-7
Apply KCL at the inverting input node of the op amp
Apply KCL at the inverting input node of the op amp
−va−0+0−(va+6)−0 = 0
10000
30000
⇒va =−1.5V
Apply KCL to the super node
corresponding the voltage source:
va −0 + va +6−0 10000 30000
+va−vb +(va+6)−vb = 0 30000 10000
⇒ 3va +va +6+va −vb +3(v +6)−v = 0
va −0 + va +6−0 10000 30000
+va−vb +(va+6)−vb = 0 30000 10000
⇒ 3va +va +6+va −vb +3(v +6)−v = 0
ab
⇒ vb =2va+6=3V
Apply KCL at node b:
⇒ v0 =8vb−4va−18=12V Apply KCL at the output node of the op amp:
i0 + v0 + v0 −vb = 0 ⇒ i0 = − 0.7 mA 30000 30000
⇒ v0 =8vb−4va−18=12V Apply KCL at the output node of the op amp:
i0 + v0 + v0 −vb = 0 ⇒ i0 = − 0.7 mA 30000 30000
vb
+vb−v0 −va−vb −(va+6)−vb = 0
30000 30000 10000
bb0abab
10000
⇒ 3v+(v−v)−(v−v)−3(v+6)−v=0
⇒ 3v+(v−v)−(v−v)−3(v+6)−v=0
6-14
P6.5-8
Apply KVL to the bottom mesh:
−i0(10000)−i0(20000)+5 = 0 ⇒ i0=1mA
6
The node voltages at the input nodes of an ideal op amp are equal. Consequently
va = 10000i0 =10 V 6
Apply KCL at node a:
va +va−v0=0⇒ v0=3va=5V
P6.5-9
KCLatnodeb: vb +12+ vb =0 ⇒ vb =−4V 40000 20000
The node voltages at the input nodes of an ideal op amp are equal, so vc = vb = −4 V . Thenodevoltagesattheinputnodesofanidealopampareequal,so v =v +0×104 =−4V.
KCLatnodeg: −vf −vg + vg =0 ⇒ v =2v
Apply KVL to the bottom mesh:
−i0(10000)−i0(20000)+5 = 0 ⇒ i0=1mA
6
The node voltages at the input nodes of an ideal op amp are equal. Consequently
va = 10000i0 =10 V 6
Apply KCL at node a:
va +va−v0=0⇒ v0=3va=5V
P6.5-9
KCLatnodeb: vb +12+ vb =0 ⇒ vb =−4V 40000 20000
The node voltages at the input nodes of an ideal op amp are equal, so vc = vb = −4 V . Thenodevoltagesattheinputnodesofanidealopampareequal,so v =v +0×104 =−4V.
KCLatnodeg: −vf −vg + vg =0 ⇒ v =2v
10000 20000
20×103 40×103 g 3 f
dc
6-15
The node voltages at the input nodes of an ideal op amp are equal, so ve = vg = 23 vf .
v −v v −v v −v
KCLatnoded: 0= d f + d e = d f +
vd −2vf
3
20×103
20×103
6
24
5
20×103 20×103 20×103
Finally,ve=vg=2vf =−16V.
f
5 d
⇒ v = v =−
V
35
By voltage division (or by applying KCL at
By voltage division (or by applying KCL at
P6.5-10
node a)
va = R0 vs
va = R0 vs
Applying KCL at node b:
vb −vs + vb −v0 = 0
R R +∆R 10
vb −vs + vb −v0 = 0
R R +∆R 10
R +R
10
⇒ R0 +∆R(vb −vs )+vb = v0
R
1
The node voltages at the input nodes of an
The node voltages at the input nodes of an
idealopampareequalso vb =va.
v = R0+∆R+1 R0 − R0+∆Rv = − ∆R v = −v R0
v = R0+∆R+1 R0 − R0+∆Rv = − ∆R v = −v R0
∆R
0 R R+R Rs R+Rs sR+RR
1 10 1 10 100
6-16
Section 6-6: Design Using Operational Amplifier
P6.6-1
Use the current-to-voltage converter, entry (g) in Figure 6.6-1.
Use the current-to-voltage converter, entry (g) in Figure 6.6-1.
P6.6-2
Use the voltage –controlled current source, entry (i) in Figure 6.6-1.
Use the voltage –controlled current source, entry (i) in Figure 6.6-1.
P6.6-3
Use the noninverting summing amplifier, entry (e) in Figure 6.6-1.
Use the noninverting summing amplifier, entry (e) in Figure 6.6-1.
6-17
P6.6-4
Use the difference amplifier, entry (f) in Figure 6.6-1.
P6.6-5
Use the inverting amplifier and the summing amplifier, entries (a) and (d) in Figure 6.6-1.
P6.6-6
Use the negative resistance converter, entry (h) in Figure 6.6-1.
Use the difference amplifier, entry (f) in Figure 6.6-1.
P6.6-5
Use the inverting amplifier and the summing amplifier, entries (a) and (d) in Figure 6.6-1.
P6.6-6
Use the negative resistance converter, entry (h) in Figure 6.6-1.
6-18
P6.6-7
Use the noninverting amplifier, entry (b) in Figure 6.6-1. Notice that the ideal op amp forces the current iin to be zero.
P6.6-8
P6.6-9
Use the noninverting amplifier, entry (b) in Figure 6.6-1. Notice that the ideal op amp forces the current iin to be zero.
P6.6-8
P6.6-9
SummingAmplifier: va =−(6v1+2v2) ⇒ v =6v+2v
InvertingAmplifier: v =−v o 1 2
oa
6-19
P6.6-10
Using superposition, vo = v1 + v2 + v3 = −9 −16 + 32 = 7 V
|
R1
|
6
|
12
|
24
|
6||12
|
6||24
|
|
R2
|
12||12||24
|
6||12||24
|
6||12||12
|
12||24
|
12||12
|
|
-vo/vs
|
0.8
|
0.286
|
0.125
|
2
|
1.25
|
|
R1
|
12||12
|
12||24
|
6||12||12
|
6||12||24
|
12||12||24
|
|
R2
|
6||24
|
6||12
|
24
|
12
|
6
|
|
-vo/vs
|
0.8
|
0.5
|
8
|
3.5
|
1.25
|
6-20
Section 6-7: Operational Amplifier Circuits and Linear Algebraic Equations
P6.7-1
6-21
P6.7-2
6-22
Section 6-8: Characteristics of the Practical Operational Amplifier
P6.8-1
The node equation at node a is:
Solving for vout:
P6.8-2
P6.8-2
vout −vos
100×103
=
vos + i
+ ib1
v =1+90×103 v +(90×103)i =10v +(90×103)i
v =1+90×103 v +(90×103)i =10v +(90×103)i
The node equation at node a is:
Solving for vo:
vos
10000
10000
=
v0 −vos
90000
10×103
b1
100×103 ( )
v = 1+ v +100×103 i =11v +100×103 i
v = 1+ v +100×103 i =11v +100×103 i
out 10×103 os
= 11 0.03×10 + 100×10 1.2×10
( −3)( 3)( −9)
= 11 0.03×10 + 100×10 1.2×10
( −3)( 3)( −9)
( )
= 0.45 mV
b1
os
b1
o 10×103os b1 os b1
=10(5×10−3)+(90×103)(.05×10−9)= 50.0045×10−3 − 50mV 6-23
=10(5×10−3)+(90×103)(.05×10−9)= 50.0045×10−3 − 50mV 6-23
P6.8-3
v1−vin+v1 +v1−v0 =0
RRR
Rin(R0−AR2)
(R+R )(R+R)+RR (1+A)
(R+R )(R+R)+RR (1+A)
1 in 2
v +Av v −v
⇒ v0
v
=−9.6078
(2×106 )(75−(200,000)(50×103 ))
(5×103 +2×106 )(75+50×103 )+(5×103 )(2×106 )(1+200,000) 2×106 (75−(200,000)(49×103 ))
=−9.6078
(2×106 )(75−(200,000)(50×103 ))
(5×103 +2×106 )(75+50×103 )+(5×103 )(2×106 )(1+200,000) 2×106 (75−(200,000)(49×103 ))
0 1+0 1=0
R0 R2
=
in
1 in 0 2 1in
P6.8-4
a)
b)
c)
a)
b)
c)
v
R
49×103
5.1×103
0 =−2 =−
v R
v0
v
=
=
=
= −9.9957
in 1
in
v
v in
v in
= −9.6037
0
(5.1×103 +2×106 )(75+49×103 )+(5.1×103 )(2×106 )(1+200,000)
6-24
P6.8-5
Apply KCL at node b:
vb = R3 (vcm−vp)
R2 +R3
Apply KCL at node a:
Apply KCL at node a:
va−v0 +va−(vcm+vn) = 0
RR
41
The voltages at the input nodes of an ideal op amp are equal so
The voltages at the input nodes of an ideal op amp are equal so
va =vb.
v =−R(v +v)+R+Rv
v =−R(v +v)+R+Rv
441
0 Rcmn Ra
0 Rcmn Ra
11
v0 =−R4(vcm+vn)+
R
1
(R+R)R(v −v)
413
R(R+R) cm p 123
R(R+R) cm p 123
R4 +1
43413134
whenR = R then(R+R)R = R ×R = R
R R R(R+R) R3+1R R
1212321
R2
so
v0 = −R4 (vcm +vn)+R4 (vcm −vp) = −R4 (vn +vp)
v0 = −R4 (vcm +vn)+R4 (vcm −vp) = −R4 (vn +vp)
RRR
111
6-25
PSpice Problems
SP6-1:
(a) vz =avw +bvx +cvy The following three PSpice simulations show
1V=vz =awhenvw=1V,vx =0Vandvy =0V 4V=vz =bwhenvw=0V,vx =1Vandvy =0V -5V=vz =cwhenvw=0V,vx =0Vandvy =1V
Therefore vz =vw +4vx −5vy (b)Whenvw=2V: vz =4vx −5vy +2
1 V = vz = a when vw= 1 V, vx = 0 V and vy = 0 V:
SP6-1:
(a) vz =avw +bvx +cvy The following three PSpice simulations show
1V=vz =awhenvw=1V,vx =0Vandvy =0V 4V=vz =bwhenvw=0V,vx =1Vandvy =0V -5V=vz =cwhenvw=0V,vx =0Vandvy =1V
Therefore vz =vw +4vx −5vy (b)Whenvw=2V: vz =4vx −5vy +2
1 V = vz = a when vw= 1 V, vx = 0 V and vy = 0 V:
6-26
4V=vz =bwhenvw=0V,vx =1Vandvy =0V:
-5 V = vz = c when vw= 0 V, vx = 0 V and vy = 1 V:
6-27
SP6-2
a) Using superposition and recognizing the inverting
and noninverting amplifiers:
v =−80v +1+80(−2)=−3.2v −8.4 o25s25 s
v =−80v +1+80(−2)=−3.2v −8.4 o25s25 s
b) Using the DC Sweep feature of PSpice produces the
plot show below. Two points have been labeled in
anticipation of c).
c) Notice that the equation predicts (−3.2)(−5)−8.4 = 7.6
c) Notice that the equation predicts (−3.2)(−5)−8.4 = 7.6
and
Both agree with labeled points on the plot.
Both agree with labeled points on the plot.
(−3.2)(0)−8.4 = −8.4
6-28
SP6-3
VOLTAGE SOURCE CURRENTS
NAME
V_V1 V_V2
V_V1 V_V2
CURRENT
-3.000E-04 -7.000E-04
-3.000E-04 -7.000E-04
v34 =−1.5−−12×10−6 ≅−1.5V
v23 = 4.5−(−1.5)= 6 V
v50 =12−0=12V
io =−7×10−4 =−0.7mA
v50 =12−0=12V
io =−7×10−4 =−0.7mA
6-29
SP6-4
V4 is a short circuit used to measure io.
The input of the VCCS is the voltage of the left-hand voltage source. (The nominal value of the input is 20 mV.) The output of the VCCS is io.
A plot of the output of the VCCS versus the input is shown below.
The gain of the VCVS is
gain= 50×10−6 −(−50×10−6) =1×10−3 A 100×10−3 −(−100×10−3 ) 2 V
(The op amp saturates for the inputs larger than 0.1 V, limiting the operating range of this VCCS.)
V4 is a short circuit used to measure io.
The input of the VCCS is the voltage of the left-hand voltage source. (The nominal value of the input is 20 mV.) The output of the VCCS is io.
A plot of the output of the VCCS versus the input is shown below.
The gain of the VCVS is
gain= 50×10−6 −(−50×10−6) =1×10−3 A 100×10−3 −(−100×10−3 ) 2 V
(The op amp saturates for the inputs larger than 0.1 V, limiting the operating range of this VCCS.)
6-30
Verification Problems
VP6-1
Apply KCL at the output node of the op
amp
io = vo +vo −(−5)=0 10000 4000
Try the given values: io =−1 mA and vo = 7 V
−1×10−3 ≠3.7×10−3 = 7 +7−(−5) 10000 4000
KCL is not satisfied. These cannot be the correct values of io and vo.
io = vo +vo −(−5)=0 10000 4000
Try the given values: io =−1 mA and vo = 7 V
−1×10−3 ≠3.7×10−3 = 7 +7−(−5) 10000 4000
KCL is not satisfied. These cannot be the correct values of io and vo.
VP6-2
va =(4×103)(2×10−3)=8V
12 ×103
vo =−10×103va =−1.2(8)=−9.6V
Sovo =−9.6Vinsteadof9.6V.
6-31
VP6-3
First, redraw the circuit as:
Then using superposition, and recognizing of the inverting and noninverting amplifiers:
v =−6−4(−3)+1+4(2)=−18+6=−12V o 2 2 2
First, redraw the circuit as:
Then using superposition, and recognizing of the inverting and noninverting amplifiers:
v =−6−4(−3)+1+4(2)=−18+6=−12V o 2 2 2
The given answer is correct.
VP6-4
First notice that ve = v f = vc is required by the ideal op amp. (There is zero current into the input
lead of an ideal op amp so there is zero current in the 10 kΩ connected between nodes e and f, hence zero volts across this resistor. Also, the node voltages at the input nodes of an ideal op amp are equal.)
The given voltages satisfy all the node equations at nodes b, c and d:
VP6-4
First notice that ve = v f = vc is required by the ideal op amp. (There is zero current into the input
lead of an ideal op amp so there is zero current in the 10 kΩ connected between nodes e and f, hence zero volts across this resistor. Also, the node voltages at the input nodes of an ideal op amp are equal.)
The given voltages satisfy all the node equations at nodes b, c and d:
nodeb:
nodec:
noded:
Therefore, the analysis is correct.
nodec:
noded:
Therefore, the analysis is correct.
0−(−5)+ 0 + 0−2 =0
10000 40000 4000
0−2 = 2−5 +0 4000 6000
2−5 = 5 +5−11 6000 5000 4000
0−2 = 2−5 +0 4000 6000
2−5 = 5 +5−11 6000 5000 4000
6-32
VP6-5
The given voltages satisfy the node equations at nodes b and e: nodeb: −.25−2+ −.25 +−.25−(−5)=0
20000 40000 40000
node e: −2.5−(−0.25) ≠ −0.25 + 0 9000 1000
Therefore, the analysis is not correct.
Notice that −2.5−(+0.25) = +0.25 + 0 9000 1000
So it appears that ve = +0.25 V instead of ve = −0.25 V.
Also, the circuit is an noninverting summer with Ra = 10 kΩ and Rb = 1 kΩ, K1 =1/ 2, K2
= 1/ 4 and K4 = 9. The given node voltages satisfy the equation
−2.5=v =K (Kv+Kv)=101(2)+1(−5)
The given voltages satisfy the node equations at nodes b and e: nodeb: −.25−2+ −.25 +−.25−(−5)=0
20000 40000 40000
node e: −2.5−(−0.25) ≠ −0.25 + 0 9000 1000
Therefore, the analysis is not correct.
Notice that −2.5−(+0.25) = +0.25 + 0 9000 1000
So it appears that ve = +0.25 V instead of ve = −0.25 V.
Also, the circuit is an noninverting summer with Ra = 10 kΩ and Rb = 1 kΩ, K1 =1/ 2, K2
= 1/ 4 and K4 = 9. The given node voltages satisfy the equation
−2.5=v =K (Kv+Kv)=101(2)+1(−5)
d1a2c24
4
None-the-less, the analysis is not correct.
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