Section 5-6: Norton’s Theorem
P5.6-1
P5.6-2
67
P5.6-3 Find v oc
1 6
KVL a i x : − 15 + i x 6 + 4 + 2i x + 3i x = 0
⇒ i x = 1 mA
∴ v oc = 3i x = 3V
Find i sc
KVL a i x + i sc : − 15 + 2i x + 10(i x + i sc ) + 3i x = 0
⇒ −15 + 15i x +10i sc = 0
(1)
KVL a i sc : −3i x + i sc = 0
(2)
Solving (1) & (2) simultaneously yields : i sc =1mA
∴ RT =
v oc
3
=
= 3kΩ
i sc
1
Norton equiv. ckt.
P5.6-4
Find v oc
by inspection i = 0
from left mesh : v1 = 3(1 3) = 1V
from KVL a: − v1 + 4i 4 Ω + v oc = 0
⇒ v oc = v1 − 4(5 2 v1 ) = −9V
Find i sc
from KVL a: − v1 + 4i sc + 10v1 + 5i sc = 0
⇒ 9v1 + 9i sc = 0
1 v
from KCL at P : − + 1 + i sc = 0
3 3
∴ RT =
v oc −9
=
= −18Ω
1
i sc
2
(1)
(2)
(1) & (2) yields
Norton equiv. ckt :
i sc =
1
A
2
68
Section 5-7: Maximum Power Transfer
∴
P5.7-1
← RT = 101Ω = RL
(b)
2
Pmax = v 2L 101 = (50) 101 = 24.75 W
P5.7-2
(a) Use source transformations to reduce ckt.
Norton equiv. where R T = 60Ω ∴want R L = 60Ω
Pmax = i 2R (R) = (30) 2 (60) = 54,000µ W= 54mW
(b)
VL = VS
P5.7-3
R "#
!R +R $
L
S
∴ PL =
VL2
RL
=
L
VS2 R L
(R S + R L ) 2
By inspection, PL is max when you vary RS to get the smallest denominator. ∴ set RS = 0
P5.7-4
Find R T using R T = v oc /i sc . First find v oc
:
KCL at P: − i x −.9 + 10i x = 0
⇒ i x = 0.1A
∴ v oc = 3 (10i x ) = 3V
Find i SC
KCL at P: − i x −.9 + 10i x = 0
∴ i sc =10i x =1 A
pL
∴ R T = v oc i sc = 3Ω = R L
=
max
1 6
1.5
VL 2
=
RL
3
.A
⇒ i x = 01
2
= 0.75 W
for max power
69
P5.7-5
(a)
For max power R L = R T . First find v oc :
KVL a i1:
− 4I 0 + 4i1 − 4 v x + 2(i1 − i 2 ) = 0
⇒ 6i1 − 2i 2 + 4 v x − 4I 0 = 0 (1)
KVL a i 2 : 2i 2 + 6i 2 − v x + 2(i 2 − i1 ) = 0
⇒ −2i1 + 10i 2 − v x = 0
(2)
also v x = 4i 2
and
(3)
v oc = 6i 2
(4)
Solving (1), (2), (3), & (4) yields v oc = I 0
Find i sc
KVL a i1 : − 4I 0 + 4i1 + 4v x + 2 (i1 − i sc ) = 0
⇒ 6i1 − 2isc + 4v x − 4I0 = 0
(1)
KVL a i x : 2i sc + 4i sc + 2(i sc − i1 ) = 0
⇒ −2i1 + 8isc = 0
also v x =−4isc
Solving (1), (2), & (3) yields
(b)
P
L max
= 54 W=
(v oc 2) 2
=I
RL
2
0
isc = 2 I 0
3
∴R T =
(2)
(3)
v oc 3
= Ω= R L
2
isc
6
⇒ I0 =18A
P5.7-6
Pmax = v T2 4 R T
Find R T ⇒ kill i source
R T = 8 + (20 + 120) (10 + 50)
= 50Ω
find v oc :
i10Ω
120 + 50
20A
120 + 50 + 20 +10
= 17A
=
∴ v10Ω = 10(17) = 170V
v 50 Ω = 50(17 − 20) = −150V
⇒
v oc = v10Ω + v 50Ω = 170 − 150 = 20V
∴ Pmax = 202 = 2 W
70
PSpice Problems
SP 5-1
Input file
V1
1
I2
2
R3
2
R4
2
R5
1
R6
3
I7
0
R8
4
.dc V1 12
.print dc
.END
0
dc
1
dc
0
3
3
4
3
dc
0
12 1
V (4)
12
2
2
1
2
1
4
2
result V(4)= v =4.952E+00V
SP 5-2
Input file
R1
1
I2
1
R3
2
R4
2
R5
1
I6
0
R7
1
V8
4
.dc V8 12
.print
dc
.END
0
2
dc
0
3
3
3
dc
4
0
dc
12 1
I(R1)
2
2
1
2
1
4
2
12
result
i = I(R1)=3.000E+00A
SP 5-3
result
Input file
V1
4
G2
0
R3
1
R4
1
R5
2
R6
2
R7
3
.dc V1 6
.print
dc
.END
0
dc
1
1
2
4
0
3
0
6 1
V(3)
6
2
500m
1
1
1
1
v = v(3) = 1.714E+00V
71
2
SP 5.4
Probe result
Input file
V1
1
R2
1
I3
2
.dc
I3 0
.probe V(2)
.end
0
2
0
2e-3
dc
10
5k
dc
1m
0.2e-3
SP 5-5
Input file
V1
1
R1
1
R2
2
G4
0
R3
3
R4
3
.dc V1 5
.print dc
.end
0
dc
2
0
3
2 0
0
0
5 1
I(R4)
5
3k
6k
4
5k
2k
result i = I(R4)= 9.524E+00A
72
SP 5-6
Input
G1
R2
I3
R4
.tf
.end
file
0
1
0
2
V(1)
1
2
2
0
I3
2
0
20
0.2
30
dc
25m
result
answer:
(
NODE
1)
V1= Voc = VT = 36V
VOLTAGE
36.0000
(
NODE
2)
VOLTAGE
24.0000
RTH = OUTPUT RESISTANCE AT V(1)=2.000E+02Ω
Verification Problems
VP 5-1
Evaluating data
Case 1:
R L = 0Ω ;
Case 2:
R L = 500Ω ; i = 43.8mA =
Solving 1+ 2 yields
v oc
Rt
(1)
v oc
R t + 500
(2)
i = I sc = 97.2 mA =
R t = 410Ω , v oc = 39.9 V
When R L = 5000Ω
So
i=
v oc
= 7.37mA
Rt +RL
not 16.5mA as recorded ∴ the data is inconsistent.
73
VP 5-2
Voc = 12 V
(line 1 of the table)
i = R 12+4K
I sc = 3mA (line 3 of the table)
so
R TH = Voc / I sc = 4 KΩ
Hence the circuit can be simplified as shown above right. (Check:
12
= 0.857 mA
10KΩ +4KΩ
as shown in line 2 of the table.)
When i = 1 mA is required
1mA =
12
12
⇒ R=
− 4kΩ = 8kΩ
R +4kΩ
1mA
I agree with my lab partner’s claim that R = 8000 causes i = 1 mA.
VP 5-3
i=
60 11
= 54.5mA
60+40
The measurement is consistent with the prelab calculations.
74
Design Problems
DP 5-1 The equation of representing the straight line in Figure DP 5-1b is v = − R t 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: R t = −
0−5
= 625 Ω and voc = 5 V.
0.008 − 0
Try R1 = R 2 = 1 kΩ . (R1 || R2 must be smaller than Rt = 625 Ω.) Then
5=
R2
R1 + R 2
vs =
1
vs
2
⇒ vs = 10 V
and
625 = R 3 +
R1 R2
= R3 + 500 ⇒ R3 = 125 Ω
R1 + R2
Now vs, R1, R2 and R3 have all been specified so the design is complete.
DP 5-2 The equation of representing the straight line in Figure DP 5-2b is v = − R t 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: R t = −
0 − ( −3)
= 500 Ω and voc = -3 V.
−0.006 − 0
From the circuit we calculate
Rt =
R 3 ( R1 + R 2 )
R1 + R 2 + R 3
so
500 Ω =
and voc = −
R 3 ( R1 + R 2 )
R1 + R 2 + R 3
R1 R 3
R1 + R 2 + R 3
and 3 V = −
is
R1 R 3
R1 + R 2 + R 3
is
Try R 3 = 1kΩ and R1 + R 2 = 1kΩ . Then R t = 500 Ω and
−3 = −
1000 R1
2000
is =
R1
2
is
⇒ 6 = R1 i s
This equation can be satisfied by taking R1 = 600 Ω and is = 10 mA. Finally, R2 = 1 kΩ - 400 Ω = 600 Ω.
Now is, R1, R2 and R3 have all been specified so the design is complete.
DP 5-3 The slope of the graph is positive so the Thevenin resistance is negative. This would require
R3 +
R1 R 2
R1 + R 2
< 0 , which is not possible since R1, R2 and R3 will all be non-negative.
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.
75
DP 5-4 The equation of representing the straight line in Figure DP 5-4b is v = − R t 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: R t = −
−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
voc =
R 2 ( d + 1)
R1 + ( d + 1) R 2
isc =
vs ,
( d + 1) v
s
R1
and
Rt =
R1 R 2
R1 + ( d + 1) R 2
Let R1 = R2 = 1 kΩ. Then
−625 Ω = R t =
and
−5 =
( d + 1) vs
d +2
1000
1000
⇒ d=
− 2 = −3.6 A/A
−625
d +2
⇒ vs =
−3.6 + 2
( − 5) = −3.077 V
−3.6 + 1
Now vs, R1, R2 and d have all been specified so the design is complete.
DP 5-5
a) Find Thev. equiv.
v oc :
Solving eqs. (1) - (3) yields v oc = v s
− v s + 800 i + v ab = 0
v ab = v oc
(2)
v oc = − (β i) (40)
(3)
(1)
1− 20
β
I sc :
Solving eqs. (4) - (6) yields
so R t =
I sc = − β i
(4)
v ab = 0
(5)
v s = 800i
(6)
I sc =−β v s 800
v oc
−800
=
and Thev. equiv.
β − 20
I sc
vs
1− 20 β
76
b) R t = R L = 400 = −800 ⇒ β = 18
β − 20
c) Max power to RL, largest vOC, largest vS, smallest Rt
PL =
and VL =
1V 6
2
L
(7)
RL
400
v oc
R total
(8)
with
R total =
−800
+ 400 (a) yields β = ±18
β − 20
d) delivering large amounts of power could melt antenna.
Max power to load : R L = R t = 50Ω
But split power equally (R L1 = R L2 = 50Ω)
DP 5-6
(R +50)(R +50)
= 50
R + 50 + R + 50
⇒ yields R=50Ω
77