Chapter 2 Properties of resistive circuits
2.1-2.2 (optional)
equivalent resistance, resistive ladder, duality, dual entities
2.3 Circuits with controlled sources
controlled sources (VCVS, CCVS, VCCS, CCVS), equivalent
resistance
2.4 Linearity and superposition
proportionality principle, superposition theorem
2.5 Thevenin and Norton networks
Thevenin’s and Norton’s theorems, source conversion
2-1
電路學講義第2章
2.1-2.2
Optional
1. equivalent two-terminal networks: two-terminal networks have
the same i-v characteristics
2. series-connected resistors: voltage divider
series equivalent resistance: Rser = R1 + R2 + ...+ RN
3. parallel-connected resistors: current divider
parallel equivalent conductance: Gpar = G1 + G2 + ...+ GN
4. resistive ladder: a network consisting entirely of series and parallel
resistors
5. series-parallel reduction method: to solve branch variables of a
resistive ladder
(1) step 1: find the equivalent resistance of the ladder network
(2) step 2: calculate the voltage or current at the network terminal
(3) step 3: use KVL, KCL, Ohm’s law to calculate branch variables
2-2
電路學講義第2章
6. Ex. 2.4 solve terminal current i, total dissipated power p, branch
variables vx, vy
(1) step 1
Req = 2 +10//15 = 8kΩ
(2) step 2
Req
40
= 0.005
8000
p = 40 × 0.005 = 200mW
i=
(3) step 3
vx = 40 − 2000× 0.005 = 30V
vy = 30×
5
= 10V
4+5+6
2-3
電路學講義第2章
7. dual networks:
i-v equation of network A= i-v equation of network B with its
i, v interchanged
Series network
KVL loop eq. v = ( R1 + R 2 + ...) i
Parallel network
KCL node eq. i = (G1 + G 2 + ...) v
equivalent resistance Rser = R1 + R2 + ... equivalent conductance G par = G1 + G 2 + ...
voltage divider
vn =
Rn
v
R1 + R2 + ...
open circuit v n = v when Rn → ∞
current divider
in =
Gn
i
G1 + G 2 + ...
short circuit in = i when G n → ∞
2-4
電路學講義第2章
8. Dual pairs
9. Ex. 2.5 dual networks
v2
+5
8
v 2 = 12 − ( 2 + 3)i1
KVL
KCL
voltage
current
resistance
conductance
loop
node
series
parallel
open ckt
short ckt
node voltage
mesh current
capacitance
inductance
impedance
admittance
i2
+5
8
i2 = 12 − ( 2 + 3) v1
電路學講義第2章
v1 =
i1 =
2-5
(dual circuit conversion)
Step 1: place a node at the center of each loop and the reference node
outside of the given circuit
Step 2: draw a line between the nodes with each line crossing an
element, and replace that element by its dual
rule of source: a voltage source that produces a positive (CW) loop
current has as its dual a current source whose reference
direction is from the ground to nonreference node
12V → i1 positive or CW → dual 12A current source directed from 0 to 1
5 A to the ground node → i2 ( negative ) = − 5 A or CCW → dual 5V voltage source
電路學講義第2章
2-6
2.3 Circuit with controlled sources
Basics
1. dependent (controlled) sources: linear element
voltage gain
(transformer, voltage amplifier)
(BJT)
transconductance
(FET)
current gain
2-7
transresistance
(transformer)
電路學講義第2章
(DC transistor circuit)
in active mode: VBE ≈ 0.7V
I C = α I E , α : CB current gain
0.98 < α < 0.999
I C = β I B , β : CE current gain
50 < β < 1000
npn transistor
DC equivalent model
method 1
loop 1: 2=100 × 10 3i1 + 200 × 10 3i2 ...(1)
β=150
loop 2: VBE =0.7=200 × 10 3i2 → i2 = 3.5uA
loop 3: − vo − 10 3 I C + 16 = 0...(2)
(1) → i1 = 13uA, node A : i1 = i2 + I B → I B = 9.5uA
I C = β I B = 1.425 mA, (2) → vo = 16 − 1.425 = 14.575V
2-8
電路學講義第2章
method 2
β=150
vo =16 − 10 3 × 150 I B ...(1)
2 − 0.7
0.7
−
= 9.5uA
100 × 10 3 200 × 10 3
(1) → vo = 16 − 10 3 × 150 × 9.5 × 10 6 = 14.575V
I B = i1 − i2 =
2-9
電路學講義第2章
2. equivalent resistance theorem (terminal behavior of a network)
For a two-terminal load network containing resistors and
controlled sources, the equivalent resistor is determined by its
terminal voltage and current.
no independent source inside
v
Req =
i
2-10
電路學講義第2章
Discussion
1. Ex.2.6 a FET with gm=5mS, vout=?
vout = iout 6 × 10 = − g m v g 6 × 10
3
1kΩ
3
vin
5
3
6 × 10 vin = − 25vin
= − gm
1+ 5
5kΩ vg
iout
6kΩ vout
gmvg
2. Ex.2.7 a VCVS to find i1(vs)
KVL : v 2 + 3v 2 − 12 i1 = 0 → v 2 = 3i1
KVL : v s − 4i − v 2 = 0 → v s = 4i + 3i1
KCL : i =
v2
3
+ i1 = i1
6
2
⇒ v s = 6i1 + 3i1 = 9i1 → i1 =
vs
9
2-11
電路學講義第2章
3. Ex.2.8 find Req
1 − gm R
v
1
KCL : i + g m v = → i = ( − g m + ) v =
v
R
R
R
v
R
Req = =
i 1 − gm R
v
1
if R = ∞ , i = − ic = − g m v → Req = = −
( active )
i
gm
2-12
電路學講義第2章
2.4 Linearity and superposition
Basics
1. Linear resistive circuit: circuit contains linear resistors and
independent sources.
2. Proportionality principle (single independent source)
if y=f(x), then f(Kx)=Kf(x)
3. Superposition theorem (multiple independent sources)
For a linear network, response of a sum of signals = sum of signal
responses
4. Controlled sources are not applied for the superposition theorem.
5. suppressed independent voltage source vs=0 → short circuit
suppressed independent current source is=0 → open circuit
2-13
電路學講義第2章
Discussion
1. Ex.2.9 (ladder network) vs =72
use proportionality principle
to find i, v2, i1
Req
step 1: assume a convenient value for a branch variable farthest
from the source
let i1 = 1 → v1 = 12
step 2: work toward source to obtain the source value x
v1
3
3
= 3 → i2 = → i = i2 + i1 =
4
6
2
v3 = 4i = 6 → vs = v3 + v2 = 9
v2 = − 3v2 + v1 → v2 =
K × 9 = 72 → K = 8
step 3: calculate the scaling factor K
step 4: calculate the actual values by the scaling factor K
i=K×
Req =
3
= 12 , v 2 = K × 3 = 24 , i1 = K × 1 = 8
2
vs
=6
i
2-14
電路學講義第2章
Ex.2.10 use superposition to find i1
O.C.
from source 1 of 30V
30
5
=
i1−1 =
6+4+2 2
from source 2 of 3A
4
i1− 2 =
×3 =1
(6 + 2) + 4
O.C.
S.C.
S.C.
from source 3 of 8A
6
i1− 3 =
× ( − 8) = − 4
6 + (2 + 4)
i1 = 2 .5 + 1 − 4 = − 0 .5
2-15
電路學講義第2章
Ex.2.11 use superposition to find i1
from source 1 of 30V
KCL → 9i1−1
KVL → 30 − 6(9i1−1 ) − (4 + 2)i1−1 = 0
i1−1 = 0.5
from source 2 of 3A
KCL → 9i1− 2
KCL →
9i1− 2 = 8i1− 2 + 3 + i4 Ω
→ i4 Ω = i1− 2 − 3
KVL → 6(9i1− 2 ) + 4(i1− 2 − 3) + 2i1− 2 = 0
→ i1− 2 = 0.2
i1 = i1−1 + i1− 2 = 0.7 A
2-16
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2.5 Thevenin and Norton networks
Basics
1. source network: a two-terminal network
containing linear resistors (controlled sources)
and at least one independent source.
2. Thevenin parameters (terminal behavior of a network): opencircuit voltage voc, short-circuit current isc, and Thevenin
resistance Rt
voc
Rt ≡
isc
2-17
電路學講義第2章
Discussion
1. Thevenin’s theorem: the equivalent circuit of a linear resistive
source network is a voltage source voc in series with a resistor Rt.
2. Norton’s theorem: the equivalent circuit of a linear resistive source
network is a current source isc in parallel with a resistor Rt.
voc ≡ v i = 0 , isc ≡ i v = 0 , Rt ≡
voc
isc
if load SC , v = 0 = voc − Rt isc
voc
if load OC , i = 0 = isc −
Rt
→ Rt =
v = voc − Rt i
i = isc −
v
Rt
2-18
voc
isc
電路學講義第2章
3. Thevenin and Norton equivalent circuits are useful to solve the
terminal behavior of a source network, but not internal branches.
4. Ex.2.12 find the Thevenin parameters
O.C . load
1 − vx
v
v − ( − 100 v x )
= x + x
250
4000
2080
→ v x = 0.0757
KCL →
v x + 100 v x
× 80 + ( − 100 v x ) = − 7.28
2080
S .C . load
1 − vx
v
v
KCL →
= x + x
250
4000 2000
→ v x = 0.842
voc =
vx
100 v x
−
= − 1.052
2000
80
v
Rt = oc = 6.92 Ω
isc
isc =
2-19
電路學講義第2章
5. Ex.2.13 use equivalent source circuits to design RL in two cases to
give v=24, and i=8
Thevinin and Norton equivalent circuits
20
= 40
5 + 20
50
40
isc =
= 10, Rt =
=4
5
10
from Tevenin equivalent circuit
RL
→ RL = 6Ω
24 = voc
Rt + RL
voc = 50 ×
from Norton equivalent circuit
Rt
→ RL = 1Ω
8 = isc
Rt + RL
2-20
電路學講義第2章
6. Thevenin resistance = the equivalent resistance of the source
network with all independent sources suppressed. (another way
to find Rt)
(independent
voltage source or
current source)
(contains only resistors
and dependent sources)
Rt =
vt voc
=
it
isc
⎧vt = voc + Rt it ,if voc = 0
vt
⎪
∵⎨
then Rt =
vt
−
=
−
=
,if
0
i
i
i
it
sc
⎪ t sc R
t
⎩
Equivalent resistance theorem is the special case of Thevenin or
Norton theorem.
7. A source network can be described by any two of the Thevenin
parameters: voc, isc, Rt
2-21
電路學講義第2章
8. Ex. 2.14 find Thevenin parameters
step 1: 40kΩ→easier to find isc
KVL → v x − 5v x = 0 → v x = 0 → isc = 0 .003
step 2: find Rt
KVL → v x − 5v x − vt = 0 → v x = − 0.25vt
KCL → it =
Rt =
vt
v
vt
+ x =−
40000 2000
10000
vt
= − 10000 (∵ controlled source) → voc = Rt isc = − 30
it
step 1:
step 2:
2-22
電路學講義第2章
9. equivalent source conversion (including dependent sources)
∵ (1) sources are off → same Rs
(2) S .C . → same is
(3) O.C . → same vs
is =
vs
Rs
10. Source conversion can ease circuit analysis.
2-23
電路學講義第2章
11. Ex. 2.15(ex. 2.9) solve v2 using equivalent source conversion
using v2
KCL → 18 =
v2
+ 0.25v2 → v2 = 24V
4 // 6 //12
When controlled sources are present, source conversion must not obliterate the
identity of any controlled variables.
2-24
電路學講義第2章
12. Ex. 2.16 (ex.2.10) find Thevenin equivalent circuit using source
conversion
step 1
step 2
step 3
2-25
RL = 2
→i=
−6
= − 0.5 A
10 + 2
電路學講義第2章