Chapter 5 Energy storage and dynamic circuits
5.1-5.2 (optional)
capacitance, displacement current, i-v relationship, parallel and
series capacitance
inductance, induced voltage, i-v relationship, parallel and series
inductance
5.3 Dynamic circuits
differential equations, natural response, forced response,
complete response
5-1
電路學講義第5章
5.1 Capacitors
Optional
dv (t )
1 t
, v (t ) = ∫ i ( λ ) d λ
1. Capacitor: i-v relation i (t ) = C
dt
C −∞
i: displacement current
2. Capacitor is an open-circuit in DC circuit, and a short-circuit as
ω=∞.
3. The voltage across a capacitor cannot change discontinuously when
the current remains finite.
4. Ex.5.3 Op-amp integrator
vout (t ) = − vc (t ) = −
1
C
1 t
=−
vin (λ ) d λ
∫
−∞
CR
∫
t
−∞
ic (λ ) d λ = −
1
C
∫
5-2
t
−∞
iin (λ ) d λ = −
1
C
vin (λ )
∫−∞ R d λ
t
電路學講義第5章
5.
capacitors in parallel :
dv
dv
i = C1
+ C2
+ ...
dt
dt
dv
dv
= (C1 + C 2 + ...)
= C par
dt
dt
→ C par = C1 + C 2 + ...
capacitors in series :
dv dv1 dv 2
=
+
+ ...
dt
dt
dt
1
1
1
=( +
+ ...) i =
i
C1 C 2
C ser
→
5-3
1
1
1
=
+
+ ...
C ser C1 C 2
電路學講義第5章
5.2 Inductors
Optional
1 t
di (t )
1. Inductor: i-v relation v (t ) = L
, i (t ) = ∫ v ( λ ) d λ
dt
L −∞
v: induced voltage
2. Inductor is a short-circuit in DC circuit, and open-circuit as ω=∞.
3. The current through an inductor cannot change discontinuously
when the voltage remains finite.
4. L and C are duals.
5.
inductors in series :
di
di
+ L2 + ...
dt
dt
di
di
= ( L1 + L2 + ...) = Lser
dt
dt
→ Lser = L1 + L2 + ...
v = L1
di1 di 2
+
+ ...
dt
dt
1
1
1
v
=( +
+ ...) v =
L1 L2
L par
inductors in parallel :
→
5-4
1
1
1
=
+
+ ...
L par L1 L2
電路學講義第5章
5.3 Dynamic circuits
Basics
1. The circuit of one energy-storage element is called a first-order
circuit. It can be described by an inhomogeneous linear first-order
differential equation as
dy (t )
+ a 0 y (t ) = f (t )
dt
y (t ) : response, f (t ) : forcing function
a1
2. The circuit with two energy-storage elements is called a secondorder circuit. It can be described by an inhomogeneous linear
second-order differential equation as
d 2 y (t )
dy (t )
a2
+
+ a 0 y (t ) = f (t )
a
1
2
dt
dt
5-5
電路學講義第5章
3. Natural response (zero-input response, or yN(t)): homogeneous
differential equation solution
a1
dy N (t )
+ a0 y N (t ) = 0, i.c. yN (0+ ) = Yo , y N (t ) : natural response
dt
a
− ot
dy N (t )
a0
dy N (t )
a0
a0
= − y N (t ) →
= − dt → d ln y N = − dt → y N (t ) = Yo e a1
(sol. 1) :
dt
a1
y N (t )
a1
a1
(sol. 2) : assume y N (t ) = Yo e st → (a1s + ao )Yo e st = 0
→ characteristic equation a1s + ao = 0 → s = −
yN (t ) = Yo e
−
ao
t
a1
ao
: natural frequency
a1
, if a0 > 0, a1 > 0 ⇒ yN (t ) → 0 as t → ∞, a stable circuit
(sol.3) : Lapalace transform (p.585), L[f (t )] = F(s),L[ f '(t )] = sF ( s ) − f (0− )
aY
Yo
a1sYN ( s ) − a1 y N (0− ) + a0YN ( s ) = 0 → YN ( s ) = 1 o =
a1s + a0 s + a0 / a1
yN (t ) = Yo e
−
ao
t
a1
, characteristic equation a1s + ao = 0
5-6
電路學講義第5章
4. Forced response (particular solution, or zero-state response, or
steady state response, yF(t)) : inhomogeneous differential
equation solution with i.c.=0,
5. Complete response (yN(t)+ yF(t)): inhomogeneous differential
equation solution with initial condition
Transient response involves both the natural and forced responses
before the steady state response (stable solution) is reached.
5-7
電路學講義第5章
Discussion
1. 1st-order circuit
RL circuit
v = Ri + L
RC circuit
di
dt
v = Ri + L
v = RC
v
dv
i = +C
:1st-order DE
R
dt
5-8
di
:1st-order DE
dt
dvC
+ vC :1st-order DE
dt
電路學講義第5章
2. 2nd-order circuit
RLC circuit
di
1
v = L + Ri +
dt
C
t
dv
d 2i
di 1
id
λ
→
=
L
+
R
+ i
∫−∞
dt
dt 2
dt C
3. Ex. 5.8
di
⎧
v
L
=
+ Ri
in
⎪⎪
LC d dvout RC dvout
dt
→ vin = −
−
⎨ dv
dv
C
dt
dt
β
β dt
out
⎪C out + β i = 0 → i = −
dt
β dt
⎪⎩
d 2 vout
dvout
⇒ LC
+
= − β vin
RC
2
電路學講義第5章
5-9
dt
dt
4. Ex.5.9 given C=300uF, R=2MΩ, vN(0-)=1000V, find vN(t) for t>0
iC
dvN vN
+
=0
dt
R
assume vN = Ae st → ( RCs + 1)vN = 0
KCL → C
characteristic equation RCs + 1 = 0 → s = −
vN (t ) = Ae
−
1
t
600
vN (t ) = 1000e
−
1
1
: natural frequency
=−
600
RC
, i.c. vN (0− ) = 1000 → A = 1000
1
t
600
,t > 0
1
t
dvN (t )
1000 − 600
,t > 0
=−
iC (t ) = C
e
600
dt
= −iR (t )
time constant
vN (600) = 1000e −1
5-10
電路學講義第5章
5. Ex. 5.10 R=4Ω, L=0.1H, v=25sin30t, find iF(t)
di
+ 4i = 25sin 30t
dt
iF (t ) = K1 cos 30t + K 2 sin 30t
KVL → 0.1
⎧ 4 K1 + 3K 2 = 0
→⎨
→ K1 = −3, K 2 = 4
3
4
25
−
K
+
K
=
1
2
⎩
→ iF (t ) = −3cos 30t + 4sin 30t
: same frequency as the source
5-11
電路學講義第5章
6. Ex.5.11 as source v(t)=10e-20t, or v(t)=10e-40t in Ex.5.10, solve iF(t)
di
+ 4i = 0 → natural response iN (t ) = Ae −40 t
dt
di
0.1 + 4i = v (t ) → forced response iF
dt
case 1:v (t ) = 10e −20 t → iF (t ) = K1e −20 t
0.1
→ 0.1( −20 K1e −20 t ) + 4 K1e −20 t = 10e −20 t
→ K1 = 5 ⇒ iF (t ) = 5e −20 t
case 2:v (t ) = 10e −40 t → iF (t ) = K1te −40 t
→ 0.1( K1e −40 t − 40 K1te −40 t ) + 4 K1te −40 t = 10e −40 t
→ K1 = 100 ⇒ iF (t ) = 100te −40 t : stable ( ∵→ 0, as t → ∞ )
5-12
電路學講義第5章
7. Ex.5.12 as source v(t)=400sin280t for t >0 with i.c. i(t)=0 for t<0
in Ex.5.10, solve iF(t)+ iN(t)
di
+ 4i = 0 → natural response iN (t ) = Ae−40t
dt
di
0.1 + 4i = 400sin 280t → forced response iF (t ) = −14 cos 280t + 2sin 280t
dt
complete response i (t ) = −14 cos 280t + 2sin 280t + Ae −40t
0.1
i.c. i (0+ ) = −14 + A = 0 → A = 14
⇒ i (t ) = −14 cos 280t + 2sin 280t + 14e −40t for t > 0
transient response
5-13
steady state
response
電路學講義第5章