Inductance and Capacitance
Inductors and capacitors cannot generate energy, they are classified as passive elements.
Inductor:
The symbol for impedance is L and measured as henrys. The relationship between the voltage
and current at the terminals of an inductor:
i
L
+
v
-
v=L
di
dt
i (t ) =
1 t
v dτ + i (t 0 )
L ∫t0
The power is:
p = vi
The energy is :
w=
1 2
Li
2
Capacitor:
It is represented by C and is measured in farads(F) The relationship between the voltage and
current at the terminals of a capacitor:
i
C
v
+
i=C
-
dv
dt
v(t ) =
1 t
i dτ + v(0)
C ∫t0
The power is:
p = vi = Cv
The energy is :
w=
1 2
Cv
2
dv
dt
Serial-parallel combination of inductance and capacitance
Inductor is serial
L1
L2
L3
v1
v2
v3
L = L1 + L2 + L3
v = v1 + v 2 + v3 = ( L1 + L2 + L3 )
di
dt
Inductors in parallel
i
L1
v
i1
L2
L3
i2
i3
Total Inductance value:
1
1
1
1
=
+
+
L L1 L2 L3
i = i1 + i2 + i3 = (
1
1
1 t
+
+ ) ∫ v tτ + i1 (t 0 ) + i 2 (t 0 ) + i3 (t 0 )
L1 L2 L3 t0
Capacitor:
Capacitors in serial:
C1
C2
C3
v1
v2
v3
Total capacitor value:
1
1
1
1
=
+
+
CT C1 C 2 C 3
Capacitors in parallel
i
v
C1
i1
C2
i2
C3
i3
Total capacitor value:
CT = C1 + C 2 + C 3
Example:
+
i
C=2F
for t < 0
for 0 ≤ t < 2
0

i (t ) = t
0

-
for t ≥ 2
For t < 0
t
t
1
1
v(t ) = ∫ i (τ )dτ = ∫ 0dτ = 0
C −∞
2 −∞
For 0 ≤ t < 0
t
t
t
1
1
1 t2
t2
v(t ) = v(0) + ∫ i (τ )dτ = 0 + ∫ τ dτ = ( ) =
C0
20
2 2 0 4
For t ≥ 0
t
v(t ) = v(2) +
 0
 t 2
v(t ) = 
4
 1
t
1
22 1
i
(
τ
)
d
τ
=
+
0 dτ = 1
C ∫2
4 2 ∫2
for t < 0
for 0 ≤ t < 2
for t ≥ 2
Remainder:
e at
e at
at
te
dt
=
(at − 1)
∫
a
a2
t n +1
n
∫ t dt = n + 1 for all n accept for n=-1
at
∫ e dt =
∫t
−1
dt = log x
(Reference: Standard Mathematical Tables CRC Press.)
Cp-7 Response of First Order RL and RC Circuits
The natural response of an RL circuit:
All the source current Is appears in the inductance before t=0. Switch is opened at t=0, then
inductance begins releasing energy. We find the v(t) for t>=0.
t=0
L
I2
Is
Ro
L
i
i
R
v
R
v
Using KVL to obtain an expression
L
di
+ Ri = 0
dt
This is a first-order differential equation. The solution of it
i (t ) = I 0 e
R
− t
L
I 0 denotes the initial current in the inductor
,t>=0 where
The voltage on the resistor is
v = I 0 Re
R
− t
L
t ≥ 0+
,
v(0 − ) = 0 and v(0) = I 0 R
The on the resistor
p = vi = i 2 R =
v2
R
or
p = I 02 Re
R
−2 t
L
Energy
w = ∫ pdx =
t
0
R
−2 t
1 2
LI 0 (1 − e L , t ≥ 0
2
We call time constant
τ=
L
R
for
t ≥ 0+
The natural response of an RC circuit
Ro
t=0
+
i
Vg
R
v
Vg
C
C
-
C
dv v
+ =0
dt R
v(t ) = V0 e − t / τ , t ≥ 0
where τ is the time constant is
i (t ) =
τ = RC .
V0 −t / τ
e , t ≥ 0+
R
V02 − 2t / τ
+
p = vi =
e
, t≥0
R
t
p = ∫ p dx =
0
1 2
V0 (1 − e − 2t / τ ) , t ≥ 0
C
The step response of an RL Circuit
R
L
v(t)
Vs
After switch is closed. KVL
Vs = Ri + L
di
dt
R
v
i (t ) =
Vs
V
+ ( I 0 − s )e − ( R / L ) t
R
R
If the initial energy is zero I0=0, then
i (t ) =
Vs Vs −( R / L ) t
− e
R R
The voltage on the inductor is
v=L
di
d V V

= L  s − s e −( R / L )t  = V s e − ( R / L ) t
dt
dt  R R

The step response of an RC circuit
Is
vc
R
C
i
Using KCL , the differential equation:
C
dvc
+ vc = I s
dt
The solution of the differential equation gives the voltage in the capacitor:
vc = I s R + (V0 − I s R)e − t / RC , t ≥ 0
The current in the capacitor yields the differential equation
V
di
1

+
i = 0 ! i = Is − 0
dt RC
R

 −t / RC
+
, t≥0
e

The Integrating Amplifier
if
C
Rs
V1
+ Vcc
-
vn
is
vo
Vs
+
-Vcc
vp
0
0
0
Let assume op amp is ideal
i f + is = 0
v p = 0 , so that i s =
vs
Rs
and
There fore
dv o
1
=−
vs
dt
Rs C
t
1
vo (t ) =
v s dτ + vo (t 0 )
Rs C t∫0
if = C
dvo
dt
Natural and Step Response of RLC Circuits
8.1 The Natural Response of a Parallel RLC Circuit
ic
C
iL
iR
L
R
+
Vo
v
-
Applying KCL to the circuit
t
v 1
dv
+ ∫ v dτ − I 0 + C
=0
R L0
dt
I0 is constant, we differentiate one to get
d 2v
1 dv
v
+
+
=0
2
RC dt LC
dt
This is second-order differentiation equation. The characteristic equation is
s2 +
s
1
+
=0
RC LC
The solution of the differential equation is
v = A1e s1t + A2 e s2t
Characteristic roots of the solution are
s1 = −α + α 2 − ω 02
s1 = −α − α 2 − ω 02
where
α=
1
and ω 0 =
2 RC
1
LC
There are three possible outcomes
1. If
ω 02 < α 2 roots are will be real. The voltage response call it overdamped.
2. If
ω 02 > α 2 roots are will be complex. The voltage response call it underdamped.
3. If
ω 02 = α 2 roots are will be real and equal. The voltage response call it critically
damped.
The Overdamped ω 02 < α 2 Voltage response
v = A1e s1t + A2 e s2t
1. Using R, L, C, find the roots s1 and s2 of the characteristic equation
2. Using circuit analysis, find
v(0 + )
v(0 + ) = A1 + A2
3. Using circuit analysis, find
dv(0 + ) / dt
dv(0 + ) iC (0 + )
=
= s1 A1 + s 2 A2
dt
C
4. Find the values of A1 and A2 from these two equation
5. Substitute the values s1, s2, A1 and A2 into the solution of the differential equation
The Underdamped
ω 02 > α 2 Voltage response
The voltage is
v(t ) = B1e −αt cos wd t + B2 e −αt sin wd t
where
wd = ω 02 − α 2
B1 and B2 values can be found by solving the following equations
v(0 + ) = V0 = B1
dv(0 + ) iC (0 + )
=
= −αB1 + ω d B2
dt
C
The critically
ω 02 = α 2 Voltage response
The voltage response is
v(t ) = D1te −αt + D2 e −αt
D1 and D2 values can be found by solving the following equations
v ( 0 + ) = V0 = D2
dv(0 + ) iC (0 + )
=
= D1 − αD2
dt
C
The Step Response of a Parallel RLC Circuit
+
I
C
L
R
Vo
v
t=0
-
Using KCL, we have
iL +
di
v
dv
+C
= I , and v = L L
R
dt
dt
or
d 2 iL
i
1 di L
I
+
+ L =
2
RC dt LC LC
dt
Solution:
function of the same form
i = If +

 as the natural response 
function of the same form
v = Vf + 

 as the natural response 
The Natural Response of a Serial RLC Circuit
R
L
I0
C
i
Applying KVL into the circuit
Ri + L
di 1 t
+
idτ + V0 = 0
dt C ∫0
which can be arranged as
V0
d 2 i R di
i
+
+
=0
2
L dt LC
dt
This is second order differential equation. The characteristic equation is
s2 +
R
1
s+
=0
L
LC
The roots of characteristic equation is
s1, 2 = −α ± α 2 − ω 02
where
α=
R
rad/s and ω 0 =
2L
1
rad/s
LC
Thus possible solution
1.
i (t ) = A1e s1t + A2 e s2t if ω 02 < α 2
2.
i (t ) = B1e −αt cos ω d t + B2 e −αt sin ω d t if ω 02 > α 2
3.
i (t ) = D1te −αt + D2 e −αt if ω 02 = α 2
The Step Response of a Serial RLC Circuit
R
L
t=0
V
C
i
vC
Applying KVL to the serial RLC circuit
V = Ri + L
i=C
di
+ vc
dt
dvc
d 2v
di
= C 2c , substitute them into the equation
, and
dt
dt
dt
d 2 vC R dvC vC
V
+
+
=
2
L dt
LC LC
dt
Three possible solution
1.
vC = V f + A1' e s1t + A2' e s2t if ω 02 < α 2
2.
vC = V f + B1' e −αt cos ω d t + B2' e −αt sin ω d t if ω 02 > α 2
3.
vC = V f + D1' te −αt + D2' e −αt if ω 02 = α 2
V f is the final value of vC .