Download Lecture 4 Review i v • Current-voltage relationships

Lecture 4 Review
• Ideal Op Amp: i1=i2=0; v1=v2
• Current-voltage relationships
— Capacitors:
dq
dv
i
C
dt
dt
t
1
v   idt  v(t0 )
C t0
•
• A capacitor is an open circuit to dc
• The voltage on a capacitor cannot change abruptly
— Inductors:
di
t
1
i   v(t )dt  i (t0 )
L t0
vL
•
dt
• An inductor is an short circuit to dc
• The current on an inductor cannot change abruptly
Lecture 5 DC Circuits
Transient Circuits – First order circuits
Contents
• First-order circuit
— The source-free R-C/R-L circuit
— Step response of an RC/RL circuit
• Review DC analysis
SF2003 Oscillations
Professor John McGilp
Objective
• Study transient behaviour (i, v) of the circuits
— Steady state
— Transient
• Practice some maths tools on circuits
—Singularity functions : to describe the input
—Differential equations: to work out a solution
Methods
• Obtain the equation for a given circuit
—Current-voltage behaviour of circuit elements
—Circuit laws: Kirchhoff's laws, Ohm’s law
—Circuit theorems
• Solve the equation
—Maths tools
• Understand the results
— current/voltage varies as a function of time
— power and energy: consume/transfer
• Applications
First-order circuits
A first-order circuit is characterized by a first-order differential
equation.
• Current/voltage as a function of time
• R-C, R-L circuits with/without sources
dv
dt
di
Inductors: v  L
dt
v
Resistors: i 
R
Capacitors: i  C
If there are capacitors and inductors
in one circuit:
To work out current for capacitors
d
d
C  v  C  vs  vR  vL 
 dt 
 dt 
d 2i
CL 2
dt
The Source-free RC circuit - Equation
Let’s first ‘guess’ the behaviour of the circuit
Initial voltage across
the capacitor
• Initial condition:
- The capacitor is charged:
v(0)  V0
Instantaneous
voltage
at time t = 0
- Energy stored in the capacitor: w(0) 
1
CV0
2
• Circuit laws: KCL ic  iR  0
• Element behaviour: ic  C
• First-order equation:
C
dv
dt
iR 
dv v
 0
dt R
v
R
The Source-free RC circuit - Solution
To solve the equation
C
dv v
 0
dt R
dv
1

dt
v
RC
(1) Rearrange the items:
v v (t )
(2) Integrating both sides:
t
dv
1


dt


v t 0 RC
v v ( 0 )
v v ( t )
Left side:
dv
v(t )
 ln v(t )  ln v(0)  ln

v
v(0)
v v ( 0 )
t
1
1
t
t 0  RC dt   RC (t  0)   RC
v(t )
t
ln


(3) Apply initial condition v(0) =V0:
V0
RC
Right side:
(4) Rearrange (taking powers of e): v(t )  V e
0

t
RC
The Source-free RC circuit – the result
• The voltage is decreasing as t increases
• How fast it decreases?
– Time constant (see next slide):   RC
• the voltage decreases towards zero
t
• Current :
• Power dissipated in the resistor
v(t ) V0  RC
iR (t ) 
 e
R
R
2t
• Energy:
V02  RC
p(t )  v(t )iR (t ) 
e
R
2t
2t



V02  RC
1
2
RC
wR (t )   p(t )dt  
e dt  CV0 1  e 
R
2
0
0


t
t
v(t )  V0e

t
RC
The Source-free RC circuit- time constant
  RC
The larger the R, …?
v(t )  V0e

t
RC
t
V0  RC
iR (t )  e
R
v( )  V0e



1
 V0e  0.368V0


v(2 )  V0e 2  e 1 V0e 1  0.368v( )
v(5 )
 0.3685  1%
v(0)
Fully discharged
To measure time constant
t

dv(t )
1

V0 e RC
dt
RC
dv(t )
1
At t  0,
  V0
dt 0

Source-free RC circuit - Example
v(t )  V0e

t
RC
The switch in the circuit has been closed for a long time, and it is
opened at t = 0.
(1) Find v(t) for t ≥ 0.
(2) Calculate the initial energy stored in the capacitor.
Thevenin Resistance seen from
the terminals of the capacitor
Ans: 8e−2t V, 5.33 J.
Step response of an RC circuit - equation
Let’s first ‘guess’ the behaviour of the circuit
• Equivalent circuit: step input
• Initial condition: 

vc (0 )  vc (0 )  V0
vc just before switching
• Circuit laws:
vc cannot change
abruptly
KVL
 Vs u (t )  vR  vC  0
KCL
ic  iR  0
• Element behaviour:
• First-order equation:
ic  C
dvc
dt
dvc vc  VS u (t )
C

0
dt
R
iR 
vR VS u (t )  vC

R
R
Step response of an RC circuit - Solution
To solve the equation
(1) For t > 0:
C
dvc vc  VS u (t )
C

0
dt
R
dvC vC  VS

0
dt
R
dvC
v  VS
 C
dt
RC
(2) Rearrange the terms:
v v (t )
dvC
1

dt
vC  VS
RC
t
dv
1
(3) Integrating both sides: 
 
dt
v  V t 0 RC
v v ( 0 ) C
v v (t )
vC (t )  VS
dv

ln
 v  VS
V0  VS
v v ( 0 ) C
Left side:
t
Right side:

t 0

initial condition
v(0) =V0 applied
1
1
t
dt  
(t  0)  
RC
RC
RC
(4) Rearrange (taking powers of e):
V0


t
v(t )  



VS  V0  VS e
t0
t0

t
v(t )  V0  VS  V0 (1  e )u (t )   RC

Step response of an RC circuit - Results
• The voltage is increasing as t increases
• How fast it increases?
V0


t
v(t )  



VS  V0  VS e
– Time constant  = RC:
 : characteristics of
The same as source-free
the circuit.
• The voltage increases towards Vs
dv(t ) C VS  V0  
i (t )  C

e
dt

• Current :
• Complete response:
Superposition:
v(t )  V0 e

t

t
t



 VS 1  e

t 0




forced response
Natural response
v(t )  VS  V0  VS e
Steady-state response

t

Steady-state response
t0
t0
To find the step response of an RC circuit - Example
v(t )  v()  [v(t0 )  v()]e

t t 0

The time constant  = RC
The final capacitor
voltage v()
Example:
The initial capacitor
voltage v(t0)
The switch has been in position A for a long
time. At t = 0, the switch moves to B.
Determine v(t) for t > 0 and calculate its
value at t = 1 s and 4 s.
Ans: v(t )  30  15e 0.5t V
v(1) = 20.9 V
v(4) = 27.97 V
RL circuits
• The source-free RL circuit
— Initial condition:
i (0)  I 0 ,
w(0) 
—Equation:
—Results
•
i (t )  I 0 e

L
1 2
LI 0
2
di
 Ri  0
dt
t
L
  
R
v(t )  RI 0e

t

p(t )  RI 02e

• Step response of an RL circuit
i(t )  i()  [i(t0 )  i()]e
The final inductor
current i()

t t 0

The initial inductor
current i(t0)
The time
constant
 = L/C
2t

2t
 
1 2
wR (t )  LI 0 1  e  
2


Appendix
Sources and input – mathematical
consideration
v(t)
• Ideal Current/Voltage
source
Vs
— v(t) = Vs; i(t) = Is
• Switching sources
— Singularity Functions
• Unit step function u(t)
• Unit impulse function (t)
• Unit ramp function r(t)
v(t)
Vs
Unit Singularity functions
• Unit step function
0 t  0
u (t )  
1 t  0
• Unit impulse function (delta function)
0
du (t ) 
 (t ) 
 Undefined
dt
0

t0
t 0
t 0
• Unit ramp function
0 t  0
r (t )   u (t )dt  
t t  0

t
0
  (t )dt  1
0
b
 f (t ) (t  t )dt  f (t )
0
a
0
Example and practice
The gate function
v(t )  10[u(t  2)  u(t  5)]
The sawtooth function
v(t )  5r (t )  5r (t  2) 10u(t  2)
Lecture 6 DC Circuits
Transient – Second-Order Circuits