Download EE42/100 Lecture 9

EE42/100 Lecture 9
Topics:
More on First-Order Circuits
Water model and potential plot for RC circuits
A bit on Second-Order Circuits
First-Order Circuits
• A circuit which contains only sources, resistors
and an inductor is called an RL circuit.
• A circuit which contains only sources, resistors
and a capacitor is called an RC circuit.
• RL and RC circuits are called first-order circuits
because their voltages and currents are
described by first-order differential equations.
R
i
i
L
vs
–
+
–
+
vs
R
C
The natural response of an RL or RC circuit is its behavior (i.e.
current and voltage) when stored energy in the inductor or
capacitor is released to the resistive part of the network
(containing no independent sources).
The step response of an RL or RC circuit is its behavior when a
voltage or current source step is applied to the circuit, or
immediately after a switch state is changed.
Natural Response of an RL
Circuit
• Consider the following circuit, for which the switch is
closed for t < 0, and then opened at t = 0:
t=0
Io
Ro
i
L
+
R
v
–
Notation:
0– is used to denote the time just prior to switching
0+ is used to denote the time immediately after switching
• The current flowing in the inductor at t = 0– is Io
Recall: The current flowing in an inductor cannot change instantly,
and the voltage across a capacitor, which is proportional to the
charge stored in the capacitor, cannot change instantly. For a
first-order circuit these are called initial values of current and
voltage. A long time after the circuit configuration changes, the
currents and voltages achieve their final, or steady-state values.
Later when we talk about second-order circuits – ones that consist of
resistors and the equivalent of two energy storage elements, like an L and
a C or two Cs – we’ll take a look at the initial and final values of these
quantities and their time derivatives.
Solving for the Current (t  0)
• For t > 0, the circuit reduces to
i
Io
L
Ro
+
R
v
–
• Applying KVL to the LR circuit:
• Solution:
i (t )  i (0)e
( R / L )t
What Does e-t/t Look Like?
e-t/t with t = 10-4
• t is the amount of time necessary
for an exponential to decay to
36.7% of its initial value.
• -1/t is the initial slope of an
exponential with an initial value of
1.
Solving for the Voltage (t > 0)
i (t )  I o e+ ( R / L ) t
Io
Ro
L
R
v
–
• Note that the voltage changes abruptly:
v (0 )  0
for t  0, v(t )  iR  I o Re

 v (0 )  I o R
( R / L ) t
Time Constant t
• In the example, we found that
i (t )  I o e  ( R / L ) t and v(t )  I o Re  ( R / L )t
L
t 
• Define the time constant
R
– At t = t, the current has reduced to 1/e (~0.37)
of its initial value.
– At t = 5t, the current has reduced to less than
1% of its initial value.
Transient vs. Steady-State Response
• The momentary behavior of a circuit (in response
to a change in stimulation) is referred to as its
transient response.
• The behavior of a circuit a long time (many time
constants) after the change in voltage or current is
called the steady-state response.
Review (Conceptual)
• Any first-order circuit can be reduced to a
Thévenin (or Norton) equivalent connected to
either a single equivalent inductor or capacitor.
RTh
RTh
L
VTh
–
+
ITh
C
– In steady state, an inductor behaves like a short circuit
– In steady state, a capacitor behaves like an open circuit
Natural Response of an RC Circuit
• Consider the following circuit, for which the switch is
closed for t < 0, and then opened at t = 0:
+
Vo 
Ro
C
t=0
+
v
–
R
Notation:
0– is used to denote the time just prior to switching
0+ is used to denote the time immediately after switching
• The voltage on the capacitor at t = 0– is Vo
Solving for the Voltage (t  0)
• For t > 0, the circuit reduces to
+
Vo 
i
+
Ro
C
v
–
• Applying KCL to the RC circuit:
• Solution:
v(t )  v(0)e
 t / RC
R
Solving for the Current (t > 0)
i
+
Vo 
+
Ro
C
v
R
–
i (0 )  0
v(t )  Vo e  t / RC
• Note that the current changes abruptly:
v Vo t / RC
for t  0, i (t )   e
R R
Vo

 i (0 ) 
R
Time Constant t
• In the example, we found that
v(t )  Vo e
t / RC
Vo t / RC
and i (t ) 
e
R
• Define the time constant t  RC
(with R in ohms and C in
farads, t is in seconds)
– At t = t, the voltage has reduced to 1/e (~0.37)
of its initial value.
– At t = 5t, the voltage has reduced to less than
1% of its initial value.
Natural Response Summary
RL Circuit
RC Circuit
i
L
+
v R
C
R
–
• Inductor current cannot
change instantaneously

•
Capacitor voltage cannot
change instantaneously

v (0 )  v (0 )
t / t
v(t )  v(0)e t /t
i ( 0 )  i (0 )
i(t )  i(0)e
L
• time constant t 
R

•
time constant

t  RC
Transient Response of 1st-Order Circuits
• We saw that the currents and voltages in RL and RC
circuits decay exponentially with time, with a
characteristic time constant t, when an applied current or
voltage is suddenly removed.
• In general, when an applied current or voltage suddenly
changes, the voltages and currents in an RL or RC
circuit will change exponentially with time, from their
initial values to their final values, with the characteristic
time constant t as follows:



x(t )  x f  x(t0 )  x f e
 ( t t 0  ) / t
where x(t) is the circuit variable (voltage or current)
xf is the final value of the circuit variable
t0 is the time at which the change occurs
This is a very useful equation!
Procedure for Finding Transient Response
1. Identify the variable of interest
•
•
For RL circuits, it is usually the inductor current iL(t)
For RC circuits, it is usually the capacitor voltage vc(t)
2. Determine the initial value (at t = t0+) of the
variable
•
Recall that iL(t) and vc(t) are continuous variables:
iL(t0+) = iL(t0) and vc(t0+) = vc(t0)
•
Assuming that the circuit reached steady state before
t0 , use the fact that an inductor behaves like a short
circuit in steady state or that a capacitor behaves like
an open circuit in steady state
Procedure (cont’d)
3. Calculate the final value of the variable
(its value as t  ∞)
•
Again, make use of the fact that an inductor
behaves like a short circuit in steady state (t  ∞)
or that a capacitor behaves like an open circuit in
steady state (t  ∞)
4. Calculate the time constant for the circuit
t = L/R for an RL circuit, where R is the Thévenin
equivalent resistance “seen” by the inductor
t = RC for an RC circuit where R is the Thévenin
equivalent resistance “seen” by the capacitor
Example: RL Transient Analysis
Find the current i(t) and the voltage v(t):
t=0
R = 50 W
i
Vs = 100 V
+

+
v
L = 0.1 H
–
1. First consider the inductor current i
2. Before switch is closed, i = 0
--> immediately after switch is closed, i = 0
3. A long time after the switch is closed, i = Vs / R = 2 A
4. Time constant L/R = (0.1 H)/(50 W) = 0.002 seconds
i(t )  2  0  2 e (t 0) / 0.002  2  2e 500t Amperes
t=0
R = 50 W
i
Vs = 100 V
+

+
v
L = 0.1 H
–
Now solve for v(t), for t > 0:
From KVL,


v(t )  100  iR  100  2  2e 500t 50
Example: RC Transient Analysis
Find the current i(t) and the voltage v(t):
R1 = 10 kW
Vs = 5 V
+

R2 = 10 kW
t=0
i
+
v
C = 1 mF
–
1. First consider the capacitor voltage v
2. Before switch is moved, v = 0
--> immediately after switch is moved, v = 0
3. A long time after the switch is moved, v = Vs = 5 V
4. Time constant R1C = (104 W)(10-6 F) = 0.01 seconds
v(t )  5  0  5 e (t 0) / 0.01  5  5e 100t Volts
R1 = 10 kW
Vs = 5 V
+

R2 = 10 kW
t=0
i
+
v
C = 1 mF
–
Now solve for i(t), for t > 0:
From Ohm’s Law,
Vs  v(t ) 5  5  5e 100t 
i (t ) 

R1
10 4
Application to Digital Integrated
Circuits (ICs)
When we perform a sequence of computations using a
digital circuit, we switch the input voltages between logic 0
(e.g. 0 Volts) and logic 1 (e.g. 5 Volts).
The output of the digital circuit changes between logic 0
and logic 1 as computations are performed.
Digital Signals
We send beautiful pulses in:
voltage
We compute with pulses.
But we receive lousy-looking
pulses at the output:
voltage
time
time
Capacitor charging effects are responsible!
• Every node in a real circuit has capacitance; it’s the charging
of these capacitances that limits circuit performance (speed)
Circuit Model for a Logic Gate
• Electronic building blocks referred to as “logic gates”
are used to implement logical functions (NAND,
NOR, NOT) in digital ICs
– Any logical function can be implemented using these gates.
• A logic gate can be modeled as a simple RC circuit:
R
+
Vin(t)
+

C
Vout
–
switches between “low” (logic 0)
and “high” (logic 1) voltage states
Logic Level Transitions
Transition from “0” to “1”
(capacitor charging)

Vout (t )  Vhigh 1  et / RC

Vout
Transition from “1” to “0”
(capacitor discharging)
Vout (t )  Vhighet / RC
Vout
Vhigh
Vhigh
0.63Vhigh
0.37Vhigh
0
time
RC
0
time
RC
(Vhigh is the logic 1 voltage level)
Vin
Sequential
Switching
What if we step up the input,
0
0
Vin
wait for the output to respond,
time
Vout
0
0
Vin
then bring the input back down?
time
Vout
0
0
time
R
Pulse Distortion
The input voltage pulse
width must be long enough;
otherwise the output pulse
is distorted.
+
+
Vin(t)
Vout
C
–
(We need to wait for the output to
reach a recognizable logic level,
before changing the input again.)
–
Pulse width = 0.1RC
6
5
4
3
2
1
0
Pulse width = RC
Pulse width = 10RC
6
5
4
3
2
1
0
Vout
Vout
Vout
6
5
4
3
2
1
0
0
1
2
Time
3
4
5
0
1
2
Time
3
4
5
0
5
10
Time
15
20
25
Example
Suppose a voltage pulse of width
5 ms and height 4 V is applied to the
input of this circuit beginning at t = 0:
t = RC = 2.5 ms
Vin
R
R = 2.5 kΩ
C = 1 nF
Vout
C
• First, Vout will increase exponentially toward 4 V.
• When Vin goes back down, Vout will decrease exponentially
back down to 0 V.
What is the peak value of Vout?
The output increases for 5 ms, or 2 time constants.
 It reaches 1-e-2 or 86% of the final value.
0.86 x 4 V = 3.44 V is the peak value
4
3.5
3
2.5
2
1.5
1
0.5
00
Vout(t) =
2
{
4
6
8
10
4-4e-t/2.5ms for 0 ≤ t ≤ 5 ms
3.44e-(t-5ms)/2.5ms for t > 5 ms
A Bit on Second-Order Circuits
A second-order circuit consists of resistors and the equivalent of two
energy storage elements (Ls, Cs). A second-order circuit is characterized
by a second-order differential equation (contains second-derivatives of time)
Example: A circuit containing R, L and C in series with a voltage source;
a circuit with R, L and C in parallel.
Initial and final values of v, i, dv/dt, and di/dt
Example: The switch in this circuit has been closed for a long time. It opens
at t = 0. Find: i(0+), v(0+), di(0+)/dt, dv(0+)/dt, i(infinite time), v(infinite time)
a. Values for t < 0
b. Values for t = 0+
c. Values for t = infinity
a. Values for t < 0
b. Values for t = 0+
c. Values for t = infinity
A 2nd Order RLC Circuit
i (t)
+
vs(t)
R
C
L
• Application: Filters
– A bandpass filter such as IF amplifier for
the AM radio.
– A lowpass filter with a sharper cutoff
than can be obtained with an RC circuit.
The Differential Equation
i (t)
vs(t)
+ vr(t) R
+
vc(t)
C
-
KVL around the loop:
vr(t) + vc(t) + vl(t) = vs(t)
t
+
vl(t) +
L
1
di(t )
Ri(t )   i( x)dx  L
 vs (t )
C 
dt
R di (t ) 1
d 2i (t ) 1 dvs (t )

i(t ) 

2
L dt
LC
dt
L dt
-
The Differential Equation
The voltage and current in a second order circuit is
the solution to a differential equation of the
following form:
2
d x(t )
dx(t )
2

2



0 x(t )  f (t )
2
dt
dt
(the forcing function –
the driving voltage or
current source)
x(t )  x p (t )  xc (t )
xp(t) is the particular solution (forced response)
and xc(t) is the complementary solution (natural
response).
The Particular Solution
• The particular solution xp(t) is usually a
weighted sum of f(t) and its first and
second derivatives.
• If f(t) is constant, then xp(t) is constant.
• If f(t) is sinusoidal, then xp(t) is sinusoidal
(with the same frequency as the source,
for a circuit of only linear elements)
The Complementary Solution
The complementary solution has the following
form:
x (t )  Ke st
c
K is a constant determined by initial conditions.
s is a constant determined by the coefficients of
the differential equation.
d 2 Ke st
dKe st
2
st

2



Ke
0
0
2
dt
dt
s 2 Ke st  2 sKe st  02 Ke st  0
s 2  2 s  02  0
Characteristic Equation
• To find the complementary solution, we
need to solve the characteristic equation:
s 2  20 s  02  0
  0
• The characteristic equation has two rootscall them s1 and s2.
xc (t )  K1e s1t  K 2 e s2t
s1  0  0   1
2
s2  0  0  2  1
Overdamped, critically damped and underdamped response of
source-free transiently excited 2nd-order RLC circuit
Assume circuit is excited by energy stored in C or L
of series RLC circuit.
Assume i(t) = K1es1t + K2es2t where (with slightly
different notation)
2
s1     2   0
s2     2   0 where
2
  R / 2L
  1/
LC
 = R/2L is called the damping factor and
 0  1 / LC is the undamped natural frequency
If  > 0  overdamped case a
If  = 0  critically damped case b
If  < 0  underdamped case c