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EE 215 Fundamentals of Electrical Engineering
Lecture Notes
First Order Circuits
8/8/01
Rich Christie
Overview:
Circuits with one capacitor or one inductor are called first order circuits, because they
give rise to first order linear differential equations.
Engineers have a love/hate relationship with differential equations. They describe the
things we do engineering with, like circuits, so we need to understand them and obtain
solutions. But we are not math majors! All we want is a solution! We don't care whether
a solution exists if we can't figure out what it is! We don't care that there are eight
different ways to solve a differential equation, we just want one way that works!
On the other hand, we DO care about what is going on in the physical object that the
equation describes. Math majors could care less about the real world. (OK, OK, this is
harsh on math majors…)
So what we'll do with first order circuits is first write the circuit equations from the
circuit. Then we'll solve them formally, just once to prove we can do it. Then we'll
develop a short cut so that all we have to do is write the form of the solution and fill in
some numbers from looking at the circuit. And that's the method you really use to solve
first order circuits.
Formal Solution:
Here's a circuit with a capacitor and a
resistor. The typical problem here is a
switch closure problem. The circuit sits
for a long time before t = 0. At t = 0, a
switch is closed. The voltages and
currents in the circuit change over time.
We want to know what the voltage v(t)
and current i(t) look like from time t = 0
to t = ∞.
t=0
vs
+
-
R
+
i
v
C
-
Note that there are actually two different circuits here: The one with the switch open,
which is around before t = 0, and the one with the switch closed.
Let's start by considering the circuit before t = 0. There is clearly no current, but what
about capacitor voltage? Well, it could be anything, really, in this case, so the problem
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has to specify this initial condition. Let v(0-) = 0. t = 0- means the time just before t = 0.
Now consider the circuit an instant after the switch closes. (This would be t = 0+.) Recall
that
i=C
dv
dt
No time goes by from 0- to 0+, so any change in voltage across the capacitor would
require infinite current, which is not possible. Therefore, capacitors cannot change
voltage instantaneously.
v (0 + )= v (0 ! )
That means all the source voltage appears across the resistor, by KVL. Then the current is
i (0 + )=
v s ! v (0 + ) v s
=
R
R
So the current did change instantaneously through the capacitor. We could calculate the
rate of change of capacitor voltage as
dv + i (0 + ) v s
(0 )= C = RC
dt
but let's try for the circuit equation instead. KCL at the capacitor is
v ! vs
dv
+C
=0
R
dt
dv
RC + v = v s and recall that v (0 + )= 0
dt
There are many, many ways to solve this equation. We'll do one formal solution before
we find a faster way…
Let's solve for the derivative, separate variables and integrate
dv v s ! v
=
dt
RC
dv
1
=!
dt note sign change
v ! vs
RC
1
1
" v ! v s dv = ! RC " dt + D where D is a constant of integration
2
t
+ D and I had to look up the integral
RC
v ! v s = Ae ! t / RC where A = e D
ln (v ! v s ) = !
v (t ) = v s + Ae ! t / RC
We can find the value of A from considering the initial condition,
so
v (0 + )= 0
v (0 + )= v s + Ae 0 = v s + A = 0
A = !v s
We can check this by considering what happens as t goes to infinity. In steady state, with
a constant source (one that does not change with time) we expect the capacitor current to
go to zero. (Think about it - initially, current (water) flows in or out, causing voltage
(pressure) to build up against the flow. Eventually the voltage (pressure) gets high or low
enough to stop the flow, and then the voltage (pressure) also stops changing.
dv
(! ) = 0
dt
Then from the circuit equation
dv
+ v = vs
dt
v = v s and i = 0
RC
Note that at t = infinity the capacitor looks like an open circuit. Replacing the capacitor
with an open circuit is an easy way to obtain steady state values. Anyway, the complete
solution is
v (t ) = v s ! v s e ! t / RC = v s (1 ! e ! t / RC )
and looks like
vs
t
3
Solution by Form of Solution:
t=0
Let's try a more complex problem.
We'll take the same approach, at
least until we have the circuit
equation.
Let's work on the initial condition, v
s
v(0+). First consider the circuit
before the switch is closed, for t <
0. Generally we assume that the
circuit has sat with the switch
open for a loooonng time. That means:
R1
R2
i
+
R3
-
+
v
C
-
dv !
(
0 )= 0 so i (0 ! )= 0
dt
The capacitor looks like an open circuit. (From the point of view of t = -∞, t = 0 looks
like t = ∞ does from t = 0…)
With the capacitor an open circuit, v(0-) is just a voltage divider.
v (0 ! )= v s
R3
R1 + R2 + R3
We could also see this as the open circuit voltage of the Thevenin equivalent seen by the
capacitor.
Now consider t = 0+, immediately after switch closure.
v (0 + )= v (0 ! )= v s
R3
is the initial condition we'll need for the solution.
R1 + R2 + R3
Before writing the circuit equation, let's find the new Thevenin equivalent seen by the
capacitor for t > 0.
Rt = R2 || R3 =
R 2 R3
R 2 + R3
R3
(note this has
R 2 + R3
gone up from v(0-).)
Rt
voc = v s
So the equivalent circuit is:
voc
+
-
i
+
v
C
-
4
Well, this looks like the first circuit we solved, so it has the same equation!
Rt C
dv
+ v = voc
dt
and the same form of solution
v (t ) = voc + Ae ! t / Rt C
but the value of A is different. At t = 0+, v = v(0+), NOT zero! Then
v (0 + ) = voc + Ae 0
A = v (0 + )! voc
v (t ) = voc + (v (0 + )! voc )e ! t / Rt C
Note that as t → ∞,
v (! ) = voc
which matches with the differential equation with dv/dt = 0.
The Real Fast Way to Solve RC Circuits (The Cookbook)
If you think about it, ANY circuit with one capacitor can be reduced to the equivalent
circuit with one resistor and one voltage source just by taking the Thevenin equivalent
seen by the capacitor. Which means that ANY circuit with one capacitor has a solution of
the form
v (t ) = voc + (v (0 + )! voc )e ! t / Rt C
Let's rewrite this slightly:
v (t ) = A + Be ! t / RtC
Of course A and B can be found from voc and v(0+), but you don't really need to remember
that. It's easier to find A and B from v(0+) and v(∞), doing a little algebra instead of
memorizing an equation.
The cookbook way of solving first order RC circuits is then:
1. Find the initial condition, v(0+), from v(0-).
2. Find the steady state solution, v(∞), using the fact that the capacitor is an open circuit
at t = ∞.
5
3. Find the value of RtC. (This value is so useful it has a special name, time constant, and
symbol, τ (lower case Greek tau)).
4. Write the solution,
v (t ) = A + Be ! t / RtC
5. Plug the two voltage values into the solution to find A and B. Use v(∞) first!
6. Make sure you answered the question being asked!
Cookbook Example:
t=0
5V
50k!
Given v(0-) = 0 V, find i(t) for 0 < t < ∞.
+
i
+
10µF
v
-
-
This problem has a twist, we are being asked for current! No fair! We know how to solve
for voltage! - Oh wait, if we know the voltage we can then find the current. OK.
1. At t = 0- v = 0 so at t = 0+ v = 0.
dv
(! ) = 0 , so i = 0, so v = vs = 5V.
2. At t = ∞,
dt
3. RC = 50k⋅10µ = 500m = 0.5 s (recall k = 103, µ = 10-6, m = 10-3).
4. v (t ) = A + Be ! t / 0.5 = A + Be !2 t
5. v(∞) = A = 5V. Then v(0+) = 5 + B = 0, B = -5.
Solution
v (t ) = 5 ! 5e !2t V
6. (Don't forget this step!)
v s ! v (t ) 5 ! 5 + 5e !2 t
i (t ) =
=
= 100e !2 t µA
50k
50 K
And we can sketch i(t) (a common request)
i
100mA
t
6
RL Circuits:
Now let's look at a circuit with an inductor instead of a capacitor. We'll start with the
simplest possible circuit:
t=0
vs
And we want to find the current i as a
function of time. (Voltage for capacitors,
current for inductors…)
R
+
i
+
v
-
L
Let's try writing the circuit equation and
hoping for a miracle…
-
From KVL:
v s ! iR ! L
L
di
=0
dt
di
+ iR = v s
dt
Hmm, compare this to
RC
dv
+ v = vs
dt
from the RC circuit. It's not quite the same - but we can make it closer…
v
L di
+i = s
R dt
R
NOW this is the same equation - which means it must have the same form of solution
(that's our miracle). Astute observers will note that the right hand side (the forcing
function in math-speak) is the Norton equivalent short circuit current, where the RC
circuit had the Thevenin equivalent open circuit voltage. Anyway, the form of the
solution is
i (t ) = A + Be " t / !
where time constant ! =
L
R
Here's an interesting thing to know about the time constant, which is useful when plotting
voltages or currents: If you take the voltage or current value at t = 0, (I'll use current,
since we’re talking about RL circuits just now) and its value at t = ∞, and its slope at t =
7
di
(0), and plot the straight line passing through i(0) with that slope, it will intersect
dt
i(∞) at t = τ!
0,
i(0)
slope
di
(0 )
dt
i(!)
t="
t=0
t
Of course, this is NOT what the graph of current would look like for the simple example
circuit. What would i(0) be for that circuit? What would the graph look like? And where
would τ be on that graph?
Inductor Example:
t=0
10K!
Find i(t) for 0 < t < ∞!
10V
+
+
i
-
v
10mH
-
Since we have essentially the same differential equation, and same form of solution, we
can use the same cookbook to find the solution.
1. Find the initial condition.
di
, the inductor current cannot change instantaneously. (It was voltage for
dt
the capacitor, now current for the inductor.)
From v = L
So, i(0+) = i(0-) = 0.
2. Find the steady state solution.
di
is zero, and inductor voltage is zero. That
dt
means the inductor is a short circuit in steady state. Then
In steady state (at t = ∞) with dc sources,
8
i (" ) =
vs
10V
=
= 1 mA
R 10 K!
3. Find the time constant
!=
L 10mH
=
= 1 µs
R 10 K"
4. Write the solution,
!6
i (t ) = A + Be ! t / " = A + Be ! t / 10 = A + Be !10
6
t
5. Plug the two boundary values (the values at t = 0 and t = ∞) into the solution to find A
and B. Use the steady state value first!
i (! ) = A = 1 mA
i (0) = A + B = 0
B = !1 mA
6t
i (t ) = 1 ! e !10 mA
6. Make sure you answered the question being asked!
Nothing sneaky in this question. Let's draw the graph, though. (Usually, I would not put
the time constant on the graph unless I was specifically asked for it.)
i(t)
1 mA
t=0
t=1µs
t
Multiple switching events:
Often we find that useful functions require switches that open and then close, possibly to
repeat in a cycle. How can we deal with this?
If the time between switching operations is "long", we can assume the circuit is in steady
state just before each switch operation. How long is "long"? Well, it depends on the time
constant τ. Usually we say the transient will settle out in 3-5 time constants.
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Of course, nothing is always that simple, and we would really like to solve circuits where
the switch changes happen less that 3-5 time constants apart. This really just requires
some attention to initial conditions and a simple modification of the form of solution.
Consider this problem:
5K!
The switch has been open for a long time.
At t = 0 the switch closes.
At t = 0.1s the switch opens(!).
+
10V
+
-
200µF
v
5K!
Sketch v(t) for 0 < t < ∞
-
The switch closure is a problem we've done before. Cookbook!
v (0 + )= v (0 ! )= 10 V
With the switch closed, the Thevenin equivalent seen by the capacitor has
5
=5V
5+5
Rt = 5 || 5 = 2.5 K!
voc = 10
So
v (! ) = 5 V
# = Rt C = 2.5K " 200µ = 500 ! 10 -3 = 0.5s
v (t ) = A + Be ! t / 0.5 = A + Be !2 t V
v (! ) = A = 5
v (0) = A + B = 10
B=5
v (t ) = 5 + 5e !2t V
Now for the switch reopening. It happens at to = 0.1s. So the form of solution has to be
modified to account for the new starting point.
v (t ) = A + Be " (t "to )/ !
Note that A, B and τ will have new values! We need the initial and steady state voltage.
The initial voltage is the voltage at t = 0.1- and we can find it from the first solution.
v (0.1" )= 5 + 5e "2!0.1 = 9.1 V
With the switch open,
10
v (! ) = 10 V
The new time constant is
" = RC = 5K ! 200µ = 1s
v (! ) = A = 10 V
v (0.1) = A + B = 9.1
B = 9.1 ! 10 = !0.9 V
v (t ) = 10 ! 0.9e ! (t !0.1)
The canonical way to express the solution is
5 + 5e %2 t V
0 < t < 0.1
% (t % to )
V 0.1 < t < $
!10 % 0.9e
#
v (t ) = "
And let's sketch the graph…
v(t)
10
9
0
0.1
t
Step Functions:
A switching change (open or close) can also be thought of as a step change in a source.
For example, in the previous problem the switch closure changes the source from 10V to
5V. A mathematical way of expressing step changes is the step function, u(t). The form of
solution to switch closure problems is sometimes called the step response of the circuit.
The step function has a magnitude of 1 (that is, the value changes from 0 to 1) and the
change occurs when the argument equals zero. Thus
#0 t < 0
u (t ) = "
!1 t > 0
1
0
t
11
An offset in the argument corresponds to a time offset in the change:
#0 t < to
u (t $ to ) = "
!1 t > to
1
0
to
t
Steps up and down are expressed by summing step functions:
v
v (t ) = u (t )! u (t ! to )
1
0
to
t
Can you figure out the step function expression of this graph?
v
2
1
0
1
2
3
4
t
-1
12
It's u (t ! 1)+ u (t ! 2 )! 3u (t ! 3)+ u (t ! 4 )
Notice how the magnitude of each step function term is the magnitude of the
corresponding change.
When a source is given as a step function, e.g.
v s = 3u (t ! 2 )! u (t ! 4 )V
Just divide the time up into segments when the source is constant, and treat each
transition point as a switch closure. Here the source has value 0 V from 0 < t < 2, 3 V
from 2 < t < 4, and 2 V for t > 4, and there are two "switch closures" to deal with, one at t
= 2 s and one at t = 4 s.
Non-Constant Sources: Exponentials:
So far we have examined sources with constant values. What happens when they change?
We need to review some math to handle this case.
When we write a differential equation for voltage, say, we put the terms with voltage in
them on the left side, and term(s) without voltage on the right. For example,
Rt C
dv
+ v = voc
dt
(Remember that voc is a constant, and the term does not contain the voltage variable v.)
Math geeks (and EEs) call the term on the right side the forcing function. It's a function
of time.
The differential equation has a natural response, (also called the homogeneous solution
in math-speak) which is the solution when the forcing function is zero, and a forced
response, (inhomogeneous solution) which is the solution as t→∞. The complete solution
is the sum of the natural response and the forced response.
Recall the form of solution with a constant forcing function,
v (t ) = A + Be " t / !
Here A is the forced response, and Be " t / ! is the natural response. From this we learn that
the form of the forced response to a constant is a constant.
13
Now we are ready to deal with forcing functions of other forms. Basically, we'll write the
form of solution for a couple of different forcing functions, and then evaluate the
coefficients of the solution. The first forcing function we will examine is the exponential
forcing function,
v s = be at
where a and b are constants. The form of the forced solution is
v f (t ) = Ae at
There is a special case when the coefficients of the exponentials are equal in the forced
and natural solution, i.e. when
a = "1 / !
In this case,
v f (t ) = Ate at
Example:
t=0
vs
Given v s (t ) = 10e !4 t u (t )V and v(0) = 0,
find v(t).
2M!
+
-
+
1µF
v
-
Don't worry about the u(t), it's just there
to keep the source equal to zero before t
= 0.
Form of solution
v (t ) = Ae at + Be " t / !
Better check τ!
" = RC = 2 M ! 1µ = 2 s
Good, not equal, the form is OK.
We find the coefficient of the forced solution by stuffing it in the differential equation.
That means we need to write the differential equation! KCL at the top capacitor node is
v ! vs
dv
+C
=0
R
dt
14
or
RC
dv
+ v = vs
dt
or
2
dv
+ v = 10e !4 t
dt
Now sub in
v = Ae !4 t
2
d
Ae !4 t + Ae !4 t = 10e !4 t
dt
2 " (! 4 )Ae !4 t + Ae !4 t = 10e !4 t
The exponentials cancel, leaving
! 8 A + A = 10
10
A=!
7
Which is an ugly number, and therefore suspect, but in this case it seems to be correct.
For those wondering what happened to the natural solution Be ! t / 2 in this process, what
happens is that the natural solution, by definition, goes to zero as t→∞, and we are
equating coefficients of the forced solution at or near t = ∞, where the natural solution is
zero and can be left out.
To evaluate coefficients in the natural solution, we have to use the initial condition at t =
0, and use the form of the complete solution (which is why we did the forced solution
first). The form of the complete solution is
10 !4 t
e + Be !t / 2
7
Plugging in t = 0,
v (t ) = !
v (0) = !
10
+B=0
7
so
15
10
7
and the complete solution is
B=
v (t ) = !
10 !4 t
(e ! e !t / 2 )V
7
Non-Constant Sources: Sinusoids:
Let's try a sinusoidal source. Sinusoidal steady state (essentially just the forced solution)
is an important operating state for many circuits, so important it had its own name,
alternating current. Here we'll look at the transient solution too.
The generic sinusoidal forcing function is
a sin("t + ! )
which can be rewritten
a1 cos !t + a 2 sin !t
using trigonometric identities. This clues us in to the form of the forced solution for
sinusoidal forcing functions,
A1 cos !t + A2 sin !t
For example, suppose that in the previous problem, the forcing function is
v s = 10 sin(10t )u (t )V
Again the u(t) just makes the forcing function zero for t < 0.
The form of the complete solution is
v (t ) = A1 cos10t + A2 sin 10t + Be ! t / 2
Let's plug the form of the forced solution into the differential equation, which is
dv
+ v = 10 sin 10t
dt
Plugging in
2
2
d
(A1 cos10t + A2 sin10t )+ A1 cos10t + A2 sin10t = 10 sin10t
dt
16
Doing the derivative
2 A1 (" sin 10t )! 10 + 2 A2 (cos10t )! 10 + A1 cos #t + A2 sin #t = 10 sin 10t
Now equate coefficients of sine and cosine.
! 20 A1 + A2 = 10
20 A2 + A1 = 0
Solve:
A1 = !20A2
! 20(! 20 A2 )+ A2 = 10
10
A2 =
401
200
A1 = !
401
And now solve for the natural response coefficient at t = 0
v (0) = !
200
10
cos(0)+
sin(0)+ Be !0 / 2 = 0
401
401
200
401
Complete solution
10
(
v (t ) = !
20 cos10t ! sin 10t ! 20e !t / 2 )V
401
B=
17