By this time you have probably studied DC circuits and looked at AC circuits.
You may have calculated how a circuit responds to a sinusoidal signal.
However, there are many situations in which the time behavior of a circuit
is important, and the signal is not a sinusoidal signal, or the circuit
is not a DC circuit. Here is an example.
In a 555 timer circuit, a DC voltage is applied to a series combination
of two resistors and a capacitor. The capacitor charges and eventually
triggers an electronic switch to change state causing the capacitor to
discharge through one of the resistors. In that way, the charging
and discharging of the capacitor controls the timing of an ever continuing
sequence of square pulses at the output of the 555 chip.
In this lesson, we will look at the time response of electrical circuits.
We will work from simple responses to more complicated response.
We will find that circuits can have several causes for the responses we
observe. We start with the simple circuit shown at the right.
We will assume that we have a capacitor that has a charge on it so that
the initial voltage across the capacitor is 10 volts. At some arbitrary
time - we will call it t = 0 - the switch is closed and current begins
to flow from the capacitor through the resistor, resulting in a discharge
of the capacitor. We want to determine - mathematically - how the
voltage across the capacitor behaves in time.
Start by writing KCL at the node between the capacitor and the resistor.
C (dVC/dt)
+ (VC /R) = 0
This equation can be rearranged to work
toward a solution.
dVC/dt
= - VC /(RC)
This expression for the rate of change
of the capacitor voltage is interesting by itself. It says that as
time goes on the capacitor voltage decreases - it tends to zero.
As the capacitor voltage gets smaller the rate of change also gets smaller.
Consider a circuit with one resistor and one capacitor. Imagine that
the circuit is suddenly connected to a constant voltage source. We
will represent the situation with a voltage source and a switch.
At some time - which we will arbitrarily call t = 0 - the voltage source
is connected to the resistor-capacitor (RC) circuit. We need to determine
what takes place in the circuit after the switch is thrown. Note
the following:
lWhat
you have learned about AC is not going to help here because there are no sinusoidal signals involved. Phasor analysis will not work. l
But, the circuit is not a DC circuit either since, something
else is going on here.
Exactly what will happen here? There is a dynamic situation here.
For purposes of discussion, we will assume that there is no charge stored
on the capacitor when the switch is thrown. Then the following will
take place.
lCurrent
is going to flow through the resistor, into
the capacitor. That happens because the voltage
across the capacitor is zero to start with,
so closing the switch puts a voltage across the
resistor causing current flow.
lAs
current continues to flow through the resistor into
the capacitor, charge builds up on the capacitor
and the voltage across the capacitor starts
to increast.
lAs
the current continues to flow, the voltage buildup
on the capacitor decreases the voltage across
the resistor, resulting in decreased current
flow through the resistor.
lThe
decreased current flow through the resistor causes
a slowdown in the voltage buildup across
the capacitor.
lAs
time goes on, the capacitor slowly approaches a steady
state voltage and current slow diminishes to zero.
lThe
steady state voltage across the capacitor will equal
the voltage of the applied DC source, and the
voltage across the resistor will be zero with zero
current flowing through the resistor.
Now, having walked through a verbal description of what happens, you should
be able to do the analysis mathematically. Let's take a shot at that.
Here is the circuit again. At some arbitrary time - call that time
t = 0 - the switch is closed. At that time, the circuit becomes the
one shown below. After the switch is closed, we can write KCL at
the output node. Here is the equation for KCL:
C (dVout/dt)
+ (Vout - Vs)/R = 0
This equation can be rearranged into the
standard form of a first order linear differential equation. Doing
that, we find:
RC (dVout/dt)
+ Vout = Vs
Since
Vs is a constant, we know that the solution of this equation
will have the form: (since the particular solution will be a constant).
vout(t)
= C1 + Homogeneous Solution
C1 is a yet-to-be-determined
constant, and we have:
vout,P(t)
= C1
The homogeneous solution is the solution of:
RC (dVout/dt)
+ Vout = 0
By several solution methods, the solution
of this equation can be found to be:
vout,H(t)
= C2 e-t/RC
Putting the particular solution and the
homogeneous solution together, we get:
vout(t)
= vout,P(t) + vout,H(t)
or:
vout(t)
= C1 + C2 e-t/RC
To finish this off, we must determine the
two constants, C1 and C2.
We know that the exponential term dies out for large time, t. When
the exponential dies out, the voltage across the capacitor must equal the
input voltage, Vs. If that is true, then:
C1
= Vs
Similarly, the initial voltage on the capacitor
must be zero:
vout(0)
= C1 + C2 e-0/RC = C1 + C2
= 0
And, that in turn implies:
C1
= - C2
or:
C2
= - Vs
so that:
vout(t)
= Vs ( 1 - e-t/RC )
This
mathematical expression looks like the figure at the right. This
plot is done for RC = .02 sec and Vs = 10 v.
In this circuit, the product of the resistance and capacitance - RC
- has the units of seconds, and is referred to as the time
constant of the circuit. You will also find time constant
behavior like this in other situations. You will find time constant
behavior in many other physical situations that satisfy a first order differential
equation like the one satisfied by this circuit. Here are a few examples.
v
A Simple Thermal System
Here is a heated Space with Insulation. In this system heat flows
into a heated space and the temperature within the heated space follows
a first order linear differential equation.
The system diagram
The system equation
Here's one more system that satisfies a first order differential equation.
v
Your memory.
Psychologists tell us that memory obeys the same kind of differential equation
as the previous two systems. If you learn information, what you retain
satisfies a first order differential equation.
v
Is this airplane a system with a time constant?
Click the button to show the airplane's path. Release the mouse button
outside of the red button to keep the path visible.
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