Time Response Of Electrical Circuits

        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.