West Virginia University
College of Engineering & Mineral Resources
Lane Department of Computer Science and Electrical Engineering
EE224
Electric Circuits Laboratory
S 2006
Experiment No.1
“Time Constants - Series RC Circuit with DC Applied”
Objective: This experiment will give students some exposure to basics in electrical circuits.
Student should design a RC circuit and measure voltage changes in a resistive-capacitive series
circuit, then using the measured voltage to plot these changes on linear graph paper, and last to
verify the experimental results by calculation.
Theoretical Background:
RC time constant circuits are used extensively in electronics for timing (setting oscillator
frequencies, adjusting delays, blinking lights, etc.) [1]. It is necessary to understand how RC
circuits behave in order to analyze and design timing circuits.
In the circuit in the figure 1, with the switch open, the capacitor is initially uncharged, and so has
a voltage, Vc, equal to 0 volts. When the single-pole single-throw (SPST) switch is closed,
current begins to flow, and the capacitor begins to accumulate stored charge. Since Q=CV, as
stored charge (Q) increases, so does the capacitor voltage (Vc). However, the growth of
capacitor voltage is not linear; in fact, it’s called an exponential growth.
Kirchoff's Loop rule states that the potential drop across the resistance, I*R, plus the potential
drop across capacitance, q/C, must equal the voltage Vs provided by the power supply; that is,
Vs = I ⋅ R + q / C
(1)
Since the current is the rate at which charges flow though the circuit,
I=
dq
dt
(2)
this expression represents a differential equation in q.
Vs =
dq
q
⋅R+
dt
C
(3)
The solution yields the charge on the capacitor as a function of time; specifically,
(
q = Qmax 1 − e − t / τ
1
)
Q = C ⋅V
,τ = R ⋅ C
(4)
Using this result, one can determine the expression for the voltage across the plates of the
capacitor as a function of time; specifically,
Vc = Vmax (1 − e − t / τ )
(5)
where Vmax is the voltage of the power supply (Vs in our case).
τ = R ⋅ C = time constant
This relation indicates that the voltage across the capacitor will start at 0 and increase quickly at
first, then more slowly. It will approach the value Vmax but never quite reach it. Over a time
interval of one time constant τ = R•C, the difference between the voltage and Vmax is decreased
by 63% = (1 - 1/e).
If you can fit an exponential curve of form
Vc = Vmax (1 − e − t / τ )
(6)
to the data values for voltage as a function of time, you can determine the time constant. If you
know the time constant and the capacitance C, you can figure out what the resistance R must be.
On the other hand, if you throw the switch to the right, then you remove the power supply from
the circuit, and complete a short circuit running in a circle around the resistor and capacitor. This
will discharge the capacitor: the charge which was stored on its plates will now run back
together, bringing the capacitor back to a neutral state. In this case, the voltage across the
capacitor ought to decrease:
Vc = Vmax (e − t / τ )
(7)
One can fit an exponential function to this data to find the time constant. It ought to agree with
the time constant measured as the capacitor was charged. The following definitions describe each
of these symbols:
Vs = the maximum possible voltage change that will occur in the circuit during a time lapse of
five time constant (in our case).
t = the lapsed time in seconds that the circuit voltages and currents have been changing.
τ = the time constant of the circuit. τ is the product of R and C (τ = RC) and the unit of τ is
seconds.
Vc = the capacitor voltage at any instant of time after the switch closes.
So, the formula which gives the instantaneous voltage across the capacitor as a function of time,
is:
vC = V S (1 - e
(− t / τ )
2
)
Figure1. Series RC Circuit
This formula describes exponential growth, in which the capacitor voltage is initially 0 volts, and
grows to a value of VS after enough time has gone by. It is important to understand each term in
this formula in order to be able to use it as a tool.
Below (Fig. 2) is a graph of the capacitor voltage in a circuit where the time constant, τ, is 1.0
second, and VS is 100 volts (for example). Notice that after about 5 seconds (i.e. 5 τ) the
capacitor voltage has grown from 0 a volt to 100 volts, and the capacitor voltage is 100 volts and
appears to have stopped changing. This is a condition called “steady-state”. While in theory it
takes an infinite time for “steady-state” conditions to be reached, in practical terms, after five
time constants using practical lab instruments it is nearly impossible to measure any changes
occurring.
Figure2. Graph of Capacitor Voltage (vc) versus Time.
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Experimental Procedure:
Construct the circuit on your proto - board for testing in the laboratory. Let C = 47 µF,
at t = 5 sec., Vs = 12 V and Vc = 7.56 V.
1. Since it takes five times constants for a capacitor to fully charge, calculate time constant and
appropriate value of resistor to be used in this experiment by using equation (5).
2. Circuit should be powered from the 12 V DC power supply (Vs). Notice the change of the
voltage at the capacitor side as time changes.
3. Calculate the resistor voltage using VR = Vs - VC. (Note: This is just Kirchhoff’s Voltage
Law applied: the sum of the resistor voltage and the capacitor voltage must equal the source
voltage).
4. Construct a single graph. Plot VR vs. time, and VC vs. time. Realize that:
•
•
The resistor voltage starts at VS and exponentially decays to nearly 0 volts.
The capacitor voltage starts at 0 volts and exponentially grows to nearly VS volts.
5. Observe the capacitor voltage on Oscilloscope
6. Simulate the same circuit in PSPICE.
Note: Please follow the format of the report given in the syllabus.
[1] Reference: www. EducatorsCorner.com Experiments.
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