Download FIRST ORDER CIRCUITS Introduction

FIRST ORDER CIRCUITS
Introduction
In this laboratory, two simple circuits consisting of only one capacitor and one resistor (RC)
and one inductor and one resistor (RL) will be examined. Voltage across the capacitor will be
observed as it charged or discharged through the resistor. The goals of this experiment are
to show that the charge/discharge follows an exponential fuction and to understand the
effect of the time constant of the circuit on the charge/discharge duration.
Theory (RC Circuits)
An uncharged capacitor can be charged by connecting its two terminals to the two terminals
of a battery. The rate at which the capacitor charges up decreases as time passes. Initially it
acts like a short circuit and when the capacitor is fully charged with its voltage essentially
equal to the voltage of the source, it acts like an open circuit. If the voltage source removed
from the circuit, the capacitor becomes the voltage source of the circuit as it discharges,
allowing current to flow until the voltage of the capacitor is zero.
The rate of the charge/discharge inversely proportional with the time constant of the circuit
which is defined as;
𝜏 =𝑅 ×𝐶
(1)
where C is the equivalent capacitance of the circuit and R is the equivalent resistance of the
circuit. Given the resistors in ohm and capacitors in farad, the time constant is defined in
seconds. Generally, 𝟓𝝉 seconds is enough for the capacitor to fully charge/discharge. Figure
1 illustrates a basic RC circuit. While the switch is at position 1, the capacitor charges. The
voltage of the capacitor, Vc, is defined by the Equation 2 given below.
1
+ Vr -
2
R
+
Vc(t)
-
Vs
Figure 1
C
𝑉𝑐 (𝑡) = 𝑉𝑠 (1 − 𝑒 −𝑡/𝑅𝐶 ) , 𝑡 ≥ 0
(2)
Vc(t)
Vs
0.63Vs
τ 2τ 3τ 4τ 5τ
t
Figure 2
In Equation 2, the effect of the exponential term decreases as time passes so the voltage of
the capacitor approaches to the source value (Vs). Figure 2 demonstrates the change of the
capacitor voltage with time. One can observe from the figure that more than the half of the
capacitor charges just in 𝜏 seconds. This behaviour of the circuit is called “transient
response”. After 𝟓𝝉 seconds, voltage and current values in the circuit became stable and the
response of the ciruit after that is called “steady state response”.
When the switch position is changed to 2, the voltage source is cancelled and the capacitor
acts like the source for the circuit, allowing the current flow over the resistor. With Vo
defining the initial voltage stored in the capacitor at the switch time, the discharge voltage
for the capacitor is given by Equation 3:
𝑉𝑐 (𝑡) = 𝑉𝑜 𝑒 −𝑡/𝑅𝐶 , 𝑡 ≥ 0
(3)
Vc(t)
V0
0.37Vo
τ 2τ 3τ 4τ5τ
t
Figure 3
As seen in Figure 3, the capacitor also discharges more than the half of the initial voltage in 𝜏
seconds.
In practice, instead of switching, square wave signal is given to circuit as the voltage source.
The capacitor charges in pulse time and discharges until the next pulse. If the pulse duration
is not long enough, the capacitor cannot be fully charged. Similarly, if the non-pulse duration
is not long enough, the capacitor cannot be fully discharged, so every time a new pulse
arrives, the capacitor will be charged more than the previous time. Sufficient time later, the
voltage change of the capacitor becomes periodic between two voltage values.
Figure 4 shows the voltage values of the source, resistor and capacitor. At the top of the
figure, T1 and T2 indicates pulse and non-pulse duration for the square wave signal. T is the
period of the wave. The capacitors voltage is shown in the middle. The voltage of the resistor
is given at the bottom. Notice the difference between maximum and minimum voltage
values.
Vs(t)
T
Vs
t
T1
T2
Vc(t)
V0
t
VR(t)
Vs
t
-V0
Figure 4
+ Vr R
+
Vc(t)
Figure 5
C
Theory (RL Circuits)
Previous analysis showed that the capacitor does not allow the voltage to change rapidly.
Inductors do the same thing to the current. Consider that in Figure 1, the capacitor is
replaced with an inductor and the switch stayed at position 2 long enough for the inductor
to release its all magnetic energy. When the switchs position changes to 1, the inductor acts
like an open circuit initially, does not allow current to flow. After a time the current flow
becomes constant and reaches to its maximum value which is determined by the resistance
of the circuit.
The time the inductor needs to store maximum current is again determined by the time
constant of the circuit. In RL circuits, the time constant is derived by dividing the inductance
of the circuit to the resistance of the circuit (Equation 4).
𝜏 = 𝐿/𝑅
(4)
At arbitrary time t, the current flows over the inductor is defined by Equation 5 given below.
Is is the maximum current of the circuit, which depends on the resistance. One can see the
similarity between Equation 5 and Equation 2. This similarity suggest that more than the half
of the maximum current flows through the inductor at 𝜏 seconds.
𝐼𝑖 (𝑡) = 𝐼𝑠 (1 − 𝑒 −𝑡𝑅/𝐿 ) , 𝑡 ≥ 0
(5)
The inductor supplies current to the circuit in the absence of an external source, e.g. while
the swith is at position 2 in Figure 1. In a situtaion like this, the current at time t of the
inductor can be calculated with the aid of Equation 6 where I o is the initial current stored in
the inductor at the switch time. This equation shows that the current of the inductor follows
the same path as the voltage of the capacitor given in Figure 3.
𝐼𝑖 (𝑡) = 𝐼𝑜 𝑒 −𝑡𝑅/𝐿 , 𝑡 ≥ 0
(6)
Preliminary work
1. For R = 1 kΩ, C = 100 µF, simulate the circuit given in Figure 5. Use 10 V p-p (0 to +10 V)
square wave at 0.5 Hz. Use these parameters in step 2,3 and 4.
2. Calculate the time constant (𝜏) for the circuit, use the value you find to calculate
voltage of the capacitor at time 𝜏 after capacitor starts charging.
3. Let i(t) be the current of the resistor. Find the expression for i(t) and draw the change
of the current for two periods.
4. Calculate the time constant graphically from the simulation. Compare the results you
found in step 2. Explain the differences, if any.
5. For the capacitor values 1 nF and 1 mF, do steps 2, 4.
6. In the circuit given in Figure 6, R = 5.6 kΩ, C1 = 4.7 µF and C2 = 47 µF. Calculate the
time constant of the circuit. Calculate the period of the square wave voltage source
that gives sufficient time to the capacitors to fully charge and discharge. For 15 Vp-p
square wave, draw the current and voltage of the resistor for two periods.
7. For the same values given in step 6, calculate the time constant of the circuit given in
Figure 7. Compare your result with the time constant you found in step 6.
8. Consider that in Figure 1 the capacitor is replaced with an inductor with a value of
100 H. R = 1kΩ and the source generates 5 Vp-p square wave. Calculate the time
constant of the circuit. Knowing that the switch stayed at position 2 long enough, so
no current flows in the circuit, at time 𝑡 = 0, the switchs position changed to 1. Draw
the voltage and current of the resistor from 𝑡 = 0 to 𝑡 = ∞. Also calculate their
values at 𝑡 = 𝜏, 𝑡 = 3𝜏, 𝑡 = 5𝜏 and show them on your plots.
9. Simulate the circuit with 5 Vp-p (0 to +5 V) square wave at 1.5 Hz and 100 H inductor.
Plot the current of the resistor for R = 100 Ω, R = 1 kΩ and R = 10 kΩ and compare
them. Explain the differences. Investigate the effect of the source amplitude by using
10 Vp-p square wave for each of three R values. Discuss the results.
10. In the circuit given in Figure 6, the capacitors are replaced with inductors L 1 = 2.2 µH,
L2 = 0.2 mH and R = 1.57 kΩ. Calculate the time constant of the circuit. Calculate the
period of the square wave voltage source that gives sufficient time to the inductors
to store maximum energy. For 12 Vp-p square wave, draw the current and voltage of
the resistor for two periods.
Experiment
1. Construct the circuit given in Figure 5 on your board. (Note: Check your signal
generator with the oscilloscope and be sure you got 0 to +10 V square wave at 10
Hz.)
2. For R = 100 kΩ, C = 100 nF, display the voltages of the resistor and capacitor on the
oscilloscope at the same time (use 2 channels). Calculate the time constant with the
aid of cursors.
3. Repeat step 2 for R = 1 kΩ, C= 100 µF. Compare the results.
4. Change the frequency of the square wave to 50 Hz. Display the voltages of the
resistor and the capacitor on the oscilloscope for the capacitor values 100 nF and 100
µF while R = 1 kΩ. Change the frequency to 200 Hz and repeat for all capacitor values.
Explain the differences in graphics.
5. Construct the circuit given in Figure 6 on your board. Use elements with the values
R = 5.6 kΩ, C1 = 4.7 µF and C2 = 47 µF. Adjust the source frequency to 5 Hz. Calculate
time constant on the oscilloscope using the cursors. Is there a difference between
theoritical (which you may already calculated at preliminary work step 6) and
practical results? If yes, what is the reason for that?
6. Construct the circuit given in Figure 7 on your board with the same values given in
step 5. Adjust the frequency of the source so that the capacitors will never be fully
stored or empty. Using the oscilloscope, verify that the frequency you select is
suitable for this criterion. Explain how did you decide the frequency.
7. Construct the circuit given in Figure 8 with 10 Vp-p square wave at 5 kHz as source.
Display the voltages of the resistor and the source on the oscilloscope for the
inductor values 10 µH and 150 µH while R = 1 Ω. Change the frequency to 500 Hz and
repeat for all ,inductor values. Explain the differences in graphics.
8. Construct the circuit given in Figure 9 on your board.For L1 = 4.7 mH, L2 = 1.5 mH
calculate the R value that allows the empty inductors to store maximum energy every
period under 10 Vp-p square wave at 200 Hz input signal. Also calculate the time
constants with the oscilloscope.
+ Vr R
C1
+
Vc(t)
-
C2
Figure 6
+ Vr R
C1
Figure 7
C2
L
+
VR(t)
-
Figure 8
L1
L2
+
VR(t)
-
Figure 9
Important Notes
Since you have to submit your report before leaving the laboratory, do not
forget to bring several blank papers to show your calculations and draw the
figures you see on the oscilloscope in the experiments.
Equipments
Resistors:
100 k, 1 k, 5.6 k, 1, 10 Ω
Capacitors:
100 nF, 100 µF, 4.7 µF, 47 µF
Inductors:
4.7 mH, 10 µH, 150 µH, 1.5 mH