SECOND ORDER CIRCUITS
Introduction
In this laboratory, natural and step responses of parallel and series RLC circuits will be
examined. Main goals of this experiment are to understand the response curves of over-,
under- and critically-damped circuits and to learn how to choose right values for R,L and C to
prevent overshooting damage.
Theory
A parallel RLC circuit is shown in Figure 1. Using Krichoff’s current law and perform time
derivative, a linear second order ordinary differential equation (ODE) can be found (Equation
1).
𝑑2 𝑣
1 𝑑𝑣
𝑑𝑡
𝑅𝐶 𝑑𝑡
2 +
+
𝑣
𝐿𝐶
=0
(1)
As the capacitor voltage cannot change abruptly and the same is true for the inductor
current, Equations 2 and 3 can be obtained.
iC
+
Vo
-
iL
C
L
ir
R
Io
+
v
-
Figure 1
𝑣(0+ ) = 𝑉𝑜
𝑣 ′ (0+ ) = −
𝐼𝑜
𝐶
−
𝑉𝑜
𝑅𝐶
(2)
(3)
Assume the solution is v(t) = Aest , where A, s are unknown constants to be solved. Substitue
into the ODE, a characteristic equation of s determined by the circuit parameters is obtained
(Equation 4).
𝑠2 +
𝑠
𝑅𝐶
+
1
𝐿𝐶
=0
(4)
The general solution of v(t) must be of the form:
𝑣(𝑡) = 𝐴1 𝑒 𝑠1𝑡 + 𝐴2 𝑒 𝑠2𝑡
(5)
where the constants A1 and A2 will be determined by the two initial conditions. In general, s
has two roots:
𝑠1,2 = −
1
2𝑅𝐶
∓ √(
1
2
) −
2𝑅𝐶
1
𝐿𝐶
= −𝛼 ∓ √𝛼 2 − 𝜔𝑜2
(6)
where 𝛼 is called the neper frequency and ωo is the resonance (natural) frequency.
The types of natural response are determined by the relative magnitudes of 𝛼 and ωo.
Table 1 summarizes this relations.
𝛼 > 𝜔𝑜
𝛼 < 𝜔𝑜
𝛼 = 𝜔𝑜
Table 1: Types of natural response
The circuit is over-damped
The circuit is under-damped
The circuit is critically-damped
Preliminary work 1: Over-damped Response
Consider that in Figure 1 C=0.2 µF, L=50 mH, R=200 Ω, V0=12 V and I0=30 mA.
a) Calculate the neper and resonance frequencies and show that the circuit is overdamped.
b) Find the expansion coefficients A1 and A2 using the equations given above.
c) Use the expansion coefficients to calculate the v(t) at t=0, 100, 250,500,750 and 1000
µs.
d) Derive the expressions for the branch currents through L and C (iL(t) and iC(t)).
Preliminary work 2: Under-damped Response
For under-damped response, the two roots of s are complex conjugate pair (Equation 7).
𝑠1,2 = −𝛼 ∓ √𝛼 2 − 𝜔𝑜2 = −𝛼 ∓ 𝑗𝜔𝑑
(6)
ωd the damped frequency defined as;
𝜔𝑑 = √ 𝜔𝑜2 − 𝛼 2
(7)
The general solution is reformulated as;
𝑣(𝑡) = 𝑒 −𝛼𝑡 (𝐵1 𝑐𝑜𝑠𝜔𝑑 𝑡 + 𝐵2 𝑠𝑖𝑛𝜔𝑑 𝑡)
(8)
a) Find the derivative of v(t).
b) Take C=0.250 µF, L=10 mH, R=2 kΩ, V0=12 V and I0=-12.25 mA. Calculate the neper
and resonance frequencies and show that the circuit is under-damped.
c) Calculate the expansion coefficients B1 and B2
d) Derive the expression of v(t).
Preliminary work 3: Critically-damped Response
The general solution and its derivative for the critically-damped response is;
𝑣(𝑡) = 𝑒 −𝛼𝑡 (𝐷1 𝑡 + 𝐷2 )
𝑣 ′(𝑡) = 𝑒 −𝛼𝑡 [(𝐷1 − 𝛼𝐷2 ) − 𝛼𝐷1 𝑡]
(9)
(10)
a) Take C=0.125 µF, L=8 H, V0=0 V and I0=-12.25 mA. Find the R value such that the
circuit is critically-damped.
b) Calculate the neper frequency.
c) Find the expansion coefficients D1 and D2.
d) Derive the expression of v(t).
The step response of the circuit is the sum of the final current and the natural response.
Consider that a current source, Is, is placed in Figure 1. The equation of the solution becomes
𝑖𝐿 (𝑡) = 𝐼𝑓 + 𝑖𝐿,𝑛𝑎𝑢𝑡𝑟𝑎𝑙 (𝑡)
(11)
where If is Is, as the inductors final current value will be equal to the current source. The
three types of natural responses are given above.
Preliminary work 4: Step response of a parallel RLC circuit
Consider that a 24 mA current source is connected to the circuit given in Figure 1 at time
t=0. For C = 10 nF, L = 10 mH and R = 625 Ω;
a) Calculate the neper and the resonance frequencies.
b) Find the complete solution.
c) Calculate the value of the inductors current at time t = 50 µs.
d) Simulate the circuit with a square wave current source (0 - 24 mA) at 2 kHz. For C =
25 nf, L = 25 mH and R = 1 kΩ, 500 Ω and 200 Ω, display the currents on the inductor.
Determine the response types of these circuits.
Io
L
i
R
+
Vo
-
C
Figure 2
Figure 2 illustrates a series RLC circuit. Using KVL, characteristic equation of s can be
obtained as Equation 12.
𝑠2 +
𝑅
𝐿
𝑠+
1
𝐿𝐶
=0
(12)
The form of s1,2 determines the solution of natural response as described above in parallel
RLC circuits. The only difference is the calculation of the neper frequency which is now R/2L
isntead of (2RC)-1.
Preliminary work 5: Natural response of a series RLC circuit
a) Consider that in Figure 2, the capacitor is connecting to the inductor via a switch. The
switch is open and the voltage stored in the capacitor is 100 V ( v c (0-) = 100 V).
b) The switch is closed at time t = 0. For C = 0.1 µF, L = 100 mH and R = 560 Ω calculate
the neper and resonance frequencies and determine the response type of the circuit.
c) Calculate the damped frequency.
d) Derive the expression of i(t) (the current flows in the circuit).
e) Calculate i(t) for t=200, 400, 600 and 1000 µs.
f) The time constant for the series RLC circuit is calculated as 𝜏 = 2𝐿/𝑅
For large 𝜏, the system is underdamped and the charge oscillates. The critical
damping condition is exist when 𝑅 = 2√𝐿/𝐶 . When the resistor is larger than the
critical value, the system is overdamped. Verify these situations by simulation
employing L = 25 mH, C = 0.1 µF and 5 V square wave source at 100 Hz. Find three R
values that satisfy the three conditions described. Display the voltages on the
capacitor. Explain how did you determine the response of the circuit with the aid of
the voltage graphics.
Experiment
1. Construct the circuit given in Figure 2 on your board. Adjust the signal generator to
supply 10 V square wave at 100 Hz. For C = 4.7 nF, L = 100 µH choose a resistor that
satisfies R2<4L/C. Observe the voltage of the capacitor with the oscilloscope. See
what will happen if you choose another resistor that not satisfies the criteria above.
Explain the results.
2. Change the components with R = 330 Ω, L = 22 µH and C = 33 pF. Set the signal
generator to produce a square wave with amplitude 0 V to +5 V. Select the frequency
large enough for the circuit to reach steady-state. Describe the damping you observe.
Change R value to 2.2 k Ω. Is the damping changed? Verify your result by calculating
the neper and resonance frequencies.
3. The steady-state value is defined as the magnitude of the voltage or the current after
the circuit has reached stability. The rise time is the period of time required for the
response to go from 10% to 90% of its steady-state value. Measure the rise time with
the aid of the cursors for both R values in step 2.
4. For underdamped circuits, the maximum and minimum values of the oscillation can
be calculated by Equation 13, where ξ is the damping ratio (𝛼/𝜔𝑜 ). Using the same
values in step 2 (with R=330 Ω), calculate first two maximum and first two minimum
values of the voltage of the capacitor. Compare the theoretical results with
oscilloscope measurements.
𝑘𝜋
𝑣𝑜 ( ) = 𝑉 − (−𝑉)𝑘 𝑒
𝜔
𝑑
−
𝑘𝜋𝜉
√1−𝜉2
, k = 0,1,2, ...
(13)
5. Construct the circuit given in Figure 3 with R = 33 Ω, L = 4.7 mH, C = 2.7 nF and RL = 1
kΩ. Note the resistor RL is necessary since the oscilloscope can only measure voltage.
Therefore, what you are going to see is the scaled version of the current flowing
through the inductor. Connect the oscilloscope probe across this resistor. Use 0 V to
+10 V square wave at 100 Hz input signal. What is the response type of the circuit?
6. Change the resistor value (R) to 2.2 kΩ and repeat the measurement. Is the response
type changed?
7. To investigate the effect of the frequency on the circuit, change the input frequency
to 1 kHz. Compare the results with the previous step and note the differences, if any.
L
R
C
RL
Figure 3
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:
330, 2.2 k, 33, 1 k
Capacitors:
4.7 nF, 33 pF, 2.7 nF
Inductors:
100 µH, 22 µH, 4.7 mH