UNIVERSITY OF NEVADA L AS VEGAS . DEPARTMENT OF ELECTRICAL AND COMPUTER ENGINEERING LABORATORIES .
Class:
EE221 Circuits II - 1002
Points
Semester:
Fall 2015
Document author:
Author's email:
Prelab 5
Document topic:
Instructor's comments:
Introduction / Theory of operation
An RC circuit is a circuit composed of resistors, capacitors and a voltage source. The charge and
discharge time of the capacitor is dependent on both the series resistance and the capacitance. The
charge rate is given by V = Vmax * (1 – e-t/(RC)) where R is the resistance and C is the capacitance.
The discharge rate is given by V = Vmax * e-5/(RC). The capacitor is approximately entirely discharge
in 5 time constants (5 * RC).
An RL circuit consists of a resistor, an inductor and a voltage source. The charge time for an
inductor is dependent on the resistance and the inductance. The charge rate is given by
I(t) = Vin/R * (1 – e-tR/L).
An RLC circuit consists of a resistance, inductance and capacitance. The response of the circuit
depends on all three elements. The circuit can have either an underdamped response, overdamped
response, or a critical response. An underdamped response will resonate naturally. A critically
damped response is just on the edge of oscillation. An overdamped response will not oscillate at
all.
Prelab main content
Analysis 1:
τ = RC = (200KΩ)(220μF) = 44s
5τ = 5 ∗ 44s = 220s
Therefore, it takes 220 seconds to charge the capacitor.
−𝑡
(𝜏)
) =
0
(
)
𝑒 44𝑠 ) =
−𝑡
(44𝑠)
𝑉𝑜𝑢𝑡 (𝑡) = 𝑉𝑖𝑛 (1 – 𝑒
𝑉𝑖𝑛 (1 – 𝑒
𝑉𝑜𝑢𝑡 (0) = 𝑉𝑖𝑛 (1 –
𝑉𝑖𝑛 (0) = 0
𝑉𝑜𝑢𝑡 (∞) = 𝑉𝑖𝑛 (1 – 𝑒
−∞
(44𝑠)
) = 𝑉𝑖𝑛 (1) = 𝑉𝑖𝑛
)
Figure 1: Analysis 1 LTSpice circuit (Left), and The simulation of the circuit (Right)
Analysis 2:
τ = RC = (200KΩ)(220μF) = 44s
5τ = 5 ∗ 44s = 220s
Therefore, it takes 220 seconds to discharge the capacitor.
−𝑡
(𝜏)
= 𝑉𝑖𝑛
0
( 𝜏)
𝑉𝑖𝑛 ∗ 𝑒 = 𝑉𝑖𝑛
−∞
(
)
𝑉𝑖𝑛 ∗ 𝑒 𝜏 = 0
𝑉𝑜𝑢𝑡 (𝑡 ) = 𝑉𝑖𝑛 ∗ 𝑒
𝑉𝑜𝑢𝑡 (0) =
𝑉𝑜𝑢𝑡 (∞) =
∗𝑒
−𝑡
(44𝑠)
Figure 2: Analysis 2 circuit (Left) and the simulation (Right)
Analysis 3:
τ=
𝐿 1.2mH
=
= 12μs
𝑅
100
−𝑡
−𝑡
𝑉𝑖𝑛
𝑉𝑖𝑛
(1 − 𝑒 τ ) =
(1 − 𝑒 12μs)
𝑅
𝑅
0
𝑉𝑖𝑛
𝐼 (0) =
(1 − 𝑒 12μs) = 0
𝑅
∞
𝑉𝑖𝑛
𝑉𝑖𝑛
(1 − 𝑒 12μs ) =
𝐼 (∞) =
𝑅
𝑅
𝐼 (𝑡 ) =
Analysis 4:
𝛼
𝑅
1
𝐿𝑒𝑡 𝜉 =
, 𝑤ℎ𝑒𝑟𝑒 𝛼 =
, 𝑎𝑛𝑑 𝜔0 =
, 𝑡ℎ𝑒𝑛 𝑡ℎ𝑒 𝑟𝑒𝑠𝑝𝑜𝑛𝑠𝑒𝑠 𝑎𝑟𝑒:
ω0
2𝐿
√𝐿𝐶
𝐶𝑟𝑖𝑡𝑖𝑐𝑎𝑙𝑙𝑦 𝑑𝑎𝑚𝑝𝑒𝑑 𝑤ℎ𝑒𝑛 𝜉 = 1
𝑂𝑣𝑒𝑟𝑑𝑎𝑚𝑝𝑒𝑑 𝑤ℎ𝑒𝑛 𝜉 > 1
𝑈𝑛𝑑𝑒𝑟𝑑𝑎𝑚𝑝𝑒𝑑 𝑤ℎ𝑒𝑛 𝜉 < 1
For this circuit,
𝛼
𝑅
100
𝜉=
=
∗ √𝐿𝐶 =
>1
ω0 2𝐿
4.32 ∗ 10−8
So, the circuit is overdamped
𝑅
2𝐿
2(1.2 ∗ 10−3 )
̅̅Ω
𝜉=
=
∗ √𝐿𝐶 = 1 → 𝑅 =
=
= 133. ̅̅
33
ω0 2𝐿
√𝐿𝐶
√1.2 ∗ 10−3 ∗ 0.27 ∗ 10−6
𝛼
This circuit will be
̅̅̅̅Ω
𝑂𝑣𝑒𝑟𝑑𝑎𝑚𝑝𝑒𝑑 𝑤ℎ𝑒𝑛 𝑅 > 133. 33
̅̅Ω
𝐶𝑟𝑖𝑡𝑖𝑐𝑎𝑙𝑙𝑦 𝑑𝑎𝑚𝑝𝑒𝑑 𝑤ℎ𝑒𝑛 = 133. ̅̅
33
̅̅
̅̅
𝑈𝑛𝑑𝑒𝑟𝑑𝑎𝑚𝑝𝑒𝑑 𝑤ℎ𝑒𝑛 𝑅 < 133. 33Ω
Overdamped circuit
Critically damped circuit
Underdamped circuit
Analysis 5:
𝛼
𝐿𝑒𝑡 𝜉 =
, 𝑤ℎ𝑒𝑟𝑒 𝛼 =
1
1
, 𝑎𝑛𝑑 𝜔0 =
, 𝑡ℎ𝑒𝑛 𝑡ℎ𝑒 𝑟𝑒𝑠𝑝𝑜𝑛𝑠𝑒𝑠 𝑎𝑟𝑒:
ω0
2𝑅𝐶
√𝐿𝐶
𝐶𝑟𝑖𝑡𝑖𝑐𝑎𝑙𝑙𝑦 𝑑𝑎𝑚𝑝𝑒𝑑 𝑤ℎ𝑒𝑛 𝜉 = 1
𝑂𝑣𝑒𝑟𝑑𝑎𝑚𝑝𝑒𝑑 𝑤ℎ𝑒𝑛 𝜉 > 1
𝑈𝑛𝑑𝑒𝑟𝑑𝑎𝑚𝑝𝑒𝑑 𝑤ℎ𝑒𝑛 𝜉 < 1
For this circuit,
1 𝐿
√ = 0.333 < 1
ω0 2𝑅 𝐶
So, the circuit is underdamped
𝜉=
𝜉=
𝛼
𝛼
ω0
=
=
1 𝐿
1 𝐿
̅̅
√ = 1 → 𝑅 = √ = 33. ̅̅
33
2𝑅 𝐶
2 𝐶
This circuit will be
̅̅Ω
𝑂𝑣𝑒𝑟𝑑𝑎𝑚𝑝𝑒𝑑 𝑤ℎ𝑒𝑛 𝑅 < 33. ̅̅
33
̅̅̅̅Ω
𝐶𝑟𝑖𝑡𝑖𝑐𝑎𝑙𝑙𝑦 𝑑𝑎𝑚𝑝𝑒𝑑 𝑤ℎ𝑒𝑛 = 33. 33
̅̅̅̅Ω
𝑈𝑛𝑑𝑒𝑟𝑑𝑎𝑚𝑝𝑒𝑑 𝑤ℎ𝑒𝑛 𝑅 > 33. 33
Overdamped circuit:
Critically damped circuit:
Underdamped circuit:
Altium:
Fig 4-1:
[
]
[
]
[
Vin
HDR1X2
Header 2
R1
AXIAL-0.3
Res1
C1
RAD-0.3
Cap
]
Fig 4-2:
[
]
[
Vin
HDR1X2
Header 2
C1
RAD-0.3
Cap
]
[
R1
AXIAL-0.3
Res1
]
Fig 4-3:
[
]
[
]
[
]
Vin
HDR1X2
Header 2
R1
AXIAL-0.3
Res1
L1
0402-A
Inductor
Fig 4-4:
[
]
[
]
[
]
[
]
[
]
[
Vin
HDR1X2
Header 2
S1
SPST-2
SW-PB
R2
AXIAL-0.3
Res1
R1
AXIAL-0.3
Res1
L1
0402-A
Inductor
c2
RAD-0.3
Cap
]
Fig 4-5:
[
]
[
]
[
]
[
]
[
]
s1
SPST-2
SW-PB
r1
AXIAL-0.3
Res1
l1
0402-A
Inductor
Iin
HDR1X2
Header 2
c1
RAD-0.3
Cap