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First-Order Circuits (RC, RL) Ulaby Ch 5; Alexander Ch 7 e170, s20, DJD Page 1 of 22
Unless noted, images in this document are from:
Circuit Analysis and Design, by Ulaby, F; Maharbiz, M and Furse, C., 2018.
http://cad.eecs.umich.edu/
First-Order Circuits
Circuits with sources, resistors, and capacitors (or inductors), but not both
caps and inductors.
RC Circuit: Resistor(s) and capacitor(s)
RL Circuit: Resistor(s) and inductors(s)
Called first-order circuits because they are modeled by first-order
differential eqs.
General idea here:
Study what happens when we go from a DC state (everything is at steady state,
or not changing), then make an abrupt change, like throwing a switch, and
model the system as it approaches a new DC state (steady state).
DC Condition ���� throw (open or close a switch) ���� New DC Condition
…. Determine the transient response (time-varying response):
v(t) , i(t), p(t), w(t) etc.
Note: steady state does not mean current is not flowing. Steady state means
nothing is changing, i.e., all currents and voltages in system are constant.
First-Order Circuits (RC, RL) Ulaby Ch 5; Alexander Ch 7 e170, s20, DJD Page 2 of 22
Two types of circuits: RC and RL
Two basic set-ups:
Natural Response/Source-Free Response
Energy initially stored in capacitor (or inductor), switch thrown so that at
“end”, no energy is stored in capacitor (inductor)
Step Response/Force Response
No energy initially stored in cap (or inductor), switch thrown so that at
“end”, energy is stored in cap (inductor)
First-Order Circuits (RC, RL) Ulaby Ch 5; Alexander Ch 7 e170, s20, DJD Page 3 of 22
Natural Response of an RC Circuit (Source-Free Response)
Situation
• Switch has been as Position 1 for a long time
(i.e., the circuit has reached a steady
state. This is for time t ≤ 0.
• Since steady state, all currents and voltages
are constant.
�� = � ����� = 0
• Cap acts like open circuit.
��� = 0�� = 0 • Voltage across cap just before t = 0 is:
��� = 0�� = �
Note: ��0�� is not necessarily Vs…. analyze
the circuit.
• Energy stored in the cap is:
��� = 0�� = 12 � ��
At t = 0, the switch is thrown to Position 2. What happens?
First-Order Circuits (RC, RL) Ulaby Ch 5; Alexander Ch 7 e170, s20, DJD Page 4 of 22
First-Order Circuits (RC, RL) Ulaby Ch 5; Alexander Ch 7 e170, s20, DJD Page 5 of 22
Adapted from: https://en.wikipedia.org/wiki/RC_circuit#/media/File:Series_RC_resistor_voltage.svg
Time, t �� ���
τ 0.368
2τ 0.135
3τ 0.050
4τ 0.018
5τ 0.007
����
� ���%
First-Order Circuits (RC, RL) Ulaby Ch 5; Alexander Ch 7 e170, s20, DJD Page 6 of 22
First-Order Circuits (RC, RL) Ulaby Ch 5; Alexander Ch 7 e170, s20, DJD Page 7 of 22
General Solution Method for RC Natural Response
1. Find Vo across cap. ��0�� = ��0�� = �
Solve DC problem just before switch is thrown.
2. Find time constant τ = RC (R = Req seen by cap, t > 0)
3. Apply equations
���� = ���� � ≥ 0 …. or ��� = ����
"��
���� = − �$ ��%
& � ≥ 0
'��� = − (�
$ ���%& � ≥ 0
���� = 12 � (��
���%& � ≥ 0
�) = 12 � (� *1 − ���%
& + � ≥ 0
Notes:
• ���� = � ,-.,% = � ����
/− 0&1 = � ����
/− 0)�1 = − 23
) ��� � ≥ 0
• Note: with one cap, the time response of all current and voltages in the
circuit vary as �^−�/6�.
• After 1 time constant τ, the capacitor voltage has gone 63% of the way to
its final value. Here, 0.37 or 37% of its initial value.
First-Order Circuits (RC, RL) Ulaby Ch 5; Alexander Ch 7 e170, s20, DJD Page 8 of 22
Example
First-Order Circuits (RC, RL) Ulaby Ch 5; Alexander Ch 7 e170, s20, DJD Page 9 of 22
Step Response of an RC Circuit (Forced Response)
Situation
• Switch has been as Position 1 for a long time
(i.e., the circuit has reached a steady
state. This is for time t ≤ 0.
• Since steady state, all currents and voltages
are constant.
�� = � ����� = 0
• Cap acts like open circuit.
��� = 0�� = 0 • Voltage across cap just before t = 0 is:
��� = 0�� = �
Note: ��0�� is not necessarily Vs1…. analyze
the circuit.
Note 2: A pure “step” has Vo = 0.
• Energy stored in the cap is:
��� = 0�� = 12 � ��
At t = 0, the switch is thrown to Position 2. What happens?
For simplicity, let’s let Vo = 0 for now.
First-Order Circuits (RC, RL) Ulaby Ch 5; Alexander Ch 7 e170, s20, DJD Page 10 of 22
First-Order Circuits (RC, RL) Ulaby Ch 5; Alexander Ch 7 e170, s20, DJD Page 11 of 22
First-Order Circuits (RC, RL) Ulaby Ch 5; Alexander Ch 7 e170, s20, DJD Page 12 of 22
General Solution Method – Any RC Circuit
1. Find Vo across cap. ��0�� = ��0�� = �
Solve DC problem just before switch is thrown.
2. Find Vf across cap at end of time. ��∞� = 8
Solve DC problem at t = ∞.
3. Find time constant τ = RC (R = Req seen by cap, t > 0)
4. Apply equations
���� = ��∞� + :��0�� − ��∞�;��%& � ≥ 0
or:
���� = 8 + < � − 8=��%& � ≥ 0
Notes:
• If � = 0, the cap voltage steps up from zero to 8.
���� = 8 >1 − ��%& ?
• If 8 = 0, the cap voltage is a natural response:
���� = ���%&
• After 1 time constant τ, the voltage has gone 63% of the way to its final
value. It is gone 63% of @ 8 − �@
• Note the cap voltage can step up A 8 > �C, or step down A 8 < �C.
• If switch is thrown at � = �E:
���� = 8 + < � − 8=��%& � ≥ 0
First-Order Circuits (RC, RL) Ulaby Ch 5; Alexander Ch 7 e170, s20, DJD Page 13 of 22
Example
2
1
First-Order Circuits (RC, RL) Ulaby Ch 5; Alexander Ch 7 e170, s20, DJD Page 14 of 22
Natural Response of an RL Circuit (Source-Free Response)
Situation
• Switch has been as Position 1 for a long time
(i.e., the circuit has reached a steady
state. This is for time t ≤ 0.
• Since steady state, all currents and voltages
are constant.
�F = G ��F�� = 0
• Inductor acts like short circuit.
�F� = 0�� = 0 • Current through inductor just before t = 0 is:
�F� = 0�� = H�
Note: ��0�� is not necessarily Is…. analyze
the circuit.
• Energy stored in the cap is:
�F� = 0�� = 12 GH��
At t = 0, the switch is thrown to Position 2. What happens?
First-Order Circuits (RC, RL) Ulaby Ch 5; Alexander Ch 7 e170, s20, DJD Page 15 of 22
General Solution Method for RL Natural Response
1. Find Io across cap. �F0�� = �F0�� = H�
Solve DC problem just before switch is thrown.
2. Find time constant 6 = F) (R = Req seen by inductor, t > 0)
3. Apply equations
�F�� = H���� � ≥ 0 …. or �F�� = H����
"��
�F�� = −$H� ��%& � ≥ 0
'��� = −$H(� ���%& � ≥ 0
�F�� = 12 GH(��
���%& � ≥ 0
�) = 12 GH(� *1 − ���%
& + � ≥ 0
Notes:
• �F�� = G ,IJ,% = GH����
/− 0&1 = GH����
/− 0F/)1 = −$H����
� ≥ 0
• Note: with one cap, the time response of all current and voltages in the
circuit vary as �^−�/6�.
• After 1 time constant τ, the capacitor voltage has gone 63% of the way to
its final value. Here, 0.37 or 37% of its initial value.
First-Order Circuits (RC, RL) Ulaby Ch 5; Alexander Ch 7 e170, s20, DJD Page 16 of 22
Example
First-Order Circuits (RC, RL) Ulaby Ch 5; Alexander Ch 7 e170, s20, DJD Page 17 of 22
Example
First-Order Circuits (RC, RL) Ulaby Ch 5; Alexander Ch 7 e170, s20, DJD Page 18 of 22
Example Ans: i1(t) =2.88e−10t mA; i2(t) =0.72e−20t mA
First-Order Circuits (RC, RL) Ulaby Ch 5; Alexander Ch 7 e170, s20, DJD Page 19 of 22
Step Response of an RL Circuit (Force Response)
Situation
• Switches have been as Position 1 for a long
time (i.e., the circuit has reached a steady
state. This is for time t ≤ 0.
• Since steady state, all currents and voltages
are constant.
�F = G ��F�� = 0
• Inductor acts like short circuit.
�F� = 0�� = 0 • Current through inductor just before t = 0 is:
�F� = 0�� = H�
Note: ��0�� is not necessarily Is…. analyze
the circuit.
• Energy stored in the cap is:
�F� = 0�� = 12 GH��
At t = 0, the switch is thrown to Position 2. What happens?
First-Order Circuits (RC, RL) Ulaby Ch 5; Alexander Ch 7 e170, s20, DJD Page 20 of 22
General Solution Method – Any RL Circuit
1. Find Io through inductor. �F0�� = �F0�� = H�
Solve DC problem just before switch is thrown.
2. Find If through inductor at end of time. �F∞� = H8
Solve DC problem at t = ∞.
3. Find time constant 6 = F) (R = Req seen by inductor, t > 0)
4. Apply equations
�F�� = �F∞� + :�F0�� − �F∞�;��%& � ≥ 0
or:
�F�� = H8 + <H� − H8=��%& � ≥ 0
Notes:
• If H� = 0, the inductor current steps up from zero to H8.
�F�� = H8 >1 − ��%& ?
• If H8 = 0, the cap voltage is a natural response:
�F�� = H���%&
• After 1 time constant τ, the current has gone 63% of the way to its final
value. It is gone 63% of @H8 − H�@
• Note the cap voltage can step up AH8 > H�C, or step down AH8 < H�C.
First-Order Circuits (RC, RL) Ulaby Ch 5; Alexander Ch 7 e170, s20, DJD Page 21 of 22
Example
First-Order Circuits (RC, RL) Ulaby Ch 5; Alexander Ch 7 e170, s20, DJD Page 22 of 22
General Response of voltages and currents through R and C (or L) elements.
K��: a voltage across any R, C or L, or current through any R, C or L.
1. Find Xo. K0�� = L� (see note below).
2. Find Xf. K∞� = K8
Solve DC problem at t = ∞.
3. Find time constant 6 4. Apply equations
K�� = K∞� + :K0�� − K∞�;��%& � ≥ 0
or:
K�� = L8 + <L� − L8=��%& � ≥ 0
Note:
• Voltage across caps must be continuous ��0�� = ��0��
• Current through inductor must be continuous �F0�� = �F0��
• Voltage across resistors and inductors can change suddenly
Solve instantaneous DC problem at t = 0+ for �)0�� and �F0��
• Current through resistors and caps can change suddenly
Solve instantaneous DC problem at t = 0+ for �)0�� and ��0��
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