First-Order Circuits

 Source-free RC/RL circuit

 Step response of RC/RL circuit
 First-Order Op Amp Circuits

First-Order Circuits
• We have considered the three passive elements (resistors, capacitors, and
inductors) and one active element (the op amp) individually, we are prepared to
consider circuits that contain various combinations of two or three of the
passive elements.
• We will carry out the analysis of RC or RL circuits by applying Kirchhoff’s Laws
• Applying Kirchhoff’s Law to pure resistive circuits generates algebraic equations
• Applying Kirchhoff’s Law to RC and RL circuits produces differential equations
• The differential equations resulting from analyzing RC and RL circuits are of the
first order

We assume that energy is initially stored in the capacitive or inductive

element. The energy causes current to flow in the circuit and is gradually
dissipated in the resistors.

2
 Source-free RC circuit
• Consider a series combination of a resistor and an initially charged capacitor
• The energy stored in the capacitor is released to the resistor
• Of course, after a very long time, all the energy in the
capacitor would be released and dissipated in the resistor.
At that time, the voltage across the capacitor would be
zero and the current in the circuit would also be zero

• But how would the circuit behave during discharge?

• Since the capacitor is initially charged, we can assume that at time t = 0, the initial
voltage is v(0) = Vo, with the corresponding energy stored w(0) = (1/2)CVo2
• Applying KCL at the top node yields iC + iR = 0
• By definition, iC = Cdv/dt and iR = v/R
dv v
• Thus, + = 0 . This is the first-order differential equation!
dt RC

Source-free RC circuit
dv 1
• To solve it, we rearrange the terms as =− dt
v RC
• Integrating both sides, we get (A is the integration constant)
v t t
• Thus, ln = − ln v = − + ln A
A RC RC
• Taking powers of e produces v(t) = Ae-t/RC
• With the initial condition v(0) = Vo, we have A = Vo
• Hence the final solution is v(t) = Voe-t/RC
• The voltage response for the RC circuit is an exponential decaying function, starting
from its initial value Vo
• The rapidity with which the voltage decreases is expressed in terms of the time
constant, denoted by τ = RC
• In terms of the time constant, we have v(t ) = Vo e−t / τ
• Since the response is due to the initial energy
stored and the physical characteristics of the
circuit and not due to some external voltage or
current source, it is called the natural response
of the circuit
4
 Source-free RC circuit
• At t = 5τ, v(t = 5τ)/Vo = 0.00674 < 1%. It is customary to assume that the capacitor
is fully discharged after 5 time constants
• Another meaning of time constant:
Draw a tangent at t=0, τ is the intercept at time axis
• Smaller the time constant (smaller resistance to
the current), the faster the response (discharge)
• Since v(t ) Vo −t / τ
iR (t ) = = e
R R
• The power dissipated in the resistor is
Vo2 −2t / τ
p (t ) = v(t )iR (t ) = e
R
• The energy absorbed by the resistor up to time t is
2 t
t Vo −2t '/ τ τ Vo2 −2t '/ τ 1
= CVo2 (1 − e−2t / τ )
t
wR (t ) = 0 p(t ')dt ' = 0 e dt ' = − e
R 2R 2
0
• As t→∞, wR(∞) →(1/2)CVo2. That is, all the energy initially stored in the capacitor is
dissipated in the resistor

Source-Free RC Circuit
Example: Let vC(0) = 15V. Find vC, vx and ix for t > 0.
• We first need to make the circuit in the original figure
conform with the standard source free RC circuit
• The 8 Ω and 12 Ω are in series and they are in parallel with
the 5 Ω resistor.
• The combined resistance is Req= (8+12)(5) / (8+12+5) = 4 Ω
• The time constant τ = ReqC = 4 (0.1) = 0.4 s
• Thus, vC(t) = vC(0)e-t/τ = 15e-t/0.4 = 15e-2.5t
• To find vx, we can use voltage division formula
12
vx (t ) = vC (t ) = 0.6(15e−2.5t ) = 9e−2.5t V
12 + 8

• Finally, ix = vx /12 = 0.75e-2.5t A

6
Source-Free RC Circuit
Example: The switch in the circuit has been closed for
a long time, and it is opened at t = 0. Find v(t) for t ≥
0. Calculate the initial energy stored in the capacitor.

Since the voltage across a capacitor cannot change

instantaneously, the voltage across the capacitor
before and immediately after the switch is open at
t=0 will be the same. That is, vC(0)=15V

For t > 0, the switch is opened, and we have the RC circuit

τ = ReqC = (9+1) × 20×10-3 =0.2s
vC(t) = vC(0)e-t/τ = 15e-t/0.2 = 15e-5tV
The initial energy stored in the capacitor
is wC(0)= ½Cv2C(0)=2.25J

Source-free RL circuit
• Consider the series connection of a resistor and an inductor
• We want to determine the response for i(t)
• We select the inductor current as the response to be determined
in order to take advantage of the idea that the inductor current
cannot change instantaneously
• At t = 0, we assume that the inductor has an initial current i(0) = Io
• Applying KVL around the loop, we have vL + vR = 0
• But vL = Ldi/dt and vR = iR. Thus
• Rearranging terms and integrating gives
i (t ) t t
di R i (t ) Rt ' i (t ) Rt
 i = −  L dt '  ln i I =−  ln =−
Io 0
o L 0 Io L

• Taking the powers of e, and define τ = L/R, we have i (t ) = I o e−t / τ

• The current response is an exponential decay of the initial current

8
 Source-free RL circuit
• Since vR(t) = iR = IoRe-t/τ , the power dissipated in the resistor is

p ( t ) = v R ( t ) i ( t ) = I o2 R ⋅ e − 2 t / τ

• The energy absorbed by the resistor is

t t t
1
w R (t ) =  p ( t ') d t ' =  I o2 R ⋅ e − 2 t '/ τ d t ' = − τ I o2 R ⋅ e − 2 t '/ τ
0 0
2 0
1
= L I o2 (1 − e − 2 t / τ )
2

• As t → ∞, wR(∞) → (1/2)LIo2. That is, the initial energy stored in the inductor
is eventually dissipated in the resistor.

Source-free RL circuit
Example: Assuming that i(0) = 10A, calculate i(t) and ix(t) in the circuit
• We have to obtain the equivalent resistance at the
inductor terminals and then apply i (t ) = I o e−t / τ
• In order to find the equivalent resistance,
• We disconnect the inductor and insert
a voltage source of 1 V at the inductor terminal
• Applying KVL to the two loops results in
2(i1 − i2 ) + 1 = 0

4i2 + 2(i2 − i1 ) − 3i1 = 0
• Simplifying and solving the equations give
i1 = −3A, io = −i1 = 3A
• Hence Req = vo/io = 1/3Ω
• The time constant is τ = L/Req = 0.5/(1/3) = 3/2s
• Thus the current through the inductor is i (t ) = i (0)e−t / τ = 10e−(2 / 3)t A
• The voltage across the inductor is v = Ldi/dt = (-10/3)e-(2/3)t V
• Since the inductor and the 2Ω resistor are in parallel, ix= v/2 = -1.6667e-(2/3)t A

10
Example: The switch in the circuit has been closed for a long time. At t = 0, the switch
is opened. Calculate i(t) for t > 0.
• For t < 0, the switch is closed, and the inductor acts
as a short circuit. The 16Ω resistor is short-circuited,
the resulting circuit is shown in Fig a)
• In Fig a), i1 = 40 / (4 || 12 + 2) = 8A
• By current division, i = 12 i1 = 6A
12 + 4
• Since the current through an inductor cannot change
instantaneously, the current before and immediately
after the switch open is i(0) = 6A
• When t > 0, the switch is open and the voltage source
is disconnected, we now have a source-free RL circuit
as shown in Fig b)
• Combining the resistors, we have Req = (12 + 4)||6 = 8Ω
• The time constant is τ = L/Req = 2/8 = 1/4s
• Thus,
i (t ) = i (0)e−t / τ = 6e−4t A

11

Example: In the circuit shown, find io, vo and i for all time, assuming that the switch
was open for a long time
• For t < 0, the switch is open. Since the inductor
acts like a short circuit to dc, the 6Ω resistor
is short-circuited, so we have the circuit shown
in Fig a)
• Hence, io = 0 and
10
i (t ) = = 2A; vo (t ) = 3i (t ) = 6V for t < 0
2+3

• For t > 0, the switch is closed, so the voltage source

is short-circuited. We now have a source-free RL
circuit as shown in Fig b)
• At the inductor terminals, Req = 3||6 = 2Ω so the time
constant is τ = L/Req = 2/2 = 1s
• Hence i(t) = i(0)e-t/τ = 2e-t A
• Since the inductor is in parallel with the 6Ω and 3Ω resistors,
di
vo (t ) = −vL = − L = −2(−2e−t ) = 4e−t V for t > 0
dt
io (t ) = vL / 6 = −(2 / 3)e−t A for t > 0 12
 Step response of an RC circuit
• Unit step function:

0, t<0
u (t ) = 
1, t >0

• When a dc source is suddenly applied, the voltage or current source can be

modeled as a step function

• The response of a circuit to such a sudden application of a dc voltage or

current source is called step response

13

Step response of an RC circuit

• Unit step function:

t<0 Singularity functions are functions

0,
u (t ) =  that either are discontinuous or
1, t >0 have discontinuous derivatives.

The unit step function delayed by 𝑡 The unit step advanced by 𝑡

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 Step response of an RC circuit
• Unit step function:
• When a dc source is suddenly applied, the voltage or current source can be
modeled as a step function

• The response of a circuit to such a sudden application of a dc voltage or current

source is called step response

15

• Consider the RC circuit shown in Fig a) which can be replaced

by the circuit in Fig b)
• We select the capacitor voltage as the circuit response to be
determined
• Assume the initial voltage of the capacitor is Vo
• Since the voltage of a capacitor cannot change
instantaneously, v(0-) = v(0+) = Vo, where v(0-) is the voltage
across the capacitor just before switching and v(0+) is its
voltage immediately after switching
• Applying KCL, we have (for t > 0)
dv Vs − v dv dt
C =  =−
dt R v − Vs RC
• Integrating both sides and introducing the initial conditions,
we have
t
v (t ) t' v(t ) − Vs t
ln(v '− Vs ) V = −  ln =−
o RC 0 Vo − Vs RC

• Taking the exponential of both sides, we get

v(t ) = Vs + (Vo − Vs )e−t / τ

16
• Combining with the capacitor voltage for t < 0, we have
Vo t<0
v(t ) =  −t / τ
Vs + (Vo − Vs )e t >0
• Special case: when Vo = 0, v(t) = Vs(1-e-t/τ ) u(t)
• The complete circuit response can be decomposed in two ways:
1. v(t ) = Vo e−t / τ + Vs (1 − e−t / τ )
    
complete natural forced
response response response

• The natural response is due to the initial stored charge, it is the same as the
source-free RC circuit response
• The forced response is due to the application of external source, it is the same
as the above special case when Vo=0
2. v(t ) = (V − V )e−t / τ +
 o 
s V
  s
complete transient steady state
response response response
• The transient response is temporary
• The steady-state response is the response as t→∞

17

• Source-free RC response: v(t ) = Vo e−t / τ

• RC circuit step response: v(t ) = Vs + (Vo − Vs )e−t / τ
• Notice that for source-free RC circuit, it is equivalent to Vs=0 in step response
• Therefore, both cases can be summarized using the response

v(t ) = v(∞) + [v(0) − v(∞)]e−t / τ

where v(0) is the initial voltage at t=0+ and v(∞) is the final or steady-state value.

Example: The switch has been in position A for a long time. At t = 0, the switch moves to
B. Determine v(t) for t > 0 and calculate its values at t = 1s and 4s.
• We have to find v(0), v(∞) and τ
• For t < 0, the switch is at position A. The capacitor
acts like an open circuit to dc, v is the same as the
voltage across the 5kΩ resistor:
5
v(0− ) = (24) = 15V
5+3
• Since the capacitor voltage cannot change instantaneously, v(0) = v(0+) = v(0-) = 15V
18
• For t > 0, the switch is in position B, the circuit becomes an simple RC circuit with step
input
• The time constant is τ = RC = 4×103 × 0.5×10-3 = 2s
• Also notice that v(∞) = 30V
• Thus, v(t ) = v(∞) + [v(0) − v(∞)]e−t / τ
= 30 + (15 − 30)e−t / 2
= 30 − 15e −0.5t
• v(1) = 30 - 15e-0.5 = 20.9 V, v(4) = 30 - 15e-2 = 27.97 V

Example: In the figure shown, the switch has been closed for a long time and is opened
at t = 0. Find i and v for all time.
• The resistor current i can be discontinuous at
t = 0, while the capacitor voltage cannot. Hence,
it is better to find v and then obtain i from v
• We have to find v(0), v(∞) and τ
• For t < 0, the switch is closed and the 30 V source has no effect (it can be replaced by a
short circuit). Furthermore, the capacitor acts like an open circuit since the switch has
been closed for a long time.
19

• So, the circuit becomes that in Fig a)

• From this circuit we obtain v = 10V, I = -v/10 = -1A
• Since the capacitor voltage cannot change instantaneously
v(0) = v(0-) = 10V
• For t > 0, the switch is opened and the 10V source is
disconnected from the circuit. The 30V source is now
operative, so the circuit becomes that in Fig b)
• After a long time, the circuit reaches steady state and the
capacitor acts like an open circuit again. By voltage division,
20
v (∞ ) = (30) = 20V
20 + 10
• In order to obtain the time constant, we have to find out the
Thevenin equivalent circuit at the capacitor terminals
20
• The Thevenin resistance at the capacitor terminals is RTH = 10 || 20 = Ω
20 1 5 3
• The time constant is τ = RTH C = ⋅ = s
3 4 3
• Thus, −t / τ
v(t ) = v(∞) + [v(0) − v(∞)]e
= 20 + (10 − 20)e−(3 / 5)t = (20 − 10e−0.6t )V
10V t<0
To obtain i, we notice from Fig b) that i = (30 - v)/10 = v(t ) = 
−0.6t
(1+e-0.6t) A (20 − 10e )V t≥0
Notice that the capacitor voltage is continuous while −1A t<0
the resistor current is not. i (t ) =  −0.6t
(1 + e )A t > 0 20
 Step response of an RL circuit
• Consider the RL circuit in Fig a), which may be replaced by the
circuit in Fig b)
• Our goal is to find the inductor current i as the circuit response
• The response can be obtained by applying the Kirchhoff’s laws
as done in the RC circuit
• We can also find the circuit response by finding the transient
response it and steady-state response iss
• We know that the transient response is always a decaying
exponential, that is, it = Ae-t/τ, where τ = L/R and A is a
constant to be determined.
• The steady-state current is iss = Vs/R
• Therefore, the complete response is in the form:
i (t ) = Ae−t / τ + Vs / R (1)
• We now need to determine the constant A form the initial
value of i

21

Step response of an RL circuit

• Assume Io be the initial current through the inductor, which may come from a source
other than Vs
• Since the current through the inductor cannot change instantaneously, I(0+) = I(0-) = Io
• Thus, at t =0, (1) becomes Io= A + Vs/R => A = Io - Vs/R
• Put the result of A back into (1) gives
V  V  −t / τ
i (t ) = s +  I o − s e
R  R 
• The above response and the source-free RL circuit response
can be summarized by the following equation:

i (t ) = i (∞) + [i (0) − i (∞)]e−t / τ

• where i(0) and i(∞) are the initial and final values of i, respectively

22
 Step response of an RL circuit
Example: Find i(t) in the circuit for t > 0. Assume that the switch has been
closed for a long time.

• We have to find i(0), i(∞) and τ

• For t < 0, the 3Ω resistor is short-circuited and the
inductor acts like a short circuit
• The current through the inductor at t = 0- (i.e., just before
t = 0) is i(0-) = 10/2 = 5A
• Since the inductor current cannot change instantaneously,
i(0) = i(0+) = i(0-) = 5A
• When t > 0, the 2Ω and 3Ω are in series so that i(∞) =
10/(2+3) = 2A
• The time constant is τ = L/R = (1/3)/(2 + 3) = 1/15 s
• Thus, i (t ) = i (∞) + [i (0) − i (∞)]e−t / τ
= 2 + (5 − 2)e−15t
= 2 + 3e−15t for t > 0

23

Example: At t = 0, switch 1 is closed and switch 2 is closed 4s later. Find i(t) for t > 0.
Calculate i for t = 2s and t = 5s
• We need to consider three time intervals
-t < 0 s
-0 ≤ t ≤ 4 s
-t > 4 s
• For t < 0, both switches S1 and S2 are open so that i = 0
• Since the inductor current cannot change instantly, i(0-) = i(0) = i(0+) = 0
• For 0 ≤ t ≤ 4, S1 is closed so that the 4Ω and 6Ω resistors are in series (remember, at
this time S2 is still open)
• Hence, assuming for now that S1 is closed for a very long time,
i(∞) = 40/(4 + 6) = 4A
• The equivalent resistance for the present RL circuit is 4 + 6 = 10Ω
• Thus, τ = L/R = 5/10 = 0.5s
• Therefore, i (t ) = i (∞) + [i (0) − i (∞)]e−t / τ
= 4 + (0 − 4)e−2t
= 4(1 − e−2t )A for 0 ≤ t ≤ 4
24
• For t > 4s, S2 is also closed. The 10V voltage source
is connected and the circuit changes. This sudden
change does not affect the inductor current because
the current cannot change instantly.
• Thus the initial current is i (4) = i (4− ) = 4(1 − e−8 ) ≈ 4A
• To find i(∞) , we have to apply KCL at node P
• Let v be the voltage at node P, using KCL, we have
40 − v 10 − v v 180
+ = => v = V
4 2 6 11
• Therefore, i(∞) = v/6 = 30/11 = 2.727A
• In order to find the time constant, we need to find the Thevenin resistance at the
inductor terminals: 4× 2 22
RTH = 4 || 2 + 6 = +6= Ω
4+2 3
• The time constant is τ = L/RTH =5/(22/3) = 15/22 s
• Hence, i (t ) = i (∞) + [i (4) − i (∞)]e−(t − 4) / τ
= 2.727 + (4 − 2.727)e−1.4667(t − 4)
= 2.727 + 1.273e−1.4667(t − 4) for t ≥ 4
• We need (t-4) in the exponential because of the time delay

25

• Putting all together,

0 for t < 0

i (t ) = 4(1 − e−2t ) for 0 ≤ t ≤ 4
 −1.4667(t − 4)
2.727 + 1.273e for t ≥ 4

• At t = 2s, i(2) = 4(1 - e-4) = 3.93A

• At t = 5s, i(5) = 2.727+1.273e-1.4667 = 3.02A

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First-Order Op Amp Circuits
• An op amp circuit containing a storage element will exhibit first-order behavior.
• As usual, we analyze op amp circuits using nodal analysis. Sometimes, the
Thevenin equivalent circuit is used to reduce the op amp circuit to one that we
can easily handle.
• RC types of op amp circuits, depending on the location of the capacitor with
respect to the op amp; that is, the capacitor can be located in the input, the
output, or the feedback loop.

Example: For the op amp circuit, find 𝑣 for t > 0, given that 𝑣(0) = 3.
If v1 is the voltage at node 1, at that node, KCL gives

27

• METHOD 2: Since v(0+) = v(0+) = 3V, we apply KCL at node 2 in the circuit of Fig.
7.55(b) to obtain

• To find we need the equivalent resistance across the capacitor terminals. If we

remove the capacitor and replace it by a 1-A current source

28
Applications: Delay Circuits
• An RC circuit can be used to provide various time delays.
• It basically consists of an RC circuit with the capacitor connected in parallel with
a neon lamp. The voltage source can provide enough voltage to fire the lamp.
When the switch is closed, the capacitor voltage increases gradually toward 110
V at a rate determined by the circuit’s time constant, (R1 + R2)C.
• The lamp will act as an open circuit and not emit light until the voltage across it
exceeds a particular level, say 70 V. When the voltage level is reached, the lamp
fires (goes on), and the capacitor discharges through it. Due to the low
resistance of the lamp when on, the capacitor voltage drops fast and the lamp
turns off. The lamp acts again as an open circuit and the capacitor recharges.

Other example: The warning blinkers commonly found on road construction sites
29

Example: Assume that R1 = 1.5MΩ, 0 < R2 < 2.5MΩ. (a) Calculate the extreme limits
of the time constant of the circuit. (b) How long does it take for the lamp to glow
for the first time after the switch is closed? Let R2 assume its largest value.
(a) The smallest value for R2 is 0 Ω and the corresponding time constant is

The largest value for R2 is 2.5 MΩ and the corresponding time constant is

(b) Assuming that the capacitor is initially uncharged, 𝑣 (0) = 0, 𝑣 (∞) = 110.

30
 Applications: Photoflash Unit
• An electronic flash unit provides a common example of an RC circuit. This
application exploits the ability of the capacitor to oppose any abrupt change in
voltage.
• It consists essentially of a high-voltage dc supply, a current-limiting large resistor
and a capacitor C in parallel with the flash lamp of low resistance When the switch
is in position 1, the capacitor charges slowly due to the large time constant.
• The capacitor voltage rises gradually from zero to while its current decreases
gradually from to zero. The charging time is approximately five times the time
constant.
• With the switch in position 2, the capacitor voltage is discharged. The low
resistance of the lamp permits a high discharge current with peak in a short
duration.

31

Example: An electronic flashgun has a current-limiting resistor and electrolytic

capacitor charged to 240 V. If the lamp resistance is find: (a) the peak charging
current, (b) the time required for the capacitor to fully charge, (c) the peak
discharging current, (d) the total energy stored in the capacitor, and (e) the
average power dissipated by the lamp.
(a) The peak charging current is

(b)
(c) The peak discharging current is

(d) The energy stored is

(a) The energy stored in the capacitor is dissipated across the lamp during the
discharging period.

32