ETB LO2 DC Transients_2019.pdf

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DEE1213
ELECTRICAL TECHNOLOGY B
Lecture #4
DC Transients
Subject Learning Outcome (SLO)
• This Lecture partially contributing to the
fulfillment of the following SLO:
– Analyze transient behavior of simple L-R and C-R
circuits

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Lecture Learning Outcomes
Upon completion of this lecture, you will be able to:
• Understand the rise or fall of current in simple series RL and RC
circuits excited with dc source.
• Understand the meaning of ‘Time Constant’ for RL and RC circuits
and explain its relationship to the performance of the circuits.
Pre-Test

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DC Transients
What happens when things change.
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Introduction
• AC Theory could be described as "The study of
electronic circuits in which things are always
changing". Voltages, currents and other quantities in
AC circuits are in a continual process of change.
• Before studying AC circuits in depth, this lecture
looks at what happens when conditions suddenly
change (called transient events) in DC circuits, so that
what is learned here can be used as a foundation for
later modules.
6

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Introduction
• Transient examples:
– When a DC voltage is applied to a capacitor C, and resistor R
connected in series, there is a short period of time
immediately after the voltage is connected during which the
current flowing in the circuit and voltages across C and R are
changing.
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Introduction
• Transient examples:
– When DC voltage is connected in series with resistance R and
inductor L, there is short period of time immediately after the
voltage is connected, during which the current flowing in the
circuit and the voltages across L and R are changing.

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Electrical Technology B
R-C Circuit Transient Response
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RC Circuit in DC Transient
Skills:
• Identify and differentiate C in charging or
discharging (considering the switch too).
• Calculate initial, instantaneous, final values of
current, voltage for R and C.
• Sketch graph…

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TRANSIENTS IN CAPACITIVE NETWORKS: THE CHARGING PHASE
VC during the charging phase.
t = 0s, vC = 0
   

/
/
1
1 t
CR
t
C e
V
e
V
v 





Revealing the short-circuit equivalent for the capacitor that
occurs when the switch is first closed.
When t = 0s, VR = E, VC = 0

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Demonstrating that a capacitor has the characteristics of an open
circuit after the charging phase has passed.
t = 0s, VR = E

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Transient Equations for a RC circuit
• The equations of the curves are:
Growth of capacitor voltage,
   

/
/
1
1 t
CR
t
C e
V
e
V
v 






/
/ t
CR
t
Ie
Ie
i 



Time constant for a C-R circuit
• The time constant,  of the circuit can be
defined as: ‘the time taken for a transient to
reach its final state if the initial rate of change
is maintained’.
• The value of the time constant is CR seconds,
i.e. for a series connected RC circuit.
time constant  = CR seconds

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TRANSIENTS CURVE: THE CHARGING PHASE
Plotting the equation VC = E(1 – e-t/) versus time (t).
t = 0s,
Vc = 0
TRANSIENTS CURVE: THE CHARGING PHASE
t = 0s
𝑰𝑪 =
𝑬
𝑹

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Universal Time Constant Chart
Universal time constant chart.
Universal Time Constant Chart
Selected values of e-x.

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Example
Example
VC versus time for the charging network
t = 0s,
Vc = 0
VC = E(1 – e-t/)

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Example
Example
iC and VR for the charging network
t = 0s
𝑰𝑪 =
𝑬
𝑹
t = 0s,
VR = E

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Example 2
A circuit of a resistor connected in series with a
0.5 F capacitor and has a time constant of
12 ms. Determine:
a) the value of the resistor, R and
b) the capacitor voltage, VC after 7 ms
connecting the circuit to a 10 V supply.
Example 2
• Solution 4
(a) The time constant  = CR, hence R = /C,
(b) The equation for the growth of capacitor
voltage is VC = V(1 - e-t/ ). Since  = 12 ms
= 12  10-3 s, V = 10 V and t = 7 ms, then






 

k
24
10
24
10
5
.
0
10
12 3
6
3
R
  4.42V
0.558)
-
10(1
1
10
1
10 583
.
0
10
12
10
7
3
3













 


 

e
e
vC

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TRANSIENTS IN CAPACITIVE NETWORKS: THE DISCHARGING PHASE
• We now investigate how to discharge a capacitor
while exerting some control on how long the
discharge time will be.
• You can, of course, place a lead directly across a
capacitor to discharge it very quickly—and possibly
cause a visible spark.
• For larger capacitors such those in TV sets, this
procedure should not be attempted because of the
high voltages involved—unless, of course, you are
trained in the maneuver.
(a)Charging network;
(b) discharging configuration.

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• For the voltage across the capacitor that is decreasing
with time, the mathematical expression is:
VC, iC, and VR for 5t switching between contacts
t = 0s,
Vc = E
t = 0s,
Vc = 0
iC = - Ie -t/
VR = -E𝒆−𝒕

VC = E𝒆−𝒕


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vC and iC for the network
VC = E𝒆−𝒕

iC = - Ie -t/
The Effect of on the Response

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The Effect of on the Response
Effect of increasing values of C
(with R constant) on the charging
curve for vC.
When C The system take longer time to
reach full charge.
Example
A capacitor is charged to 100 V and then discharged
through a 50 k resistor. If the time constant of the
circuit is 0.8 s, determine:
a) the value of the capacitor;
b) the time for the capacitor voltage to fall to 20 V;
c) the current flowing when the capacitor has been
discharging for 0.5 s; and
d) the voltage drop across the resistor when the
capacitor has been discharging for one second.

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Example
Given V = 100 V,  = 0.8 s, R = 50 k.
(a) Since time constant,  = CR,
(b) vC = Ve-t/ ;20 = 100e-t/0.8, t = 1.29s.
(c) The initial current flowing,
i =- Ie -t/ = -2e(-0.5/0.8) = -1.07mA
(d)
F
16
10
50
8
.
0
3




C
mA
2
10
50
100
3




R
V
I
V
7
.
28
287
.
0
100
100 25
.
1
1
/





 


e
Ve
v
v t
C
R

36

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Electrical Technology B
R-L Circuit Transient Response
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Transient Response in RL circuit
t = 0s,
VR = 0
t = 0s,
IL = 0
t = 0s,
VL = V
vL = Ve (-t/)
vR = V(1-e-t/ ) i = I(1-e-t/ )

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Time constant for an RL circuit
The time constant,  of a series connected L-R
circuit is defined in the same way as the time
constant for a series connected C-R circuit. Its
value is given by:
time constant,  =
𝑳
𝑹
seconds
Transient Curves Equations
• Decay of induced voltage,
vL = Ve(-Rt/L) = Ve (-t/)
• Growth of resistor voltage,
vR = V(1-e-Rt/L) = V(1-e-t/ )
• Growth of current flow,
i = I(1-e-Rt/L) = I(1-e-t/ )

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Example
A coil of inductance 0.04 H and resistance 10  is
connected to a 120 V, d.c. supply. Determine:
a) the final value of current,
b) the time constant,  of the circuit,
c) the value of current after a time equal to the time
constant from the instant the supply voltage is
connected,
d) the expected time for the current to rise to within
1% of its final value. (99% of I)
Example
(a) Final steady current,
I = (V/R) = 120/10 = 12 A.
(b) Time constant of the circuit,
 = (L/R) = 0.04/10 = 0.004 s or 4 ms.
(C) Given that: t = 
IL = I(1-𝑒
−
𝑡
)
IL = 12A(1-𝑒−
𝑡
𝑡)
IL = 12A(1-𝑒−1
)

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IL = 12A(0.6321)
IL = 7.59A
43
d) 𝐼𝐿 = 99%I
𝐼𝐿 =0.99I
 = 0.004 s
𝐼𝐿 =I 1 − 𝑒
−
𝑡

0.99I = I 1 − 𝑒−
𝑡
0.004
0.99 = 1 − 𝑒−
𝑡
0.004
44
𝑒−
𝑡
0.004 = 0.01
-
𝑡
0.004
= −4.60517
t = 0.0184s
 t = 18.4 ms

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RL Transient Response: Current Decay
When a series-connected L-R circuit is
connected to a d.c. supply as shown with S in
position A, a current I = V/R flows after a short
time, creating a magnetic field
(  I) associated with the inductor.
When S is moved to position B:
• Thus vL = -vR The current decays exponentially
to zero and since vR is proportional to the current
flowing, vR decays exponentially to zero. The
curves representing these transients are similar to
these shown in Figure below.

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decay of voltages, vL = -vR = -Ve(-t/)
• The equations representing the decay transient
current curves are:
decay of current, i = Ie (-t/)

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• The equations representing the decay transient
current curves are:
decay of resistor voltage, VR = Ee (-t/)
Example
An inductor has a negligible resistance and an inductance of 200
mH and is connected in series with a 1 k resistor to a 24 V, d.c.
supply. Determine the time constant of the circuit and the
steady-state value of the current flowing in the circuit.
Determine:
a) the current flowing in the circuit at a time equal to one time
constant,
b) the voltage drop across the inductor at a time equal to two
time constants, and
c) the voltage drop across the resistor after a time equal to three
time constants.

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Example
The time constant,  = (L/R) = 0.2/1000 = 0.2 ms.
The steady-state current, I = (V/R) = 24/1000 = 24 mA.
(a) The transient current, i = I(1 - e -t/ ) and t = 1.
Working in mA units gives,
i = 24(1 - e -(1/) ) = 24(1 - e-1) = 24(1 - 0.368) = 15.17 mA
(b) The voltage drop across the inductor,
vL = Ve -t/ = 24e-2/ = 24e-2 = 3.248 V
(c) The voltage drop across the resistor,
vR = V(1 - e-t/ ) =24(1 - e-3/ ) = 24(1 - e-3) = 22.81 V
Post-Test
52

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53
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