W09 314 HW08 extra credit Ind kick

W09 314 HW08 extra credit Ind kick - 
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 EECS
314
Winter
2009

 HW
08
Extra
credit
problem
on
the
Inductive
Kick
 Extra
Credit
Problem
on
the
Inductive
Kick
(50
points)
 
 An
abrupt
change
of
inductor
current
(for
example,
an
interruption
of
current
when
 the
switch
is
suddenly
open)
causes
an
abrupt
change
of
inductor
voltage,
known
as
 the
inductor’s
kickback
(which
can
damage
the
switch
and
can
be
dangerous
for
the
 human
operator).
Note
that
an
inductor
has
many
loops
of
wire
around
a
magnetic
 core,
and
an
electric
motor
has
many
loops
of
wire
around
a
magnetic
core
thus
 motors
act
as
inductive
loads,
in
particular,
motors
exhibit
the
kickback.

 
 Consider
the
circuit
shown
on
the
 diagram.
Here
the
battery
is
shown
as
a
 combination
of
the
voltage
source
VB
 and
resistance
RB;
the
DC
motor
is
 shown
as
a
series
combination
of
the
 inductance
L
and
resistance
RM,
and
RS
 is
the
shunt
resistance.

 
 Due
date:
Tuesday
March
24,
2009,
before
the
lecture
 Consider
the
role
of
the
shunt
resistor
RS:
if
it
is
large,
the
inductive
kick
at
time
t=0+
 gets
high
and
dangerous
for
the
operator
and
the
circuit;
if
RS
is
small,
it
consumes
 too
much
power
at
time
t<0
(due
to
current
division).

 
 In
other
words,
the
shunt
resistance
should
be
high
when
the
switch
is
closed
but
it
 should
be
very
low
when
the
switch
gets
open.
This
requirement
cannot
be
fulfilled
 with
a
resistor
but
a
semiconductor
diode
can
be
used
to
avoid
the
inductive
kick.
 

 In
the
circuit
shown
on
the
second
 diagram,
the
shunt
resistor
is
replaced
 with
a
semiconductor
diode,
which
is
 reverse‐biased
(not
conducting)
at
DC
 steady
state
when
the
switch
is
closed
 and
the
motor
running.
Basically,
the
DC
 source
pushes
the
current
through
the
 diode
in
the
direction
opposite
to
the
arrow
of
the
diode
symbol;
the
diode
does
not
 conduct
and
thus
does
not
absorb
any
power.

 
 When
the
switch
is
open
and
the
DC
source
disconnected,
the
inductor
 pushes
the
current
through
the
diode
in
the
direction
shown
by
the
 arrow
of
the
diode
symbol
thus
the
diode
conducts
and
the
voltage
 drop
across
it
does
not
exceed
the
offset
voltage
VDO
(assuming
the
 offset
diode
model).
 
 ©
2009
Alexander
Ganago

 Page
1
of
4
 
 
 EECS
314
Winter
2009

 HW
08
Extra
credit
problem
on
the
Inductive
Kick
 Part
1:
The
circuit
with
a
shunt
resistor
 
 Given:
VB
=
50
V,
RB
=
0.5
Ω,
RS
=
16
Ω,
RM
=
0.4
Ω.

 DC
steady‐state
conditions
are
reached
at
t
=
0‐
then
the
switch
is
open.

 In
the
given
circuit,
at
DC
steady‐state
conditions,
the
power
absorbed
by
the
shunt
 resistor
RS
is
only
2.5%
of
the
power
absorbed
by
the
motor
resistance
RM
.

 
 Assume
that
VB
and
RM
are
given
above,
and
RB=0
(an
ideal
battery).

 If
needed,
assume
L
=
1
H.

 
 (5
points)
Calculate
the
magnitude
|VM(t
=
0+)|
in
the
given
circuit.

 
 Your
answer:
|VM(t
=
0+)|
=
_________________
V
 
 (5
points)
Calculate
the
shunt
resistance
RS
such
that
|VM(t
=
0+)|=
50
V.
 
 Your
answer:
RS
=
_________________
Ω
 
 (5
points)
Calculate
the
powerPS
consumed
by
this
new
RS
under
DC
steady‐state
 conditions
(in
watts
and
as
%
of
that
absorbed
by
RM).

 
 Your
answer:
PS
=
________________
W,
or
_____________
%
of
that
absorbed
by
RM
 
 Show
your
work. ©
2009
Alexander
Ganago

 Page
2
of
4
 
 
 EECS
314
Winter
2009

 HW
08
Extra
credit
problem
on
the
Inductive
Kick
 Part
2:
The
circuit
with
a
semiconductor
diode
 
 Problem
 Given:
VB
=
50
V,
RB
=
0.5
Ω,
VDO
=
0.7
V,
RM
=
0.4
Ω.
 
 (5
points)
Under
DC
steady‐state
conditions
at
t=0‐,
calculate
the
power
P
supplied
 by
the
source.
If
the
data
given
in
the
problem
are
not
sufficient
for
this
calculation,
 make
a
clear
statement
about
it.

 
 Your
answer:
P
=
_________________
W
 

 (5
points)
Calculate
the
voltage
VM
across
the
motor
at
t
=
0+
 
 Your
answer:
VM
=
_________________
V
 
 (5
points)
Calculate
the
voltage
VL
across
the
inductor
at
t
=
0+
 
 Your
answer:
VL
=
_________________
V
 

 (5
points)
Calculate
the
power
PD
absorbed
by
the
diode
at
t=0+
 
 Your
answer:
PD
=
_________________
W
 
 Show
your
work.

 ©
2009
Alexander
Ganago

 Page
3
of
4
 
 
 EECS
314
Winter
2009

 HW
08
Extra
credit
problem
on
the
Inductive
Kick
 Part
3:
The
circuit
with
a
semiconductor
diode
 
 An
Old
Exam
Problem
 

 In
the
circuit
on
the
top
diagram,
 the
battery
is
shown
as
a
 combination
of
the
voltage
source
 VB
and
resistance
RB;
the
DC
 motor
is
shown
as
a
series
 combination
of
the
inductance
L
 and
resistance
RM,
and
RS
is
the
 shunt
resistance.

 
 Assume
VB
=100
V;
RB
=
0.
 
 In
an
attempt
to
reduce
the
inductive
kick
at
time
t
=
0+
(right
after
the
switch
is
 opened),
the
shunt
resistor
is
replaced
with
a
semiconductor
diode:
see
the
diagram
 below.
All
other
components
are
the
same
in
both
circuits.

 
 Assume
the
offset
 model
for
the
diode
 (see
the
sketch)
 with
VD0
=
0.7
V.

 
 
 
 In
the
circuit
with
the
diode:

 
 A. When
the
switch
is
closed,
the
motor
will
work
as
it
does
in
the
circuit
with
 the
shunt
resistor.

 B. When
the
switch
is
closed,
the
motor
will
not
run
because
the
voltage
across
 it
will
be
exactly
zero.
 C. When
the
switch
is
closed,
the
motor
will
hardly
work,
because
VM
=
0.7
V
 instead
of
100
V.

 D. When
the
switch
is
open,
the
diode
will
conduct
and
therefore
reduce
the
 inductive
kick
to
0.7
V.
 E. When
the
switch
is
open,
the
diode
will
not
conduct
thus
the
inductive
kick
is
 reduced
to
0.7
V.
 
 (5
points)
Choose
the
correct
statement.
Explain
why
it
is
correct.

 (10
points)
Explain
why
each
of
the
other
statements
is
wrong.

 
 Show
your
work.
Use
separate
pages
as
needed.

 ©
2009
Alexander
Ganago

 Page
4
of
4
 
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This note was uploaded on 10/28/2009 for the course EECS 314 taught by Professor Ganago during the Winter '07 term at University of Michigan.

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