Question

Need Help for Lab 6
please ...

Answer 1

me too stuck in the same ..

Answer 2

1. The
final
output
voltage
of
the
inverter
when
its
input
is
high
was
measured
to
be
~193mv.
The
equivalent
circuit
for
the
inverter
when
the
MOSFET
switch
is
on
is
a
voltage
divider
with
a
pullup
resistance
of
RL
and
a
pulldown
resistance
of
RON.
The
voltage-­‐divider
equation
for
the
output
voltage
is
.193
=
VSRON/(
RON
+
RL)
=
3RON/(RON
+
10000)
Solving
for
RON
gives
689Ω.
2. VTH
=
VSRON/(RON
+
RL)
=
.193V
RTH
=
RON
||
RL
=
643Ω
3. Equation
10.66
is
vC (t) =VTH +(VS −VTH )e−t/RTHCGS which,
after
substituting
the
known
values
becomes.25 =.193+(3−.193)e−t/(643⋅2⋅10−13 ) .
Solving
for
t
gives
t = −(643)(2 ⋅10−13 ) ln
.25−.193
3−.193
#
$ %
&
' (
=.5011⋅10−9 ,
i.e.,
about
0.5ns.
4. The
measured
value
for
tpd,0→1
is
0.6ns.
5. The
measured
value
for
tpd,1→0
is
3.45ns.
Substituting
the
known
values
into
Equation
10.71
gives
2.5 = 3+(.193−.3)e−t/(10000⋅2⋅10−13 ) ,
which,
when
solved
for
t
gives
3.4505ns.
The
plot
from
the
transient
analysis
of
the
ring
oscillator
is
shown
below
with
the
start
and
stop
times
of
one
cycle
as
indicated
(measurements
at
any
given
point
in
two
successive
cycles
gives
a
similar
result).
The
estimate
for
both
transitions
is
(31.375ns
–
15.125ns)/9
=
1.8ns.
From
questions
4
and
5:
tpd,0→1
+
tpd,1→0
=
4.15ns which
is
the
conservative
worst-­‐case
time
for
two
transitions,
almost
a
factor
3
longer!

Answer 3

for The circuit below contains an inverter designed to be used in a system where VS=3V, VOL=0.25V and VOH=2.5V. The input to the inverter is hooked to a voltage source that makes a 0→1 transition at t=0. The performance of the inverter is measured as it drives a 200fF capacitive load, which represents the parasitic capacitance of the wiring and the inputs of other logic gates hooked to the output of the inverter.

Answer 4

thanks a lot for your help :)