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18
01
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1011
(101112--123)
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<
START OF
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DATA
SELECT
.
INPUTS
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⑨
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Q
④
:
Q
①
configuration
The
signal
as
a
of
Q
one
is
edge
of this
divider
flip-flop
( BY 2)
fed back into
of the clock
.
it to
allows
This
is
function
because the
D and switches
signal ( positive
or
only
inverse
on
negative)
.
:
:
:
:
;
O
O
O
O
O
O
O
O
①
LET
Re×t=390kR
1-
0.5s
-
w
fw=I
CEXT
.
I
→
51408ns
RE✗T( Ext
b- ✗ 108ns
=
1 /
✗
.
390hr
1.166×10-6
Cext ⇐ 1.2µF
⑤
R ,+2Rz=
¥4
36000
=
R
12,1-2122=36000
R,
+
Ra
=
+
,
R
,
Rz
+2122=008
0.8
36000
+122=28800
R,
R -28800
-
,
Let
Rz⇐6.8kR
solve for R
Rz
R -28800-6.81×52--22.1=52
,
→
,
R ,=22kR
,
22k +6.8k
221<+2×6.86
Rz=6.8kR
✗
100%
=
80.899%