Transcript Presentasi Bab7 PENCACAH

Mata Kuliah
Teknik Digital
7. PENCACAH
Pencacah Reguler
Tabel 7.1. Tabel keadaan pencacah biner berurutan.
A
0
0
0
0
0
0
0
0
1
1
1
1
1
1
1
1
B
0
0
0
0
1
1
1
1
0
0
0
0
1
1
1
1
C D A+ B+ C+ D+
0 0
0 0 0 1
0 1
0 0 1 0
1 0
0 0 1 1
1 1
0 1 0 0
0 0
0 1 0 1
0 1
0 1 1 0
1 0
0 1 1 1
1 1
1 0 0 0
0 0
1 0 0 1
0 1
1 0 1 0
1 0
1 0 1 1
1 1
1 1 0 0
0 0
1 1 0 1
0 1
1 1 1 0
1 0
1 1 1 1
1 1
0 0 0 0
(a)
Pencacah naik
A
0
0
0
0
0
0
0
0
1
1
1
1
1
1
1
1
B
0
0
0
0
1
1
1
1
0
0
0
0
1
1
1
1
C
0
0
1
1
0
0
1
1
0
0
1
1
0
0
1
1
D
A+ B+ C+ D+
0
1 1 1 1
1
0 0 0 0
0
0 0 0 1
1
0 0 1 0
0
0 0 1 1
1
0 1 0 0
0
0 1 0 1
1
0 1 1 0
0
0 1 1 1
1
1 0 0 0
0
1 0 0 1
1
1 0 1 0
0
1 0 1 1
1
1 1 0 0
0
1 1 0 1
1
1 1 1 0
(b )
Pencacah turun
Pencacah dengan flip-flop T
AB
C
Pencacah Naik.
A
0
0
0
0
1
1
1
1
B
0
0
1
1
0
0
1
1
A+ B+ C+
0 0 1
0 1 0
0 1 1
1 0 0
1 0 1
1 1 0
1 1 1
0 0 0
C
0
1
0
1
0
1
0
1
00
01
11
1
1
0
TA TB TC
0 0 1
0 1 1
0 0 1
1 1 1
0 0 1
0 1 1
0 0 1
1 1 1
1
TA= BC
AB
C
00
01
11
10
1
1
1
1
0
1
TB= C
C
T
B
T
C
10
A
T
B
A
TC= 1
Pencacah dengan flip-flop T
AB
Pencacah Turun.
C
A
0
0
0
0
1
1
1
1
B
0
0
1
1
0
0
1
1
A+ B+ C+
1 1 1
0 0 0
0 0 1
0 1 0
0 1 1
1 0 0
1 0 1
1 1 0
C
0
1
0
1
0
1
0
1
TA TB TC
1 1 1
0 0 1
0 1 1
0 0 1
1 1 1
0 0 1
0 1 1
0 0 1
00
0
01
11
1
1
1
TB= BC
AB
C
0
TC= 1
00
01
11
10
1
1
1
1
1
TB= C
C
T
B
T
C
10
A
T
B
A
Pencacah dengan flip-flop T
Pencacah Naik-Turun
Up/Dn= M
M= 0 Down
M= 1 Up
Up/Dn= M
C
P
T
B
T
C
TC= P
A
T
B
TB= MPC + MPC
A
TA= MPBC + MPBC
Pencacah tak beraturan
ABC
000
001
010
011
100
101
110
111
A+B+C+
0 1 1
- - 1 0 0
0 1 0
1 0 1
0 0 0
- - - - -
BC
A
00
01
11
10
00
0
1
0 1
1
x
x
1 x
A+
BC
A 0 1
00 1 1
01 x
11 x
10
x
C+
AB
AB
C
BC
A 0 1
00 1
01 x
11 1 x
10
x
B+
x
01
11
1
x
x
10
1
TA = BC + BC
= B+C
C
AB
01
11
0
1
1
x
0
1
1
x
x
1
x
TB= AC
10
C
00
00
01
1
11
10
x
1
x
1
TC= B + C
Pencacah tak beraturan:
flip-flop T, Diagram Rangkaian
B
C
P
T
T
C
TC= P(B + C)
A
T
B
TB= P(A + C)
A
TA= P(B + C)
Diagram waktu pencacah
irreguler
P
A
0
0
0
1
1
1
1
0
0
0
1
0
0
1
0
B
0
C
0
TA
TB
TC
0
Pencacah dengan flip-flop RS
ABC
000
001
010
011
100
101
110
111
A+B+C+ RA SA RB SB RC SC
0 1 1
x 0 0 1 0 1
- - x x x x x x
1 0 0
0 1 1 0 0 x
0 1 0
x 0 0 x 1 0
1 0 1 0 x 0 x 0 1
0 0 0
1 0 x 0 1 0
- - x x x x x x
- - x x x x x x
Q
0
0
1
1
Q+
0
1
0
1
R
x
0
1
0
S
0
1
0
x
Peta-K Pencacah dengan RS
A
BC
00
01
11
10
A
0
A
0 1 BC 0 1
0 1 BC 0 1
x
00 1
x
00 1 1
x 1
01 x
x x
01 x
x x
11 x x
x
11
x
x
10
x
1 x
10
x
SB
RB
SC RC
RA= C
SB = AB
RB = BC
SC = BC
(c)
1
x
x
1
x
x
SA
RA
SA = BC
AB
C 00
0
0
01
1
11
x
10
1
1
0
x
0
x
AB
A+
C
0
1
x
1
x
1
x
x
RC = C
AB
0
00 01 11 10
1
0 x 0
1
x
1
B+
x
0
C
00 01 11
0 1 0 x
1 x
C+
0
x
10
1
0
Pencacah dengan flip-flop JK
Peta Keadaan Berikut
ABC
000
001
010
011
100
101
110
111
A+B+C+ JA KA JB KB JC KC
0 1 1
0 x 1 x 1 x
- - x x x x x x
1 0 0
1 x x 1 0 x
0 1 0
0 x x 0 x 1
1 0 1
x 0 0 x 1 x
0 0 0
x 1 0 x x 1
- - x x x x x x
- - x x x x x x
Q Q+
0
0
1
1
J
0
1
0
1
K
0
1
x
x
x
x
1
0
Peta-K Pencacah dengan JK
A
BC
00
01
11
10
A
0
x
1
A
0 1 BC 0 1
0 1 BC 0 1
x
00 1
x x
00 1 1
x 1
01 x
x x
01 x x
x x
11 x x
x
11 x x
x x
10 x x
1 x
10
x
JB
KB
JC K C
KA= C
JB = A
KB = C
JC = B
1
x
x
x
x
JA
KA
JA = BC
AB
C 00
0
0
01
1
11
x
10
1
1
0
x
0
x
AB
A+
C
0
x
x
1
x
1
x
1
x
x
KC = 1
AB
0
00 01 11 10
1
0 x 0
1
x
1
B+
x
0
C
00 01 11
0 1 0 x
1 x
C+
0
x
10
1
0
Peta-K Pencacah dengan ff D
AB
C 00
0
0
01
1
11
x
10
1
1
0
x
0
x
AB
C
AB
0
00 01 11 10
1
0 x 0
1
x
1
x
A+
B+
DA= AC + BC
DB= A B + AC
0
C
00 01 11
0 1 0 x
1 x
0
C+
DC= B C
x
10
1
0
Pencacah dalam Rangkaian Terpadu
Input
A
14
QA
13
QD
12
GND
QC
10
11
QA QD
>A
R0(1)
R0(2)
R0(1)
QB
9
Input
B
8
QC QB
R0(1) R0(2) QD QC QB QA
H
H
L L L
L
L
x
Count
x
L
Count
B<
R0(2)
R0(2)
NC
VCC
NC
NC
NC
K
>CK
J
R0(1)
>CK
7
K
6
J
5
>CK
4
K
3
J
2
J
>CK
K
1
Q
Q
Q
Q