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74F283SJ データシート(PDF) 2 Page - Fairchild Semiconductor

部品番号 74F283SJ
部品情報  4-Bit Binary Full Adder with Fast Carry
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メーカー  FAIRCHILD [Fairchild Semiconductor]
ホームページ  http://www.fairchildsemi.com
Logo FAIRCHILD - Fairchild Semiconductor

74F283SJ データシート(HTML) 2 Page - Fairchild Semiconductor

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Functional Description
The 74F283 adds two 4-bit binary words (A plus B) plus the
incoming Carry (C0). The binary sum appears on the Sum
(S0–S3) and outgoing carry (C4) outputs. The binary weight
of the various inputs and outputs is indicated by the sub-
script numbers, representing powers of two.
20 (A0 + B0 + C0) + 2
1 (A
1 + B1)
+ 22 (A
2 + B2) + 2
3 (A
3 + B3)
= S
0 + 2S1 + 4S2 + 8S3 + 16C4
Where (
+) = plus
Interchanging inputs of equal weight does not affect the
operation. Thus C0, A0, B0 can be arbitrarily assigned to
pins 5, 6 and 7 for DIPS, and 7, 8 and 9 for chip carrier
packages. Due to the symmetry of the binary add function,
the 74F283 can be used either with all inputs and outputs
active HIGH (positive logic) or with all inputs and outputs
active LOW (negative logic). See Figure 1. Note that if C0 is
not used it must be tied LOW for active HIGH logic or tied
HIGH for active LOW logic.
Due to pin limitations, the intermediate carries of the
74F283 are not brought out for use as inputs or outputs.
However, other means can be used to effectively insert a
carry into, or bring a carry out from, an intermediate stage.
Figure 2 shows how to make a 3-bit adder. Tying the oper-
and inputs of the fourth adder (A3, B3) LOW makes S3
dependent only on, and equal to, the carry from the third
adder. Using somewhat the same principle, Figure 3 shows
a way of dividing the 74F283 into a 2-bit and a 1-bit adder.
The third stage adder (A2, B2, S2) is used merely as a
means of getting a carry (C10) signal into the fourth stage
(via A2 and B2) and bringing out the carry from the second
stage on S2. Note that as long as A2 and B2 are the same,
whether HIGH or LOW, they do not influence S2. Similarly,
when A2 and B2 are the same the carry into the third stage
does not influence the carry out of the third stage. Figure 4
shows a method of implementing a 5-input encoder, where
the inputs are equally weighted. The outputs S0, S1 and S2
present a binary number equal to the number of inputs I1
I5 that are true. Figure 5 shows one method of implement-
ing a 5-input majority gate. When three or more of the
inputs I1–I5 are true, the output M5 is true.
Active HIGH: 0
+ 10 + 9 = 3 + 16
Active LOW: 1
+ 5 + 6 = 12 + 0
FIGURE 1. Active HIGH versus Active LOW Interpretation
FIGURE 2. 3-Bit Adder
FIGURE 3. 2-Bit and 1-Bit Adders
FIGURE 4. 5-Input Encoder
FIGURE 5. 5-Input Majority Gate
C0
A0
A1
A2
A3
B0
B1
B2
B3
S0
S1
S2
S3
C4
Logic Levels
L
L
H
L
H
H
L
L
H
H
H
L
L
H
Active HIGH
00
101
100
111
001
Active LOW
1
1
010
011
000
110


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