74LS283 (7483): the 4-bit binary full adder chip, live

The 74LS283 adds two 4-bit binary numbers, A4 A3 A2 A1 and B4 B3 B2 B1, plus a carry in C0, and outputs the 4-bit sum S4 S3 S2 S1 (the datasheet writes sigma) and a carry out C4. Bits are numbered 1 to 4, with bit 1 the least significant. The older 7483 does the same addition with a different pinout. This live circuit adds like the chip.

TI's datasheet describes the 74LS283 as performing "the addition of two 4-bit binary words", with full internal look-ahead across all four bits that generates the carry in about 10 ns, and it notes that the '283 is electrically and functionally identical to the '83A, "only the arrangement of the terminals has been changed": the 74LS283 puts VCC and GND on the corner pins (16 and 8), which the 7483 did not. This version chains four gate-level full adders (two XORs, two ANDs and an OR each), a ripple-carry adder, so its carry takes four steps where the chip's look-ahead (see the carry-lookahead adder) takes one; the outputs are the same. Two chips make an 8-bit adder: C4 of the low chip into C0 of the high chip. Put XOR gates (a 7486) on the B inputs and the same signal on C0, and it becomes an adder-subtractor. All 512 combinations of A, B and C0 are checked against real addition before the page is built.

Big circuit: pinch, scroll or tap + to zoom in, then drag to move around.

Live circuit: click the inputs to toggle them and watch the outputs update. It runs the real simulator, right here on the page.

Open this circuit in the lab →

What you need

8× XOR: A XOR B for each bit, and that XOR the carry in, the sum bit S
8× AND: A AND B (generate) and (A XOR B) AND carry (propagate) for each bit
4× OR: the carry into the next bit, generate OR propagate

Step by step

Place the inputs A1 to A4, B1 to B4 and C0, and probes S1 to S4 and C4.
Build a full adder for bit 1 from A1, B1 and C0.
Chain three more: each full adder takes its bit's A and B and the carry from the bit below.
The last carry is C4. Try 0111 + 0101 (7 + 5) with C0 = 0: S = 1100 (12), C4 = 0. Then 1111 + 0001: S = 0000 and C4 = 1.

Examples

A (A4..A1)	B (B4..B1)	C0	S (S4..S1)	C4	Decimal
0011	0100	0	0111	0	3 + 4 = 7
0111	0101	0	1100	0	7 + 5 = 12
1001	0110	1	0000	1	9 + 6 + 1 = 16
1111	1111	0	1110	1	15 + 15 = 30
1111	1111	1	1111	1	15 + 15 + 1 = 31

74LS283 pinout

Pin	Name	Pin	Name
1	S2 (sum 2)	16	VCC
2	B2	15	B3
3	A2	14	A3
4	S1 (sum 1)	13	S3 (sum 3)
5	A1	12	A4
6	B1	11	B4
7	C0 (carry in)	10	S4 (sum 4)
8	GND	9	C4 (carry out)

The value is A + B + C0, with C4 worth 16. The pinout (16-pin DIP and SOIC) and the facts above are from TI's SN74LS283 datasheet (SDLS095A), checked 2026-10-25; it lists typical times of 25 ns to add two 8-bit words and 45 ns for two 16-bit words with the 74LS283. The 7483 has the same function but VCC and GND in the middle of the package, so the two are not pin compatible. The datasheet's own bit names, 1 to 4, differ from the 0 to 3 used on the other adder pages here.

You can also read the full theory in the lesson on multi-bit adders. Prefer to build it yourself from scratch? Open the digiwleea lab and place the gates on the canvas, or start from the logic gate simulator.

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Each of these is one step from a transistor to a working CPU. Build every gate from CMOS transistors and work all the way up in the free digiwleea lab, with the free theory reference alongside.

Open the 74LS283 4-bit adder in the lab →

Frequently asked

What does the 74LS283 do?

It adds two 4-bit binary numbers and a carry in, giving a 4-bit sum and a carry out, with internal carry look-ahead so the carry does not ripple bit by bit.

What is the difference between the 7483 and the 74LS283?

The addition is the same; TI's datasheet says the 283 is electrically and functionally identical to the 83A with only the arrangement of the terminals changed. The 74LS283 has VCC and GND on the corner pins 16 and 8.

How do you make an 8-bit adder with 74LS283s?

Use two chips: the low four bits in one and the high four bits in the other, with the carry out C4 of the low chip wired to the carry in C0 of the high chip.

Can the 74LS283 subtract?

Yes, with help: invert B with XOR gates controlled by a SUB signal and feed the same SUB into C0. A plus NOT B plus 1 is A minus B in two's complement.

Why are the 74LS283 bits numbered 1 to 4?

That is the datasheet's naming: A1 and B1 are the least significant bits, C0 is the carry into bit 1 and C4 is the carry out of bit 4.