To build an eight-bit adder from two four-bit adders, which connection is required between the low-order and high-order devices?
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Athe high-order carry-in to ground
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Bthe low-order carry-out to the high-order carry-in
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Cthe high-order carry-out to ground
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Dthe low-order sum to the high-order data input
Answer
Correct Answer: the low-order carry-out to the high-order carry-in
Explanation
Introduction / Context:Cascading smaller adders to form wider adders is a standard digital design technique. Ensuring correct carry propagation between the lower and higher nibbles determines correctness for multi-bit operations.
Given Data / Assumptions:
- Two 4-bit adders are used to create one 8-bit adder.
- The low-order adder handles bits 3..0; the high-order adder handles bits 7..4.
- Each adder provides a carry-out and may accept a carry-in.
Concept / Approach:Ripple chaining requires that the carry generated by the lower block become the carry-in of the upper block. This preserves arithmetic correctness so that sums exceeding 15 in the low nibble increment the high nibble appropriately.
Step-by-Step Solution:
Compute low nibble sum and produce Cout_low.Feed Cout_low into Cin_high of the upper adder.Compute high nibble sum using Cin_high to include propagated carry.Form final 8-bit sum by concatenating high and low sums.Verification / Alternative check:Timing analysis shows a carry chain from LSB to MSB; functional simulation confirms correct results across all input combinations with this connection.
Why Other Options Are Wrong:
- High-order carry-in to ground: Ignores carries from the lower nibble.
- High-order carry-out to ground: Not a valid cascading connection.
- Low-order sum to high-order data input: Data buses are independent; only carry must propagate.
Common Pitfalls:Leaving Cin_high floating; forgetting to account for initial global Cin for full-adder chains.
Final Answer:the low-order carry-out to the high-order carry-in