GATE EC 2016 Set 1 — Question 54
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Digital Circuits → Computer Organization → Machine Instructions & Addressing
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Question
In an 8085 system, a PUSH operation requires more clock cycles than a POP operation. Which one of the following options is the correct reason for this?
Correct answer
(C) The stack pointer needs to be pre-decremented before writing registers in a PUSH, whereas a POP operation uses the address already in the stack pointer.
Solution
In the 8085 microprocessor, a PUSH instruction (e.g.,
PUSH B) takes 12 T-states, while a POP instruction (e.g., POP B) takes 10 T-states. The difference lies in the opcode fetch machine cycle. For PUSH, the opcode fetch cycle is 6 T-states because it includes internal operations to decrement the stack pointer (SP) before the first memory write. For POP, the opcode fetch cycle is 4 T-states because the SP is used as is for the first memory read, and incremented afterwards. Thus, the pre-decrement requirement in PUSH adds extra clock cycles.Continue learning with Success Tracker
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