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?

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?

For POP, the data transceivers remain in the same direction as for instruction fetch (memory to processor), whereas for PUSH their direction has to be reversed.
Memory write operations are slower than memory read operations in an 8085 based system.
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.
Order of registers has to be interchanged for a PUSH operation, whereas POP uses their natural order.

LohaHridoy

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2025-01-04 04:42

1 Answers

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AhmedNazir · Accepted Answer · 2022-09-04T04:29:57+06:00

Option 3 : 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.

PUSH is pre decremented operation

Since it is 16 bit operation it takes 6 clock cycles for Fetch and decoding operation.

For PUSH operation:

T₁	T₂	T₃
6 clock cycles	3 clock cycles	3 clock cycles
Fetch and Decode	Write operation	Write operation

Total no of clock cycles = 6+3+3= 12 clock cycles

Whereas POP is a post incremented operation and it takes only 4 clock cycles in fetch and decode operation.

For POP:

T₁	T₂	T₃
4 clock cycles	3 clock cycles	3 clock cycles
Fetch and Decode	Write operation	Write operation

Total no of clock cycles = 4+3+3= 10 clock cycles