Test: Microprocessors - 2

Therefore required number of chips

Address of last memory location can always be find using this formula last location’s address
= Initial address + Hexadecimal equivalent of memory size - (0001 )_H

This is a indirect addressing mode because the address of the data which is to be transferred to register B is given by register pair HL.
It’s execution involves 2 machine cycles opcode fetch cycle and memory read cycle and total 7-T states.

SUB A when this instruction is executed content of A is subtracted from content of A so content of accumulator after execution of this instruction is

So

therefore, CY = 1,Z = 0 and in compare instruction content of accumulator remains unchanged. Therefore content of A = (30)H.
So, correct option is (c).

Initial content of A = (07)_H = (7)₁₀. So this set of instructions multiply the content of A by 10.

Before execution first RRC,

In both cases D₇ becomes carry flag, but in case of RAL carry before execution of this instruction goes to D₀ while in case of RLC, D₇ goes to D₀. So when D₇ and CY will be same then D₀ in both cases will be same and result will be same.

Now LDXA B → load the accumulator by the content at memory location pointed by BC register pair.
i.e. loaded by content at memory location
BC = 20FFH

Carry flag is set because carry is generated at MSB of the addition. Now conditional jump instruction JC display will be satisfied (condition is true) so control goes to display.
Therefore, output at port 01H = (FE)_H.

the conditional jump JNC LOOP will transfer the control to LOOP if condition is true i.e. carry CY = 0. But there CY = 1 so condition is false so it comes out of loop. So loop will be executed only once.

So function of above subroutine is to reset all the flags.

Since push instruction decreamented the stack pointer by 2. While POP operation increamented the stack pointer by 2.
Since there are one PUSH and one POP so overall effect is cancelled and content of SP remains unchanged.
SP = 209F

For finding call location we treat it as RST 4.5.
So, call location = Hexadecimal equivalent of 8 x4.5
= (36)₁₀ = (0024)H

SIM instruction interprets the accumulator content as

LDAX D → load accumualtor indirect by the content at memory location pointed by DE pair.

The loop includes four instructions DCX, MOV, ORA, and JNZ takes 24 clocks (6 + 4 + 4 + 10) for execution. The loop is repeated 2384H times. Which is converted to decimal as

it will be executed 2383H times completely and 1 time partially as condition becomes false f = 2 MHz.

LXI H, XX65H → HL = XX65H MVI M, FFH
Memory location pointed by HL pair will be loaded by the immediate data that is data at

INR M    data at memory location pointed by HL pair is increamented by 1 so it becomes