A. MOV C, A
B. MVI A, 32H
C. LDA 2050H
D. JMP 2085H
E. ADD B
Choose the correct answer from the options given below :
To determine which 8085 instructions are 2-byte, we need to analyze the size of each instruction based on its opcode and operands.
The size of an instruction is determined by the number of bytes it occupies in memory. This includes the opcode byte and any immediate data or address bytes.
| Instruction | Type | Size (Bytes) | Explanation |
|---|---|---|---|
| A. MOV C, A | Register Transfer | $1$ byte | Opcode only. |
| B. MVI A, 32H | Immediate Load | $2$ bytes | Opcode + $1$ byte immediate data. |
| C. LDA 2050H | Memory Load (Direct) | $3$ bytes | Opcode + $2$ byte address. |
| D. JMP 2085H | Jump (Unconditional) | $3$ bytes | Opcode + $2$ byte address. |
| E. ADD B | Register Arithmetic | $1$ byte | Opcode only. |
Based on the analysis:
Therefore, the correct option includes only instruction B.
For the series diode configuration of the given figure, determine $V_D$ and $I_D$.
For the given emitter-bias network, determine the values of $I_C$ and $V_{CE}$. Assume $\beta = 100$.