Question 1

8051 assembly control flow Will the following code execute only once or loop repeatedly? STAT: MOV A, #01H JNZ STAT

Accepted Answer

Introduction / Context: This question checks understanding of the 8051 conditional jump instruction JNZ (jump if A != 0) and how immediate moves affect the accumulator A. Given Data / Assumptions: MOV A, #01H loads accumulator with 0x01 (decimal 1). JNZ LABEL branches when accumulator is nonzero. No instructions modify A between the move and the jump. Concept / Approach: Because A is explicitly loaded with 1 before each JNZ, the zero condition is never met. Therefore, the branch is always taken, forming an infinite loop at STAT. Step-by-Step Solution: 1) Execute MOV A, #01H → A = 1.2) Execute JNZ STAT → since A != 0, jump to STAT.3) Repeat the same two instructions forever because A is reloaded with 1 each iteration.4) No path clears A to zero, so the loop never exits. Verification / Alternative check: Single-step in a simulator: the program counter toggles between the two instructions indefinitely. Why Other Options Are Wrong: True: Incorrect—code does not execute only once. It depends on the initial PSW flags: JNZ depends on A, not PSW flags. Only if interrupts are disabled: Interrupts are unrelated to JNZ behavior here. Common Pitfalls: Confusing JNZ with conditional flags; on 8051, JNZ checks accumulator zero directly. Final Answer: False — the sequence loops forever.

Question 2

8051 loop behavior with JNZ Will the following code lock the CPU in a loop? LOOP: MOV A, #00H JNZ LOOP

Accepted Answer

Introduction / Context: Understanding conditional jumps is essential in 8051 assembly. JNZ branches only when the accumulator is nonzero. This snippet deliberately loads zero before testing, so you must reason about what JNZ will do. Given Data / Assumptions: MOV A, #00H sets accumulator A = 0. JNZ LOOP branches if A != 0. No intervening instructions alter A. Concept / Approach: Immediately after loading 0, the condition for JNZ is false. Therefore, the jump is not taken and control falls through to the next instruction after JNZ, not back to LOOP. Step-by-Step Solution: 1) Execute MOV A, #00H → A = 0.2) Execute JNZ LOOP → since A == 0, do not jump.3) Program continues linearly; there is no infinite loop.4) A loop would require A to be nonzero at the JNZ. Verification / Alternative check: Run on an 8051 simulator and watch the program counter: it does not branch to LOOP. Why Other Options Are Wrong: True: Incorrect because A equals zero at the test. Only if A was nonzero before MOV: Prior value is overwritten by MOV. Only when the stack is empty: Stack state is irrelevant to JNZ here. Common Pitfalls: Assuming JNZ checks a generic zero flag; it specifically checks the accumulator value. Final Answer: False — the code executes once and then falls through.

Question 3

8051 indirect addressing semantics What does the instruction MOV A, @R1 do in the 8051?

Accepted Answer

Introduction / Context: Indirect addressing allows access to internal RAM using a register as a pointer. This question checks your understanding of the 8051 syntax @R1, which dereferences R1 to obtain the target byte. Given Data / Assumptions: @R1 means “address contained in R1”. MOV A, @R1 reads from internal data memory at that address. R0 or R1 are valid indirect address registers in the 8051. Concept / Approach: Dereferencing R1 provides a byte from internal data memory 0x00–0x7F (lower 128) or indirectly from 0x00–0xFF on variants; the byte is then moved into A. No flags are affected by the addressing mode itself. Step-by-Step Solution: 1) Fetch R1 → contains an 8-bit address n.2) Read internal RAM at address n → value m.3) Load accumulator A with m.4) R1 remains unchanged; only A receives the byte. Verification / Alternative check: Compare with MOV @R1, A which writes A to the addressed memory, confirming the direction of transfer is reversed in that form. Why Other Options Are Wrong: Copy R1 to A: That would be MOV A, R1. Copy A to R1: That would be MOV R1, A. Copy A to memory at R1: That would be MOV @R1, A. Common Pitfalls: Confusing register direct (R1) with register indirect (@R1) addressing, and mixing up source/destination order. Final Answer: copy the contents of memory whose address is in R1 to the accumulator

Question 4

8051 memory transfer instructions Is it true that data transfer from I/O to external data memory can only be done with MOV?

Accepted Answer

Introduction / Context: The 8051 has distinct instructions for internal data memory, external data memory, and code memory spaces. This question probes whether MOV is the only way to move data involving external data memory and I/O ports. Given Data / Assumptions: MOV operates on internal RAM, SFRs, and port SFRs. MOVX is the dedicated instruction for external data memory (xdata). I/O ports are accessed via SFR addresses (for example, P0, P1, P2, P3). Concept / Approach: External data memory requires MOVX @DPTR, A or MOVX A, @DPTR (or @Ri). Data can be staged through A between port SFRs and external RAM using MOV and MOVX in sequence; therefore, MOV is not the only instruction involved when xdata is accessed. Step-by-Step Solution: 1) Read from a port: MOV A, P1 (from I/O SFR to A).2) Write to external RAM: MOVX @DPTR, A.3) Or the reverse: MOVX A, @DPTR then MOV P1, A to output to a port.4) Conclusion: External data memory access specifically uses MOVX, not only MOV. Verification / Alternative check: Consult any 8051 instruction set summary: MOVX is listed as the instruction for xdata; MOV accesses internal/SFR spaces. Why Other Options Are Wrong: True: Incorrect because MOVX is required for external data memory. True, but only in mode 0 / EA LOW: Operating mode or EA pin does not change MOVX requirement for xdata. Common Pitfalls: Assuming MOV can directly access all spaces; on 8051, memory spaces are distinct and require the correct instruction. Final Answer: False — external data memory transfers use MOVX, often combined with MOV to/from ports.

Question 5

8051 program (code) memory capacity What is the maximum external code memory that can be interfaced to a standard 8051?

Accepted Answer

Introduction / Context: Classic 8051 architectures separate program memory (code) and data memory. Knowing the maximum addressable external program space is a fundamental device capability question. Given Data / Assumptions: Program counter is 16 bits in the baseline 8051. External code memory is addressed over a 16-bit bus via PSEN. EA pin selects internal versus external code fetch at reset and beyond. Concept / Approach: With a 16-bit program counter, the address range is 0x0000 to 0xFFFF, which is 65,536 bytes = 64 KB. Therefore, the maximum external code memory space is 64K bytes for a standard 8051 (without banking or extended addressing in derivatives). Step-by-Step Solution: 1) Bits in PC = 16 → unique addresses = 2^16 = 65,536.2) Bytes of addressable code space = 65,536 bytes.3) Express in kilobytes: 65,536 / 1024 = 64 KB.4) Conclude: 64K is the maximum external program memory size. Verification / Alternative check: Manufacturer datasheets and 8051 architecture references consistently state 64K external code memory for the standard core. Why Other Options Are Wrong: 32K: Understates the 16-bit space. 128K / 256K: Exceeds baseline 8051 PC width; applies only to enhanced derivatives with banking or paging. Common Pitfalls: Confusing external code size with external data (xdata) size or with internal code ROM sizes in specific derivatives. Final Answer: 64K

Question 6

8051 internal memory map understanding Are special function registers (SFRs) located in the next 128 addresses immediately above the lower 128 bytes of internal RAM?

Accepted Answer

Introduction / Context: The 8051 has a distinctive internal address map: lower internal RAM and upper special function registers share the same 8-bit address space but map to different functional regions. This question probes recognition of the address ranges involved. Given Data / Assumptions: Lower internal RAM occupies addresses 0x00–0x7F (128 bytes). Special function registers are mapped at 0x80–0xFF. Direct addressing to 0x80–0xFF targets SFRs, not general RAM. Concept / Approach: Conceptually, SFRs reside in the “next 128 locations” above the lower RAM block in the direct address space: 0x80–0xFF. These are not general-purpose RAM bytes; they control peripherals, ports, timers, serial, and CPU status. Indirect addressing cannot access SFRs; it wraps within lower RAM (and upper RAM on 8052), reinforcing the separation. Step-by-Step Solution: 1) Identify lower RAM: 128 bytes at 0x00–0x7F.2) Identify SFR region: 128 byte-wide address map at 0x80–0xFF.3) Direct addressing to 0x80–0xFF accesses SFRs (for example, P0 = 0x80, TMOD = 0x89).4) Conclude that SFRs lie in the next 128 addresses above lower RAM in direct address terms. Verification / Alternative check: Check any 8051 memory map: SFRs occupy addresses 0x80–0xFF. Why Other Options Are Wrong: False: Disagrees with the standard 8051 map. Only on 8052: 8052 adds upper 128 RAM via indirect addressing, but SFR map remains 0x80–0xFF on both. Only when EA is HIGH: EA relates to code memory selection, not SFR mapping. Common Pitfalls: Assuming SFRs are normal RAM, or trying to access SFRs with indirect addressing. Final Answer: True

Question 7

8051 reset vector and code source On reset with EA (External Access) pin held HIGH, where does the 8051 start fetching instructions from (PC = 0000H)?

Accepted Answer

Introduction / Context: The EA pin determines whether code fetches are from on-chip ROM/Flash or from external program memory when the address is within the internal block. Understanding reset behavior and the reset vector is a core 8051 concept. Given Data / Assumptions: Reset vector is at address 0000H. EA = 1 selects internal code for addresses mapped on-chip. EA = 0 forces external code fetches for all addresses. Concept / Approach: When EA is HIGH during reset, the CPU fetches the first instruction from internal code memory at 0000H (if present). Subsequent fetches within the internal code range also come from internal memory unless accesses exceed the internal range, where external fetch may occur in some derivatives. Step-by-Step Solution: 1) Assert reset; PC loads 0000H.2) EA = HIGH → select on-chip code for base addresses.3) CPU fetches the first opcode from internal code memory at 0000H.4) Execution proceeds from internal memory unless configured otherwise. Verification / Alternative check: Datasheets specify EA behavior and the reset vector mapping explicitly. Why Other Options Are Wrong: External code memory: Chosen when EA = 0. Internal/external data memory: Data memories are not used for opcode fetch. Common Pitfalls: Confusing EA pin function with the PSEN pin or with data memory spaces. Final Answer: internal code memory

Question 8

8051 timers and registers Is it necessary to use both TL0 and TL1 to start Timer 0?

Accepted Answer

Introduction / Context: The 8051 provides two timers: Timer 0 and Timer 1. Each timer uses its own pair of high and low byte registers. This question checks recognition of which registers belong to which timer. Given Data / Assumptions: Timer 0 uses TH0 (high) and TL0 (low). Timer 1 uses TH1 (high) and TL1 (low). TMOD configures modes; TCON starts/stops timers via TR0/TR1. Concept / Approach: Starting Timer 0 requires configuring TMOD for Timer 0 and setting TR0 in TCON. Only TH0 and TL0 are involved for count/reload depending on mode; TL1 is unrelated and belongs to Timer 1. Step-by-Step Solution: 1) Configure TMOD for Timer 0 mode (for example, mode 1, 16-bit).2) Load TH0/TL0 as needed.3) Set TR0 = 1 in TCON to run Timer 0.4) TL1 and TH1 remain unused for Timer 0 operation. Verification / Alternative check: Examine SFR map: TL0/TH0 at distinct addresses from TL1/TH1; no cross-dependency exists. Why Other Options Are Wrong: True: Incorrect; Timer 0 does not require Timer 1 registers. Only in auto-reload / external clock: These modes still use only TH0/TL0. Common Pitfalls: Confusing the two timers or mixing their SFRs when writing initialization code. Final Answer: False — Timer 0 uses TH0 and TL0 only.

Question 9

Microcontroller building blocks Which of the following do microcontrollers typically contain on-chip?

Accepted Answer

Introduction / Context: Unlike general-purpose microprocessors that often require many external components, microcontrollers integrate multiple subsystems on a single chip to enable compact embedded designs. This question tests recognition of what is typically integrated. Given Data / Assumptions: Microcontroller = CPU core plus essential memory and peripherals. Target applications include control and real-time tasks. Integration reduces board size, cost, and power. Concept / Approach: A typical microcontroller integrates a CPU, volatile memory (RAM) for variables and stack, and nonvolatile program memory (ROM/Flash). Many also include timers, GPIO, ADC/DAC, serial interfaces, and more, forming a complete embedded system on chip. Step-by-Step Solution: 1) CPU executes instructions and controls data flow.2) RAM holds working data and stacks during runtime.3) ROM/Flash stores firmware that persists across resets and power cycles.4) Combining these enables standalone operation without large external memories. Verification / Alternative check: Review any mainstream MCU family block diagram (for example, 8051 derivatives, AVR, PIC, ARM Cortex-M): CPU + RAM + Flash are standard. Why Other Options Are Wrong: CPUs / RAM / ROM individually are components; microcontrollers typically include all, not just one. Common Pitfalls: Confusing microcontrollers with microprocessors that may require external RAM/ROM and peripherals. Final Answer: all of the above

Question 10

8051 microcontroller — Port 3 alternate functions What is the alternate (special) function of port pin P3.4 on the 8051 family devices?

Accepted Answer

Introduction / Context: Each pin of Port 3 on the 8051 serves a dual role: a general-purpose I/O and a dedicated peripheral signal. Knowing these alternate functions is essential for interfacing timers, serial ports, and external memory correctly. Given Data / Assumptions: P3.4 is one of the Port 3 pins with an alternate function. Standard 8051 pin mapping is assumed (P3.0–P3.7 have fixed special roles). We are identifying the specific peripheral function attached to P3.4. Concept / Approach: In the classic 8051, Port 3 signals are: P3.0 = RXD, P3.1 = TXD, P3.2 = INT0, P3.3 = INT1, P3.4 = T0, P3.5 = T1, P3.6 = /WR, P3.7 = /RD. Therefore, P3.4 is the external input for Timer 0 (T0). Step-by-Step Solution: 1) Recall the standard mapping of 8051 Port 3 alternate functions.2) Locate P3.4 in the list: P3.4 corresponds to T0.3) T0 is the external clock/input for Timer 0.4) Conclude the correct option is Timer 0. Verification / Alternative check: Cross-reference any canonical 8051 pinout; P3.4 is universally shown as T0 (Timer 0 input). Why Other Options Are Wrong: Timer 1: mapped to P3.5 (T1), not P3.4. interrupt 0: mapped to P3.2 (INT0). interrupt 1: mapped to P3.3 (INT1). Common Pitfalls: Memorizing P3.x functions out of order or confusing T0/T1 with INT0/INT1. Always remember the ascending pairings: INT0 (P3.2), INT1 (P3.3), T0 (P3.4), T1 (P3.5). Final Answer: Timer 0

Question 11

8051 timers — counting resources How many 16-bit counter/timer modules are provided in a standard 8051 microcontroller?

Accepted Answer

Introduction / Context: The classic 8051 architecture includes hardware timers used for time delays, event counting, and baud-rate generation. Recognizing how many and what width these timers are is a fundamental specification question. Given Data / Assumptions: Baseline 8051 (not extended variants). Each timer is 16-bit wide in the most common modes. Naming convention: Timer 0 and Timer 1. Concept / Approach: Standard 8051 devices implement two 16-bit timer/counter modules: Timer 0 and Timer 1. These support multiple modes (including 13-bit, 16-bit, and auto-reload modes), and can be driven by internal clocks or external pins T0 and T1 (P3.4, P3.5). Step-by-Step Solution: 1) Recall the baseline peripheral set of 8051.2) Identify timers: T0 and T1 are present.3) Each can operate as a 16-bit counter/timer in standard mode.4) Therefore, the total count is two. Verification / Alternative check: Any 8051 datasheet block diagram clearly shows two timer/counter modules labeled Timer 0 and Timer 1. Why Other Options Are Wrong: 1, 3, or 4: do not match the standard 8051 core (some derivatives add Timer 2, but that is not the classic 8051). Common Pitfalls: Confusing classic 8051 with later derivatives that include Timer 2; always read the exact variant's datasheet. Final Answer: 2

Question 12

8051 external data memory interface (MOVX space) What is the maximum external data memory capacity that can be interfaced to the 8051?

Accepted Answer

Introduction / Context: The 8051 can address on-chip RAM plus external data memory through the MOVX instruction. Understanding the external memory address space is important for system design and memory mapping. Given Data / Assumptions: Classic 8051 external data memory bus with 16-bit addressing. Separate program and data spaces (Harvard-like architecture). Focus is on external data memory capacity. Concept / Approach: With a 16-bit address for external data memory, the addressable range is 2^16 bytes = 65,536 bytes = 64 KB. This is accessed via MOVX using DPTR or Ri (for paged modes on some variants). Step-by-Step Solution: 1) External data space uses 16 address lines.2) Capacity = 2^16 bytes = 65,536 bytes.3) 65,536 bytes ≈ 64 KB.4) Therefore, the correct capacity is 64K. Verification / Alternative check: Memory interfacing examples in 8051 textbooks consistently show 64 KB external data space with latches to demultiplex address/data lines (ALE). Why Other Options Are Wrong: 32K: half the actual address space. 128K or 256K: exceed the 16-bit addressing limit of the standard external data space. Common Pitfalls: Confusing external data memory size with external program memory size; both are 64 KB separately, but they are distinct spaces in the classical architecture. Final Answer: 64K

Question 13

8031 vs. 8051 — key architectural difference Which statement correctly distinguishes the 8031 from the 8051?

Accepted Answer

Introduction / Context: The 8031 and 8051 are closely related; however, one is intended for systems with external program memory while the other includes on-chip program ROM. Recognizing this difference helps when selecting parts for cost or board-space constraints. Given Data / Assumptions: Baseline Intel 8051 family definitions. 8031 and 8051 share CPU core, timers, serial port, and interrupts. Difference centers on on-chip program memory. Concept / Approach: The classic 8051 contains on-chip program ROM (originally mask ROM, later flash in derivatives). The 8031 is the ROM-less counterpart, requiring external program memory connected to the PSEN, ALE, and Port 0/Port 2 buses. Step-by-Step Solution: 1) Identify whether each part has internal program memory.2) 8051: has internal program ROM; 8031: no internal program ROM.3) Therefore, the correct statement is that the 8031 is ROM-less.4) Other features (timers/interrupts) are otherwise comparable. Verification / Alternative check: Any standard pinout diagram shows the 8031 using external program memory signals; 8051 can also use external program memory but additionally includes internal ROM. Why Other Options Are Wrong: No interrupts: incorrect; 8031 has interrupts. The 8051 is ROM-less: opposite of the truth. Extra 64 bytes claim: not the key differentiator between these part numbers. Common Pitfalls: Confusing RAM size differences in derivatives with the presence/absence of on-chip program memory. Final Answer: The 8031 is ROM-less.

Question 14

8051 instruction — moving the value from Port 3 to register R2 Which instruction correctly transfers the current value present at Port 3 into general-purpose register R2?

Accepted Answer

Introduction / Context: 8051 supports register-to-direct, direct-to-register, and accumulator-based data moves. Knowing valid addressing forms helps avoid assembler errors and unintended side effects. Given Data / Assumptions: We need to read a port's latch/pin value (Port 3) into a CPU register. Direct addressing of SFR P3 is permitted. Target register is R2 (in the current active register bank). Concept / Approach: The instruction format MOV Rn, direct is valid: it copies the byte at a direct address (such as SFR P3) into a working register Rn. Thus MOV R2, P3 transfers the Port 3 value into R2 in a single instruction. Step-by-Step Solution: 1) Determine source operand: SFR P3 (direct address).2) Determine destination: R2 (register in current bank).3) Use MOV R2, P3 to perform direct-to-register transfer.4) Assemble and verify addressing mode is supported. Verification / Alternative check: Referencing the 8051 instruction set shows opcodes MOV Rn, direct (A8–AF opcodes) are legal and execute in 2 bytes, 2 machine cycles on many variants. Why Other Options Are Wrong: MOV P2, R3: writes R3 to Port 2, wrong direction and wrong port. MOV R3, P2: reads Port 2, not Port 3, and puts it in R3, not R2. MOV 3P, R2: invalid mnemonic/syntax. Common Pitfalls: Forgetting that ports are SFRs and can be accessed via direct addressing; mixing up source/destination order in MOV. Final Answer: MOV R2, P3

Question 15

8051 register banks — count of general-purpose data registers In total, how many general-purpose data registers (R0–R7 across all four register banks) are available in the 8051's internal RAM?

Accepted Answer

Introduction / Context: The 8051 implements four banks of eight registers each, mapped in the lower internal RAM. Understanding the total number of GPRs (across banks) versus the simultaneously active set (one bank) is key for low-level programming and context switching. Given Data / Assumptions: Four register banks exist in the standard 8051. Each bank contains R0–R7 (8 registers). Only one bank is active at a time, selected by PSW bits RS1:RS0. Concept / Approach: Total defined GPRs = number of banks * registers per bank = 4 * 8 = 32. Although only 8 are actively addressed at once, all 32 physically exist in internal RAM (addresses 00h–1Fh). Step-by-Step Solution: 1) Banks: 4 (Bank 0..3).2) Each bank: 8 registers (R0..R7).3) Total = 4 * 8 = 32.4) Therefore, 32 general-purpose registers are implemented. Verification / Alternative check: Memory map shows 00h–1Fh allocated to register banks: 32 bytes total. Why Other Options Are Wrong: 8: counts only one active bank. 16 or 64: do not match the defined four-bank structure in classic 8051. Common Pitfalls: Confusing the total implemented registers with the concurrently accessible bank; the PSW selects which set responds to Rn opcodes. Final Answer: 32

Question 16

8051 polling code — does it check bit 2 of Port 1? Given the following program (formatted with real newlines), determine whether it checks if bit 2 of P1 is HIGH and, if so, writes FFH to Port 3. READ: MOV A, P1 ANL A, #02H CJNE A, #02H, READ MOV P3…

Accepted Answer

Introduction / Context: This question tests precise bit masking and comparison in 8051 assembly. Correctly identifying which bit is being tested requires understanding hexadecimal masks and the CJNE instruction semantics. Given Data / Assumptions: Code reads Port 1 into A. Mask used is #02H (binary 0000 0010). Loop repeats until CJNE finds equality with #02H. Concept / Approach: Bit values are: bit0 = 01H, bit1 = 02H, bit2 = 04H, etc. Applying ANL A, #02H zeroes all bits except bit1. If bit1 is high, A becomes 02H; if bit1 is low, A becomes 00H. The CJNE A, #02H, READ loops until A equals 02H (bit1 set). Therefore, the code checks bit1, not bit2. Step-by-Step Solution: 1) After MOV A, P1, accumulator contains P1 value.2) ANL A, #02H keeps only bit1.3) If bit1 = 1, A = 02H; else A = 00H.4) CJNE A, #02H, READ branches back unless A equals 02H.5) When A = 02H, the loop exits and MOV P3, #0FFH executes. Verification / Alternative check: Had the intention been to test bit2, the mask should be #04H and the comparison should be #04H accordingly. Why Other Options Are Wrong: True: incorrect because the mask checks bit1, not bit2. Common Pitfalls: Off-by-one errors in bit masks (confusing bit numbering with 1-based human counting). Always map bit n to 1 << n: bit2 corresponds to 04H. Final Answer: False

Question 17

8051 logical AND practice — accumulator result Compute the accumulator after executing these two instructions (showing real newlines): MOV A, #0BH ANL A, #2CH

Accepted Answer

Introduction / Context: This exercise checks understanding of bitwise AND on hexadecimal values in 8051 assembly. Being fluent in hex-to-binary conversion and bitwise operations is crucial for low-level I/O and mask manipulation. Given Data / Assumptions: MOV A, #0BH loads A with 0x0B. ANL A, #2CH computes A = A AND 0x2C. Binary conversions are performed to visualize the result. Concept / Approach: 0x0B = 0000 1011, 0x2C = 0010 1100. The AND operator keeps only positions where both operands have 1s. The outcome is 0000 1000 (0x08). Step-by-Step Solution: 1) Convert 0x0B → 0000 1011.2) Convert 0x2C → 0010 1100.3) AND bitwise: result → 0000 1000.4) Express in binary as 00001000 (matches the option). Verification / Alternative check: Compute in hex: 0x0B AND 0x2C = 0x08, which equals binary 00001000. Why Other Options Are Wrong: 11010111, 11011010, 00101000: do not match the correct AND result of the given operands. Common Pitfalls: Misaligning bits during conversion or confusing AND with OR/XOR. Always write both operands in full 8-bit binary before applying the operator. Final Answer: 00001000

Question 18

8051 internal RAM map — general-purpose storage region Is the statement true or false? “The last 96 locations in the internal data memory are reserved for general-purpose data storage and stack.”

Accepted Answer

Introduction / Context: Understanding the internal RAM layout of the 8051 is vital for stack placement, register bank usage, and bit-addressable operations. The lower 128 bytes are divided into specific regions, not all of which are general-purpose. Given Data / Assumptions: Classic 8051 lower internal RAM is 128 bytes (00h–7Fh). Upper 128 bytes (80h–FFh) are SFRs in the original 8051. General-purpose RAM and the stack usually occupy a subset of 00h–7Fh. Concept / Approach: Lower RAM map: 00h–1Fh = register banks (32 bytes), 20h–2Fh = bit-addressable area (16 bytes), 30h–7Fh = general-purpose RAM (80 bytes). The stack typically grows within this GPR area unless relocated by the programmer. Therefore, the “last 96 locations” claim is incorrect; it is actually 80 bytes (30h–7Fh). Step-by-Step Solution: 1) Compute GPR size: 0x7F − 0x30 + 1 = 0x50 = 80 bytes.2) Recognize stack resides in GPR region by default (SP starts at 07h).3) Conclude the statement claiming 96 bytes is false. Verification / Alternative check: Refer to any 8051 memory map diagram showing the partitions of lower RAM and SFR space. Why Other Options Are Wrong: True: contradicts the canonical 80-byte size of the GPR region in classic 8051. Common Pitfalls: Confusing lower RAM size (128 bytes) with the GPR subset; miscounting the bit-addressable area as general-purpose. Final Answer: False

Question 19

8051 logical OR practice — OR with zero (immediate) Compute the accumulator after these two instructions (shown with real newlines). Assume immediate OR with zero (#00H): MOV A, #2BH ORL A, #00H

Accepted Answer

Introduction / Context: This checks understanding of the ORL instruction and the difference between immediate and direct addressing. ORing any value with 00H (immediate) leaves the value unchanged; ORing with a direct address would depend on that RAM location's content. Given Data / Assumptions: The code sequence is interpreted as immediate OR: ORL A, #00H. MOV A, #2BH loads 0x2B (binary 0010 1011) into A. OR with 0x00 immediate should not modify the accumulator. Concept / Approach: OR rules: x OR 0 = x. Immediate addressing (prefixed with #) uses the literal constant. Therefore, A remains 0x2B after ORL A, #00H. Step-by-Step Solution: 1) Initial A = 0x2B.2) Execute ORL A, #00H → A = 0x2B OR 0x00 = 0x2B.3) Convert to the expected answer format: “2B H”.4) Select the option matching 2B H. Verification / Alternative check: Binary check: 0010 1011 OR 0000 0000 = 0010 1011 (unchanged). Why Other Options Are Wrong: 1B H, 3B H, 4B H: these would require setting or clearing specific bits, which does not happen when ORing with zero. Common Pitfalls: Confusing immediate with direct addressing. Without the #, ORL A, 00H would OR with RAM location 00H (R0 in Bank 0), yielding an unknown result unless that memory value is specified. Final Answer: 2B H

Question 20

Embedded Systems — Microcontrollers and Fixed Firmware A microcontroller is most useful in systems where the program is non-variable (stored in ROM/Flash) and the application is dedicated (single purpose). State whether this statement is true or fal…

Accepted Answer

Introduction / Context: Microcontrollers integrate CPU, RAM, nonvolatile program memory, and peripherals on a single chip. They are designed to run a fixed application reliably for long periods, which is why they dominate embedded products like appliances, meters, and controllers. Given Data / Assumptions: The statement refers to nonvariable programs stored in ROM or Flash. The target use case is a dedicated or single purpose application. General microcontroller examples include 8051, AVR, PIC, ARM Cortex-M. Concept / Approach: For dedicated tasks, microcontrollers offer low cost, low power, deterministic timing, and tight peripheral coupling. Nonvariable firmware avoids the overhead of dynamic loading, large operating systems, and complex memory hierarchies, making MCUs ideal for cost and reliability constrained designs. Step-by-Step Solution: 1) Identify the workload: single function control, sensing, and actuation.2) Map requirements to MCU strengths: integrated timers, ADCs, GPIO, UART, I2C, SPI.3) Fixed firmware enables simple boot, quick start, and minimal memory footprint.4) Result: microcontrollers are the natural fit for nonvariable, dedicated systems. Verification / Alternative check: Compare with microprocessors plus external chipset: those excel when applications vary, memory is large, and an operating system is needed. For fixed tasks, the simpler MCU platform proves more efficient and cheaper. Why Other Options Are Wrong: False: Contradicts the embedded design rationale for MCUs. Only true for 32-bit MCUs: Bit width does not change the principle. True only when an operating system is present: Most MCU designs are bare metal or use a small RTOS, not a full OS. Common Pitfalls: Assuming microprocessors and microcontrollers are interchangeable; confusing firmware upgradability with variability of the mission. Firmware can be updated, yet the application remains dedicated. Final Answer: True

The 8051 Microcontroller Questions