An Embedded systems is a machine framework with a dedicated capacity inside a bigger mechanical or electrical framework, frequently with real time-computing constraints. It is implanted as a feature of a complete gadget often including equipment and mechanical parts. By contrast, an universally useful machine, for example, a (PC), is intended to be adaptable and to meet an extensive variety of end-client needs. Embedded systems control numerous gadgets in common use today. Here are for you the common embedded systems interview questions.
Now with experience in embedded systems there are immense opportunities for every person or professional. But even after that one has to face a interview. So here are some common questions asked in the embedded systems interview questions for advance level recruitment, usually with experience.
Embedded systems interview questions with answers:
1. What is read modify write technique?
• Read modify write is a technique used to access the ports.
• Here in a single instruction all the three actions are combined.
• Here initially the data is read from the port then modified .
• At last the value is written back on the port.
• This feature saves many lines of code and makes the process easier.
2. In which addressing mode is the DPTR register used?
• Data pointer register is used in the indexed addressing mode.
• It is used in accessing data from tne look-up table entries stored in ROM.
• SYNTAX: MOVC A, @A+DPTR
• Here c means the code (shows data elements are stored in code space of ROM).
• The contents of A are added to 16 bit DPTR to form the 16 bit address of the needed data.
3. Which registers are used for register indirect addressing mode if data is on-chip?
• R0 and R1 are the only registers used for register indirect addressing mode.
• These registers are 8 bit wide.
• Their use is limited to accessing only internal RAM.
• When these registers hold addresses of RAM ,they must be preceded by a @ sign.
• In absence of this sign it will use the contents of register than the contents of memory location pointed by the same register.
4. Of the 128-byte internal RAM how many bytes are bit addressable?
• Only 16 bytes of the 128 bytes of RAM are bit addressable.
• The bit addressable RAM locations are 20H to 2FH.
• They are addressed as 0 to 127 (decimal) or 00 to 7F.
• Also the internal RAM locations 20 to 2FH are both byte and bit addressable.
• These 16 bytes can be by single bit instructions using only direct addressing mode.
5. Explain parallel address space.
• The two physically separate memories using the same addresses form the parallel address space.
• Here the two memories are accessed using different access modes.
• This parallel addressing is generally used in 8052(enhanced version of 8051) having extra 128 bytes of RAM with addresses 80 to FFH and memory is called as upper memory.
• This distinguishes it from lower 128 bytes 00to 7FH.
• To access lower bytes direct addressing mode is used and for higher bytes indirect addressing mode is used.
6. Which port in 8051 performs a dual role?
• The port 3 can be used as simple input /output port or provides signals like interrupts.
• P3.0 and p3.1 are used for serial communication.
• P3.2and P3.3 are used as external interrupts.
• P3.4and P3.5 are used for timers.
• P3.6and P3.7 are used as read write signals of external memory.
7. Explain high-end embedded processor.
• The microcontroller & processor can be interchangeably used here.
• It forms a general purpose processor .
• High-end stands for system of greatest power.
• Here when a microcontroller cannot do a specific task the processor does the same with higher power
• Example: ADM 64
8. Which 8051 version is using UV-EPROM?
• Ultraviolet EPROM is used in the version 8751.
• Here it becomes easier to erase the data using uv rays.
• It takes around 20 minutes to erase the data.
• To overcome this short coming flash ROM versions of 8751 available.
9. Which 8051 version uses Flash ROM?
• Atmel Corporation is manufacturing flash ROM
• At89C51 chip of atmel is using flash ROM.
• Here no Rom eraser is needed.
• Erasing process is done by the PROM burner itself.
• To avoid use of PROM burner the 8052 version support in-system programming.
10. In which registers the immediate values cannot be directly loaded?
• The general purpose registers named A,B, R0-R7 can not accept the immediate data.
• To indicate immediate value pound sign is used before it.
• These register accept 8 bit data preceding a pound sign.
• There are registers present in 8051 used to store data temporarily.
11. Explain the difference between statements.
MOV A, #17H -a
MOV A, 17H -b
• Statement a indicates immediate data is copied into the register A.
• # in statement a indicates 17 is an immediate data which is moved to the destination.
• Here one copies the immediate data and other copies the data present at the specified address.
• In statement b the value present in 17H memory location is copied into A register.
• Absence of pound sign does not cause any error but required operation fails.
12. Which register is considered as the destination register for any arithmetic operation.
• The ‘A’ register called the accumulator is used as the destination register.
• Example : ADD R2, # 12H.
This is invalid as 12 is to be added to value present in R2 and data is also stored in R2.
• ADD A,# 12H
This is valid as 12 is to be added to the value present in accumulator.
• Foregoing discussion explains why register A is used as accumulator.
13. Explain LCALL.
• It is called long call and is a 3 byte instruction.
• Here 1st byte is used for opcode &2nd & 3rd bytes are used for address of target subroutine.
• Call subroutine can be used anywhere within 64 K- byte address space.
• Processor automatically saves on stack the address of instruction immediately below LCALL.
• After execution instruction return (RET) transfers control back to next instruction.
14. How much time is required by an instruction for execution in 8051?
• Time required depends on the number of clock cycles used to execute instruction.
• These clock cycle are called machine cycle.
• Length of machine cycle of an instruction depends on frequency of crystal oscillator of controller.
• Also one machine cycle lasts 12 oscillator period so machine cycle is 1 \12of crystal frequency.
• Crystal frequency 8051 = 11.0592MHz
Machine cycle = 11.0592MHz\12=921.6KHz=1.085micro second.
These are the most common embedded systems interview questions. Read these embedded systems interview questions for better opportunity and an edge to crack the interviews.
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