B. FFH C. 7BH D. 7DH 26. For the 8051, what is the content of the program counte
ID: 3724911 • Letter: B
Question
B. FFH C. 7BH D. 7DH 26. For the 8051, what is the content of the program counter (PC) upon A. PC = FFFFH reset? B. PC = 0000H D. PC value unknown 27. What is the result of NOT (OxAA)? A. OXAA 0x55 C. Ox00 D. 0x58 28. What is the magnitude of the unsigned integer data type? A.-128 to +127 B. 0 to 255 C. 0 to 65,535 D.-32,768 to +32,767 29. What is the frequency of the clock that is being used as the clock source for the timer? a) Externally applied frequency f b) Controller's crystal frequency /12 c) Controller's crystal frequency d) Externally applied frequency /12 30. What is the function of the TMOD register? a) TMOD register is used to load the count of the timer. b) TMOD register is used to set timers or counter's to their appropriate modes c) Is the destination or the final register where the result is obtained after the operation of the timer d) is used to interrupt the timer 31. What is the maximum delay that can be generated with the crystal frequency of 24MHz? a) 2.9 sec b) 32.768 m sec c) 0.11 m sec d) 24 m secExplanation / Answer
26. we know the program counter(PC) which is 2 byte address, Whenever the 8051 is initialised or say reseted then PC always starts from 0000H and accordingly as the instruction starts executing. hence B is the ans for this
27. 0x55 is enabled to be reset if the value comes in next is 0xAA. As 0x55 is used to reset the value of 0xAA hence this is known to be NOT of 0xAA. Hence the answer to the question is B.
28. it is pretty clear that the unsigned int which takes 16 bit hence range will be rom 0 to ((2^16)-1) which is 0 - 65535. hence the answer is C.
29.according to the defination the frequency of the clock is equal to crystal frequency/12, hence the answer is C.
30. As we know there are different timers or counters need to be set to perticular modes whcih can be used for operating timer or counter these things are done by TMOD registers. Hece the answer is A.
31. as in the question it is given that we need to get the maximum delay we need to get the maximum counts and time period required to execute one cycle.
65536 * (1/24MHz) = 0.002730625 sec which is approximately 2.73 m sec but here there is no option given for the answer.
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