B D.70H 26 For the 8051, what is the content of the program counter (PC) upon re

Question

B D.70H 26 For the 8051, what is the content of the program counter (PC) upon reset? A PC= FFFFH B PC 0000H D. PC value unknown 27. What is the result of NOT (OXAA)? C 0x00 D.0x55 28. What is the magnitude of the unsigned integer data type? A. 0 to 255 B.-128 to +127 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 Simer? a) some externally applied frequency f b) controller's orystal frequency f c) controllers crystal frequency 12 d) exdenally applied frequency 12 30. What is the function of the TMOD register? a) TMOD register is used to set dflerent timer's or counter's to their appropriate modes b) TMOD register is used to load the count of the timer c) Is the destination or the final register where the resut 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) 0.11 m sec c) 32.768 m sec d) 24 m sec

Explanation / Answer

Ans 26. The contents of the PC (program counter) upon RESET of the 8051 is  

B). 0000 H(as we restart it starts with this value only)

Ans 27.Result of NOT(0XAA)

0xAA and 0XAA gives same decimal value so (A) might be the answer but not sure about

the question.According to me the answer should be FF55 which is not there in the

option.

Ans 28. Magnitude of unsigned Integer data type is

C). 0 -65535(Unsigned type of integers don't have sign bits that is the reason they

have this specefic range )

Ans 29. frequency of the clock that is being used as the clock source for the timer

C).controller's crystal frequency f/12

Ans 30.Function of the TMOD register

A).TMOD register is used to set different timer’s or counter’s to their appropriate

modes (in this way it tells the user that what mode is being used when operating any

of the timer or the counter).

Ans 31.maximum delay that can be generated with the crystal frequency of 24MHz

A).2.9sec( Formula is multiply the maximum number of counts with the time period

required to execute one machine cycle( 65536*1/24MHz=2.7sec approx) so here 2.9 is

the nearest one.)

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