26. For the 8051, what is the content of the program counter (PC) upon reset? A.

Question

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)? A. OXAA B. 0x55 C. 0x00 D. 0x58 28. What is the magnitude of the unsigned integer data type? A. -128 to +127 B. 0 to 2545 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) Controllers 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 sec

Explanation / Answer

26
After a reset, the program counter is loaded with 0000H but the content of on-chip RAM is not affected

27
0xAA = 0b10101010
NOT 0xAA = 0b01010101 = 0x55

28
Unsigned integer generally ranges from 0 to 65535 (this is true when the size if int is 2bytes)

29
The size of the crystal frequency attached to the 8051 also decides the speed at which the 8051 timer ticks. The frequency for the timer is always 1 12th the frequency of the crystal attached to the 8051. i.e. 8051 TIMERS USE 1/12 OF XTAL FREQUENCY

30
Both timers 0 and 1 use the same register, called TMOD, to set the various timer operation modes. TMOD is an 8-bit register in which the lower 4 bits are set aside for Timer 0 and the upper 4 bits for Timer 1. In each case, the lower 2 bits are used to set the timer mode and the upper 2 bits to specify the operation.i.e. 'b'

31
XTAL = 24MHz
So timer freq = 2MHz
time period = 0.0005 ms
max timer counter value = 65535

hence max time = max timer counter value * time period = 65535 * 0.0005ms = 32.768ms

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