Find the delay generated by Timer 0 in the following code of 8051 microcontroller : clock period = 1.085 µs. MOV TMOD, #01H MOV TLO, #3DH MOV THO, #0A2H. SET B TRO Loop : J N B TFO, LOOP CLR TRO CLR TFO
26.043 ms
Read the code.
\(\texttt{MOV TMOD,\#01H}\) — the low nibble configures Timer 0; 0001 selects Mode 1, the 16-bit timer mode (GATE = 0, C/T = 0, so it is a timer clocked internally, started purely by software).
\(\texttt{MOV TL0,\#3DH}\ /\ \texttt{MOV TH0,\#A2H}\) — load the initial count. \(\texttt{SETB TR0}\) starts it; the JNB TF0, LOOP line simply spins until the overflow flag is set; CLR TR0 stops the timer.
How an 8051 timer measures a delay. The timer counts up from the loaded value, one increment per machine cycle, and sets the overflow flag TF0 when it rolls over from FFFFH to 0000H. So the number of counts is the distance from the loaded value to 65536.
Step 1 — form the 16-bit initial value. TH0 is the high byte, TL0 the low byte:
\(TH0{:}TL0 = A23D_{16}\)
Step 2 — convert to decimal.
\(A23D_{16} = (162 \times 256) + 61 = 41472 + 61 = 41533\)
Step 3 — counts to overflow.
\(N = 65536 - 41533 = 24003\ \text{counts}\)
Step 4 — multiply by the machine-cycle time. One count = one machine cycle = 1.085 µs (this corresponds to the usual 11.0592 MHz crystal, since 12/11.0592 MHz = 1.085 µs):
\(t = 24003 \times 1.085\ \mu s = 26043.3\ \mu s\)
\(t = 26.043\ \text{ms}\)
Shortcut for such problems. Counts = 65536 − (TH:TL); an equivalent trick is to take the 2's complement of the loaded value, i.e. \(FFFF_{16}-A23D_{16}+1 = 5DC3_{16} = 24003\). To design for a desired delay you invert the procedure: divide the delay by 1.085 µs, subtract from 65536, and load the hexadecimal result. Mode 1's longest single delay is \(65536 \times 1.085\ \mu s \approx 71.1\ \text{ms}\), so longer delays need a repeat loop.
Hence, the delay generated by Timer 0 is 26.043 ms.
Similar Questions
Given below are two statements :
Statement I : In 8051 micro-controller register banks and stack reserves 32 bytes from locations 00 to 1F Hex
Statement II : Bit-Addressable RAM locations are 20 H to 2 F H.
In the light of the above statements, choose the most appropriate answer from the options given below :
In "JZ Next" instruction of 8051 microcontroller, which register's content is checked to see if it is zero?
Using 8051 assembly language programming numbers 54 H and 98 H are added. After addition the status of CY, AC and P flags are respectively:
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Advantage of segmented memory in 8086 is
A. Allows the memory capacity to be 1M byte, although the actual address is of 16 bit size.
B. Allows the placing of code and data portions of the same program in different parts of memory.
C. Allows the memory capacity to be 2M byte, although the actual address is of 16 bit size.
D. Allows the placing of code and data portion of the same program in same parts of memory.
E. Allows the memory capacity to be 1M byte, although the actual address is of 20 bit size
Choose the correct answer from the options given below :
Which of the following statements are correct for a microcontroller ?
A. A microcontroller has on-chip I/O ports
B. A microcontroller has a fixed amount of RAM on the chip
C. The flag register in the 8051 is called program standard word
D. Auxiliary carry flag is set when there is a carry from D3 to D4 during an ADD or SUB operation
E. Overflow flag is only used to detect errors in unsigned arithmetic operations.
Choose the correct answer from the options given below :
Match List I with List II
LIST I (instruction of 8051 microcontroller) LIST II (Addressing mode) A. MOV A, # 32 H I. Register addressing mode B. MOV R 6, A II. Register Indirect addressing mode C. MOV R4, 7 FH III. Immediate Addressing mode D. MOV A, @ RO IV. Direct Addresing mode Choose the correct answer from the options given below:
The ALE line of an 8085 microprocessor is used to
Contents of ‘A’ register after execution of the following 8085 microprocessor program is
MVI A, 55 h
MVI C, 25 h
ADD C
DAA
Word 20 contains 40
Word 30 contains 50
Word 40 contains 60
Word 50 contains 70
Which of the following instructions loads 60 into the accumulator ?
(a) load immediate 60
(b) load direct 30
(c) load indirect 20
(d) load indirect 30
Which of the above statements are correct ?
Important Questions from Microprocessors
Register which is used to store values of arithmetic and logical operations is termed:
How many bytes of bit addressable memory is present in 8051 based microcontrollers?
A single instruction to clear the lower four bits of the accumulator in 8085 assembly language is-
In Microprocessor 8085 Address/Data buffer is a/an _______ buffer.
Microprocessor 8085 operates on a clock cycle with ______ duty cycle.
