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Chapter 9 8051 Timer Programming in Assembly and C

Chapter 9 8051 Timer Programming in Assembly and C. Objective. Embedded Systems with Timer/Counter 紅綠燈、電子鐘 碼錶 (count up) 、計時器 (count down) 計數器、計步器、跑步機等運動器材。 8051 有 2 個 Timer/Counter 的機制，我們在這一章裡要瞭解： 什麼是 Timer ，什麼是 Counter。 如何來使用 Timer。 如何來使用 Counter。. Sections.

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Chapter 9 8051 Timer Programming in Assembly and C

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  1. Chapter 9 8051 Timer Programming in Assembly and C

  2. Objective • Embedded Systems with Timer/Counter • 紅綠燈、電子鐘 • 碼錶(count up)、計時器(count down) • 計數器、計步器、跑步機等運動器材。 • 8051 有 2 個 Timer/Counter 的機制，我們在這一章裡要瞭解： • 什麼是 Timer，什麼是 Counter。 • 如何來使用 Timer。 • 如何來使用 Counter。

  3. Sections 9.1 Programming 8051 timers 9.2 Counter programming 9.3 Programming timers 0 and 1 in 8051 C

  4. Section 9.1Programming 8051 Timers

  5. Inside Architecture of 8051 External interrupts On-chip ROM for program code Timer/Counter Interrupt Control Timer 1 On-chip RAM Counter Inputs Timer 0 CPU Serial Port Bus Control 4 I/O Ports OSC P0 P1 P2 P3 TxD RxD Address/Data Figure 1-2. Inside the 8051 Microcontroller Block Diagram

  6. Timers /Counters • The 8051 has 2 timers/counters: timer/counter 0 and timer/counter 1. They can be used as • The timer is used as a time delay generator. • The clock source is the internal crystal frequency of the 8051. • An event counter. • External input from input pin to count the number of events on registers. • These clock pulses cold represent the number of people passing through an entrance, or the number of wheel rotations, or any other event that can be converted to pulses.

  7. Timer • 8051 timers use 1/12 of XTAL frequency as the input of timers, regardless ofmachine cycle. • Because the input of timer is a regular, fixed-periodic square wave, we can count the number of pulses and calculate the time delay. 8051 ÷ 12 Timer XTAL oscillator P1 to LCD TH0 Set Timer 0 TL0

  8. Counter • Count the number of events • External input from Tx input pin (x=0 or 1). • We use Tx to denote T0 or T1. • External input from T0 input pin (P3.4) for Counter 0 • External input from T1 input pin (P3.5) for Counter 1 8051 TH0 P1 to LCD TL0 Vcc P3.4 a switch T0

  9. Vcc P1.0 1 40 P0.0(AD0) P1.1 2 39 P0.1(AD1) P1.2 3 38 P0.2(AD2) P1.3 4 37 8051 (8031) P0.3(AD3) P1.4 5 36 P0.4(AD4) P1.5 6 35 P0.5(AD5) P1.6 7 34 P0.6(AD6) P1.7 8 33 P0.7(AD7) RST 9 32 (RXD)P3.0 EA/VPP 10 31 (TXD)P3.1 ALE/PROG 11 30 PSEN (INT0)P3.2 12 29 P2.7(A15) (INT1)P3.3 13 28 (T0)P3.4 P2.6(A14) 14 27 (T1)P3.5 P2.5(A13) 15 26 P2.4(A12) (WR)P3.6 16 25 P2.3(A11) (RD)P3.7 17 24 P2.2(A10) XTAL2 18 23 P2.1(A9) XTAL1 19 22 P2.0(A8) GND 20 21 Figure 4-1. 8051 Pin Diagram PDIP/Cerdip

  10. XTAL oscillator ÷12 C/T = 1 C/T = 0 T0 Pin Pin 3.4 TR0 Gate INT0 Pin Pin 3.2 Figure 9-8: Timer/Counter 0 timer input TH0 TL0 counter input 1:start 0:stop TF0 1. monitor by JNB 2. interrupt hardware control Sec 9.2 

  11. XTAL oscillator ÷12 C/T = 0 C/T = 1 T1 Pin Pin 3.5 TR1 Gate INT1 Pin Pin 3.3 Figure 9-9: Timer/Counter 1 timer input TH1 TL1 counter input 1:start 0:stop TF1 1. monitor by JNB 2. interrupt hardware control

  12. Registers Used in Timer/Counter • TH0, TL0 (Timer 0 registers) • TH1, TL1 (Timer 1 registers) • TMOD (Timer mode register) • TCON (Timer control register) • You can see Appendix H (pages 607-611) for details. • Since 8052 has 3 timers/counters, the formats of these control registers are different. • T2CON (Timer 2 control register), TH2 and TL2 used for 8052 only.

  13. Basic Registers of the Timer • Both Timer 0 and Timer 1 are 16 bits wide. • Each 16-bit timer can be accessed as two separate registers of low byte and high byte. • Timer 0: TH0 & TL0 • Timer 0 high byte, timer 0 low byte • Timer 1: TH1 & TL1 • Timer 1 high byte, timer 1 low byte • These registers stores • the time delay as a timer • the number of events as a counter

  14. TH0 TH1 TL0 TL1 D15 D15 D14 D14 D13 D13 D12 D12 D11 D11 D10 D10 D9 D9 D8 D8 D7 D7 D6 D6 D5 D5 D4 D4 D3 D3 D2 D2 D1 D1 D0 D0 Timer Registers Timer 0 Timer 1

  15. (MSB) (LSB) TF1 TR1 TF0 TR0 IE1 IT1 IE0 IT0 Timer 1 Timer0 for Interrupt TCON Register (1/2) • Timer control register: TCON • Upper nibble for timer/counter, lower nibble for interrupts • TR (run control bit) • TR0 for Timer/counter 0; TR1 for Timer/counter 1. • TRx is set by programmer to turn timer/counter on/off. • TRx=0: off (stop) • TRx=1: on (start)

  16. (MSB) (LSB) TF1 TR1 TF0 TR0 IE1 IT1 IE0 IT0 Timer 1 Timer0 for Interrupt TCON Register (2/2) • TF(timer flag, control flag) • TF0 for timer/counter 0; TF1 for timer/counter 1. • TFx is like a carry. Originally, TFx=0. When TH-TL roll over to 0000 from FFFFH, the TFx is set to 1. • TFx=0 : not reach • TFx=1: reach • If we enable interrupt, TFx=1 will trigger ISR.

  17. TF1 TR1 TF0 TR0 IE1 IT1 IE0 IT0 Table 9-2: Equivalent Instructions for the Timer Control Register TCON: Timer/Counter Control Register

  18. (MSB) (LSB) GATE C/T M1 M0 GATE C/T M1 M0 Timer 1 Timer 0 TMOD Register • Timer mode register: TMOD MOV TMOD,#21H • An 8-bit register • Set the usage mode for two timers • Set lower 4 bits for Timer 0 (Set to 0000 if not used) • Set upper 4 bits for Timer 1 (Set to 0000 if not used) • Not bit-addressable

  19. (MSB) (LSB) GATE C/T M1 M0 GATE C/T M1 M0 Timer 1 Timer 0 Figure 9-3. TMOD Register GATE Gating control when set. Timer/counter is enabled only while the INTx pin is high and the TRx control pin is set. When cleared, the timer is enabled whenever the TRx control bit is set. C/T Timer or counter selected cleared for timer operation (input from internal system clock). Set for counter operation (input from Tx input pin). M1 Mode bit 1 M0 Mode bit 0

  20. C/T (Clock/Timer) • This bit is used to decide whether the timer is used as a delay generator or an event counter. • C/T = 0 : timer • C/T = 1 : counter

  21. Gate • Every timer has a mean of starting and stopping. • GATE=0 • Internal control • The start and stop of the timer are controlled by software. • Set/clear the TR0 (or TR1) for start/stop timer. • GATE=1 • External control • The hardware way of starting and stopping the timer by softwareand an external source. • Timer/counter is enabled only while the INT0 (or INT1) pin has an 1 to 0 transition and the TR0 (or TR1) control pin is set. • INT0: P3.2, pin 12; INT1: P3.3, pin 13.

  22. M1, M0 • M0 and M1 select the timer mode for timers 0 & 1. M1 M0 Mode Operating Mode 0 0 013-bit timer mode 8-bit THx + 5-bit TLx (x= 0 or 1) 0 1 116-bit timer mode 8-bit THx + 8-bit TLx (x= 0 or 1) 1 0 28-bit auto reload 8-bit auto reload timer/counter; THx holds a value which is to be reloaded into TLx each time it overflows. 1 1 3Split timer mode

  23. (MSB) (LSB) GATE C/T M1 M0 GATE C/T M1 M0 Timer 1 Timer 0 Example 9-1 Indicate which mode and which timer are selected for each of the following. (a) MOV TMOD,#01H (b) MOV TMOD,#20H (c) MOV TMOD,#12H Solution: (a) TMOD = 00000001, mode 1 of timer 0 is selected. (b) TMOD = 00100000, mode 2 of timer 1 is selected. (c) TMOD = 00010010 mode 2 of timer 0, and mode 1 of timer 1 are selected. timer 1timer 0

  24. Find the timer’s clock frequency and its period for various 8051- based systems, with the following crystal frequencies. (a) 12 MHz (b) 16 MHz (c) 11.0592 MHz Solution: (a) 1/12 × 12 MHz = 1 MHz and T = 1/1 MHz = 1 s (b) 1/12 × 16 MHz = 1.333 MHz and T = 1/1.333 MHz = 0.75 s (c) 1/12 × 11.0592 MHz = 921.6 KHz; T = 1/921.6 KHz = 1.085 s ÷12 XTAL oscillator Example 9-2 

  25. (MSB) (LSB) GATE C/T M1 M0 GATE C/T M1 M0 Timer 1 Timer 0 Example 9-3 Find the value for TMOD if we want to program timer 0 in mode 2, use 8051 XTAL for the clock source, and use instructions to start and stop the timer. Solution: TMOD= 0000 0010 Timer 1 is not used. Timer 0, mode 2, C/T = 0 to use XTAL clock source (timer) gate = 0 to use internal (software) start and stop method.

  26. Timer Mode 1 • In following, we all use timer 0 as an example. • 16-bit timer (TH0 and TL0) • TH0-TL0 is incremented continuously when TR0 is set to 1. And the 8051 stops to increment TH0-TL0 when TR0 is cleared. • The timer works with the internal system clock. In other words, the timer counts up every 12 clocks from XTAL. • When the timer (TH0-TL0) reaches its maximum of FFFFH, it rolls over to 0000, and TF0 is raised. • Programmer should check TF0 and stop the timer 0.

  27. XTAL oscillator ÷12 C/T = 0 TL TH TF TF goes high when FFFF 0 overflow flag TR Mode 1 Programming like MC for 89C51 Start timer

  28. Steps of Mode 1 (1/3) • Chose mode 1 timer 0 • MOV TMOD,#01H • Set the original value to TH0 and TL0. • MOV TH0,#0FFH • MOV TL0,#0FCH • You had better to clear the flag to monitor: TF0=0. • CLR TF0 • Start the timer. • SETB TR0

  29. FFFC FFFD FFFE FFFF 0000 Steps of Mode 1 (2/3) • The 8051 starts to count up by incrementing the TH0-TL0. • TH0-TL0= FFFCH,FFFDH,FFFEH,FFFFH,0000H TR0=1 TR0=0 TH0 TL0 Start timer Stop timer Roll over TF0=0 TF0=0 TF0=0 TF0=0 TF0=1 TF0 Monitor TF01 until TF0=1

  30. Steps of Mode 1 (3/3) • When TH0-TL0rolls over from FFFFH to 0000, the 8051 set TF0=1. • TH0-TL0= FFFEH, FFFFH, 0000H (Now TF0=1) • Keep monitoring the timer flag (TF) to see if it is raised. • AGAIN: JNB TF0, AGAIN • Clear TR0 to stop the process. • CLR TR0 • Clear the TF flag for the next round. • CLR TF0

  31. Initial Count Values • The initial count value = FFFC. • The number of counts = FFFFH-FFFCH+1 = 4 • we add one to 3 because of the extra clock needed when it rolls over from FFFF to 0 and raises the TF flag. • The delay = 4 MCs for 89C51 • If MC=1.085 ms, then the delay = 4.34 ms • Figure 9-4 show a formula for delay calculations using mode 1 of the timer for a crystal frequency of XTAL=11.0592 MHz. • Examples 9-4 to 9-9 show how to calculations the delay generated by timer.

  32. (a) in hex (FFFF – YYXX + 1) × 1.085 swhere YYXX are TH, TL initial values respectively. Notice that values YYXX are in hex. (b) in decimal Convert YYXX values of the TH, TL register to decimal to get a NNNNN decimal number, then (65536 – NNNNN) × 1.085 s Figure 9-4. Timer Delay Calculation for XTAL = 11.0592 MHz 

  33. Find Timer Values • Assume XTAL = 11.0592 MHz . • How to find the inter values needed for the TH, TL? • Divide the desired time delay by1.085 s. 20ms ÷ 1.085 ms = 18433 • Perform65536 –n, where n is the decimal value we got in Step 1. 65536-18433=47103=B7FFH • Convert the result of Step 2 to hex, where yyxx is the initial hex value to be loaded into the timer’s registers. • Set TH = yy and TL = xx. TH=B7H, TL=FFH • Example 9-10

  34. Example 9-4 (1/4) In the following program, we are creating a square wave of 50% duty cycle (with equal portions high and low) on the P1.5 bit. Timer 0 is used to generate the time delay. Analyze the program. ;each loop is a half clock MOV TMOD,#01 ;Timer 0,mode 1(16-bit) HERE: MOV TL0,#0F2H ;Timer value = FFF2H MOV TH0,#0FFH CPL P1.5 ACALL DELAY SJMP HERE P1.5 50% 50% whole clock

  35. FFF2 FFF3 FFF4 FFFF 0000 TF0 = 0 TF0 = 0 TF0 = 0 TF0 = 0 TF0 = 1 Example 9-4 (2/4) ;generate delay using timer 0 DELAY: SETB TR0 ;start the timer 0 AGAIN:JNB TF0,AGAIN CLR TR0 ;stop timer 0 CLR TF0 ;clear timer 0 flag RET

  36. Example 9-4 (3/4) Solution: In the above program notice the following steps. 1. TMOD = 0000 0001 is loaded. 2. FFF2H is loaded into TH0 – TL0. 3. P1.5 is toggled for the high and low portions of the pulse. 4. The DELAY subroutine using the timer is called. 5. In the DELAY subroutine, timer 0 is started by the “SETB TR0” instruction. 6. Timer 0 counts up with the passing of each clock, which is provided by the crystal oscillator. As the timer counts up, it goes through the states of FFF3, FFF4, FFF5, FFF6, FFF7, FFF8, FFF9, FFFA, FFFB, FFFC, FFFFD, FFFE, FFFFH. One more clock rolls it to 0, raising the timer flag (TF0 = 1). At that point, the JNB instruction falls through.

  37. Example 9-4 (4/4) 7. Timer 0 is stopped by the instruction “CLR TR0”. The DELAY subroutine ends, and the process is repeated. 8. Remember to clear TF0 by the instruction “CLR TF0”. Notice that to repeat the process, we must reload the TL and TH registers, and start the timer again (in the main program).

  38. Example 9-5 In Example 9-4, calculate the amount of time delay in the DELAY subroutine generated by the timer. Assume that XTAL = 11.0592 MHz. Solution: The timer works with the internal system clock. frequency of internal system clock = 11.0592/12 = 921.6 KHz machine cycle = 1 /921.6 KHz = 1.085 s (microsecond) The number of counts = FFFFH – FFF2H +1 = 14 (decimal). The delay = number of counts × 1.085 s = 14 × 1.085 s = 15.19 s for half the clock. For the entire period of a clock, it is T = 2 × 15.19 s = 30.38 s as the time delay generated by the timer.

  39. Example 9-6 (1/2) In Example 9-5, calculate the accurate frequency of the square wave generated on pin P1.5. Solution: In the time delay calculation of Example 9-5, we did not include the overhead due to instructions in the loop. To get a more accurate timing, we need to add clock cycles from Table A-1 in Appendix A, as shown below.

  40. Example 9-6 (2/2) HERE: MOV TL0,#0F2H 2 MOV TH0,#0FFH 2 CPL P1.5 1 ACALL DELAY 2 SJMP HERE 2 ;----------delay using timer 0 DELAY: SETB TR0 1 AGAIN: JNB TF0,AGAIN 14 CLR TR0 1 CLR TF0 1 RET 2 0 Total 28 T = 2 × 28 × 1.085 s = 60.76 s, and F = 16458.196 Hz.

  41. Example 9-7 (1/2) Find the delay generated by timer 0 in the following code, using both of the methods of Figure 9-4. Do not include the overhead due to instructions. CLR P2.3 MOV TMOD,#01 ;Timer 0,mode 1(16-bit) HERE: MOV TL0,#3EH ;Timer value=B83EH MOV TH0,#0B8H SETB P2.3 SETB TR0 ;start the timer 0 AGAIN:JNB TF0,AGAIN CLR TR0 CLR TF0 CLR P2.3 P2.3

  42. Example 9-7 (2/2) Solution: (a) or (b) is OK. (a) (FFFF – B83E + 1) = 47C2H = 18370 in decimal and 18370 × 1.085 s = 19.93145 ms. (b) Since TH-TL = B83EH = 47166 (in decimal ) we have 65536 – 47166 = 18370. This means that the timer counts from B83EH to FFFF. This plus rolling over to 0 goes through a total of 18370 clock cycles, where each clock is 1.085 s in duration. Therefore, we have 18370 × 1.085 s = 19.93145 ms as the width of the pulse.

  43. Example 9-8 (1/2) Modify TL and TH in Example 9-7 to get the largest time delay possible. Find the delay in ms. In your calculation, exclude the overhead due to the instructions in the loop. Solution: TH0=TL0=0 means that the timer will count from 0000 to FFFF, and then roll over to raise the TF0 flag. As a result, it goes through a total of 65536 states. Therefore, we have delay = (65536 – 0) × 1.085 s = 71.1065 ms.

  44. Example 9-8 (2/2) CLR P2.3 ;clear P2.3 MOV TMOD,#01 ;Timer 0,mode1(16-bit) HERE: MOV TL0,#0 ;TL0=0, the low byte MOV TH0,#0 ;TH0=0, the high byte SETB P2.3 ;SET high P2.3 SETB TR0 ;start timer 0 AGAIN: JNB TF0,AGAIN ;monitor timer Flag 0 CLR TR0 ;stop timer 0 CLR TF0 ;clear timer 0 flag CLR P2.3

  45. Example 9-9 (1/2)  The following program generates a square wave on pin P1.5 continuously using timer 1 for a time delay. Find the frequency of the square wave if XTAL = 11.0592 MHz. In your calculation do not include the overhead due to instructions in the loop. MOV TMOD,#10H ;timer 1, mode 1 AGAIN:MOV TL1,#34H ;timer value=3476H MOV TH1,#76H SETB TR1 ;start BACK: JNB TF1,BACK CLR TR1 ;stop CPL P1.5 ;next half clock CLR TF1 ;clear timer flag 1 SJMP AGAIN ;reload timer1

  46. Example 9-9 (2/2)  Solution: In mode 1, the program must reload the TH1, TL1 register every timer if we want to have a continuous wave. FFFFH – 7634H + 1 = 89CCH = 35276 clock count Half period = 35276 × 1.085 s = 38.274 ms Whole period = 2 × 38.274 ms = 76.548 ms Frequency = 1/ 76.548 ms = 13.064 Hz. Also notice that the high portion and low portion of the square wave are equal. In the above calculation, the overhead due to all the instructions in the loop is not included.

  47. Example 9-10 (1/2)  Assume that XTAL = 11.0592 MHz. What value do we need to load into the timer’s registers if we want to have a time delay of 5 ms (milliseconds)? Show the program for timer 0 to create a pulse width of 5 ms on P2.3. Solution: XTAL = 11.0592 MHz  MC = 1.085 s. 5 ms / 1.085 s = 4608 MCs. To achieve that we need to load into TL0 and TH0 the value 65536 – 4608 = 60928 = EE00H. Therefore, we have TH0 = EE and TL0 = 00.

  48. Example 9-10 (2/2)  CLR P2.3 MOV TMOD,#01 ;Timer 0,mode 1 HERE: MOV TL0,#0 MOV TH0,#0EEH SETB P2.3 ;SET high P2.3 SETB TR0 ;start AGAIN: JNB TF0,AGAIN CLR TR0 ;stop CLR TF0 CLR P2.3 5ms P2.3

  49. Example 9-11 (1/2)  Assuming that XTAL = 11.0592 MHz, write a program to generate a square wave of 2 kHz frequency on pin P1.5. Solution: This is similar to Example 9-10, except that we must toggle the bit to generate the square wave. Look at the following steps. • The period of square wave = 1 / frequency = 1 / 2 kHz = 500 s. (b) The half period = 500 s /2 = 250 s. (c) 250 s / 1.085 s = 230 65536 – 230 = 65360 = FF1AH. (d) TL1 = 1AH and TH1 = FFH

  50. Example 9-11 (2/2)  MOV TMOD,#10H ;timer 1, mode 1 AGAIN:MOV TL1,#1AH ;timer value = FF1AH MOV TH1,#0FFH SETB TR1 ;start BACK: JNB TF1,BACK CLR TR1 ;stop CPL P1.5 CLR TF1 ;clear timer flag 1 SJMP AGAIN

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