*Chapter 9 8051 Timer Programming in Assembly and C

*Objective8051 2 Timer/Counter Timer Counter Timer Counter

*Sections9.1 Programming 8051 timers9.2 Counter programming9.3 Programming timers 0 and 1 in 8051 C

*Section 9.1Programming 8051 Timers

*Figure 4-1. 8051 Pin DiagramPDIP/Cerdip

*Inside Architecture of 8051CPUOn-chip RAMOn-chip ROM for program code4 I/O PortsTimer 0Serial PortFigure 1-2. Inside the 8051 Microcontroller Block DiagramOSCInterrupt ControlExternal interruptsTimer 1Timer/CounterBus ControlTxD RxDP0 P1 P2 P3Address/DataCounter Inputs

*Timers /CountersThe 8051 has 2 timers/counters: timer/counter 0 and timer/counter 1. They can be used asThe 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.

*TimerSet the initial value of registersStart the timer and then the 8051 counts up.When the registers equal to 0, the 8051 sets a bit to denote time out8051 timers use 1/12 of XTAL frequency, regardless of machine cycle.toLCDP18051TL0TH0SetTimer 0XTALoscillator 12Timer

*CounterCount the number of eventsShow the number of events on registersExternal input from T0 input pin (P3.4) for Counter 0External input from T1 input pin (P3.5) for Counter 1External input from Tx input pin. We use Tx to denote T0 or T1.

T0toLCDP3.4P18051a switchTL0TH0Vcc

*Registers Used in Timer/CounterTH0, TL0, TH1, TL1 (Timer 0 and 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.

*Basic Registers of the TimerBoth Timer 0 and Timer 1 are 16 bits wide.These registers stores the time delay as a timerthe number of events as a counter Timer 0: TH0 & TL0Timer 0 high byte, timer 0 low byteTimer 1: TH1 & TL1Timer 1 high byte, timer 1 low byteEach 16-bit timer can be accessed as two separate registers of low byte and high byte.

*Timer RegistersTimer 0Timer 1

*TCON Register (1/2)Timer control register: TCONUpper nibble for timer/counter, lower nibble for interruptsTR (run control bit)TR0 for Timer/counter 0; TR1 for Timer/counter 1.TR is set by programmer to turn timer/counter on/off.TR=0: off (stop)TR=1: on (start)

*TCON Register (2/2)TF (timer flag, control flag) TF0 for timer/counter 0; TF1 for timer/counter 1.TF is like a carry. Originally, TF=0. When TH-TL roll over to 0000 from FFFFH, the TF is set to 1.TF=0 : not reach TF=1: reach If we enable interrupt, TF=1 will trigger ISR.

*Table 9-2: Equivalent Instructions for the Timer Control RegisterTCON: Timer/Counter Control Register

For timer 0SETB TR0=SETB TCON.4CLR TR0=CLR TCON.4

SETB TF0=SETB TCON.5CLR TF0=CLR TCON.5For timer 1SETB TR1=SETB TCON.6CLR TR1=CLR TCON.6

SETB TF1=SETB TCON.7CLR TF1=CLR TCON.7

*TMOD RegisterTimer mode register: TMODMOV TMOD,#21HAn 8-bit registerSet 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

*Figure 9-3. TMOD RegisterGATE 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 1M0 Mode bit 0

*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 : timerC/T = 1 : counter

*GateEvery timer has a mean of starting and stopping.GATE=0Internal controlThe start and stop of the timer are controlled by software.Set/clear the TR for start/stop timer.GATE=1External controlThe hardware way of starting and stopping the timer by software and an external source.Timer/counter is enabled only while the INT pin is high and the TR control pin is set (TR).INT0: P3.2, pin 12; INT1: P3.3, pin 13.

*M1, M0M0 and M1 select the timer mode for timers 0 & 1.

M1 M0 Mode Operating Mode 0 0 0 13-bit timer mode 8-bit THx + 5-bit TLx (x= 0 or 1)0 1 1 16-bit timer mode 8-bit THx + 8-bit TLx 1 0 2 8-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 3 Split timer mode

*Example 9-1Indicate which mode and which timer are selected for each of the following. (a) MOV TMOD,#01H (b) MOV TMOD,#20H (c) MOV TMOD,#12HSolution:

(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 1 timer 0

*Example 9-2Find the timers 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

*Example 9-3Find 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.

*Timer Mode 1In 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.

*Mode 1 Programminglike MC for 89C51Start timer

*Steps of Mode 1 (1/3)Chose mode 1 timer 0MOV TMOD,#01HSet the original value to TH0 and TL0.MOV TH0,#0FFHMOV TL0,#0FCHYou had better to clear the flag to monitor: TF0=0.CLR TF0Start the timer.SETB TR0

*Steps of Mode 1 (2/3)The 8051 starts to count up by incrementing the TH0-TL0.TH0-TL0= FFFCH,FFFDH,FFFEH,FFFFH,0000HTF = 0TF = 0TF = 0TF = 0TF = 1TH0TL0Start timerStop timerMonitor TF until TF=1TR0=1TR0=0TF

*Steps of Mode 1 (3/3)When TH0-TL0 rolls 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 TR0Clear the TF flag for the next round.CLR TF0

*Initial Count ValuesThe 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 89C51If MC=1.085 ms, then the delay = 4.34 msFigure 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.

*Figure 9-4. Timer Delay Calculation for XTAL = 11.0592 MHz(a) in hex

(FFFF YYXX + 1) 1.085 s where 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

*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 50%50%whole clockP1.5

*Example 9-4 (2/4);generate delay using timer 0DELAY: SETB TR0 ;start the timer 0AGAIN:JNB TF0,AGAIN CLR TR0 ;stop timer 0 CLR TF0 ;clear timer 0 flag RET

*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.

*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).

*Example 9-5In 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.

*Example 9-6 (1/2)In Example 9-5, calculate the 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.

*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 0DELAY: SETB TR0 1AGAIN: JNB TF0,AGAIN 14 CLR TR0 1 CLR TF0 1 RET 2 0 Total 28T = 2 28 1.085 s = 60.76 s, and F = 16458.196 Hz.

*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 0AGAIN:JNB TF0,AGAIN CLR TR0 CLR TF0 CLR P2.3P2.3

*Example 9-7 (2/2)Solution:

(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.

*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.

*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 0AGAIN: JNB TF0,AGAIN ;monitor timer Flag 0 CLR TR0 ;stop timer 0 CLR TF0 ;clear timer 0 flag CLR P2.3

*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 1AGAIN:MOV TL1,#34H ;timer value=3476H MOV TH1,#76H SETB TR1 ;startBACK: JNB TF1,BACK CLR TR1 ;stop CPL P1.5 ;next half clock CLR TF1 ;clear timer flag 1 SJMP AGAIN ;reload timer1

*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 msFrequency = 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.

*Find Timer ValuesAssume that we know the amount of timer delay and XTAL = 11.0592 MHz .How to find the inter values needed for the TH, TL?Divide the desired time delay by 1.085 s. Perform 65536 n, where n is the decimal value we got in Step 1.Convert the result of Step 2 to hex, where yyxx is the initial hex value to be loaded into the timers registers.Set TH = yy and TL = xx.Example 9-10

*Example 9-10 (1/2)Assume that XTAL = 11.0592 MHz. What value do we need to load into the timers 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 TL and TH the value 65536 4608 = 60928 = EE00H. Therefore, we have TH = EE and TL = 00.

*Example 9-10 (2/2) CLR P2.3 MOV TMOD,#01 ;Timer 0,mode 1HERE: MOV TL0,#0 MOV TH0,#0EEH SETB P2.3 ;SET high P2.3 SETB TR0 ;startAGAIN: JNB TF0,AGAIN CLR TR0 ;stop CLR TF0 CLR P2.3 P2.35ms

*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

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

*Example 9-12 (1/2)Assuming XTAL = 11.0592 MHz, write a program to generate a square wave of 50 Hz frequency on pin P2.3.

Solution:Look at the following steps.(a) The period of the square wave = 1 / 50 Hz = 20 ms.(b) The high or low portion of the square wave = 10 ms.(c) 10 ms / 1.085 s = 9216 65536 9216 = 56320 in decimal = DC00H in hex.(d) TL1 = 00H and TH1 = DCH.

*Example 9-12 (2/2) MOV TMOD,#10H ;timer 1, mode 1AGAIN: MOV TL1,#00 ;Timer value = DC00H MOV TH1,#0DCH SETB TR1 ;startBACK: JNB TF1,BACK CLR TR1 ;stop CPL P2.3 CLR TF1 ;clear timer flag 1 SJMP AGAIN ;reload timer since ;mode 1 is not ;auto-reload

*Generate a Large Time DelayThe size of the time delay depends on two factors:They crystal frequencyThe timers 16-bit register, TH & TLThe largest time delay is achieved by making TH=TL=0. What if that is not enough? Example 9-13 show how to achieve large time delay.

*Example 9-13Examine the following program and find the time delay in seconds. Exclude the overhead due to the instructions in the loop. MOV TMOD,#10H MOV R3,#200 AGAIN: MOV TL1,#08 MOV TH1,#01 SETB TR1 BACK: JNB TF1,BACK CLR TR1 CLR TF1 DJNZ R3,AGAIN Solution:TH TL = 0108H = 264 in decimal 65536 264 = 65272.One of the timer delay = 65272 1.085 s = 70.820 msTotal delay = 200 70.820 ms = 14.164024 seconds

*Timer Mode 0Mode 0 is used to compatible with MSC-48Mode 0 is exactly like mode 1 except that it is a 13-bit timer instead of 16-bit.8-bit TH0 + 5-bit TL0The counter can hold values between 0000 to 1FFF in TH0-TL0.213-1= 2000H-1=1FFFHWe set the initial values TH0-TL0 to count up.When the timer reaches its maximum of 1FFFH, it rolls over to 0000, and TF0 is raised.

*Timer Mode 28-bit timer. TL0 is incremented (or roll over) continuously when TR0=1.Auto-reloadingTH0 is loaded into TL0 automatically when TL0=00H.TH0 is not changed .You need to clear TF0 after TL0 rolls over.In the following example, we want to generate a delay with 200 MCs on timer 0.See Examples 9-14 to 9-16

*Mode 2 programming

*Steps of Mode 2 (1/2)Chose mode 2 timer 0MOV TMOD,#02HSet the original value to TH0.MOV TH0,#38HClear the flag to TF0=0.CLR TF0Start the timer.SETB TR0After TH0 is loaded with the 8-bit value, the 8051 gives a copy of it to TL0.TL0=TH0=38H

*Steps of Mode 2 (2/2)The 8051 starts to count up by incrementing the TL0.TL0= 38H, 39H, 3AH,....When TL0 rolls over from FFH to 00, the 8051 set TF0=1. Also, TL0 is reloaded automatically with the value kept by the TH0.TL0= FEH, FFH, 00H (Now TF0=1)The 8051 auto reload TL0=TH0=38H. Go to Step 6 (i.e., TL0 is incrementing continuously).Note that we must clear TF0 when TL0 rolls over. Thus, we can monitor TF0 in next process.Clear TR0 to stop the process.

*Example 9-14 (1/2)Assuming that XTAL = 11.0592 MHz, find (a) the frequency of the square wave generated on pin P1.0 in the following program(b) the smallest frequency achievable in this program, and the TH value to do that. MOV TMOD,#20H ;Timer 1,mode 2 MOV TH1,#5 ;not load TH1 again SETB TR1 ;start (no stop TR1=0) BACK:JNB TF1,BACK CPL P1.0 CLR TF1 ;clear timer flag 1 SJMP BACK ;mode 2 is auto-reload

*Example 9-14 (2/2)Solution:(a) First notice that target address of SJMP. In mode 2 we do not need to reload TH since it is auto-reload. Half period = (FFH 05 +1) 1.085 s = 272.33 s Total period = 2 272.33 s = 544.67 s Frequency = 1.83597 kHz.(b) To get the smallest frequency, we need the largest period and that is achieved when TH = 00. Total period = 2 256 1.085 s = 555.52 s Frequency = 1.8kHz.

*Example 9-15 Find the frequency of a square wave generated on pin P1.0.Solution: MOV TMOD,#2H ;Timer 0,mode 2 MOV TH0,#0 AGAIN:MOV R5,#250 ;count 250 times ACALL DELAY CPL P1.0 SJMP AGAINDELAY:SETB TR0 ;startBACK: JNB TF0,BACK CLR TR0 ;stop CLR TF0 ;clear TF DJNZ R5,DELAY ;timer 2: auto-reload RETT = 2 (250 256 1.085 s) = 138.88 ms, and frequency = 72 Hz.

*Assemblers and Negative ValuesSince the timer is 8-bit in mode 2, we can let the assembler calculate the value for TH.For example (AT89C51), if we want to generate a time delay with 200 MCs, then we can use MOV TH1,#38H orMOV TH1,#-200Way 1: 256-200 = 56 = 38HWay 2: -200 = -C8H 2s complement of 200 = 100H C8H = 38 H

*Example 9-16Assuming that we are programming the timers for mode 2, find the value (in hex) loaded into TH for each of the following cases.(a) MOV TH1,#-200 (b) MOV TH0,#-60 (c) MOV TH1,#-3(d) MOV TH1,#-12 (e) MOV TH0,#-48Solution:Some 8051 assemblers provide this way.-200 = -C8H 2s complement of 200 = 100H C8H = 38 H

Decimal2s complement (TH value)-200 = - C8H38H- 60 = - 3CHC4H- 3FDH- 12F4H- 48D0H

*Example 9-17 (1/2)Find (a) the frequency of the square wave generated in the following code, and (b) the duty cycle of this wave.

Solution:MOV TH0,#-150 uses 150 clocks. The DELAY subroutine = 150 1.085 s = 162 s. The high portion of the pulse is twice that of the low portion (66% duty cycle). The total period = high portion + low portion = 325.5 s + 162.25 s = 488.25 s Frequency = 2.048 kHz.

*Example 9-17 (2/2) MOV TMOD,#2H ;Timer 0,mode 2 MOV TH0,#-150 ;Count=150AGAIN:SETB P1.3 ACALL DELAY ACALL DELAY CLR P1.3 ACALL DEALY SJMP AGAIN

DELAY:SETB TR0 ;startBACK: JNB TF0,BACK CLR TR0 ;stop CLR TF0 ;clear TF REThigh periodlow period

*Section 9.2Counter Programming

*CounterThese timers can also be used as counters counting events happening outside the 8051.When the timer is used as a counter, it is a pulse outside of the 8051 that increments the TH, TL.When C/T=1, the counter counts up as pulses are fed fromT0: timer 0 input (Pin 14, P3.4)T1: timer 1 input (Pin 15, P3.5)

*Table 9-1: Port 3 Pins Used For Timers 0 and 1

PinPort PinFunctionDescription14P3.4T0Timer/Counter 0 external input15P3.5T1Timer/Counter 1 external input

*Counter Mode 116-bit counter (TH0 and TL0)TH0-TL0 is incremented when TR0 is set to 1 and an external pulse (in T0) occurs.When the counter (TH0-TL0) reaches its maximum of FFFFH, it rolls over to 0000, and TF0 is raised.Programmers should monitor TF0 continuously and stop the counter 0.Programmers can set the initial value of TH0-TL0 and let TF0 as an indicator to show a special condition. (ex: 100 people have come).

*Figure 9-5. (a) Timer 0 with External Input (Mode 1)

*Figure 9-5. (b) Timer 1 with External Input (Mode 1)

*Counter Mode 28-bit counter. TL0 is incremented if TR0=1 and external pulse occurs.Auto-reloadingTH0 is loaded into TL0 when TF0=1.It allows only values of 00 to FFH to be loaded into TH0.You need to clear TF0 after TL0 rolls over.See Figure 9.6, 9.7 for logic viewSee Examples 9-18, 9-19

*Figure 9.6: Timer 0 with External Input (Mode 2)

*Figure 9.7: Timer 1 with External Input (Mode 2)

*Example 9-18 (1/2)Assuming that clock pulses are fed into pin T1, write a program for counter 1 in mode 2 to count the pulses and display the state of the TL 1 count on P2.Solution: MOV TMOD,#01100000B ;mode 2, counter 1 MOV TH1,#0 SETB P3.5 ;make T1 input portAGAIN:SETB TR1 ;startBACK: MOV A,TL1 MOV P2,A ;display in P2 JNB TF1,Back ;overflow CLR TR1 ;stop CLR TF1 ;make TF=0 SJMP AGAIN ;keep doing it

*Example 9-18 (2/2)We use timer 1 as an event counter where it counts up as clock pulses are fed into pin3.5. Notice in the above program the role of the instruction SETB P3.5. Since ports are set up for output when the 8051 is powered up , we must make P3.5 an input port by making it high. P2 is connected to 8 LEDs and input T1 to pulse.

*Example 9-19 (1/3)Assume that a 1-Hz frequency pulse is connected to input pin 3.4. Write a program to display counter 0 on an LCD. Set the initial value of TH0 to -60.Solution:Note that on the first round, it starts from 0 and counts 256 events, since on RESET, TL0=0. To solve this problem, load TH0 with -60 at the beginning of the program.

*Example 9-19 (2/3) ACALL LCD_SET_UP ;initialize the LCD MOV TMOD,#00000110B ;Counter 0,mode2 MOV TH0,#-60 SETB P3.4 ;make T0 as inputAGAIN:SETB TR0 ;starts the counterBACK: MOV A,TL0 ; every 60 events ACALL CONV ;convert in R2,R3,R4 JNB TF0,BACK ;loop if TF0=0 CLR TR0 ;stop CLR TF0 SJMP AGAIN

*Example 9-19 (3/3);converting 8-bit binary to ASCIICONV: MOV B,#10 ;divide by 10 DIV AB MOV R2,B ;save low digit MOV B,#10 ;divide by 10 once more DIV AB ORL A,#30H ;make it ASCII MOV R4,A MOV A,B ORL A,#30H MOV R3,A MOV A,R2 ORL A,#30H MOV R2,A ;ACALL LCD_DISPLAY here RETR4R3R2

*A Digital ClockExample 9-19 shows a simple digital clock.If we feed an external square wave of 60 Hz frequency into the timer/counter, we can generate the second, the minute, and the hour out of this input frequency and display the result on an LCD. You might think that the use of the instruction JNB TF0,target to monitor the raising of the TF0 flag is a waste of the microcontrollers time.The solution is the use of interrupt. See Chapter 11.In using interrupts we can do other things with the 8051. When the TF flag is raised, interrupt service routine starts.

*GATE=1 in TMODAll discuss so far has assumed that GATE=0.The timer is stared with instructions SETB TR0 and SETB TR1 for timers 0 and 1, respectively.If GATE=1, we can use hardware to control the start and stop of the timers.SETB TR0 and SETB TR1 for timers 0 and 1 are necessary.INT0 (P3.2, pin 12) starts and stops timer 0INT1 (P3.3, pin 13) starts and stops timer 1This allows us to start or stop the timer externally at any time via a simple switch.

*Example for GATE=1 The 8051 is used in a product to sound an alarm every second using timer 0. Timer 0 is turned on by the software method of using the SETB TR0 instruction and is beyond the control of the user of that product.The timer is turned ON/OFF by the 1 to 0 transition on INT0 (P3.2) when TR0 =1 (hardware control). However, a switch connected to pin P3.2 can be used to turn on and off the timer, thereby shutting down the alarm.

*Figure 9-8: Timer/Counter 0

*Figure 9-9: Timer/Counter 1

*Timer/Counter Mode 3For Timer/Counter 0As a Timer:Two 8-bit timers (TH0/TL0)As a Counter:8-bit counter. (TL0)For Timer/Counter 1Not available

*Section 9.3Programming Timers 0 and 1 in 8051 C

*8051 Timers in CIn 8051 C we can access the timer registers TH, TL, and TMOD directly using the reg51.h header file.See Example 9-20Timers 0 and 1 delay using mode 1See Example 9-22 and Example 9-25Timers 0 and 1 delay using mode 2See Examples 9-23 and 9-24Look by yourself

*Example 9-20 (1/2)Write an 8051 C program to toggle bits of P1 continuously with some time delay. Use Timer 0, 16-bit mode.Solution:#include void T0Delay(void);void main(void) { while(1) { //repeat forever P1=0x55; T0Delay(); //time delay P1=0xAA; T0Delay(); }}

*Example 9-20 (2/2)Assume XTML=11.0592MHz for the AT89C51 FFFFH-3500H+1=CB00H=51968 1.085ms 51968 56.384ms void T0Delay() { TMOD=0x01; //Timer 0, Mode 1 TL0=0x00; TH0=0x35; //initial value TR0=1; //turn on T0 while (TF==0); //text TF to roll over TR0=0; //turn off T0 TF0=0; } //clear TF0

*Example 9-25 (1/3)A switch is connected to P1.7. Write an 8051 C program to monitor SW and create the following frequencies on P1.5. SW=0: 500 Hz; SW=1: 750 Hz. Using Timer 0, mode 1.Assume XTML=11.0592MHz for the AT89C51

Solution:#include sbit mybit=P1^5;sbit SW=P1^7;void T0Delay(void);

*Example 9-25 (2/3)void main(void) { SW=1; //make P1.7 an input pin while(1) { //repeat forever mybit=~mybit; //toggle if (SW==0) T0Delay(0); //500Hz time delay else T0Delay(1); //750Hz time delay }}

*Example 9-25 (3/3)void T0Delay(unsigned char c) { TMOD=0x01; //Timer 0, Mode 1 if (c==0) { TL0=0x67; TH0=0xFC }; // FFFFH-FC67H+1=921, about 999.285ms, 500 Hz else { TL0=0x9A; TH0=0xFD }; // FFFFH-FD9AH+1=614, about 666.19ms, 750 Hz TR0=1; while (TF==0); TR0=0; TF0=0;}

*Time Delay with Different ChipsAlthough the numbers of clocks per machine cycle are vary with different versions, the frequency for the timer is always 1/12 the frequency of the crystal.To maintain compatibility with the original 8051The same codes put on AT89C51 and DS89C420 will generate different time delay.The factors of C compiler and MC of others instructionsThe C compiler is a factor in the delay size since various 8051 C compilers generate different hex code sizes. So, we still have approximate delay only.See Examples 9-21

*Example 9-21 (1/3)Write an 8051 C program to toggle bits P1.5 continuously every 50 ms. Use Timer 0, mode 1 for (a) AT89C51 and (b) DS89C420.Solution:#include void T0Delay(void);sbit mybit=P1^5;void main(void) { while(1) { //repeat forever mybit=~mybit; T0Delay(); }}

*Example 9-21 (2/3)(a) Assume XTML=11.0592MHz for the AT89C51 FFFFH-4BFDH+1=B403H=46083 1.085ms 46083 50ms void T0Delay() { TMOD=0x01; //Timer 0, Mode 1 TL0=0xFD; TH0=0x4B; //initial value TR0=1; //turn on T0 while (TF0==0); //BACK: JNZ TF0,BACK TR0=0; //turn off T0 TF0=0; } //clear TF0

*Example 9-21 (3/3)(b) Assume XTML=11.0592MHz for the DS89C4x0 FFFFH-4BFDH+1=B403H=46083 1.085ms 46083 50ms void T0Delay() { TMOD=0x01; //Timer 0, Mode 1 TL0=0xFD; TH0=0x4B; //initial value TR0=1; //turn on T0 while (TF0==0); //BACK: JNZ TF0,BACK TR0=0; //turn off T0 TF0=0; } //clear TF0

*8051 Counters in CExternal pulses to T0 (P3.4) and T1 (P3.5).See Examples 9-26 to 9-29.

*Example 9-26 (1/2)Assume that a 1-Hz external clock is being fed into T1 (P3.5). Write a C program for counter 1 in mode 2 to count up and display the state of the TL1 count on P1. Start the count at 0H.

Solution:P1 is connected to 8 LEDs.T1 is connected to 1Hz external clock.T1P3.5P11HzTL1TH1LEDs

*Example 9-26 (2/2)#include sbit T1=P3^5;void main(void) { T1=1; //make T1 an input pin TMOD=0x60; //counter 1, mode 2 TH1=0; //reload value while(1) { //repeat forever do {TR1=1; P1=TL1;} while (TF1==0); TR1=0; TF1=0; //clear flags}}

*You are able to List the timers of the 8051 and their associated registersDescribe the various modes of the 8051 timersProgram the 8051 timers in Assembly and C to generate time delaysProgram the 8051 counters in Assembly and C as event counters

*HomeworkChapter 9 Problems9,20,22,26,31,36,44,47,48,53

*, pass, !, , ., , , !

***Timer 0 is turned on by the software method of using the SETB TR0 instruction and is beyond the control of the user of that product. TR0 user control (INTx) .