杭州 • 浙江大学 • 2014
Principle and Interface
Techniques of Microcontroller
--8051 Microcontroller and Embedded Systems Using Assembly and C
LI, Guang (李光)
Prof. PhD, DIC, MIETWANG, You (王酉)
PhD, MIETChapter 10
Timer Programming
Tuesday, December 09, 2014
Outline
§10-1 Programming Timer
§10-2 Counter Programming
§10-3 Programming Timers in C
§10-1 Programming Timer
The 8051 has two timers/counters, they can be used either as
Timers to generate a time delay or as
Event counters to count events happening outside the microcontroller
Both Timer 0 and Timer 1 are 16 bits wide
Since 8051 has an 8-bit architecture, each
16-bits timer is accessed as two separate
registers of low byte and high byte
Structure of Timer0
Timer 0 & 1 Registers
Accessed as low byte and high byte
The low byte register is called TL0/TL1 and
The high byte register is called TH0/TH1
Accessed like any other register
MOV TL0,#4FH
MOV R5,TH0
TMOD Register
Both timers 0 and 1 use the same register, called TMOD (timer mode), to set the
various timer operation modes TMOD is a 8- bit register
The lower 4 bits are for Timer 0
The upper 4 bits are for Timer 1
In each case,
The lower 2 bits are used to set the timer mode
The upper 2 bits to specify the operation
TMOD Register
Gating control when set.
Timer/counter is enable 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
Timer or counter selected
Cleared for timer operation (input from internal system clock)
Set for counter operation (input from Tx input pin)
Example 10-1
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:
We convert the value from hex to binary. From Figure 9-3 we have:
(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.
TMOD Register
Example 10-2
Find the timer’s clock frequency and its period for various 8051-based
system, with the crystal frequency 11.0592 MHz when C/T bit of TMOD is 0.
Solution:
1/12 × 11.0529 MHz = 921.6 MHz;
T = 1/921.6 kHz = 1.085 us XTAL
oscillator ÷12 If C/T = 0, it is used as a timer for time delay generation.
The clock source for the time delay is the crystal frequency of the 8051
Timers of 8051 do starting and stopping by either software or hardware control
In using software to start and stop the timer where GATE=0
The start and stop of the timer are controlled by way of software by the TR (timer start) bits TR0 and TR1
– The SETB instruction starts it, and it is stopped by the CLR instruction
– These instructions start and stop the timers as long as GATE=0 in the TMOD register
The hardware way of starting and stopping the timer by an external source is achieved by making GATE=1 in the TMOD register
TMOD Register
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.
TMOD = 0000 0010
• Timer 0, mode 2
• C/T = 0 to use XTAL clock source
• gate = 0 to use internal (software) start and stop method.
Mode 0
13-bit timer mode
8-bit timer/counter THx with TLx as 5-bit
prescaler
Mode 1 Programming
The following are the characteristics and operations of mode1:
1. It is a 16-bit timer; therefore, it allows value of 0000 to FFFFH to be loaded into the timer’s register TL and TH
2. After TH and TL are loaded with a 16-bit initial value, the timer must be started
This is done by SETB TR0 for timer 0 and SETB TR1 for timer 1
3. After the timer is started, it starts to count up
It counts up until it reaches its limit of FFFFH
When it rolls over from FFFFH to 0000, it sets high a flag bit called TF (timer flag)
– Each timer has its own timer flag: TF0 for timer 0, and TF1 for timer 1
– This timer flag can be monitored
When this timer flag is raised, one option would be to stop the timer with the instructions CLR TR0 or CLR TR1, for timer 0 and timer 1, respectively
4. After the timer reaches its limit and rolls over, in order to repeat the process
TH and TL must be reloaded with the original value, and TF must be reloaded to 0
Mode 1 Programming
Steps to Mode 1 Program
To generate a time delay
1. Load the TMOD value register indicating which timer
(timer 0 or timer 1) is to be used and which timer mode (0 or 1) is selected
2. Load registers TL and TH with initial count value 3. Start the timer
4. Keep monitoring the timer flag (TF) with the JNB TFx, target instruction to see if it is raised
Get out of the loop when TF becomes high
5. Stop the timer
6. Clear the TF flag for the next round
7. Go back to Step 2 to load TH and TL again
Example 10-3
In the following program, we create 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
MOV TMOD,#01 ;Timer 0, mode 1(16-bit mode) HERE: MOV TL0,#0F2H ;TL0=F2H, the low byte
MOV TH0,#0FFH ;TH0=FFH, the high byte CPL P1.5 ;toggle P1.5
ACALL DELAY SJMP HERE
In the above program notice the following step.
1. TMOD is loaded.
2. FFF2H is loaded into TH0-TL0.
3. P1.5 is toggled for the high and low portions of the pulse.
DELAY:
SETB TR0 ;start the timer 0
AGAIN: JNB TF0,AGAIN ;monitor timer flag 0 until it rolls over CLR TR0 ;stop timer 0
CLR TF0 ;clear timer 0 flag RET
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, and so on until it reaches FFFFH. One more clock rolls it to 0, raising the timer flag (TF0=1).
At that point, the JNB instruction falls through.
7. Timer 0 is stopped by the instruction CLR TR0. The DELAY subroutine ends, and the process is repeated.
Notice that to repeat the process, we must reload the TL and TH registers, and start the process is repeated
Example 10-4
In Example 10-3, calculate the amount of time delay in the DELAY subroutine generated by the timer. Assume XTAL = 11.0592MHz.
Solution:
The timer works with a clock frequency of 1/12 of the XTAL frequency; therefore, we have 11.0592 MHz / 12 = 921.6 kHz as the timer frequency. As a result,
each clock has a period of T = 1/921.6kHz = 1.085us. In other words, Timer 0 counts up each 1.085us resulting in delay = number of counts × 1.085us.
The number of counts for the roll over is FFFFH – FFF2H = 0DH (13 decimal).
However, we add one to 13 because of the extra clock needed when it rolls over from FFFF to 0 and raise the TF flag. This gives 14 × 1.085us = 15.19us for half the pulse. For the entire period it is T = 2×15.19us = 30.38us as the time delay generated by the timer.
(a) in hex(FFFF – YYXX + 1) ×1.085 us, where YYXX are TH, TL initial values respectively.
Notice that value YYXX are in hex.
(b) in decimal Convert YYXX values of the TH, TL register to decimal to get a NNNNN decimal, then
(65536 - NNNN) ×1.085 us
Example 10-5
In Example 10-4, calculate the frequency of the square wave generated on pin P1.5.
Solution:
In the timer delay calculation of Example 9-5, we did not include the
overhead due to instruction in the loop. To get a more accurate timing, we need to add clock cycles due to this instructions in the loop. To do that, we use the machine cycle from Table A-1 in Appendix A, as shown below.
Cycles
HERE: MOV TL0,#0F2H 2
MOV TH0,#0FFH 2
CPL P1.5 1
ACALL DELAY 2
SJMP HERE 2
DELAY: SETB TR0 1
AGAIN: JNB TF0,AGAIN 14
CLR TR0 1 CLR TF0 1 RET 2 Total 28 T = 2 × 28 × 1.085 us = 60.76 us and F = 16458.2 Hz
Example 10-6
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 instruction.
CLR P2.3 ;Clear P2.3
MOV TMOD,#01 ;Timer 0, 16-bitmode HERE: MOV TL0,#3EH ;TL0=3Eh, the low byte MOV TH0,#0B8H ;TH0=B8H, the high byte SETB P2.3 ;SET high timer 0
SETB TR0 ;Start the timer 0 AGAIN: JNB TF0,AGAIN ;Monitor timer flag 0 CLR TR0 ;Stop the timer 0
CLR TF0 ;Clear TF0 for next round CLR P2.3
Solution:
(a) (FFFFH – B83E + 1) = 47C2H = 18370 in decimal and 18370 ×1.085 us = 19.93145 ms
(b) Since TH – TL = B83EH = 47166 (in decimal) we have
65536 –47166 = 18370. This means that the timer counts from B38EH to FFFF. This plus Rolling over to 0 goes through a total of 18370 clock cycles, where each clock is 1.085 us in duration. Therefore, we have 18370 × 1.085 us = 19.93145 ms as the width of the pulse.
Example 10-7
Modify TL and TH in Example 10-6 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:
To get the largest delay we make TL and TH both 0. This will count up from 0000 to FFFFH and then roll over to zero.
CLR P2.3 ;Clear P2.3
MOV TMOD,#01 ;Timer 0, 16-bitmode 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 the timer 0
CLR TF0 ;Clear timer 0 flag CLR P2.3
Making TH and TL both zero means that the timer will count from 0000 to FFFF, and then roll over to raise the TF flag. As a result, it goes through a total Of 65536 states. Therefore, we have delay =(65536 - 0)×1.085 us = 71.1065ms.
Example 10-8
The following program generates a square wave on 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,#10 ;Timer 1, mod 1 (16-bitmode) AGAIN: MOV TL1,#34H ;TL1=34H, low byte of timer MOV TH1,#76H ;TH1=76H, high byte timer SETB TR1 ;start the timer 1
BACK: JNB TF1,BACK ;till timer rolls over CLR TR1 ;stop the timer 1
CPL P1.5 ;comp. p1. to get hi, lo CLR TF1 ;clear timer flag 1
SJMP AGAIN ;is not auto-reload
Solution:
Since FFFFH – 7634H = 89CBH + 1 = 89CCH and 89CCH = 35276
clock count and 35276 × 1.085 us = 38.274 ms for half of the square wave.
The frequency = 13.064Hz.
Also notice that the high portion and low portion of the square wave pulse are equal. In the above calculation, the overhead due to all the instruction in the loop is not included.
Finding the Loaded Timer Values
To calculate the values to be loaded into the TL and TH registers, look at the following example
Assume XTAL = 11.0592 MHz, we can use the following steps for finding the TH, TL registers’ values
1. Divide the desired time delay by 1.085 us
2. Perform 65536 – n, where n is the decimal value we got in Step1
3. Convert the result of Step2 to hex, where yyxx is the
initial hex value to be loaded into the timer’s register
4. Set TL = xx and TH = yy
Example 10-9
Assume that XTAL = 11.0592MHz. What value do we need to load the timer’s register 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:
Since XTAL = 11.0592 MHz, the counter counts up every 1.085 us. This means that out of many 1.085 us intervals we must make a 5 ms pulse. To get that, we divide one by the other. We need 5 ms / 1.085 us = 4608
clocks. To Achieve that we need to load into TL and TH the value 65536 – 4608 = EE00H. Therefore, we have TH = EE and TL = 00.
CLR P2.3 ;Clear P2.3
MOV TMOD,#01 ;Timer 0, 16-bitmode HERE: MOV TL0,#0 ;TL0=0, the low byte MOV TH0,#0EEH ;TH0=EE, 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 the timer 0 CLR TF0 ;Clear timer 0 flag
Example 10-10
Assume 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 Example10-9, except that we must toggle the bit to generate the square wave. Look at the following steps.
(a) T = 1 / f = 1 / 2 kHz = 500 us the period of square wave.
(b) 1 / 2 of it for the high and low portion of the pulse is 250 us.
(c) 250 us / 1.085 us = 230 and 65536 – 230 = 65306 which in hex is FF1AH.
(d) TL = 1A and TH = FF, all in hex. The program is as follow.
MOV TMOD,#01 ;Timer 0, 16-bitmode
AGAIN: MOV TL1,#1AH ;TL1=1A, low byte of timer MOV TH1,#0FFH ;TH1=FF, the high byte SETB TR1 ;Start timer 1
BACK: JNB TF1,BACK ;until timer rolls over CLR TR1 ;Stop the timer 1 CLR P1.5 ;Clear timer flag 1 CLR TF1 ;Clear timer 1 flag SJMP AGAIN ;Reload timer
Example 10-11
Assume XTAL = 11.0592 MHz, write a program to generate a square wave of 50 kHz frequency on pin P2.3.
Solution:
Look at the following steps.
(a) T = 1 / 50 = 20 ms, the period of square wave.
(b) 1 / 2 of it for the high and low portion of the pulse is 10ms.
(c) 10 ms / 1.085 us = 9216 and 65536 – 9216 = 56320 in decimal, and in hex it is DC00H.
(d) TL = 00 and TH = DC (hex).
MOV TMOD,#10H ;Timer 1, mod 1
AGAIN: MOV TL1,#00 ;TL1=00,low byte of timer MOV TH1,#0DCH ;TH1=DC, the high byte SETB TR1 ;Start timer 1
BACK: JNB TF1,BACK ;until timer rolls over CLR TR1 ;Stop the timer 1 CLR P2.3
SJMP AGAIN ;Reload timer mode 1 isn’t auto-reload
Example 10-12
Examine the following program and find the time delay in seconds. Exclude the overhead due to the instructions in the loop.
MOV TMOD,#10H ;Timer 1, mod 1 MOV R3,#200
AGAIN: MOV TL1,#08H ;TL1=08,low byte of timer MOV TH1,#01H ;TH1=01,high byte
SETB TR1 ;Start timer 1
BACK: JNB TF1,BACK ;until timer rolls over CLR TR1 ;Stop the timer 1 CLR TF1 ;clear Timer 1 flag
DJNZ R3,AGAIN ;if R3 not zero then reload timer
Solution:
TH-TL = 0108H = 264 in decimal and 65536 – 264 = 65272.
Now 65272 × 1.085 μs = 70.820 ms, and for 200 of them we have 200 ×70.820 ms = 14.164024 seconds.
Mode 2 Programming
The following are the characteristics and operations of mode 2:
1. It is an 8-bit timer; therefore, it allows only values of 00 to FFH to be loaded into the timer’s register TH
2. After TH is loaded with the 8-bit value, the 8051 gives a copy of it to TL
Then the timer must be started
This is done by the instruction SETB TR0 for timer 0 and SETB TR1 for timer 1
3. After the timer is started, it starts to count up by incrementing the TL register
It counts up until it reaches its limit of FFH
When it rolls over from FFH to 00, it sets high the TF (timer flag)
4. When the TL register rolls from FFH to 0 and TF is set to 1, TL is reloaded automatically with the original value kept by the TH register
To repeat the process, we must simply clear TF and let it go without any need by the programmer to reload the original value This makes mode 2 an auto-reload, in contrast with mode 1 in which the programmer has to reload TH and TL
Mode 2 Programming
Steps to Mode 2 Program
To generate a time delay
1. Load the TMOD value register indicating which timer (timer 0 or timer 1) is to be used, and the timer mode (mode 2) is selected
2. Load the TH registers with the initial count value 3. Start timer
4. Keep monitoring the timer flag (TF) with the
JNB TFx , target instruction to see whether it is raised
Get out of the loop when TF goes high
5. Clear the TF flag
6. Go back to Step4, since mode 2 is auto-reload
Example 10-13
Assume XTAL = 11.0592 MHz, find the frequency of the square wave generated on pin P1.0 in the following program
MOV TMOD,#20H ;T1/8-bit/auto reload MOV TH1,#5 ;TH1 = 5
SETB TR1 ;start the timer 1
BACK: JNB TF1,BACK ;till timer rolls over CPL P1.0 ;P1.0 to hi, lo
CLR TF1 ;clear Timer 1 flag
SJMP BACK ;mode 2 is auto-reload
Solution:
First notice the target address of SJMP. In mode 2 we do not need to reload TH since it is auto-reload. Now (256 - 05)×1.085 us = 251×1.085 us =
272.33us is the high portion of the pulse. Since it is a 50% duty cycle square wave, the period T is twice that; as a result T = 2×272.33 us = 544.67us and the frequency = 1.83597 kHz
Example 10-14
Find the frequency of a square wave generated on pin P1.0.
Solution:
MOV TMOD,#2H ;Timer 0, mod 2 (8-bit, auto reload) MOV TH0,#0
AGAIN: MOV R5,#250 ;multiple delay count ACALL DELAY
CPL P1.0 SJMP AGAIN
DELAY: SETB TR0 ;start the timer 0
BACK: JNB TF0,BACK ;stay timer rolls over CLR TR0 ;stop timer
CLR TF0 ;clear TF for next round DJNZ R5,DELAY
RET
T = 2 ( 250 × 256 × 1.085 us ) = 138.88ms, and frequency = 72 Hz
Example 10-15
Assuming 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,#-48
Solution:
You can use the Windows scientific calculator to verify the result provided by the assembler. In Windows calculator, select decimal and enter 200.
Then select hex, then +/- to get the TH value. Remember that we only use the right two digits and ignore the rest since our data is an 8-bit data.
Decimal 2’s complement (TH value) -3 FDH
-12 F4H -48 D0H -60 C4H -200 38H The number
200 is the timer count till the TF is set to 1
The advantage of using negative values is that you don’t need to calculate the value loaded to THx
§10-2 Counter Programming
Timers can also be used as counters counting events happening outside the 8051
When it is used as a counter, it is a pulse outside of the 8051 that increments the TH, TL registers
TMOD and TH, TL registers are the same as for the timer discussed previously
Programming the timer in the last section also applies to programming it as a counter
Except the source of the frequency
C/T Bit in TMOD Register
The C/T bit in the TMOD registers decides the source of the clock for the timer
When C/T = 1, the timer is used as a counter and gets its pulses from outside the 8051
The counter counts up as pulses are fed from pins 14 and 15, these pins are called T0 (timer 0 input) and T1 (timer 1 input)
Port 3 pins used for Timers 0 and 1
Example 10-16
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 TL1 count on P2, which connects to 8 LEDs.
Solution:
MOV TM0D,#01100000B ;counter 1, mode 2, ;C/T=1 external pulses MOV TH1,#0 ;clear TH1
SETB P3.5 ;make T1 input AGAIN: SETB TR1 ;start the counter BACK: MOV A,TL1 ;get copy of TL MOV P2,A ;display it on port 2 JNB TF1,Back ;keep doing, if TF = 0 CLR TR1 ;stop the counter 1 CLR TF1 ;make TF=0
SJMP AGAIN ;keep doing it
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 make P3.5 an input port by making it high. In other words, we must configure (set high) the T1 pin (pin P3.5) to allow pulses to be fed into it.
TCON Register
TCON (timer control) register is an 8-bit register
TCON: Timer/Counter Control Register
The upper four bits are used to store the TF and TR bits of both timer 0 and 1
The lower 4 bits are set aside for controlling the interrupt bits
TCON register is a bit-addressable register
TCON Register
Equivalent instruction for the Timer Control Register
TCON Register
Case of GATE = 1
If GATE = 1, the start and stop of the timer are done externally through pins P3.2 and P3.3 for timers 0 and 1, respectively
This hardware way allows to start or stop the
timer externally at any time via a simple switch
§10-3 Programming Timers in C Accessing Timer Registers
Example 10-17
Write an 8051 C program to toggle all the bits of port P1 continuously with some delay in between. Use Timer 0, 16-bit mode to generate the delay.
Solution:
#include <reg51.h>
void T0Delay(void);
void main(void){
while (1) { P1=0x55;
T0Delay();
P1=0xAA;
T0Delay();
} }
void T0Delay(){
TMOD=0x01;
TL0=0x00;
TH0=0x35;
TR0=1;
while (TF0==0);
TR0=0;
TF0=0;
}
FFFFH – 3500H = CAFFH
= 51967 + 1 = 51968
51968×1.085μs = 56.384ms is the approximate delay
Calculating Delay Length Using Timers
To speed up the 8051, many recent versions of the 8051 have reduced the number of clocks per
machine cycle from 12 to four, or even one
The frequency for the timer is always 1/12
ththe frequency of the crystal attached to the 8051,
regardless of the 8051 version
Times 0/1 Delay Using Mode 1 (16-bit Non Auto-reload)
Example 10-18
Write an 8051 C program to toggle only bit P1.5 continuously every 50ms.
Use Timer 0, mode 1 (16-bit) to create the delay. Test the program on the (a) AT89C51 and (b) DS89C420.
Solution:
#include <reg51.h>
void T0M1Delay(void);
sbit mybit=P1^5;
void main(void) {
while (1) {
mybit=~mybit;
T0M1Delay();
} }
void T0M1Delay(void) {
TMOD=0x01;
TL0=0xFD;
TH0=0x4B;
TR0=1;
while (TF0==0);
TR0=0;
TF0=0;
}
FFFFH – 4BFDH = B402H
= 46082 + 1 = 46083 46083×1.085 μs = 50ms
Example 10-19
Write an 8051 C program to toggle all bits of P2 continuously every 500ms.
Use Timer 1, mode 1 to create the delay.
Solution:
//tested for DS89C420, XTAL = 11.0592 MHz
Times 0/1 Delay Using Mode 1 (16-bit Non Auto-reload)
#include <reg51.h>
void T1M1Delay(void);
void main(void) {
unsigned char x;
P2=0x55;
while (1) {
P2=~P2;
for (x=0;x<20;x++) T1M1Delay();
} }
void T1M1Delay(void) {
TMOD=0x10;
TL1=0xFE;
TH1=0xA5;
TR1=1;
while (TF1==0);
TR1=0;
TF1=0;
}
A5FEH = 42494 in decimal 65536 – 42494 = 23042
23042×1.085μs = 25ms and 20×25ms = 500ms
Example 10-20
A switch is connected to pin P1.2. Write an 8051 C program to monitor SW and create the following frequencies on pin P1.7:
SW=0: 500Hz
SW=1: 750Hz, use Timer 0, mode 1 for both of them.
Solution:
Times 0/1 Delay Using Mode 1 (16-bit Non Auto-reload)
#include <reg51.h>
sbit mybit=P1^5;
sbit SW=P1^7;
void T0M1Delay(unsigned char);
void main(void){
SW=1;
while (1) {
mybit=~mybit;
if (SW==0) T0M1Delay(0);
else
T0M1Delay(1);
} }
void T0M1Delay(unsigned char c){
TMOD=0x01;
if (c==0) { TL0=0x67;
TH0=0xFC;
}
else {
TL0=0x9A;
TH0=0xFD;
}
TR0=1;
while (TF0==0);
TR0=0;
TF0=0;
}
FC67H = 64615
65536 – 64615 = 921
921×1.085 μs = 999.285μs 1 / (999.285μs×2) = 500Hz
Times 0/1 Delay Using Mode 2 (8-bit Auto-reload)
Example 10-21
Write an 8051 C program to toggle only pin P1.5 continuously every 250ms.
Use Timer 0, mode 2 (8-bit auto-reload) to create the delay.
Solution:
#include <reg51.h>
void T0M2Delay(void);
sbit mybit=P1^5;
void main(void){
unsigned char x,y;
while (1) { mybit=~mybit;
for (x=0;x<250;x++)
for (y=0;y<36;y++) //we put 36, not 40 T0M2Delay();
} }
void T0M2Delay(void){
TMOD=0x02;
TH0=-23;
TR0=1;
while (TF0==0);
TR0=0;
TF0=0;
}
Due to overhead of the for loop in C, we put 36 instead of 40
256 – 23 = 233
23×1.085μs = 25μs and 25μs×250×40 = 250ms
Example 10-22
Write an 8051 C program to create a frequency of 2500 Hz on pin P2.7. Use Timer 1, mode 2 to create delay.
Solution:
#include <reg51.h>
void T1M2Delay(void);
sbit mybit=P2^7;
void main(void){
unsigned char x;
while (1) { mybit=~mybit;
T1M2Delay();
} }
void T1M2Delay(void){
TMOD=0x20;
TH1=-184;
TR1=1;
while (TF1==0);
TR1=0;
TF1=0;
}
1/2500Hz = 400μs 400μs /2 = 200μs 200 μs / 1.085μs = 184
C Programming of Timers as Counters
Example 10-23
Assume that a 1-Hz external clock is being fed into pin T1 (P3.5). Write a C program for counter 1 in mode 2 (8-bit auto reload) to count up and display the state of the TL1 count on P1. Start the count at 0H.
Solution:
#include <reg51.h>
sbit T1=P3^5;
void main(void){
T1=1;
TMOD=0x60;
TH1=0;
while (1) { do {
TR1=1;
P1=TL1;
}
while (TF1==0);
TR1=0;
TF1=0;
} }
Example 10-24
Assume that a 1-Hz external clock is being fed into pin T0 (P3.4). Write a C program for counter 0 in mode 1 (16-bit) to count the pulses and display the state of the TH0 and TL0 registers on P2 and P1,respectively.
Solution:
#include <reg51.h>
void main(void){
T0=1;
TMOD=0x05;
TL0=0 TH0=0;
while (1) { do {
TR0=1;
P1=TL0;
P2=TH0;
}
while (TF0==0);
TR0=0;
TF0=0;
} }
