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7
Segment Display
INTRODUCTION
For the seven segment display you can use the LT-541 or LSD5061-11 chip.
Each of the segments of the display is connected to a pin on the 8051 (the
schematic shows how to do this). In order to light up a segment on the the
pin must be set to 0V. To turn a segment off the corresponding pin must be
set to 5V. This is simply done by setting the pins on the 8051 to '1' or
'0'.
LED displays are
But they are cheaper than LCD display
7-SEG Display are available in two types -1. Common anode & 2. common
cathode , but command anode display are most suitable for interfacing
with 8051 since 8051 port pins can sink current better than sourcing it.
CREATING DIGIT
PATTERN
For displaying
Digit say 7 we need to light segments -a ,b, c. Since we are using Common
anode display , to do so we have to to provide Logic -0 (0 v) at anode
of these segments.
so need to clear
pins- P1.0 ,P1.1,P1.2. that is 1 1 1 1 1 0 0 0 -->F8h .
Connection
Hex Code
|
Segment number |
8051
pin number |
|
a |
P1.0 |
|
b |
P1.1 |
|
c |
P1.2 |
|
d |
P1.3 |
|
e |
P1.4 |
|
f |
P1.5 |
|
g |
p1.6 |
|
h(dp) |
P1.7 |
|
Digit |
Seg. h |
Seg. g |
Seg. f |
Seg. e |
Seg. d |
Seg. c |
Seg. b |
Seg. a |
HEX |
|
0 |
1 |
1 |
0 |
0 |
0 |
0 |
0 |
0 |
C0 |
|
1 |
0 |
0 |
0 |
0 |
0 |
1 |
1 |
0 |
06 |
|
2 |
1 |
0 |
1 |
0 |
0 |
1 |
0 |
0 |
A4 |
|
3 |
1 |
0 |
1 |
1 |
0 |
0 |
0 |
0 |
B0 |
|
4 |
1 |
0 |
0 |
1 |
1 |
0 |
0 |
1 |
99 |
You can also do this
for some characters like A ,E .. but not for D or B because it will be
same as that of 0 & 8 . So this is one of limitation of 7-seg display.
Since we can Enable
only one 7-seg display at a time ,we need to scan these display at fast
rate .The scanning frequency should be high enough to be flicker-free. At
least 30HZ .Therefore – time one digit is ON is 1/30 seconds
INTERFACING
Note that I am
using Common Anode display. so the common Anode pin is tied to 5v .The
cathode pins are connected to port 1 through 330 Ohm resistance (current
limiting).
 
Common Anode display
CODE EXAMPLE
Connection
- a:h to port
p1.0:p1.7 , D0:D1 to p3.0:p3.1.
To Display
- Consider example of
vending machine where we want to display number of soft drink bottles
on display entered by customer. Suppose he enter 3 (03) bottles then we will use
lookup table to see DIGIT PATTERN of these keys.
So DIGI[1]=c0 (hex
code for '0') &
DIGI[2]=bo(hex
code for '3').
Note: I have taken
values for DIGI[1] & DIGI[2] directly in code.
|
Algorithm |
|
C
LANGUAGE (SPJ SIDE51) |
|
|
start
: disable [D0:D1]
again : enable D0
[a:h] - pattern for Digit1
delay
disable D0. Enable D1
[a:h] - pattern for Digit2
delay
Goto again
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#include
<Intel\8052.h>
#include <standard.h>
#include<stdio.h>
#include<etc.h>
void main( ){
unsigned char DIGI[2];
unsigned char right,cnt;
P1 = 0; /* initialize all P1 outputs to be zero*/
P3 = 0;
DIGI[1]=0xc0;/*(hex code for '0')*/
DIGI[2]=0xb0;/*(hex code for '3')*/
while(1) {
for (cnt=1;cnt<3;cnt++)
{
P3 = cnt; /*Enable D0 and then D1*/
P1=DIGI[cnt];/*P1=c0 when D0 is enabled & P1=b0 when d1 is enabled*/
delay_ms(20);
}
}
}
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