S
sommes
Guest
I would like to ask Is 7 segment decoder/driver necessary to connect 7
segment display with Atmel 90S8535 microcontoller.
segment display with Atmel 90S8535 microcontoller.
D
David L. Jones
Guest
sommes wrote:
You can do the 7-seg decoding in software using a lookup table.
And if you drive the LEDs at a low enough current (5mA for example),
you can drive them direcly via a resistor.
Dave
No.
You can do the 7-seg decoding in software using a lookup table.
And if you drive the LEDs at a low enough current (5mA for example),
you can drive them direcly via a resistor.
Dave
J
Jasen Betts
Guest
On 2006-03-21, sommes <jlj@kl.com> wrote:
you can drive the display directly. just like if it was it's 7 LEDs)
it'll take 7 IO pins to do this (8 if you want the dot too) instead of 4
(or fewer if using I2C etc) for the 7-segment decoder/driver.
--
Bye.
Jasen
no. if you use a common-anode 7 segment display (and apropriate resistors)
you can drive the display directly. just like if it was it's 7 LEDs)
it'll take 7 IO pins to do this (8 if you want the dot too) instead of 4
(or fewer if using I2C etc) for the 7-segment decoder/driver.
--
Bye.
Jasen
S
sommes
Guest
Thanks Dave
Would you mind to explain some more if 4 digi 7 segment display are used in
this case.
Cheers
Sommes
Would you mind to explain some more if 4 digi 7 segment display are used in
this case.
Cheers
Sommes
S
Simone Merrett
Guest
These usually work by fast (, ~1/30 sec) 'time-multiplexing', so only is
digit is lit at any time.
In this case, there is something like an 'address' bus (turn on the wire /
line corresponding to the digit you are lighting).
If each of the 7-segs in the display draws 1mA, then you should be able to
attach directly to the microcontroller, (which should be at least 5mA
drive/sink capable).
Of course, if the display contains no 'internal ' resistors, then you will
have to provide these to get the correct current.
(You can make them as bright as you like, provided you dont exceed the limit
of the microcontroller).
For example, a 5V microcontroller output pin lighting up a red LED at 2mA
will need a 1900 ohm resistor in series.
This is based on ohms-law, where the voltage drop across the LED being 1.2 V
{ (5-1.2)/0.002}, which is typical of red LEDs. (use ~1.7V for Green.... or
refer to LED / 7-seg specification for exact dropping voltage).
Dave Merrett
"sommes" <ljk@jlk.com> wrote in message news:dvq7c8$7q1$1@emma.aioe.org...
digit is lit at any time.
In this case, there is something like an 'address' bus (turn on the wire /
line corresponding to the digit you are lighting).
If each of the 7-segs in the display draws 1mA, then you should be able to
attach directly to the microcontroller, (which should be at least 5mA
drive/sink capable).
Of course, if the display contains no 'internal ' resistors, then you will
have to provide these to get the correct current.
(You can make them as bright as you like, provided you dont exceed the limit
of the microcontroller).
For example, a 5V microcontroller output pin lighting up a red LED at 2mA
will need a 1900 ohm resistor in series.
This is based on ohms-law, where the voltage drop across the LED being 1.2 V
{ (5-1.2)/0.002}, which is typical of red LEDs. (use ~1.7V for Green.... or
refer to LED / 7-seg specification for exact dropping voltage).
Dave Merrett
"sommes" <ljk@jlk.com> wrote in message news:dvq7c8$7q1$1@emma.aioe.org...
J
Jasen Betts
Guest
On 2006-03-22, sommes <ljk@jlk.com> wrote:
cheat sheet
So you could drive the 8 LEDs in 4 7 segment displays and have three I/0 pins
left over.
if the chip is anything like the 2313 the pins can sink upto 20mA which could
give a very bright 7 segment display.
If you're wanting to use fewer IO pins consider matrixing the displays.
this'd mean using 4 I/O pins to drive transistors to select a one of the 4
7-segment units and using 8 other I/O pins to drive the individual segments
and then writing a program to light the individual displays one at a time,
very fast (eg 100 times per second) this way the light output will be 1/4
of what you'd ge driving the display full-time, but only 8 + 4 = 12 I/O pins
would be needed leaving 23 for other purposes...
Bye.
Jasen
IIRC the chip you were proposing to use has 35 I/O pins according to my
cheat sheet
So you could drive the 8 LEDs in 4 7 segment displays and have three I/0 pins
left over.
if the chip is anything like the 2313 the pins can sink upto 20mA which could
give a very bright 7 segment display.
If you're wanting to use fewer IO pins consider matrixing the displays.
this'd mean using 4 I/O pins to drive transistors to select a one of the 4
7-segment units and using 8 other I/O pins to drive the individual segments
and then writing a program to light the individual displays one at a time,
very fast (eg 100 times per second) this way the light output will be 1/4
of what you'd ge driving the display full-time, but only 8 + 4 = 12 I/O pins
would be needed leaving 23 for other purposes...
Bye.
Jasen
S
sommes
Guest
Thanks Jasen
P
Peter Jakacki
Guest
Jasen Betts wrote:
maximum current that you can run through the chip substrate, usually in
the vicinity of 100 to 200ma absolute maximum.
to do is to increase the peak current 4 times to boost the average
current back to "normal".
Another possibility is to simply use high brightness displays and run at
much lower current.
*Peter*
Hmmm, I think you'd better consult the datasheet again, there is a
maximum current that you can run through the chip substrate, usually in
the vicinity of 100 to 200ma absolute maximum.
If you buffer the segment drive as well as the commons then all you need
to do is to increase the peak current 4 times to boost the average
current back to "normal".
Another possibility is to simply use high brightness displays and run at
much lower current.
*Peter*