matrix keypad
Archiver
Posts: 46,084
Hi,
As an addition to Tracy Allen's matrix schematic consider this:
Normal 3x4 matrix:
} | |
---o
o
o--- r1 P0
| | |
---o
o
o--- r2 P1
| | |
---o
o
o--- r3 P2
| | |
---o
o
o--- r4 P3
| | | switches between row and column at o character
c1 c2 c3 requires 7 I/O lines for 12 switches
P7 P6 P5
Adding 4 diodes
/ / / /
\ \ \ \ 4x pullup resistor 10k
/ / / /
} | | |
1} 2| 3| D1/|
---o
o
o
+--- r1 P0
| | | |
4} 5| D2/| 7|
---o
o
+
o--- r2 P1
| | | |
6} D3/| 8| 9|
---o
+
o
o--- r3 P2
| | | |
D4/} 0| #| *|
---+
o
o
o--- r4 P3 diodes between row (cathode) and column
(anode) at /
| | | | switches between row and column at o
character
c1 c2 c3 c4 requires 4 I/O lines for 12 switches
Now only the rows are connected to the stamp. Make one row a low output
(others are input) and
check the remaining inputs. Then make the next row a low output etc.
r1 low output -> c4 pulled low -> keys 7,9,* give low input on P1,P2,P3
r2 low output -> c3 pulled low -> keys 3,8,# give low input on P0,P2,P3
r3 low output -> c2 pulled low -> keys 2,5,0 give low input on P0,P1,P3
r4 low output -> c1 pulled low -> keys 1,4,6 give low input on P0,P1,P2
Example: r3 ia low output: press 2 -> r1 low input
press 5 -> r2 low input
press 0 -> r4 low input
This scheme could very easily be expanded to support more switches while
reducing the
required I/O lines. It just takes a few extra diodes (e.g 1N4148).
Greetings peter
As an addition to Tracy Allen's matrix schematic consider this:
Normal 3x4 matrix:
} | |
---o
o
o--- r1 P0
| | |
---o
o
o--- r2 P1
| | |
---o
o
o--- r3 P2
| | |
---o
o
o--- r4 P3
| | | switches between row and column at o character
c1 c2 c3 requires 7 I/O lines for 12 switches
P7 P6 P5
Adding 4 diodes
/ / / /
\ \ \ \ 4x pullup resistor 10k
/ / / /
} | | |
1} 2| 3| D1/|
---o
o
o
+--- r1 P0
| | | |
4} 5| D2/| 7|
---o
o
+
o--- r2 P1
| | | |
6} D3/| 8| 9|
---o
+
o
o--- r3 P2
| | | |
D4/} 0| #| *|
---+
o
o
o--- r4 P3 diodes between row (cathode) and column
(anode) at /
| | | | switches between row and column at o
character
c1 c2 c3 c4 requires 4 I/O lines for 12 switches
Now only the rows are connected to the stamp. Make one row a low output
(others are input) and
check the remaining inputs. Then make the next row a low output etc.
r1 low output -> c4 pulled low -> keys 7,9,* give low input on P1,P2,P3
r2 low output -> c3 pulled low -> keys 3,8,# give low input on P0,P2,P3
r3 low output -> c2 pulled low -> keys 2,5,0 give low input on P0,P1,P3
r4 low output -> c1 pulled low -> keys 1,4,6 give low input on P0,P1,P2
Example: r3 ia low output: press 2 -> r1 low input
press 5 -> r2 low input
press 0 -> r4 low input
This scheme could very easily be expanded to support more switches while
reducing the
required I/O lines. It just takes a few extra diodes (e.g 1N4148).
Greetings peter
Comments
of diodes, and pulling each pin high or low to each other pin, and pulling
each pin high or low (5v or 0v externally). Pulling 2 lines or 3 lines low /
high in all combinations gives 72 possibilities. It would be code hungry,
and messy to make but it can be done.
Chris
Original Message
From: Peter Verkaik <peterverkaik@b...>
To: <basicstamps@yahoogroups.com>
Sent: Friday, June 29, 2001 6:47 PM
Subject: [noparse][[/noparse]basicstamps] RE: matrix keypad
> Hi,
>
> As an addition to Tracy Allen's matrix schematic consider this:
>
> Normal 3x4 matrix:
>
> } | |
> ---o
o
o--- r1 P0
> | | |
> ---o
o
o--- r2 P1
> | | |
> ---o
o
o--- r3 P2
> | | |
> ---o
o
o--- r4 P3
> | | | switches between row and column at o character
> c1 c2 c3 requires 7 I/O lines for 12 switches
> P7 P6 P5
>
>
> Adding 4 diodes
>
> / / / /
> \ \ \ \ 4x pullup resistor 10k
> / / / /
> } | | |
> 1} 2| 3| D1/|
> ---o
o
o
+--- r1 P0
> | | | |
> 4} 5| D2/| 7|
> ---o
o
+
o--- r2 P1
> | | | |
> 6} D3/| 8| 9|
> ---o
+
o
o--- r3 P2
> | | | |
> D4/} 0| #| *|
> ---+
o
o
o--- r4 P3 diodes between row (cathode) and column
> (anode) at /
> | | | | switches between row and column at o
> character
> c1 c2 c3 c4 requires 4 I/O lines for 12 switches
>
>
> Now only the rows are connected to the stamp. Make one row a low output
> (others are input) and
> check the remaining inputs. Then make the next row a low output etc.
> r1 low output -> c4 pulled low -> keys 7,9,* give low input on P1,P2,P3
> r2 low output -> c3 pulled low -> keys 3,8,# give low input on P0,P2,P3
> r3 low output -> c2 pulled low -> keys 2,5,0 give low input on P0,P1,P3
> r4 low output -> c1 pulled low -> keys 1,4,6 give low input on P0,P1,P2
>
> Example: r3 ia low output: press 2 -> r1 low input
> press 5 -> r2 low input
> press 0 -> r4 low input
>
> This scheme could very easily be expanded to support more switches while
> reducing the
> required I/O lines. It just takes a few extra diodes (e.g 1N4148).
>
> Greetings peter
>
>
>
> To UNSUBSCRIBE, just send mail to:
> basicstamps-unsubscribe@yahoogroups.com
> from the same email address that you subscribed with. Text in the Subject
and Body of the message will be ignored.
>
>
> Your use of Yahoo! Groups is subject to http://docs.yahoo.com/info/terms/
>
>
>It is actually possible to have 72 switches read by 4 io pins, using a heap
>of diodes, and pulling each pin high or low to each other pin, and pulling
>each pin high or low (5v or 0v externally). Pulling 2 lines or 3 lines low
/
>high in all combinations gives 72 possibilities. It would be code hungry,
>and messy to make but it can be done.
>Chris
Would you care to elaborate a bit more?
Following your scheme I calculate:
1 output + 3 inputs give 4 possibilities to detect 3 switches -> 12
switches (my scheme)
2 outputs + 2 inputs give 6 possibilities to detect 2 switches -> 12
switches
3 outputs + 1 input give 4 possibilities to detect 1 switch -> 4
switches
So I come no further than 28.
How many diodes would be required for your 72 switches and how would they
electrically connect using only 4 I/O lines?
Please elaborate.
Greetings peter
high in all combinations, this gives heaps more. You can then pull 2 inpits
high / low in all combinations. You can then pull each pin high (to external
5v) in all combinations (ie, 2 and three at a time) gives 15, then pull them
low gives another 15.
If I get time, I will try to come up with a clear explanation. A freind and
I were challenging each other recently and we came up with approx 72.
Like I said, it sure would be messy though, especially with code.
Regards, Chris.
Original Message
From: Peter Verkaik <peterverkaik@b...>
To: <basicstamps@yahoogroups.com>
Sent: Sunday, July 01, 2001 6:33 PM
Subject: [noparse][[/noparse]basicstamps] RE: matrix keypad
> Hi Chris,
>
> >It is actually possible to have 72 switches read by 4 io pins, using a
heap
> >of diodes, and pulling each pin high or low to each other pin, and
pulling
> >each pin high or low (5v or 0v externally). Pulling 2 lines or 3 lines
low
> /
> >high in all combinations gives 72 possibilities. It would be code hungry,
> >and messy to make but it can be done.
> >Chris
>
> Would you care to elaborate a bit more?
> Following your scheme I calculate:
> 1 output + 3 inputs give 4 possibilities to detect 3 switches -> 12
> switches (my scheme)
> 2 outputs + 2 inputs give 6 possibilities to detect 2 switches -> 12
> switches
> 3 outputs + 1 input give 4 possibilities to detect 1 switch -> 4
> switches
> So I come no further than 28.
>
> How many diodes would be required for your 72 switches and how would they
> electrically connect using only 4 I/O lines?
>
> Please elaborate.
>
> Greetings peter
>
>
>
>
> To UNSUBSCRIBE, just send mail to:
> basicstamps-unsubscribe@yahoogroups.com
> from the same email address that you subscribed with. Text in the Subject
and Body of the message will be ignored.
>
>
> Your use of Yahoo! Groups is subject to http://docs.yahoo.com/info/terms/
>
>
>I agrre with your message, but you can use 1 switch to pull 3 inputs low or
>high in all combinations, this gives heaps more. You can then pull 2 inpits
>high / low in all combinations. You can then pull each pin high (to
external
>5v) in all combinations (ie, 2 and three at a time) gives 15, then pull
them
>low gives another 15.
>If I get time, I will try to come up with a clear explanation. A freind and
>I were challenging each other recently and we came up with approx 72.
>Like I said, it sure would be messy though, especially with code.
>Regards, Chris.
This clearifies somewhat. Instead of looking at single inputs to detect one
switch
you look at the inputs as a group. So fot the 1 output with 3 inputs you get
8 possibilities
(3 bits equals 8 states) to detect switches. The all-ones state (i.e 111) is
however
used to indicate that no switch is pressed, so I am left with 7.
Following this scheme I calculate:
1 output + 3 inputs give 4 possibilities to detect 7 switches -> 28
switches
2 outputs + 2 inputs give 6 possibilities to detect 3 switches -> 18
switches
3 outputs + 1 input give 4 possibilities to detect 1 switch -> 4
switches
So I come no further than 50.
Also you did not say how many diodes are necessary and how they connect.
I am not that interested in the code, but a schematic would be nice.
Greetings peter
I believe the following is also part of your scheme:
0 outputs + 4 inputs give 1 possibility to derect 15 switches -> 15 switches
This would give a total of 65 switches.
Anxious to see your schematic.
Greetings peter
combinations (control-alt-delete!. So with four keys, each with two
states (open and closed):
4 single keys
6 two-key combinations
4 three-key
1 four-key
---
15 total
There is also of course one way to get zero keys, giving 16 binary
states. A 16 to 4 encoder chip (available in standard CMOS or TTL)
can do that logic function, to map 16 individual keys into 4 bits at
the output. Or it can be done with diodes. But I don't see how to
get 72 out of it!
Unless, if each switch can have more than 2 states, then the
possibilities multiply. Say, each switch is a SPDT with center off,
so it can be either high, low or open. Then there are 9 combinations
with 2 of these, 27 with three, and 81 with four.
SPDT center off
+5
o
<---o--/\/\---P0 stamp
com
o 10k
I can imagine that there would be a device, call it a "81 to 4
ternary decoder" (using diode?) that would be able to decode all of
those states (call the device a "microprocessor"?!).
The Stamp itself can test ternary switch positions, using the fact
that the capacitance of a floating input will hold a charge when a
pin is switched from output to input. To test for the three
different states, the stamp pin is made a LOW output, then
immediately back to an input. If the input reads high, then the
switch is in the upper position. If input reads low, then another
test is needed. Make the pin output HIGH briefly and then back to
input. If the pin reads low, the switch is in the lower position; if
it reads high, the switch is in the middle, open position.
FWIW, a 4x4 keypad (each key with two states) allows 120 2-key combinations.
-- best regards
Tracy Allen
electronically monitored ecosystems
http://www.emesystems.com
mailto:tracy@e...
>Hi Chris,
>
>>I agrre with your message, but you can use 1 switch to pull 3 inputs low or
>>high in all combinations, this gives heaps more. You can then pull 2 inpits
>>high / low in all combinations. You can then pull each pin high (to
>external
>>5v) in all combinations (ie, 2 and three at a time) gives 15, then pull
>them
>>low gives another 15.
>>If I get time, I will try to come up with a clear explanation. A freind and
>>I were challenging each other recently and we came up with approx 72.
>>Like I said, it sure would be messy though, especially with code.
>>Regards, Chris.
>
>This clearifies somewhat. Instead of looking at single inputs to detect one
>switch
>you look at the inputs as a group. So fot the 1 output with 3 inputs you get
>8 possibilities
>(3 bits equals 8 states) to detect switches. The all-ones state (i.e 111) is
>however
>used to indicate that no switch is pressed, so I am left with 7.
>Following this scheme I calculate:
>
>1 output + 3 inputs give 4 possibilities to detect 7 switches -> 28
>switches
>2 outputs + 2 inputs give 6 possibilities to detect 3 switches -> 18
>switches
>3 outputs + 1 input give 4 possibilities to detect 1 switch -> 4
>switches
>So I come no further than 50.
>
>Also you did not say how many diodes are necessary and how they connect.
>I am not that interested in the code, but a schematic would be nice.
>
>Greetings peter
>There is also of course one way to get zero keys, giving 16 binary
>states. A 16 to 4 encoder chip (available in standard CMOS or TTL)
>can do that logic function, to map 16 individual keys into 4 bits at
>the output. Or it can be done with diodes. But I don't see how to
>get 72 out of it!
The encoder solution does not allow for simultaneuos keypresses as it is a
priority encoder. I figured
the same out with diodes, but the chip would be better. All this however
does not involve a
matrix keypad, which was the starting point: many switches without (complex)
hardware.
If I want many switches I could also hook up a PS/2 keynoard with on eof
those PAKs.
I mentioned the diode solution to save 3 or 4 I/O lines which are more
valuable than the 4 diodes.
I too don't see how Chris can get 72 switches in a matrix circuitry using
only 4 I/O lines.
I await his schematic.
Greetings peter
time to do one) ?
Regards,
Chris
Original Message
From: Peter Verkaik <peterverkaik@b...>
To: <basicstamps@yahoogroups.com>
Sent: Monday, July 02, 2001 7:29 PM
Subject: [noparse][[/noparse]basicstamps] RE: matrix keypad
> Hi Chris,
>
> >I agrre with your message, but you can use 1 switch to pull 3 inputs low
or
> >high in all combinations, this gives heaps more. You can then pull 2
inpits
> >high / low in all combinations. You can then pull each pin high (to
> external
> >5v) in all combinations (ie, 2 and three at a time) gives 15, then pull
> them
> >low gives another 15.
> >If I get time, I will try to come up with a clear explanation. A freind
and
> >I were challenging each other recently and we came up with approx 72.
> >Like I said, it sure would be messy though, especially with code.
> >Regards, Chris.
>
> This clearifies somewhat. Instead of looking at single inputs to detect
one
> switch
> you look at the inputs as a group. So fot the 1 output with 3 inputs you
get
> 8 possibilities
> (3 bits equals 8 states) to detect switches. The all-ones state (i.e 111)
is
> however
> used to indicate that no switch is pressed, so I am left with 7.
> Following this scheme I calculate:
>
> 1 output + 3 inputs give 4 possibilities to detect 7 switches -> 28
> switches
> 2 outputs + 2 inputs give 6 possibilities to detect 3 switches -> 18
> switches
> 3 outputs + 1 input give 4 possibilities to detect 1 switch -> 4
> switches
> So I come no further than 50.
>
> Also you did not say how many diodes are necessary and how they connect.
> I am not that interested in the code, but a schematic would be nice.
>
> Greetings peter
>
>
>
> To UNSUBSCRIBE, just send mail to:
> basicstamps-unsubscribe@yahoogroups.com
> from the same email address that you subscribed with. Text in the Subject
and Body of the message will be ignored.
>
>
> Your use of Yahoo! Groups is subject to http://docs.yahoo.com/info/terms/
>
>
>
See message from Russ Miller: No more attachments.
Preferable drawing formats are BMP, TIF or GIF.
Greetings peter