 |
 |
| View previous topic :: View next topic |
| Author |
Message |
Sean7326
Joined: 28 Nov 2003
Posts: 1
|
| Question on DIP Switch Combinations |
 Posted: Sun Nov 30, 2003 10:56 pm |
 |
|
Can someone please check my logic with this - I have a 12 position DIP switch that has on, off, and float - marked +, -, and 0. I am interfacing that to 12 pins on a PIC to provide different options.
Would I be correct that 3 possible positions, times 12 switches, would equal 3^12 or 531,441 possible options? |
|
 |
wedilo
Joined: 16 Sep 2003
Posts: 71
Location: Moers, Germany
|
|
| Posted: Mon Dec 01, 2003 2:36 am |
|
|
Hello,
I think that is correct, but how do you determine 3 states on a digital I/O pin? 
Sorry, I hope I understand your question correctly.
73 Sven |
|
|
Ttelmah
Guest
|
| Re: Question on DIP Switch Combinations |
| Posted: Mon Dec 01, 2003 4:07 am |
|
|
| Sean7326 wrote: | Can someone please check my logic with this - I have a 12 position DIP switch that has on, off, and float - marked +, -, and 0. I am interfacing that to 12 pins on a PIC to provide different options.
Would I be correct that 3 possible positions, times 12 switches, would equal 3^12 or 531,441 possible options? |
Yes.
To detect the third state though, requires some circuitry, if it is to be done safely. One solution, is to have a small capacitor in series with a small resistor between ground, and the wiper of each switch, and a larger resistor between the wiper and the PIC's input pin. You can then drive the pin as an output 'high', switch to using it as an input, and read the result. If you see '1', then the switch is either in the 'open' position, or set to 1. Repeat by driving low, and you can then know where the switch is positioned. The small resistor in series with the capacitor, is to reduce te current experienced by the switch when discharging the capacitor. The resistor between the wiper and the PIC input, is to limit the current when testing, and trying to drive the PIC 'high', when the switch is set 'low' (and vice versa)..
Other methods are to use a few of the switches together with resistors for form a voltage divider tree, and sense the voltage with an A/D input. With this you could reduce the number of inputs used, but don't try to decode too many combinations with one 'tree', or the limits of voltage reference accuracy, and A/D converter accuracy, will 'sneak out' to hit you.
Best Wishes |
|
|
wedilo
Joined: 16 Sep 2003
Posts: 71
Location: Moers, Germany
|
|
| Posted: Mon Dec 01, 2003 5:35 am |
|
|
Hey Ttelmah,
Wooh, I'm enthusiastic about that fine idea.
Please more... 
73 Sven |
|
|
Remko
Joined: 24 Sep 2003
Posts: 14
|
|
| Posted: Mon Dec 01, 2003 5:57 am |
|
|
Mail Microchip and ask for the little Tips 'n Tricks book, it is a give-a-way book. This trick and more are explained.
Greetingz
RFM |
|
|
wedilo
Joined: 16 Sep 2003
Posts: 71
Location: Moers, Germany
|
|
| Posted: Mon Dec 01, 2003 9:23 am |
|
|
Hello Remko,
thank you for that information 
73 Sven |
|
|
burnsy
Joined: 18 Oct 2003
Posts: 35
Location: Brisbane, Australia
|
| 4 dipswitches per analog pin |
| Posted: Sat Dec 06, 2003 11:04 pm |
|
|
I've been connecting 4 dipswitches on an0 for ages using a simple voltage divider like sean said.
On the processor side of the dipswitch common the dipswitch pins to a 10k pullup and connect it to an0. On the other side of the dipswitch, connect a 10k, 5k1, 2k4 and 1k2 resistor to each of the pins and common them to gnd.
It even works with normal switches, but make sure you use 1-2% resisitors to reduce errors.
If anyone wants the code, email me at jason@jtekelectronics.com.au
cheers,
Jason
_________________
This is the last code change until its ready.... |
|
|
KerryW
Guest
|
| Another way to do that |
| Posted: Sun Dec 07, 2003 8:40 am |
|
|
Ground the common pin of each switch. Run the N.O. and N.C. sides of each switch to the inputs of a 74LS244 (or 74HC245, or whatever) Use 3 to do the 24 inputs. Use pull-ups on the 74LS244 inputs. Run 3 outputs to the enable lines of the 74LS244s. Tie all 3 output 0s together to 1 input pin, all outputs 1 to another input pin, etc.
This would use 3 20 pin ICs, 3 9 pin sips, 8 input lines and 3 output lines. If this method interestes you, I can whip up a quick schematic for you if you need it. |
|
|
Bruno
Guest
|
|
| Posted: Sat Mar 18, 2006 2:52 am |
|
|
Does someone knows if Mr. Jason is still on the same e-mail.
Im asking this becuse i have e-mail to him, but i didnt get the replay.
If someone has some infos or maybe link to the code, please be so kind and pass it to this forum.
Thank you ! |
|
|
|
|
You cannot post new topics in this forum
You cannot reply to topics in this forum
You cannot edit your posts in this forum
You cannot delete your posts in this forum
You cannot vote in polls in this forum
|
Powered by phpBB © 2001, 2005 phpBB Group
|
FAQ 
Search
Register
Profile 
Log in to check your private messages 
Log in
