SX Chip Input and Switches
Green Phantom
Posts: 21
I have a question on the SX Chip's I/O pins when set as inputs.
By what I saw in the manuals so far, you connect 1 to 10K-ohm pull-up resistors to the pins set as inputs to VDD when using switches for input, where a 0 in the input· (the switches connected to GROUND) will give a result depending on the program code of your project.· This is called "active-low."
For instance: if you use a standard logic TTL chip (like the 7400 2-input NAND gate), the two inputs at the HIGH state gives you a 0 (LOW) state at the output.· But when you make one of the two inputs a 0 (LOW,) the output becomes a 1 (HIGH.)· This is what is known as "negative-edge" OR gate or "active-low" OR gate.
Can you use pull-up resistors connected to GROUND at each SX pin (Ports A, B and C) used as inputs and then use the switches (tactile, etc.,) each connected to VDD, to input a 1 (HIGH) state at each pin?
The thing is, that the Port A, B and C pins, any one made inputs are normally set a 0 with pull-up resistors and then set to 1 using switches.
I ask this question because some of the schematics I saw on the Internet of some projects for the SX chip have me a little confused.
I appreciate your replies.
Green Phantom
By what I saw in the manuals so far, you connect 1 to 10K-ohm pull-up resistors to the pins set as inputs to VDD when using switches for input, where a 0 in the input· (the switches connected to GROUND) will give a result depending on the program code of your project.· This is called "active-low."
For instance: if you use a standard logic TTL chip (like the 7400 2-input NAND gate), the two inputs at the HIGH state gives you a 0 (LOW) state at the output.· But when you make one of the two inputs a 0 (LOW,) the output becomes a 1 (HIGH.)· This is what is known as "negative-edge" OR gate or "active-low" OR gate.
Can you use pull-up resistors connected to GROUND at each SX pin (Ports A, B and C) used as inputs and then use the switches (tactile, etc.,) each connected to VDD, to input a 1 (HIGH) state at each pin?
The thing is, that the Port A, B and C pins, any one made inputs are normally set a 0 with pull-up resistors and then set to 1 using switches.
I ask this question because some of the schematics I saw on the Internet of some projects for the SX chip have me a little confused.
I appreciate your replies.
Green Phantom
Comments
You can do it either way. But you can use the internal pull-ups if you connect the switch to ground.
There are no internal pull-downs.
Bean.
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www.iElectronicDesigns.com
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Using internal pull-up resistors from the SX chip itself is a good idea for simplifying hardware connections on a circuit board.· Having even pull-up resistors in an SX chip is a terrific advantage over fitting external pull-ups on a circuit board!· Only that the inputs are at logic 1 state and you use a switch to make one of the inputs a 0 (Ground) for input from a keypad!· I'll explain to you later about a project I am developing.· I am challenging myself to make one single SX chip into a handheld device, that's right: build a hand-held device with a keyboard and LED display with only one SX chip.· I'm talking about a touchtone-phone-style 12-button keypad with 12 of the SX chip's I/O inputs used (RA and RB) and RC for output.· But not a 1-digit 7-segment LED display, but an 8-digit LED display!· I ordered an Avago HCMS-2913 8-digit serial dot-matrix LED display which lets me control one of 320 individual LEDs on the display and create a custom character display set built-in the SX!· The SX chip is already a micro-computer in a chip, but what's the use if there is no keypad and display to "input" data and "see" what is happening in the chip on the "display?"
The internal pull-up resistors in SX chip do give a hardware circuit construction advantage and they really do work!
I just got the hang of it!
Thaks again!
Green Phantom
[noparse][[/noparse]edit] I removed all the white space from the bottom of your post. (Bean)
Post Edited By Moderator (Bean (Hitt Consulting)) : 1/23/2008 12:32:06 PM GMT