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Let's say I am trying to make a chat program like in the YouTube video XBee Basics - Lesson 2 - Simple Chat Program Between Two XBees in AT Mode.

In this video, he explained that one XBee is a sender which sends "Hello world" to the another XBee using Arduino.

However, is the XBee able to send and receive data at the same time? Are the routers able to communicate to the coordinator and the other way around?

Additionally, I want to write my own software to receive and send data and not using Arduino software. Are there any libraries and APIs?

Articles or videos would be nice.

Peter Mortensen
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COLD ICE
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1 Answers1

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Yes, it is possible to send and receive data at the same time. When XBee modules are configured in "AT mode" (also referred to as "Transparent Serial"), you configure the DH and DL parameters to identify the destination node that will receive the serial data.

You should definitely spend some time at examples.digi.com for step-by-step tutorials on setting up various examples like a two way chat. The Quick Reference page includes a long list of libraries in various languages, including the Digi XBee ANSI C Library for C and xbee-api for Java.

tomlogic
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  • How do I use 2 series 1 xbee for a 2 way chat – timberlake Dec 13 '13 at 13:15
  • @timberlake, the examples.digi.com site has just what you're looking for: http://examples.digi.com/get-started/basic-xbee-802-15-4-chat/ – tomlogic Dec 13 '13 at 23:04
  • I have a xbee conected to the computer with a explorer and another xbee the sits on a shield. The xbeesheild has a jumper that switches between xbee/usb. When I have it on the xbee side it can receive msgs and when on usb it can send msgs. I need to do bothat the same ttime. So should I use different pins like rx tx for sending and 3,4 for receiving? – timberlake Dec 14 '13 at 07:47
  • @timberlake, maybe you should ask that as a separate question here on StackOverflow, instead of having it stuck in these comments. I don't have direct experience with Arduino and its shields, but someone else who sees your question might. – tomlogic Dec 16 '13 at 17:29
  • ahh thanks.. but i managed to solve the problem. Thanks for your comments – timberlake Dec 18 '13 at 16:20