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From the product manual / reference guide it looks like SIM900 / SIM900A is already equipped with 12 GPIOs.

But could not find information on how to use these GPIOs to control relays directly, instead of routing them through additional MCUs.

Most examples on the net (such as this) use the SIM900 just for read/write, completely ignoring its GPIO capabilities.

Any pointers on how to use the GPIOs of SIM900 as control signals directly - please help.

Thanks in advance.

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The GPIOs in sim900 module is meant to be controlled by AT commands through serial port ofcourse.

if u see the AT command manual of sim900 you will find the following at command to control the GPIOs

AT+SGPIO=<operation>,<GPIO>,<Function>,<level>

following is the Description to the above function!

enter image description here

P.S. i know this question is about 20days old and he might have already got the answer.I'm only sharing because it may help other.!

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Yes, but you will need to blow away the included AT-command virtual machine and write your own firmware layer, to be loaded in the SIM900 flash.

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  • \$\begingroup\$ Thank you. Know any examples or tutorials anywhere on this subject? \$\endgroup\$ – Gopalakrishna Palem Feb 3 '15 at 14:58
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Yes it is possible. Refer "SIM900_Customer Application Building Tutorial_Application Note_V1.00.pdf"

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  • \$\begingroup\$ Welcome to EE.SE! Usually very short answers that simply provide a link to resources are discouraged. In the future (when you have enough reputation), a post like this may be better served as a comment. \$\endgroup\$ – Daniel Feb 9 '18 at 5:54
  • \$\begingroup\$ Hi, I don't have enough reputation to add comment,hence added answer \$\endgroup\$ – Bahubali Feb 9 '18 at 5:57
  • \$\begingroup\$ No problem! Just a tip for the future. Again, welcome. \$\endgroup\$ – Daniel Feb 9 '18 at 5:58
  • \$\begingroup\$ Thanks Daniel,Why I am not getting "add answer" option to this question?(electronics.stackexchange.com/questions/340637/…) \$\endgroup\$ – Bahubali Feb 9 '18 at 6:01
  • \$\begingroup\$ It has been marked as a duplicate question, and cannot be answered. \$\endgroup\$ – Daniel Feb 9 '18 at 6:05

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