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I got a SIM800L, same as the one under this link, Quickstart SIM800 (SIM800L) With Arduino, and played around with an Arduino Nano and Uno. I had a few problems, at the start, not getting it work properly. It turned out that my Nano 5V pin wasn't enough for the SIM unit. It works fine with Uno though.

There is the code I tested the GSM (same link) and got it working somehow, I got simple AT commands and managed to send an SMS. But after I dismantled the connections and the next day I tried again... Nothing. I can see the SIM unit connects to network as the LED blinking slows down to once every 3 seconds but when I type commands in nothing happens in the serial monitor. It doesn't give errors or anything, just nothing happens on the serial monitor.

I did wire it incorrectly, once with the Tx/Rx, but that shouldn't damage the unit, should it?

I tried to upload different programs to my Arduino and they work fine so that shouldn't be the problem. Is there a way to reset the SIM800 unit? Or am I missing something here?

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  • If you turn around the Rx and Tx pins, nothing would be damaged. Try connecting a "USB to serial TTL cable" directly, so you can speak from with terminal from PC (USB) to the device directly.
    – Paul
    Mar 22, 2016 at 12:01
  • These modules cannot tolerate a 5v supply. It is theoretically possible for a carrier board to have a 5v to 4.2v regulator, but there is no indication of the existence of one at your link. the LM317 mentioned there is extremely unsuitable - insufficient current handling, and too high a dropout voltage.
    – Chris Stratton
    May 25, 2016 at 21:26
  • It looks like it has a 5V pin right at the top left doesn't it? Also kinda curious - what disqualifies the LM317 from candidacy?
    – Malachi
    May 25, 2016 at 23:00

2 Answers 2

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The module needs a good 5v supplier as it might have a peak of 1A or 2A, when it registers to the network or connects to 3G.

You need to make sure that you share the ground between the power supplier and you use some 2 pins on your arduino to connect the GSM module. Dont try to get a fast baudrate as software serial doesnt work, try with 9600 to start then see if it response to the AT command.

SIM800 is a very simple module, try with an FTDI in case you want to make sure you didnt break your Arduino.

For more help, share your schematic how you have connected the arduino.

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  • While you are correct about the high current requirement, the device cannot tolerate a 5v supply unless there is an appropriate regulator somewhere in the system, which would be less than trivial to select (the link's mention of an LM317 is entirely unsuitable)
    – Chris Stratton
    May 25, 2016 at 21:23
  • Isn't he using a breakout board which has an onboard regulator?
    – Malachi
    May 25, 2016 at 22:57
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I have put in a lot of time with the SIM808, and I always have trouble delivering enough power to it. Note that the SIM808 will alert you with an "undervoltage warning" sometimes as a response to AT commands. I bet the SIM800 would do the same.

As Max above is alluding to - try powering the SIM800L via a separate 5V 2A adapter. When doing this don't forget to still hook up ground between the Uno and your SIM800L board.

If you accidentally hooked 5V to the VDD pin on your module, you may have damaged it.

FTDI is a tool every embedded dev should have. Get a few!

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  • Keep in mind that the SIM800/808 cannot tolerate a 5v supply - you hint at this, but it's not clear what regulator you imagine would provide an appropriate connection to such a supply.
    – Chris Stratton
    May 25, 2016 at 21:24
  • You're right. I didn't make it clear - I am using a FONA 808 shield module which is designed to plug in to an Uno 5V directly
    – Malachi
    May 25, 2016 at 22:57
  • The FONA shield accepts a 5v input, yes, but it is to recharge the required lithium cell. It is actually the cell rather tha the input which provides the peak power required by the module. That means that a high current 5v supply is not really required, nor even likely to be taken advantage of, since the charge current is set for a relatively small cell.
    – Chris Stratton
    May 25, 2016 at 23:21

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