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I bought a SIM808 module to use it with a PIC microcontroller. Everything works great but now I'm trying to reduce the module power consumption. The module I have is this one from DFRobot. When the module is in power down mode it draws about 60mA!

The datasheet says it should draw 50uA at max in power down mode. I know I'm in power down mode because the network activity LED is off and I haven't powered it up using the PWRKEY. I removed all the LEDs from the board and the active GPS antenna that also draws current. I even removed the buck converter and powered it from a 4V power supply! But I can't go lower than 60mA. The same happens when I put the module in sleep mode, 60mA is the lowest current.

Do I have a faulty module or I'm missing something on the schematic that is drawing so much current? You can download the schematic from here.

enter image description here

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  • \$\begingroup\$ After you (believe that you) are in low power mode, what response do you get to this command: AT+CGNSPWR? (Yes, the final ? is part of the command and must be included.) \$\endgroup\$
    – SamGibson ♦
    Dec 2, 2017 at 3:04
  • \$\begingroup\$ @SamGibson It returns +CGNSPWR:0 if I send it just before setting DTR=1, after entering low power mode I get no response from the module as expected. \$\endgroup\$
    – Andres
    Dec 2, 2017 at 3:36
  • \$\begingroup\$ OK, thanks for checking :-) I didn't know you were putting it into low power using DTR, rather than the AT command (...FUN... if I remember correctly), which affects response to AT commands, as you say. Have you felt each component with your finger, after giving plenty of time for it to "warm up", to see if you can find one that is warmer than the others (or use a non-contact IR or FLIR sensor)? \$\endgroup\$
    – SamGibson ♦
    Dec 2, 2017 at 3:51
  • \$\begingroup\$ @SamGibson I will leave it connected all night and will check tomorrow for any warm component but as far as I remember everything stays cool. \$\endgroup\$
    – Andres
    Dec 2, 2017 at 4:05
  • \$\begingroup\$ Something is consuming 60mA x 4V = 240mW. If most of that consumption is by just one component, then I hope & expect it to become warmer than the others, especially if it is a small component. Include all components, even capacitors, when you are testing for something warm. Answers to this previous question - Overvoltage failure modes/PCB repair - may give you some ideas, but that was for a "short circuit" situation, whereas yours is much more subtle (only 240mW is being consumed "somewhere"). Good luck - hope you make progress! \$\endgroup\$
    – SamGibson ♦
    Dec 2, 2017 at 4:19

2 Answers 2

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Ok I finally could get the sleeping current advertised on the datasheet.

First, I removed the buck converter and supplied 4V directly to the module so I get better efficiency and removed the supply current needed for the buck converter. That allowed me to strip down a few milliamps, but the true cut down in power consumption was upgrading the module firmware. After that, the power consumption in all states went down dramatically. I had to flash a firmware without Bluetooth support to get the lowest power (even disabling Bluetooth I couldn't get to the sleeping current advertised).

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I am not going to take the time to attempt to download a schematic from some unknown web site from an unknown part of the world.

But I would venture to guess that you have some signal lines from the PIC microcontroller that are feeding into the powered down chips on the module. If such signals are at a output high drive level from the PIC they will forward bias the input pins on the powered down silicon and cause current to flow. This will prevent the silicon from fully shutting down.

The work around for this type of problem is to ensure that all signal pins that are driving out of the PIC are set to a low (near GND) level before attempting to put the module into the powered down state.

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  • \$\begingroup\$ I attached the schematic as an image. The current I'm measuring is with no MCU connected to the module, just the board itself. \$\endgroup\$
    – Andres
    Dec 2, 2017 at 1:36

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