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Trouble using Launchpad with battery

Dylan Williams
Prodigy 80 points

Hello, 

I have my launchpad setup interfaced with a serial LCD using the UART pins.  If I power the launchpad with 3.3V and unplug the usb cord, my program no longer works.  I have tried removing the VCC jumper.

-Thanks

Dylan

  • 0
    Guru 92251 points

    Try removing the RST jumper. When the emulator part isn't powered from the USB it is probably holding RST low which prevents the program was running.

    (haven't got a launchpad to try it, but did have the same problem on a fraunchpad)

  • 0 in reply to Chester Gillon
    Prodigy 80 points

    I'm afraid that didnt work.  I tried removing the RST, VCC, and TEST jumpers with no luck.  If I apply 3.3V to J6, I'm only reading about 2.0V at the J1 VCC and with the VCC jumper removed I only read about 1.5V.  The emulator is not plugged in or anything.  But as soon as I put the jumpers back in and plug the emulator in, the program kicks in.

  • 0 in reply to Dylan Williams
    Guru 227245 points

    Dylan Williams said:
    If I apply 3.3V to J6, I'm only reading about 2.0V at the J1 VCC and with the VCC jumper removed I only read about 1.5V.

    According to the schematics, J6.1 (ext_PWR V+ in) and J1.1 (VCC) are directly connected. If you really put 3.3V into J6, you must read 3.3V on VCC. And if you twisted V+ and GND, you shouldn't read more than ~0.5V until you fried the clamp diodes inside the MSP. Well, possibly the MSP will still work even if you did.

    But maybe your batteries are just sucked dry so that the small current of the MSP lets the voltage break down to 2V

  • 0 in reply to Jens-Michael Gross
    Prodigy 80 points

    This cannot be the case because I am powering it from a power supply at the moment.  Really, I am applying 5V to an LCD and voltage dividing down to 3.3V for the launchpad.  At my voltage divide point I can read 3.3V without the launchpad plugged in. But, when I use this point to power the launchpad, its pulled down to the ~2V.  Does the UART have anything to do with the emulation section?  Out of the Vcc, Rst, and Test jumpers on the launchpad, which ones should I remove to use it without usb.  Also a note, when I plug the launchpad into usb I see a power light on the emulation side. But, when I use battery power I dont see this light with or without vcc jumper in.

    Thanks,

    -Dylan

  • 0 in reply to Dylan Williams
    Genius 3330 points

    How are you "dividing down"?

  • 0 in reply to eltury
    Prodigy 80 points

    Im using a simple voltage divider.  2 resistors in series from Supply Voltage to GND.  My voltage sense point or supply to launchpad is in between these resistors. The resistor values were chosen such that the center point is 3.3V when Supply is 5V.  

  • 0 in reply to Dylan Williams
    Guru 92251 points

    What is the value of the resistor values? The current drawn from the launchpad will cause a voltage drop at the "voltage sense point". Consider using a linear voltage regulator instead.

  • 0 in reply to Chester Gillon
    Genius 3330 points

    That was my guess.

  • 0 in reply to Chester Gillon
    Prodigy 80 points

    Thanks guys. I hooked up a linear regulator and it solved the problem.  

  • 0 in reply to Dylan Williams
    Guru 227245 points

    Dylan Williams said:
    Thanks guys. I hooked up a linear regulator and it solved the problem.  

    Just to clarify to you what happened: since the MSP draws some current, it acts like a 8variable) resistor too, so it is in parallel to your voltage dividers bottom resistor. As a result, the bottom resistor becomes (significantly) smaller and the voltage drops accordingly.

    Using a voltage divider is only for applications that have a significantly (at least by a magniture) higher impedance than the divider itself.

    So if you use a 2Ohms/3Ohms divider to break down 5V to 3V, an MSP that draws 10mA will pull the 3V down by only 0.03V. If you use 2k/3k, the voltage will drop to 0.3V.
    However, in first case, you'll be consuming 1A cross-current (burning 5W on the resistors).

    The linear regulator you now use is more or less a voltage divider with an active impedance changer. So the voltage divider doesn't get loaded. You can see it as a voltage divider with high impedance and an (stripped-down) OpAmp following it.

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