Shop OBEX P1 Docs P2 Docs Learn Events
excessive battery consumption — Parallax Forums

excessive battery consumption

turboEK2000turboEK2000 Posts: 18
edited 2006-08-04 16:24 in BASIC Stamp
I have a problem with a BS2 and some other circuitry with its power consumption, to be exact I’m using a EDE-1144 keypad encoder and some other stuff to open a door, the circuit works perfectly for about a day then my 4 brand new AAA batteries are dead in less then 24 hours. With a 6v supply its pulls about 7mA, how bad is this, what should the current be for the batteries to last a few months.
The code i'm using is based off an example Chris Savage posted a few months ago...if this means any thing

Comments

  • Beau SchwabeBeau Schwabe Posts: 6,568
    edited 2006-08-01 02:03
    Can you post a schematic?

    ▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔
    Beau Schwabe

    IC Layout Engineer
    Parallax, Inc.
  • Jason-WIJason-WI Posts: 35
    edited 2006-08-01 03:32
    7 ma can't be right as the average AAA alkyline battgery is about 1100 mAh. Are you using a low dropout regulator like the LM2940 series form National Semiconductor. A plain jane 7805 is usually 1.5 V droupout depending on the load. If you supply drops below 6.5 volts your output voltage from the regulator will also drop causing the stamp to reset/shutdown.

    Jason

    ▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔
    He who dies with the most toys wins!
  • Chris SavageChris Savage Parallax Engineering Posts: 14,406
    edited 2006-08-01 17:23
    Well the EDE1144 saves current (and EMI) by not contiuously scanning the keypad (like the 74C922 does).· If you have a BASIC Stamp, and the EDE1144 that (7 mA) should be about right...No display though?·

    ▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔
    Chris Savage
    Parallax Tech Support
    csavage@parallax.com
  • allanlane5allanlane5 Posts: 3,815
    edited 2006-08-01 18:27
    Yes. So a 7 mA circuit should drain a 1100 mA-hour battery pack in -- 1100 / 7 == 157 hours.

    But he's seeing the pack drained in less than 24 hours. So there's a lot of current leaving the battery that he's not measuring somehow. Or, the battery pack isn't 1100 mA-Hours.
  • turboEK2000turboEK2000 Posts: 18
    edited 2006-08-03 02:08
    it is 7mA, I still cant figure out why its killing these batteries, there brand new alkaline. I'm using a L7805 5v regulator, think this is the problem??? When the circuit stops working the batteries are not completely dead there just weak... what would be a better 5v regulator that will not kill the batteries as fast???


    ps. no display, just a LED, a beeper and a 3v relay, its a pretty simple circuit

    Post Edited (turboEK2000) : 8/3/2006 2:11:41 AM GMT
  • LarryLarry Posts: 212
    edited 2006-08-03 03:28
    you'll get much better results with a low drop out regulator. The 7805 just won't cut it with 4 alkalines. As you have discovered empirically, the thing will stop working almost as soon as the batteries drop their voltage even a little. Add another cell to the battery pack for better results.


    You still won't get past the issue that resitive regulators waste power as heat dirrectly related to the voltage they are scrubbing while they are regulating. For better results long term, use a switching regulator. Sounds exotic, but they are pretty simple to make (and cheap).

    Go here: www.romanblack.com/smps.htm

    Two transistors will do it.

    ▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔
  • Chris SavageChris Savage Parallax Engineering Posts: 14,406
    edited 2006-08-03 04:16
    Are you measuring the 7 mA on the input side of the regulator or the output side, because the regulator itself will consume current dropping the voltage down to 5V.· This will be disappated as heat.

    ▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔
    Chris Savage
    Parallax Tech Support
    csavage@parallax.com
  • turboEK2000turboEK2000 Posts: 18
    edited 2006-08-04 00:10
    yep 7.7mA right at the input to the regulator with fresh batteries, the output is at 4.9V

    now I just measured the output of the regulator with the half dead batteries and its at 4.4V, so this would cause the Stamp and EDE1144 not to work????

    I'm going to·throw another cell in there as Larry suggested...
  • Chris SavageChris Savage Parallax Engineering Posts: 14,406
    edited 2006-08-04 05:18
    4.4V?· Yeah, the BASIC Stamp will be in a RESET state at that voltage.

    ▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔
    Chris Savage
    Parallax Tech Support
    csavage@parallax.com
  • NateNate Posts: 154
    edited 2006-08-04 11:12
    If you are drawing 7ma continuous from AAA batteries, you are going to be replacing/recharging batteries like a madman no matter what you do.· Time to look into the 'NAP' function to save current draw.·

    If you were to take the leap to the·SX, it has a 'SLEEP' function which can be interrupted by a keypress interrupt.· The SX's current draw in it's 'SLEEP' mode is 1uA or less.· You can also easily integrate the keypad scanning into the SX programming·to cut down on current draw and parts count at the same time.· (With the SX, a LED, a beeper and a 3V relay, you wouldn't even need the voltage regulator (run off batt. direct) - talk about simple!)

    Nate
  • Tracy AllenTracy Allen Posts: 6,664
    edited 2006-08-04 16:24
    Not only low dropout, but also low quiescent current is necessary in the regulator. The 7805 ix a bad choice. With that regulator and a 6 volt battery pack, you can't get anywhere near the full amp-hour capacity out of the batteries, because of the high dropout voltage and because of the tendency for the regulator itself to draw more current as it goes into dropout.

    The LM2940 that is used on the BOE is low dropout but _not_ low quiescent current.

    The LM2936 that is on the BS2 module and the LT1121 that is on the multibank BS2s are both low dropout (0.2 volt) and low quiescent current (<100 microamps). The LM2936 can supply up to 50 milliamps, which must include the 3 ma for the Stamp itself but can also include 40ma of external loads. If your whole system draws only 7 ma, you might dispense with the external regulator and just use the Stamp's Vdd.

    You asked, "what should be the current to last for a few months?" The battery capacity of AAA alkalines is 1200 ma-h. In 4 months there are about 1464 hours. Divide:
    1200 / 1464 = 0.82 milliamp.
    data.energizer.com/ <-- great info and tutorials

    I'm not familiar with the EDE-1144, but to reach that lower current consumption your system will have to do some napping or tricks.

    ▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔
    Tracy Allen
    www.emesystems.com
Sign In or Register to comment.