Arduino uno with official Wifi Shield take 5 Volts. Arduino lights up as it should when powered but the WiFi shield is dead the l9 doesn't light up but works fine when arduino is connected to usb cable.

I use 5 battery holder 5x1.5V = 7.5 V ; 2000 mAH

50 mA - Arduino 
300 mA - WiFi shield

350 mA * 5h = 1750 mAh

5xAA = 7.5V ; 2000 mAh

Am I missing here something? It should last in the worst case scenario 5 hours with new batteries at 2000 mAH (which I think they have more but I do not have them on me right now).

  • We are not the arduino help desk.
    – Olin Lathrop
    May 14, 2016 at 19:36
  • 2
    @OlinLathrop I'm asking if I did the calculations correctly? if that makes sense
    – Higeath
    May 14, 2016 at 19:38
  • 1
    Ignore Olin. He's stuck in the fifties.
    – Passerby
    May 14, 2016 at 19:41
  • What does the 5V rail show when on batteries and the wifi shield connected?
    – Passerby
    May 14, 2016 at 19:43
  • 1
    @davetweed not true. The 5V reg could handle a few hundred mA, closer to 700 before issues, with a low VIN. The 3.3V reg is the weak one.
    – cde
    May 14, 2016 at 22:13

2 Answers 2


As a general answer, yes. AA alkaline should be more around 2500~3000 mAh depending on quality. 5x AA in series would result in 7.5 Volts and 2500 mAh.

But that's only for almost fully charged batteries. AA batteries can range from 1.6V to 0.8V "dead for most purposes". As they drain, the voltage drops.

What may be happening is that you are powering the Arduino through its power jack, which goes through a diode, causing a 0.7V drop. Then it goes to the linear regulator which has a very high drop out voltage. Or even via the VIN pin, bypassing the diode. But as the battery drains, the voltage drops below the regulator ' s limit.

The Arduino microcontroller can easily work below 5V. The shield might not.

  • thank you for reply! I definitely use jack I will try going straight for VIN pin tomorrow is there a simple way I could put a connector on a jack to get positive and negative? So I won't have to cut off the connector?
    – Higeath
    May 14, 2016 at 20:42
  • @higeath there are plenty of pigtails on ebay. If I recall correctly, you need a 2.5mm female dc jack.
    – Passerby
    May 14, 2016 at 20:56

4xAA = 6.0V connected to a standard diode like 1n4001 = ~5.4V (within the specs for an AVR), connect directly to Vdd (+5V on arduino's). Unless there's a reason you need to use a 7.5V pack. Bypass the protection diode & Vreg = less wasted energy on heat/diode drops. IIRC, the WIFI/Ethernet shields can spike to 500mA or so, so that may also be causing extra drain. If your code constantly uses the WIFI radio, it will cause more current spikes as opposed to using the WIFI less in your code. Your avg current will decrease and battery life will go up. You'll save batteries by using 4 instead of 5 and not cause the 5V regulator to drop out and cause power issues by bypassing its input at Vin and connecting the 6V pack with a series diode to drop it to ~5.4V and connecting to +5V on an Arduino. It will also help with Vdrop when the WIFI has a current spike.

Also note that as any battery discharges, it's more of a slope than straight line and then dead, even for lithium packs. look at this link, good info for you to understand. http://www.powerstream.com/AA-tests.htm

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