I have purchased this lithium polymer battery (11.1V, 5500mAh), to power a robotics project. The project uses a Raspberry Pi, three continuous-rotation servo motors, a couple of LEDs and a Sparkfun Geiger Counter.

I've also set up a voltage regulation circuit, as shown with the battery below.

The battery is behaving very strangely. Firstly, very loud cracking noises were made when I accidentally short circuited it. Is this normal? Secondly, when testing it with a multimeter, both the (fairly thick) wires got very hot. The ammeter setting (when I turned the dial to the one with the 200 on it) showed a ridiculously high reading - something in the range of 26A. When I plugged the battery into the voltage regulation circuit and turned it on, the LED (5V), came on dimly briefly and then faded out quickly. I thought that the regulator had failed and the LED was burnt out, but when I checked by connecting a power pack, it was fine.

What's gone wrong? Do I need to get a smaller battery? Is my circuit for voltage regulation wrong? Do I have a faulty regulator?

Thanks for your help, Archie


simulate this circuit – Schematic created using CircuitLab

  • \$\begingroup\$ As suggest below by Leo, you likely had the meter on the setting to measure current when you tried to measure the voltage. Consequentially, the (very small) internal resistance of the meter was the only impedance seen by the battery and this effectively shorted it out. \$\endgroup\$
    – sherrellbc
    Jun 27, 2015 at 19:12

1 Answer 1


The amp meter has ideally a zero internal resistance, so when you connect it directly without anything in series you're basically shorting the battery. Hence the high current reading. Since you have a high current the wires get hot because of all the power they are dissipating.

Best case scenario, you're battery simply need to be recharged. I wouldn't be surprised if it was damaged though. Maybe somebody with more experience on batteries can add on that.

As a general rule, remember that an amp meter always goes in series with the component which current you want to measure. If nothing else is there, you will short the circuit.

  • \$\begingroup\$ Right. So, how should one read the voltage left in a battery? I will try recharging it now. \$\endgroup\$ Jun 27, 2015 at 19:16
  • \$\begingroup\$ Set the meter to read volts, not amps. Make sure you do this before you connect the meter to the battery. And check to make sure your battery is connected to the volts input, not the current input. Most meters use different input jacks for the two functions, just to prevent customers from doing what you did. \$\endgroup\$ Jun 27, 2015 at 20:36
  • \$\begingroup\$ With the multimeter set to volt reading, not amps. \$\endgroup\$
    – Leo
    Jun 27, 2015 at 20:36
  • \$\begingroup\$ I've just given it a fresh charge, and it's just done the same thing. I think that it might be a capacitor issue - rather than smoothing out the supply, it's just storing it all up, except for the initial 'spike' in voltage? \$\endgroup\$ Jun 27, 2015 at 21:14
  • \$\begingroup\$ Did you put a 0.33uF cap at the input of the voltage regulator, as the datasheet suggests? \$\endgroup\$
    – Leo
    Jun 28, 2015 at 13:02

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