# What informations can we obtain with these voltage and current measurements of a little electronic device?

I powered a little electronic device with 2 AA batteries and here are the results measured with a multimeter:

• When using a first set of 2 AA batteries:

• Voltage when device unplugged: 2.45 V
• Voltage when device plugged / ON: 2.10 V, consumption: 0.30 A
• When using a second set of 2 brand new AA batteries:

• Voltage when device unplugged: 3.10 V
• Voltage when device plugged / ON: 3.00 V, consumption: 0.20 A

Just being curious: what informations can I deduce about the device or about the batteries with these measurements?

• Try posting a picture of the device and get a better response instead of guesses or plainly basic information like the device's input port resistance. Do you know ohms law BTW? – Andy aka Sep 18 at 11:28
• is this electronics puzzle question you got as homework :P – Mitu Raj Sep 18 at 11:47
• @Andyaka It's a Zoom pedal like this: sweetwater.com/store/detail/… It can be powered by 2 AA batteries or 9V DC adapter or 5V USB. Pretty convenient, so I was curious about how it works! And yes I know Ohm's law – Basj Sep 18 at 11:58
• @MituRaj No no ;) It's a device I have, and since it can be powered by 2 AA batteries or 5V USB or 9V DC adapter, I was curious about its powering system :) – Basj Sep 18 at 11:58
• I was curious about its powering system - This may be obvious enough to not warrant explanation, but the device has a voltage regulator internally that converts any of those three power sources to the correct voltage for the actual circuit to function. – dwizum Sep 18 at 20:24

2.1V * 0.3A = 0.63W

3.1V * 0.2A = 0.62W

The device seems to use constant power no matter what the input voltage is, which most likely indicates it is internally powered by a switching DC-DC converter. It could also be a flashlight with a switching LED driver.

• Ok thank you! Is there something we can say about the batteries? What does the voltage drop (0.35V for the first set of batteries, 0.10V for the second set) tell about them? – Basj Sep 18 at 12:00
• Fresh batteries have lower internal resistance – peufeu Sep 18 at 12:06
• Because the first batteries are getting a bit older ;) , internal resistance increases with usage over time – Mitu Raj Sep 18 at 12:08
• @MituRaj can we estimate the internal resistances with these measurements? – Basj Sep 18 at 12:28
• Ofcourse, its just the dropped value 0.35V divided by the current drawn .... – Mitu Raj Sep 18 at 12:31
*•  When using a first set of 2 AA batteries:
•   Voltage when device unplugged: 2.45 V
•   Voltage when device plugged / ON: 2.10 V, consumption: 0.30 A


Battery ESR = 0.35V/0.30A = 1.17 Ohms, load = 0.63W.
Battery loss= 0.3^2* 1.17= 0.1W

•   When using a second set of 2 brand new AA batteries:
•   Voltage when device unplugged: 3.10 V
•   Voltage when device plugged / ON: 3.00 V, consumption: 0.20 A*


Battery ESR = 0.1V/0.2A = 0.5 Ohms, load = 0.60W Battery loss = 0.2^2* 0.5= 0.02W Cells may or may not be balanced when reading 2.1/3/1 = 67% of no load fresh cell voltage and ESR always rises which DoD.

Load appears to be constant power yet 5% more efficient with fresh batteries while battery losses reduce from 0.1W to 0.02W with low DoD fresh cells.

So supply inefficiency rises from 0.02/0.60W= 3.3% to 0.1W/0.63W = 16% additional losses when drained. This explains more rapid decay in storage energy remaining when batteries get low with a constant power load.

• Interesting: Measuring Internal Resistance of Batteries. – Basj Sep 18 at 20:30
• Batteries are just like batteries with a low ESR but exceptionally high C value. My answer assumes you know Ohm’s Law and ESR of dielectric ~ electrode interface exists for every battery and every capacitor. There is also a double-electric layer effect, giving 2 time constants, the longer one appears like “memory effect” – Tony Stewart Sunnyskyguy EE75 Sep 18 at 20:35