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The rechargeable battery info: 1900 mAH (1 battery)

A device that have input of 3V DC and output of 5 V / 500mA

If one rechargeable battery is inserted into the device, I can expect that it can provide me (1900 mAH / 500 mA of 3.8 hours) of output. (or it can last me for 3.8 hours)

So, if two rechargeable batteries are inserted in parallel, I can expect 3.8 hours * 2 = 7.6 hours of output

So, if the device only provide me 1 hour of output with 2 batteries inserted, can I say that the circuit is faulty?
Suggestive that the device provide 3800 mA of output in 1 hour?

Also, a side note - the device becomes hot (temperature around 40 to 45 degree Celsius)

The 2 battery is insert into the following device:

enter image description here

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  • 1
    \$\begingroup\$ What device, and could you perhaps share photos of how you have the batteries hooked up? \$\endgroup\$ – Anindo Ghosh Feb 6 '13 at 3:29
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    \$\begingroup\$ Depends on the batteries and what you consider to be your "discharged" voltage - in general the higher the discharge current, the lower the capacity obtained. \$\endgroup\$ – Oli Glaser Feb 6 '13 at 4:13
  • \$\begingroup\$ @AnindoGhosh I have added the photo of the device. \$\endgroup\$ – Jack Feb 6 '13 at 6:20
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First, stating the background based on the device now shown in the question:

The emergency charger shown is designed to provide 5 Volts at 500 mA, from 2 AA cells.

It is not clear from the device photo alone, and without specifications being provided, whether the device uses the batteries in series (will not work with a single cell inserted), or parallel as the question implies (will work even with a single cell inserted).

The circuit inside would be a boost converter, generating 5 Volts from a 1 to 3 Volt input. Typical boost converter efficiency is anywhere from 75 to 90 percent, depending on design and component selection.

Assumptions:

  • 85% efficiency of boost converter for calculations
  • Eneloop HR-3UTGA NiMH rechargable AA cells
  • The device uses the AA cells in parallel (series calculations also follow)

Output power: 500 mA @ 5 Volts = V x I = 2.5 Watts
Required input power:  2.5 / 0.85 = 2.941 Watts (at 85% efficiency)
Required min current: 2.45 Amperes (P / V, fully charged NiMh 1.2 Volt battery)
Required max current: 2.94 Amperes (with depleted battery, at 1 Volt)

Assume roughly equal split of current draw...

Min current from each: 1.225 Amperes
Max current from each: 1.47 Amperes
Actual capacity: ~ 1800 mAh per AA (Discharge graph in datasheet, between 1 and 2 A)
Capacity of 2xAA: ~ 3600 mAh
Time to Discharge: Capacity / Input Power = 3600 / 2941 =  ~1.224 hours

So, even in optimistic conditions, the charger is expected to deliver less than 1 hour 14 minutes of power, from the above calculations. In reality, the current drawn would rise as voltage drops, thus causing faster depletion. Also, whichever cell has lower internal resistance due to manufacturing and life-cycle differences, would discharge first, loading the second into a quicker depletion.

Thus the observed 1 hour of operation is not unexpected.


Now, if the batteries were connected in series, as is possible, the following calculations change, the rest remain as above:

Required min current: 1.226 Amperes (fully charged, 2 x 1.2 = 2.4 Volts)
Required max current: 1.47 Amperes (with depleted battery, 2 x 1 Volt)
Actual capacity: ~ 1800 mAh per AA (Discharge graph in datasheet, between 1 and 2 A)
Capacity of 2xAA: ~ 3600 mAh
Time to Discharge: Capacity / Input Power = 3600 / 2941 =  ~1.224 hours

Thus, operation time would be about the same, but single battery operation would not be an option.


Additional notes:

  • Typical boost converter efficiency reduces with increase in gap between input and output voltage. This impacts both faster depletion towards the end of battery charge, and lower efficiency for a parallel battery arrangement.
  • Thus, realistically the parallel battery option is likely to last much less than with batteries in series.
  • The probable reason the device heats up more with rechargeable cells than with standard 1.5 Volt cells, is that the lower voltage causes greater efficiency losses. Such losses are typically dissipated as heat in electronic devices.
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  • \$\begingroup\$ Probably, you would like to answer the part 2 of this question as here - electronics.stackexchange.com/questions/57262/… ? \$\endgroup\$ – Jack Feb 6 '13 at 8:23
  • \$\begingroup\$ @Jack Answered that as well. \$\endgroup\$ – Anindo Ghosh Feb 6 '13 at 8:45
  • \$\begingroup\$ Yes, I have marked as correct. Would be great (and I will upvote) if you can write down the calculation like the way you answer this question. \$\endgroup\$ – Jack Feb 6 '13 at 8:54
  • \$\begingroup\$ @Jack Added estimation calculations there as well. \$\endgroup\$ – Anindo Ghosh Feb 6 '13 at 11:08
  • \$\begingroup\$ Thanks. Please see my comments at the part 2 of this question. \$\endgroup\$ – Jack Feb 7 '13 at 1:01

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