Electronic – Battery pack with 12 NiMH 1.2 V AA batteries

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.

You speak of dropout voltages, which makes me think you are planning to use a pair of linear regulators for this, such as a 7812 and a 7805.

At idle, a Pi draws about 0.4 A.¹ P=VI, so you're talking about burning up 7 × 0.4 = 2.8 W across the second regulator.² That's a fair bit of waste.

A bare TO-220 regulator has a 50°C/W thermal resistance, and 50 × 2.8 comes to 140°C over ambient. Ambient temperature will only be as low as 20-25°C indoors, and then only if you're running this device naked on the workbench with a fan blowing on it. Unless your final application will have this thing running in a ventilated case outside in winter, count on the actual ambient temperature to be higher as the electronics heat up the air inside the case.

The second regulator is almost certain to destroy itself if you don't put a fairly big heat sink on it.

But it gets worse.

The first regulator will also be producing heat, up to about 5 W worth of it.³ The heat sinks on both regulators will have to be bigger than you calculate for each alone, since they will each be heating up the air inside the electronics' enclosure. The dual regulator system will come to an equilibrium higher than if you were running the two regulators in separate enclosures.

A far superior solution is to use one of the many off-the-shelf DC-DC converters that will put out both 5 and 12 V from a single input supply. You will find that there are models covering a wide range of input voltages. Any battery that can put out enough current for your job can be made to work.

When you start stringing many cells together into a high-voltage battery, you make it much easier for the battery to kill itself due to cell reversal. The fewer the cells you can use in series, the better. Another of the advantages of the DC-DC converter solution is that you can find types that will put out more voltage than they take in,⁴ so you could get away with just 1-4 cells, which is less likely to self-destruct.

You might be able to avoid the requirement for a regulated 12 V supply. LEDs don't really want constant voltage, they want constant current. The EE 101 way to get that is to drop a constant voltage across a fixed resistor, but there are a whole lot of ways to make a constant current source/sink. So, you'd put the CCS either between the raw supply voltage and the LED array or between the LED array and ground, then run the Pi from a 5 V DC-DC converter.

Footnotes:

1. It can go way up from there.

2. The second regulator only sees the 12-to-5 V drop, due to the way you've drawn your system in the question. Thus, we can ignore the battery voltage question for this part of the analysis.

3. The first regulator throws off even more heat than the first, for a couple of reasons.

   First, the battery can't drop below 12 V+V_d, the regulator's dropout voltage, if we're going to meet your "stable voltage" requirement.

   To get the full use out of the battery, you want to divide a single cell's lowest useful voltage into the lowest battery voltage you can tolerate to get the minimum number of cells. Then you multiply that by the highest voltage of the battery to get the peak battery voltage.

   NiMH cells are still the most DIY-friendly sort. NiMH cells range from about 1.35 V when fresh off the charger down to about 0.8 V when nearly dead. If we use a regulator with a 2 V dropout voltage, we divide 0.8 into (12+2) V, which comes to 17.5 cells, which we have to round up to 18 × 1.35 V = 24.3 V. That means the first regulator could be throwing off another 12.3 × 0.4 = 5 W!

   As the battery voltage drops, heat thrown off the the first regulator will also drop, unlike heat produced by the second regulator, which sees a constant voltage. It only drops to about 4 W, though, so it does't really change our conclusions.

   Lithium rechargeables differ on behavior from NiMH a fair bit, but if you run the same calculations on them, the conclusions don't change much.

   But all of that only considers the heat due to the Pi's power draw. You also have to add the current required by your LED matrix, which you haven't specified. If it's another 400 mA, you double the power wasted in the first regulator.

4. At a cost of higher input current.