# The number of the cells required for supplying power to an embedded system

I'm designing a battery powered embedded system. The system consists of mainly the Raspberry Pi zero & Raspberry Pi Camera V2 and some other sensor modules. This system is supposed to capture still images in sequence for 4 hours with certain time interval and measures some environmental observables. An important condition is that the ambient temperature would be around -40$^{\circ}$C. I wanted to ask you if my calculation of the number of battery cells required for this application is valid.

For this calculation I would suppose that the TL5930 from Tadiran will be used. Also a boost converter(switching regulator) with an efficiency of 90% will be assumed. The Raspberry Pi requires its power supply to generate 5V output. The maximum power consumption of the Raspberry Pi, Raspberry Pi camera and others are measured by myself, and I would like to assume here the devices consume 3.0W(5V*600mA) in total continuously. (Please assume this value)

Then from the power consumption of 3.0W and the conversion efficiency of 90%. I find the power that should be supplied to the voltage regulator is $$P_{in} = P_{out} \times \frac{1}{\eta} = 3.0W \times \frac{100}{90} = 3.3W \qquad \quad [1]$$

Now I need to know how many batteries and which configuration I need in order to supply 3.3W for about 4 hours. I looked at the plots 'Voltage vs Temperature' and 'Capacity vs Current' in the datasheet of the TL5930 and found one needs to make assumption of the discharge rate to estimate the terminal voltage and the capacity of the battery. I suppose that the current load would be 230mA. Then in the condition of $T=-40^{\circ}$ and $I_{d}=230mA$, the terminal voltage of the cell would be $V_{BAT}=2.2V$, and the capacity would be $Q_{BAT}=0.5Ah$. Assuming that I connect cells in parallel and convert 2.2V to 5V. The total current draw from the batteries to the voltage regulator is $$I_{in}=3.3W/2.2V=1.43A \qquad \quad (2)$$ Since I already made an assumption of the discharge rate and derived voltage as well as capacity, here I again assume that the discharge rate for each battery will be $I_{d}=230mA$ $$0.23A\times N = 1.43A. \qquad \quad (3)$$ $$N=6.22. \quad (4)$$

If I have less than 6 cells, the current would exceed 230mA. This is not recommended, because the maximum recommended continuous current of the TL5930 is 230mA. Therefore I need 7 cells connected in parallel.

To me the calculation seems odd. The discharge rate would be less than 230mA this time, and it is inconsistent with my initial assumption. I would need to change my initial assumption of the terminal voltage and the capacity. Then I will get a different discharge rate in the end. Could you please let me know what is wrong in my calculation?

• At a 200 mA current and a temperature of -40 your battery will have around 18% of it's 0 C current capability and around 15% of it's power capability. You'd get a longer battery life if you used some of that battery power to run a heater to keep that battery at 10 to 20C Aug 23, 2017 at 15:55
• The whole idea of parallel batteries is wrong in general. Your choice of battery is quite wrong for a device with 4-hour run time. The TL5930 is designed for 4mA discharge rate in an equipment designed to run for years. Get two 18650 cells in series, and forget all your calculations. Aug 23, 2017 at 16:38
• @Andrew thanks for your comment. Is there some general tendency of the temperature dependence? Could you please point me to some references(article, post and etc)? It's impressive that you can keep temperature at 10C, even though the air temperature is -40C. Aug 23, 2017 at 17:36
• @AliChen thank you very much. So you mean one shouldn't use lithium-thionyl chloride? I was interested in this type, because of its high capacity and low temperature operation. Could you lease suggest me other battery type that one can use at low temperature? Also do you mean that I should use lots of batteries connected in series together with a step down regulator? 2 batteries don't seem to be enough to provide 3W for 4 hours, isn't it? Aug 23, 2017 at 17:40
• Under normal conditions, a 18650 rechargeable Li-ion cell has typical capacity of 10 Wh, so two should be plenty. I don't know much about low temperatures, but did you look into sealed lead-acid batteries? They are used in Arctic/Antarctic areas... Aug 23, 2017 at 18:37

The ESR of frozen electrolytic capacitors goes through the roof, if you have any in your design, please check for potential problems.

Whether any of the components in a Raspberry Pi is actually rated for -40°C (most likely none) is your problem. Considering the price of a rPi, I would buy some extra, freeze them, and keep the ones which still work.

The battery works much better warm than frozen.

I'm not gonna do the math for this, but I would consider putting the device, or at least the battery, inside an insulated enclosure, 5cm thick PIR foam for example. We will assume whoever puts it down outside in -40°C weather got there in a nice comfortable heated car, so the device can be warm inside when it starts. You only want 4 hours runtime, which is quite low.

You could even go:

"Dear customer, we can sell you the $200 battery which works at -40°C, or you could crack a handwarmer, throw it in the enclosure and close the lid, and in this case the$5 battery will give you the same runtime."

• Thank you for the answer. (1) The raspberry pi zero and raspberry pi camera v2 work fine at -40C. I tested it over 2 weeks, and there was no noticeable problem. It could boot at -40C and could read values from sensor modules every seconds and reading/writing with microSD card worked tool. In addition, the photos from the picamera look just fine. Over the 2 weeks there was no crash. (2) Unfortunately the enclosure of the whole device is quite small. Thus I'm not sure how effective insulation I could achieve. Aug 24, 2017 at 4:37
• Is it a wearable? Aug 24, 2017 at 11:25
• No, it's not a wearable device. Aug 25, 2017 at 2:56

Try to approach this from a different angle. The job needs 600 mA at 5 V for 4 hours. This is 12 W*hr total.

Now, since there are obvious limitations in term of discharge current per cell, the power supply topology must be chosen as buck converter, not a boost from low voltage. Therefore the cells should be connected in series.

If a battery has a 200 mA limit, you need a 15V battery downconverted to 5V. Assuming 3V cut-off and conversion losses, 5 cells should be enough. With 12 Wh total requirements, each cell should have ~2.5 Wh capacity, or only ~1 A*h for a 3V cell technology.

Therefore, you might go for a smaller cells from Tadrian, say C-size TLH-5920, which has 1 A*H at 100 mA discharge and -30C. This might be marginal, so the D-size will likely do the required job. The uncertainty comes from the fact that Tadrian doesn't offer any characterization at 200 mA discharge, nor at -40C. So you might need to do some experimental research on your own.

An alternative would be a sealed lead-acid 12-V 1 A*h battery. I heard that they work fine in polar areas of Earth.

• Thank you very much for the detailed answer. Could I please ask some questions? (1) Can't one use a boost converter in order to decrease the current draw on each cell? I think either serial or parallel connection work in principle. With the batteries connected in series, you would get a lower current draw on the battery pack, but the discharge rate of each cell is same as the total current draw. In contrast, the parallel connection decreases the discharge rate of each cell. Aug 24, 2017 at 6:09
• 12Wh is the total energy to be transferred for 4 hours. But we also need to take into account the energy transferred every second. Thus starting point would be 3W. Considering the conversion efficiency of 90%, You need an input power of 3.3W. When you want to have the batteries connected in series, you need a 3.3W/0.2A=16.5V battery pack. Aug 24, 2017 at 6:40
• Then we need to make an assumption of the terminal voltage and the capacity of each battery, since the values are not provided by Tadiran. This is same for both TL5930 and for TLH5920. When you start making the assumption, you are in the same situation as me. To assume certain discharge rate and get the terminal voltage and capacity, and it would result in a discharge rate which can be inconsistent with your initial assumption. Aug 24, 2017 at 6:43
• If I understood correctly, you suggested the same calculation for the batteries connected in series. But the point of my original questions is, whether it makes sense to make such initial assumption of the discharge rate of each cell. Aug 24, 2017 at 6:47