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I'm designing a small telemetry system that will take measurements every few hours and send the data through 3G at the end of the day. The target battery life is 1 year.

The GSM modules is an XBee 3G. The module draws about 700mA at 3.3V when transmitting. That is the averaged current, I don't know how is current waveform composed.

  • My first idea was to use 4 AA alkaline batteries and two 50F supercapacitors with a current limited charging circuit, so I won't draw more than 150mA from the batteries and allow the GSM modules to run just from the energy stored in the supercapacitors. The problem with this is the cost of the supercapacitors and circuit, the large space it ocuppies and the time is takes to load the caps (about half an hour if I'm not wrong).

  • My second idea was to use 4 AA Energizer Lithium batteries. This batteries allow me to draw up to 1A without significant capacity reduction and they are well known so you can buy them almost everywhere.

  • My third idea was to use a ER34615 battery. The energy density is really good and I could easily achieve more than one year of battery life. The problem is they are also rated for currents up to around 300mA which is very low. I have seen some brands like Ultralife which offers this ER34615 with currents up to 2A which is perfect for me but they seem to be rare and I can't find them in my country.

I really like the idea to use a ER34615 cell. Is there any way I could use it without destroying the battery capacity due to the large currents? I though about using a smaller supercapacitor in parallel with the battery but if the average current is 700mA there's not much a parallel capacitor can do. How are GSM modules usually powered in this applications where battery current capability is so limited?


EDIT 1:

Ok, to properly test the current drawn from the XBee 3G I connected to a 3G network and covered the antenna in aluminium foil and powered the module with 3.5V to get a -90dBm signal strength.

I sent 1000 of data (plus HTTP header overhead) with a POST request to an echo server, so I receive 1000 bytes (plus HTTP header overhead) of data back.

During the data transfer the current was around 320mA, no big current spikes as in 2G connections. The 320mA were pretty much constant during a few seconds, after that, the module sits in idle at around 80mA also with no current spikes. I don't really know what to think, because Digi has this power consumption table:

enter image description here

I don't have a way to know the transmission power but having such a low signal level and being at 320mA makes me think how they measured those 700mA.


EDIT 2:

I dropped the voltage to the absolute minimum of 3V and covered the antenna even more so I get -95dBm now. The current stays about the same but only sometimes the current goes to about 800mA during the transmission (a seconds or less). So this 700mA advertised by Digi seems to be the worst case scenario. With relatively good signal levels it's hard to go above 400mA, at least in my network conditions.

The ER34615 are rated for 300mA maximum while dropping the capacity to 9Ah, although this is rated at a continuous 300mA during the battery lifespan. In worst cases a 800mA current drawn from the battery for a about a second can lead to significant voltage drops unless I add some beefy capacitor? Is it possible to calculate how much energy I would be wasting from the battery due to the internal resistance when draining say 800mA during a second?

I think the Energizer Lithium may end up being a better choice because as the ER34615 discharges I may be using more than 300mA due to the boost converter and this will reduce even more the battery lifespan.

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  • \$\begingroup\$ Any time I have analysed the energy cost of the energizer lithiums , they have looked very poor. i.e the small advantage doesn't offset the huge cost. The only advantage seems to be when you have something that already uses AA's and you will pay anything to get a single set of batteries to run longer. \$\endgroup\$ – Henry Crun Mar 20 '18 at 2:47
  • \$\begingroup\$ @HenryCrun what do you suggest instead of the energizer Lithium? I'm looking for non-rechargable batteries capable of supplying enough current for 3G modems. \$\endgroup\$ – Andres Mar 20 '18 at 3:02
  • \$\begingroup\$ I suggest big alkalines eg D cells. Cheap, long shelf life, available everywhere. Below I suggest a way to keep step-up switch mode currents low, using a small lipo (ipod battery) as a cheap,big super cap. An alternative is to step down. Peak current = output current, and battery current = Iout*Vout/Vin \$\endgroup\$ – Henry Crun Mar 20 '18 at 3:37
  • \$\begingroup\$ An alternative is to step down. Peak current = output current, and battery current = Iout*Vout/Vin. You need at least 4 cells as alkaline batteries end life at 0.9 V. But 6 or 8 cells will reduce the current. With 8 cell, you would only draw 100mA in for 300mA out. You might want to look closely at the module. the current specs suggest it already has a step-down switch mode in it. Perhaps it will just run off 9V already? \$\endgroup\$ – Henry Crun Mar 20 '18 at 3:43
  • \$\begingroup\$ Yes, D alkaline cells are cheap, but running from a single cell can prove difficult because of the very low voltage I have to step-up so I would need at least two cells which takes up a lot of space and are pretty heavy. The problem is not only peak current, the module runs from 3 to 5V and the 300mA (800mA on poor signal conditions) are not peak current but a constant and steady current that lasts a few seconds. Energizer Lithium batteries price is not such a big problem in my case and they have great advantages like current capacity, weight and size. \$\endgroup\$ – Andres Mar 20 '18 at 3:57
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You should get a scope across a 0.1 ohm resistor in series with the module, and see what the actual current drain waveform looks like. Cellular systems are time-sliced, so a requirement for 700mA (peak) may be quite different to the 1 second average. Perhaps an 1000uF electro will be all you need.

Clever Li-Thionyl batteries are un-obtainium whenever you really need a replacement, I avoid them like the plague, unless they let you have a lifetime battery.

D cells have better $/J, and higher current than AA, if you don't mind the size. 75kJ - 21Wh, about 1/2 the cost of AA's here. Varta D cells give energy at 11Wh@600mA or 20Wh@150mA, so probably D cells can run 700mA. (but you get twice as much energy from the battery if you keep the current down)

You could consider a small lipo battery that can power your radio, and float charged by a step up converter from 2x D cells, or even 1 x Dcell. In this case something like a 300maH lipo is acting as a cheap super-cap. It always gets recharged slowly by the low-current switchmode between cycles. Its capacity only needs to be small - you just need it to have good peak current - but this is exactly what lipos have.

You can recharge alkaline to a limited extent, especially if you never discharge it.

Assuming 20J per upload (3V,700mA,10secs), you could do 10 uploads a day for a year from 1 D cell.

For an outdoor application, a small solar panel may significantly increase the lifetime of the alkalines. 75kJ = 21Wh. At 250 sun days*5hrs/day, you need a 17mW solar panel to keep the battery charged. A 40mW solar panel is 1" square

Ah-ha! I hear you say, but won't this just make the switchmode division of TI rich? Well you can just use you micro for the very low current flyback convertors needed if you want.

schematic

In this case, software is generating the flyback convertor pulses, and using the ADC to check Vsolar, Valkaline and VDD, and thus do the regulation. The L1,D2 switchmode efficiency doesn't really matter. The L2,D3 switch mode efficiency does matter, so probably D3 should be schottky and the current kept low so Von of P1 is small.

The key here is that the currents are very low. If the transmit energy is 20J, the L2,D3 switchmode only has to deliver 2mA for 1hour to recover the energy (and the alkaline battery is only delivering 5mA). L1,2 are very small and cheap.

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  • \$\begingroup\$ Thank you Henry for your answer! AFAIK 2G connections are time slotted but 3G connections are continuos so that should mean a continuos 700mA current consumption. In this Duracell PDF (goo.gl/iNrCig) draining 1 amp from a D cell gives about 10.5Wh. The 3G modules takes up most 60 seconds from power on to finishing transmitting the data. So, saying I have an 80% efficiency boost converter to elevate from 1.5V that is about 1.93mW used by the 3G module averaged in 24 hours. That gives me (10.5Wh)/(1.93mW) = 226 days. Not taking into account any other component in the system. \$\endgroup\$ – Andres Mar 17 '18 at 3:50
  • \$\begingroup\$ You also see the considerable benefit of getting the enrgy from a capacitor/lipo : the battery energy goes from 11Wh ->21Wh \$\endgroup\$ – Henry Crun Mar 17 '18 at 5:05
  • \$\begingroup\$ If 3G has a 2Mb/s data rate, and speech needs 10kb/s, is the radio transmitting all the time?Is the radio transmitting at the same time as the receiver is receiving? Probably the transmitter is not running continuously at all. You need to measure it. And you need to measure the integrated energy for a cycle - there will be a lot of waiting, not using much power I expect. \$\endgroup\$ – Henry Crun Mar 17 '18 at 5:10
  • \$\begingroup\$ you need to measure the current with a poor signal. Cellular systems lower the transmit power so all phones give the same signal at the base station, so if you are right next to a base station it won't be using much current. \$\endgroup\$ – Henry Crun Mar 17 '18 at 5:15
  • \$\begingroup\$ Yes, getting the energy from capacitors helps a lot but I wanted to see if there was another solution I was missing because the caps and the charging system is kind of expensive for the product (around 20 USD). Tomorrow I will measure the power consumption to see a real-world scenario. Thank you for all the help Henry :) \$\endgroup\$ – Andres Mar 17 '18 at 5:21

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