How to power a GSM Modem from a Li-Po battery?

Question

I want to integrate an AirPrime SL6087 GSM modem into a design, this is my first try at working with a GSM module. The datasheet states that the modem draws 2A for 1mS every 4mS (max) at 3.6V, the problem is that I intend to run this project from a single Li-Po battery and want to use a buck-boost regulator so I can keep the unit alive for as long as possible (running it down all the way to 3V). The only regulator I can find capable of this is the LTC3113 but there doesn't seem to be stock anywhere. The LTC3112 is also a possibility but it's 2.5A output current is rated at 5V input not down to 3V.

My question is this, there's a lot of small GSM products on the market, how do these go about keeping the power supply stable while running on a single Li-Po battery? I have this nagging feeling that I'm over complicating this...

The complete datasheet for the module is only available after logging in, registration is free however. Here's a quick recap (page 20):

In connected mode, the RF Power Amplifier current (2.0A peak in GSM/GPRS mode) flows with a ratio of 2/8 of the time (around 1154uS every 4.615mS)

Answer 1

My experience (across four different cellular modems, both GSM and CMDA), is that they are intended to be run directly off the battery. For the AirPrime SL808x series (as an example as I haven't worked with the SL6087), VCC_3V6 (which is used for the power amplifier), is rated at 3.3v min to 4.3v max, with a typical value of 3.6v -- this closely matches a nominal 3.7v Li-Po battery which may reach 4.2 volts on a full charge.

The digital section of the cell module (logic levels), on the other hand, runs off a 1.8v rail which is internally generated. It is also output on the VREF_1V8 pin (1 ma max).

So you don't really need a high power buck-boost regulator at all. If you need more than 1 ma to power your logic level conversion circuitry (assuming 3.3v logic levels elsewhere), you will need to provide an LDO regulator to generate 1.8v from the 3.3v rail.

Answer 2

In summary, you need to go from

Vin = 2.7 V to 4.23 V

to

Vout = 3.6 V  
Iout_peak = 2 A  
Iout_average = 500 mA.

If you use a big capacitor at the output, able to provide the 2 A peak without loosing too much voltage, you'll be able to use a buck-boost regulator that provides 500 mA average, at the output. For instance, the LTC3536 could work for you (which may deliver up to 1 A at 2.7 V input), and it's in stock.

Choose the output capacitor this way:

\$ C=\dfrac{I·\Delta t}{\Delta V}=\dfrac{2·0.0012}{\Delta V}\$,

where \$\Delta V\$ is the voltage drop you allow yourself to have, at that 3.6 V output. The smaller the drop you permit, the higher the capacitance you will need. For instance, to have a drop of only 0.2 V, you will need 12 mF (=12000 uF). Also, the capacitor must have a low ESR, to be able to deliver 2 A to the load.

Since the LTC3536 has an adjustable output voltage, you probably want to aim at a Vout slightly higher than 3.6 V, to compensate for the drop in the capacitor, or even at a clearly higher Vout, and use a 3.6 V LDO (but able to provide 2 A peak) between the big cap and the load, to completely hide the drop at that capacitor. That will allow you to use lower capacitances.

Answer 3

There are many regulators which are capable of buck-boost regulation. Don't limit your search to just buck-boost type regulators, you can also do this with a SEPIC or flyback topology. This is also easy to do with an inverting topology: Connect the battery such that it provides -3 to -4.2V, and you can output most any positive voltage (including 3.6V) which you'd like.

If you're interested in an LT part, it looks like there are a lot of options. I did a switching regulator search for 3V minimum input, 5V maximum input, 3.6V output, and 2.2A max current which produced 644 results - and those offerings are all from one company!

These offerings notwithstanding, you have some large and fast current spikes from 2mA of sleep current to 2.2A of active current. It's likely that both your battery and your regulator will have a hard time adjusting that quickly. Therefore, I'd recommend boosting (always, so you just have a simple boost regulator) to 5V, putting a great big capacitor (say 220 uF) on this rail, and using an LDO to output 3.6V. The capacitor will still droop under the sudden load, but the LDO will stabilize the output and the switcher will kick in before the capacitor droops too far. It's only a 1.4V drop, so your efficiency is still pretty good and power dissipation isn't that bad.

Since this is your first try, it's probably a good idea to make this one a little overbuilt and later you can iterate the design to get the smallest, most efficient high-frequency buck-boost converters that can handle this load behavior.

Answer 4

For a similar project (battery powered only, gsm, autonomy time ~10days) I'm thinking of suppling the gsm (3,2 to 4,8V input, 2A spikes) direct from the batteries (3AA backtoback=4,5volts & ~3Ah) with only a big capacitor. This way I won't have any consumption from any LDOs, the battery is able to supply as much current as it has and the large capacitor (maybe supercasitor) will help. Great attention will be given to sleep modes.

Answer 5

100% Practically working Solution:-

Actually GSM modules are very compatible with single cell li-ion / li-polymer batterris as there operating voltage is 3.6V-4.2V but not 5/3.3. You can directly connect Battery to GSM MODULE.But here important thing is you have to charge/dischage battery.Simply you can use MCP73833/4 Stand-Alone Linear Li-Ion / Li-Polymer Charge Management Controller(see datasheet http://www.microchip.com/TechDoc.aspx?type=datasheet&product=mcp73833). It will work very easily.Keep in mind Vbat track should be enough broad with PTH.