# How to provide enough current for my GSM circuit, and still use a LiOn battery?

I am very new to electronics.

I am building a simple GPS tracker prototype circuit that uses a 3.7 volt 500mah lithium ion rechargeable battery.

The circuit has 2 major components: a simple microcontroller that interfaces serially with a SIM908 GSM & GPS chip. When the SIM908 chip is active but not transmitting or receiving data, it uses only 80ma of current. But when I instruct the microcontroller to request the SIM908 to send or receive data, the SIM908 requires 2 amps of current during that period of radio transmission, then it goes back to utilizing only 80ma of current.

My idea is to send GPS location every 5 minutes to a website using the GSM cellular data. This solution is all working fine already, but the problem is that it drains the battery in about 2 hours.

I understand that could go for a much more powerful battery, but my project requires a small form factor, so the 500mah is the largest I can go, and thus, I need to stick with it.

So, maybe I am asking something impossible here, but my question is: does anyone know what would take to keep the same 500mah battery, still provide the circuit something between 3.3 and 3.7 volts, but have a way to increase amps to 2A during those current peaks for a lot lot lot more than 2 hours??... maybe 24 hours... would it be possible?

Hope this question makes sense, and someone can direct me to either a better solution or a more efficient way to provide power and current for many hours.

• Have you tried turning the GSM chip off? Aug 20, 2014 at 23:57
• What (circuit) is deciding when the battery is "dead"? 2A from a 0.5Ah battery is 4C discharge rate, albeit in short bursts. Could be the battery voltage is dipping below a comparator threshold (or similar) under such heavy load, and being turned off when it's only really too low a voltage for tens of milliseconds and recovers between bursts. Some adjustment (slowdown) to the low battery detector may be in order. Aug 21, 2014 at 3:02
• 80mA from 500mAh battery gives estimated runtime of 500/80=6.25 hours. Getting to 24 hours runtime will take either 4x larger battery, or reduction in average current consumption by using or creating lower power modes, as in Dan Laks answer. Aug 21, 2014 at 3:06
• You can reduce the impact of high-current spikes on the battery by adding a supercapacitor in parallel with the battery. However, capacitors tend to be very large and expensive compared to battery on a per energy stored basis. You might try one to see if it helps, but you may find it increases the size beyond what you have allowable, or that using the same volume increase for a larger battery gives better results for the money. Aug 21, 2014 at 3:09