Electronic – Measuring GameBoy power consumption

The calculation seems correct to me, but some points:

• Assume your device is going to use more than you expect. Depending on the circuit you're going to use, not only the modules consume power. It might be insignificant though, I'm not sure.

• The power consumption of the microcontroller is dependent on the oscillator frequency and the supplied voltage.

  Given is the consumption at 1MHz. For your application, this may or may not be too slow, depending on how much instructions you're going to need to read from the GPS module, process the data, send SMS, read SMS, etc. My guess is that 1MHz is not going to be enough, but it depends on your code and speed requirements. However, you cannot just adjust the frequency.

  On page 22 of the datasheet you linked to, you can see different power consumption graphs for different oscillator frequencies and supply voltages:

  

  ^{(http://www.ti.com/lit/ds/symlink/msp430g2553.pdf, p. 22)}

  In this graph you can also see that you cannot use 16MHz with any supply voltage. This is clarified on page 21:

  

  ^{(http://www.ti.com/lit/ds/symlink/msp430g2553.pdf, p. 21)}

  So as you can see, if you want to run the microcontroller at 16MHz, you're going to need 3.3V (see the "Recommended Operating Conditions" table on the same page), and you will be consuming about 4.15mA instead of 0.230uA * 16 = 3.68mA.

  Another note: it depends on the characteristics of the microcontroller that you're using (I don't know it), but running the microcontroller at 2.2V and the GPS/GSM at a different voltage might cause problems, since the input/output of the modules might not be compatible with the input/output of the microcontroller anymore. It's always safe to have both on the same supply to make sure you don't have this kind of issues, but it will also consume some more power.

• You calculation supposes a perfect voltage source. Real-world batteries however don't have a fixed voltage. The voltage will change over time, and alkaline doesn't have the best specs:

  

  ^{(http://www.stefanv.com/electronics/using_nimh.html)}

  Now this graph is an example for a 2Ah battery, but you see the idea. You're going to use two alkaline's to get the necessary 3V for the GSM module, however, directly after using the batteries the voltage will drop and the module won't work anymore. A better option would be to use three NiMH's, which provide 3.6V together and will keep the device running until the batteries are almost completely discharged.

• You say you want to turn off the GPS when unused. Yes, this is possible, but remember that the module needs some time before it has a fixed position, as it needs to look for satellites. I don't know the details of your application, but say you want to send an SMS every 10 seconds, then you cannot turn the GPS off in between.

  How much time you need to get a fix depends on your module and environmental circumstances like buildings and maybe the weather, I'm not sure. I built a device with a fairly old GPS module once, and that one needed about 4 minutes in open field to get a fix. Nowadays modules should be much better.

  Even though you can use the GPS module within a certain voltage range, that doesn't mean you get the same results. When I was building that device, I noticed I could get a fix slightly faster, in about 3 minutes, when using a supply voltage about 0.5V higher than the minimum. (In the end I didn't use it since I didn't need the minute and it would mean adding an extra battery, for which I didn't have space.)

On a very basic level you have it correct(or at least a reasonable estimate), but we can get more detailed and in depth.

First of all, 12v * 40 Ah gives 480 Wh NOT watts. Watt-hours, I.E. 480 watts for 1 hour or 240 watts for 2 hours etc. A technicality that doesn't change your calculations, but might make future calculations less confusing.

Your first assumption about lead acid batteries and not discharging more than 50-60% is, in general, a correct assumption. However, you call it a deep cycle battery, and a good quality deep cycle lead acid battery may tolerate up to 80% depth of discharge. Less discharge is always better, but the better deep cycle batteries do allow a lower discharge and still have a reasonable cycle life. The only way to know is to look at the data sheet from the manufacturer.

Your second assumption is that your battery will simply charge in a linear fashion. Although your calculations for how long it will take to charge your battery are an OK estimate, in reality a lead acid battery charges to 70% in 5-7 hours (if your solar panel can provide enough current) and the remaining 30% in the following 7-10 hours. Lead acid batteries follow a 3 stage charge algorithm (maybe 4 stages depending on your charge controller).

So in your battery if you discharged it 75% = 10a remaining. To get up to 70% need 18a (10+18=28a). 18A÷5h=3.6a/h which your panel is capable of and you said you have at least 6 hours of sunlight and it will take 5 hours to 70% so in the last hour the battery will continue to charge past 70%. So if you only have 6 hours of sunlight and your controller strictly follows the 5 hours to charge to 70% algorithm, then you will not fully charge the battery each day IF you discharge to 75%! In all likelihood your charge controller will take advantage of your panel's ability to provide more than the 3.6 amps and charge the battery a little quicker, but not much - there are limits.

Finally your router. Does it use AC power? If so, then you need an inverter which will convert the battery power to 120 volt AC which you will plug in your router. All inverters are somewhat inefficient and you will lose at least 15% of your solar power to the inverter. So add this to your calculation and you will find out how long your router will be powered.