Radio test with graph

[kevm14]

Using the Battlog program, I ran benchmark tests on the Sprint 1.11.00F, USC 1.07.06F and Telus 1.04.05V. I do not yet have a complete set of data but I think I can conclude a few things. Signal strength varied between 95 and 97% and did not seem to correlate to a particular radio. I show only watt-hours because that takes into account battery voltage, assuming it was calculated correctly. Current draw can be misleading without knowing voltage but if someone would rather see the average current draw (or mAH) I will make a graph with those values instead.



One bug I noted in Battlog is that if you don't save the log right when it finishes, the phone data becomes invalid, as it continues to compute the values and includes them in the saved log. That is what happened to the second Telus test and why I didn't include the phone portion.

Are my values in the ballpark, particularly among people with "Excellent" battery life?

[kevm14]

Not sure why I can't edit. I meant to add the following:

I also read somewhere that the current usage goes up as the battery drains and the thought at the time was some kind of defect. It's not. It's just the way electricity works. For a given power level, you multiply voltage by current. If the voltage drops (like because of a draining battery) the current must increase to sustain the required power output.

[Rusty Duck]

Interesting but I am not sure I understand your graph.

[kevm14]

What do you not understand? How the data was collected or the way it is displayed in the graph?

[kevm14]

Maybe move this to the CDMA upgrades section.

[mixxy]

wow, dont take this the wrong way but someone was boredd =p hehe

[LoneSnark]

Did you repeat the tests and average the results? If so, then your results do seem significant. However, if you did not then it seems likely that the significance was lost in the system noise.

Your views on voltage and current are absolutely correct. The phone compensates for voltage drops by drawing more current in accordance with P = I x V. While transmitting, the transmitter chooses a power level, P, to transmit at and the dynamic regulator chooses an I and V to satisfy it.

That said, the system is not as simple as that. dynamic (or switching) regulators are rarely that simple as they introduce their own efficiency on top of their energy conversion. And it just so happens that dynamic regulators tend to become less efficient as input voltage drops. What this means is, as battery voltage drops the power drawn from the battery increases, not just the current.

Psystem = Pbattery x %efficiency

So, if the system is consuming 1 watt and the efficiency is 90% then the battery is supplying 1.11 watt. So, if the battery voltage is 3.8V then the current supplied would be 292mA.

However, if the battery drains and voltage drops to 3.7V and the efficiency drops to 80% then the battery is supplying 1.25 watt, or 338mA, just to supply the system with that same 1 watt.

[kevm14]

Each test is displayed singularly. However, I could average the 3 1.07.06 results as well as the Telus results. You can kinda see what it would be just by looking at it, though. I need to do more tests with 1.11.00F but I also want to flash the stock 1.03 back and test that 3 times.

[kevm14]

New graph with averages.

[izanagi]

label your axes!

BAD STUDENT!