And that's why I said half charge. Small cycles don't count almost at all. Also, N900 charges to about 98 then doesn't charge until approx 86. Standby and self-discharge doesn't take that much overnight in normal circumstances. So, actually, N900 only chages one cycle overnight. Should it take longer, it tops it off again. My phone still says battery full if standby overnight.

And yes, battery does go away if phone is left plugged in for extended periods.

As for how much a cycle counts for a chemistry, I already linked battery university.

He does, just not under normal circumstances.

We already went through this in another thread. Also, if you decide to use a different metric, the battery module versus Nokia's, the we can't talk numbers, now can we? The module you use does not make the charge decision, bme does.

And that's why we don't use that module for reference. If bme says 90 percent and you plug it in, N900 says "battery full" and doesn't charge. If you want to see how bme works, ask bme via lshal.

You have long lost me. In my post I said that IF IT REALLY ATE 10 PERCENT in a few seconds, dissipated power would be in the order of 100s watts. Nothing above has anything to do with my post. You seem to be on about what it really eats, my post was about what it would eat if.

If you still think I'm off, please quote the segment so we are on the same page.

[QUOTE=Estel;1004643]Then, if we want to calculate how much power could be drained if powercycle would use 3% of battery capacity, we need to keep in mind that Wh = Ch*V, so (Wh = Ch*V) / h -> W=C*V - we need to divide by h, not multiply by h as you did.[QUOTE]

Power multiplied by time is capacity/energy. If capacity is the same, small time equals big power. If you used C as discharge factored as capacity per hour, then the equation is:

Capacity = power multiplied by time.
Capacity = volt times ampere times time.

if you factor hour, you get WATT equals Amp times Volt. Which is what we already know. You substituted C oddly. There is no Ch. C is an abstract-ish concept, in which a battery is drained in one hour, so A equals Ah. No operations using h is allowed.

Wh = W * h
Wh= V * A * h
5 Wh = 3.8V * 1.3A * 1h

If you make 1 to be 0.1, or 6 minutes, and volts is the same, then Amp jumps tenfold.

5 Wh = 3.8V * 13A * 0.1h

Same bucket spilled in half the time means twice the intensity.

So, if instead of 3600s you do 36 seconds, that's 100 times current, and 1.3 A becomes 130 amps, so, you know, some 500W.

You missted the h again. 5 Wh is 5 * 3600 watts second. That's 18 kWs. Notice the s, it's not plural. Same battery can power a high power drill for a second not for an hour.

1 Ws is 1 watt for a second, 1 kW for a miliwecond or a wris****ch for a year.

Assuming zero internal resistance.

ETA: Haha, forum censored wris****ch. Wrist Watch. Timepiece. Portable clock. Heh.