Hey, that simple threat becomes very interesting battery knowledge source!

Most of ndi post is very useful for me, but I think that shadowjk got a point here. If you let the phone charge overnight, in fact it is just charged to full in few hours, then discharged to 95%, then charged again to 100, etc... It's true that RGB diode is green all the time, also battery-eye show it as "full" part of cycle. Unfortunately, N900 doesn't use power directly from charger when plugged - it just use the battery, charging in the same time. When battery reaches "full", power is cut off until battery drop to 95% or so. You can easily monitor it with bq27200 kernel-power module, by:

cat /sys/class/power_supply/bq27200-0/current_now

...when plugged in, you got value under 0 (for example, "-2475") which is exactly (charge current)-(current used by phone to "live"). When battery reach "full", current_now report just positive (i.e used actually) current - exactly the same as if plug is disconnected = non charging.

So, that would mean You got 10-20 cycles every night...

Also, after short logic check i totally disagree with Your calculations about powercycle drain. Precisely, your calculations are correct, but upside down - it is not 180W in couple of seconds, but 0.180W = 180mW!

Look ndi. N900 battery rounded up got 5Wh - You're right at that. It's capable to safe deliver for our needs 1315 mA constant current @ 3.8V (rounded from 3.3V-4.2V, doesn't matter for our calculations) for time of 1 hour. Also, it can deliver 657 mA constant current for 2 hours, etc. Every fluctuation of current change total time, but after C (current)*V (voltage)=5W or higher, battery is already at 3.3V and is discharged, aka "dead".

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.

Using known values, we want to get 3% power during, lets say, 15 seconds of bootup - it is W*3/100=C*V, V is constant four our calculation (3.8V), W after /h is just 5. 5*3/100=0.15.

We want C (current), so C=W/V (or, if you prefer that, Ch=Wh/V, doesn't matter at all for us now), we get 0.039=0.15/3.8. Our 0.039 is in A, so it is 39mA. When we don't use /h, we get 3% of battery power = 39mAh. If You want to check for mistakes, 39mAh *33.33 = 1299.87 mAh - thats our 1315 mAh capacity rounded down.
@ 3.8V it is 5Wh.

So, we already know and proofed (if anyone understand my not so self-explanatory calculations ), that 3% of battery power is 150 mW (yea, it's logical, but we wanted hard proofs, doesn't we?). 10%, then, is 500 mW, 10% battery capacity use is just 500mW. No need for blue smoke and other things.

---

Also, if we treat serious Your calculations about 10% beign 180W, that would mean we got 1800W at 100% - real beast. That would allow us for powering up high-power driller for an hour, just using our little battery, no need for wall plug I got nothing against that idea, but for now we must be happy with 5W thing...

I tried to explain calculations as easy as it can be, but I'm again after sleepless night, so sorry for any hard to understand shortcuts. We can pull here dr_frost from "3000mAh battery mod" thread, he use high W batteries in his RC machines, so maybe he will be able to explain it more user-friendly

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.

It is true that the battery module does not control charging. This is a good thing, because that means any observations we make through it do not change bme's behaviour.

bme's charge control actually seems to run independently of bme's battery meter in a number of ways.
First, when charging it will display "battery full" before charging actually stops or reaches full. Then it will display green forever even if it charges and stops charging.
Secondly, if the battery is near full, like 90-100% in real terms (not the percentages reorted by bme, hal and so on), and you connect charger, it will often go straight to "Battery full" green led. This doesn't mean it's not charging, it does charge, and typically the charge rate is 2 - 3 times higher than when green light is triggered during a charge started from lower initial state of charge.

How "full" the battery is charged depends on a few factors:

Before PR1.3:
CC/CV @ 950mA to 4.2V, Charge termination at 50mA during CV stage.
The charge chip current sensor can not distinguish between system load and current going into battery. (Check leaked schematics and you'll see it).
Additionally there are some PCB losses between charging chip and system+battery amounting to about 0.125 Ohms.

Let's investigate two scenarios here: Idle phone, and not quite idle phone.
An idle N900 has a power consumption of about 6mA.
This decreases the termination current (at the battery) to 44mA, while the charger chip is outputting 50mA @ 4200mV.
By ohms law, The voltage drop across .125 Ohm is 6.25mV. Let's call it 6. The voltage at the battery will then be 4194mV.

So, charge ended at: 4194mV, 44mA.
Manufacturer recommended charge is CC/CV to 4200mA @ 0.7C with C/20 cutoff. Or, 924mA, 66mA.
The end voltage was lower, but so was the termination current, meaning more power was put into the battery.
Very close to the specs recommended by battery manufacturers anyway.

For the highly loaded N900, let's take 350mA system load. This is typical of heavy browsing on edge or wifi, or heavy IM activity on 3g, screen on.

Charge will never terminate, as 350mA > 50mA.
Voltage drop due to losses in path between charger and system: 0,35A * 0,125 Ohm = 43,75mV. Let's call it 44mV.
The battery voltage eventually stabilizes at 4200mV - 44mV, 4156mV, with the charger chip float charging it indefinitely.

The end voltage is lower, but then it has floated at that voltage for much longer than in the previous example.

After PR1,3 (but not with some community kernels) there's a minimum system load of about 60-80mA, which means that the charge never ends.


From a battery longevity point of view, time spent at 3.7-3.9V degrade the battery the least. Time spent over 4.1V accelerate aging. Time spent at 4.15 trickle charging (charging never stops) accelerates aging even more.
Likewise, time spent at low voltages also accelerate aging.

What this mean is that the battery is happiest when it's at 40-80%.

Anyway, after 100 cycles (where a partial cycle from 60-40-60 would be counted as one fifth of a cycle) and spending 8-12 hours a day on the charger, my battery tests out as 1261mAh (was 1300 as new). Seems to hold up much better than N8x0 BP-4L and BP-5L batteries

I actually regret not having a spare charger I can destroy by hooking up a multimeter.

Agree with shadowjk, i will only clarify that indeed i used bx module only to measure, not to control charging - as joerg mentioned in other thread, bx module can't be used for that caused it occupy hardware doesn't allowing other monitors (also bme) to work in parallel.

sorry, but i can;t agree with logic of that sentence - i feel that either You or me is very wrong in understanding what other mean.

Calculations i posted on last post were NOT for actual capacity use, but for our teoretical 3% or 10% capacity (whatever, i calculated for both). I proved that using 10% of battery power during 15 seconds boot time WILL NOT produce 150W or any other similiar high value, but 150 mW.

Also - maybe I fail to understand it, what is possible - battery that is 5Wh can deliver 5 W after a 1 hour of work with 1C drain. That doesn't mean that it can deliver 10 W for 0.5 hour! You multiply it when time decrease, what is wrong. Battery CAN work for an 2 hour if C is 0.5 etc. In Your calculation, when time decrease, C remain the same, what is wrong.

Thats also why battery charge for an hour if charged with 1C (with 0 internal resistance etc), for 0.5 hour if charged with 2C, etc...You can not just multiply Wattage, when time decrease, ignoring C decreasing. Thats why in your calculations 10% of battery power would produce 150-180W energy during 15 seconds, what is wrong. Just because You don't decreased C in calculations.

That is just plain wrong. Battery is 1250mWh so 10% is 125mWh so 125 mW PER HOUR. So 15 seconds would be 1/240 hours, 0.00416667 hours. now to get the Wattage used during that time when 125mWh was spent, we divide away time, which is 0.00416667 hours. in so we get 125 / ~0.00416667 mW which is 30000 mW or 30W.

EDIT: erm ok that ^^ is also just plain wrong. but if you replace all W's with A's it will be correct and multiply with some nice V, 3.9 or so, to get the W.

http://talk.maemo.org/showpost.php?p...&postcount=657

Post by joerg-rw, currently working on jrbme that is replacement for closed-source bme, for HEN project.

He stated that 1W won't exceed 30% of battery capacity... Because all capacity is 5W, You can call it 5Wh if You want to count with mAh, but that doesn't matter

I link this only because joerg-rw is very respected community member, and maybe that will help sceptics to understand that our belowed BL-5J is beast capable of store high W values In fact, that citation is not needed, cause basic math skills tell you than when on both side of "=" you got same multiplier - h in this case - You can drop it by dividing both sides /h.

sorry to call basic math skills - i don't want to offend anyone, but I;m just running out of ideas how to prove basic things that has been proved by calculation itself...

BL-5J can deliver max POWER of 5 W (no matter of time which it was used) what is CAPACITY of 5Wh independent of voltage (theoretically, You can use transformer to obtain 5Wh with for example, 12V, in 0 power loss during transforming, what is of course theoretical only), or 1300 mAh at 3.7.

Power != Capacity

Capacity is constant (unless our battery degrade ;P )

10% of 5Wh battery is damn 0.5Wh, and using that 10% of capacity during 15 seconds is using damn 0.5W = 500mW in 15 seconds.

I don't know how to write it in simpler words.

i have just been browsing through an can't really be sure what has been concluded here but some ref here from someone that uses battery's to a greater extent then normal

the 1320mAh battery can in no way supply over 24W they are not built for high power delivery but high capacity and the 24W is based on 5C which they cannot deliver anyway, the circuit can defiantly not deliver much more the 5-10W

On the charge cycles, well you will already feel a slight drop (very low tho) from 5 cycle, and after 300 cycles it will have lost some % capacity, and here it really depends on the quality of the battery.

I have not been nice to my EV Ciao battery through the last year an it has lost a little over 20%, next time (in the near future) im getting LiFe instead.

Ok, everyone:

@dr_frost_dk: "the 1320mAh battery can in no way supply over 24W they are not built for high power delivery"

You are correct, I said ideal battery on purpose back there, even mentioned resistance.

The proposition was, booting the battery decreased charge by 10% as shown by the indicator. I said it can't be, because bootup is 30 seconds and if you killed 10% in 30 seconds that would be many-a-watt, so it can't happen. For many reasons, one being battery can't deliver, another being the phone would fry. Nobody is claiming it can happen. Au contraire.

@mece
That took about 10 minutes to understand, but yes, you are correct. Roughly 30A, or ~100W, in 15 seconds, that's 180W for 10 seconds.

@shadowjk
My measurements disagree with your last part, but I have a custom kernel. I'm fairly sure that my phone doesn't charge battery at all if left connected after end-of-charge, because the consumption slope is identical when connected and not connected.

Will dig further.

@Estel:
"A candle that burns twice as bright burns half as long". We can all agree to that. 5Wh is 5Wh and nobody is changing that (we wish).

5Wh = 5W in 1h
5 Wh = 10 W in 30 minutes

You said:
"10% of 5Wh battery is damn 0.5Wh, and using that 10% of capacity during 15 seconds is using damn 0.5W = 500mW in 15 seconds."

You again mix hours and seconds like they are the same.

10% of 5Wh battery is damn 0.5Wh(1), and using that 10% of capacity during 15 seconds is using damn 0.5W(2)(see what you did here?) = 500mW in 15 seconds.

Herein lies your problem. You said that 10% of 5Wh is 0.5W. It's not. It's 0.5 Wh. Since 10% of a kilometer is not 0.1 oranges. You should keep the measurements to the end of the equation.

You used 5Wh (at 1), and said that 10% is 0.5W (2). It's actually 0.5Wh, which is 0.5W FOR AN HOUR. Not 15 seconds. 0.5W for an hour, but delivered in 10/15 seconds.

If you still have issues with this, I suggest you take the simplistic approach to physics. Go with metric basic measurements. They are volt, ampere and second. I'll use the extended Watt as V*A.

A 5Wh battery delivers 5W for one hour, that's 5W for 3600 seconds. The measurements is watt hour, NOT watt per hour, mind you, it's not a division. Watt per hour implies flux of energy, watt hour is capacity.

A faucet delivers water at 1 liter per hour, meaning that left alone for an hour, it produces a liter. That allows for a bucket that gets filled in one hour to be one liter (1l/h*1h simplifies to liter). So, Wh is capacity, and Wh per time (hour) is W.

We simplify to metric, so the equation becomes:

5W * 1h = 5W * 3600 seconds = 18000 Ws

Note that when you do math with measurements, measurements keep the operands. So, two apples times two oranges makes four apple*orange. We write that as apple orange.

When you multiply watts and seconds, the result is watt second, or Ws. A grand total of 18,000 Ws. That's it. 18000 watts second. If you take 18000 seconds (5 hours) that's 1W. Makes sense. Also, if you take one second, it's 18000W. Whole battery.

10% is 1800Ws. If it boots in 10 seconds, it's 180W dissipated power, vast majority in heat. It'll glow like a light bulb (well ... -ish).

My original post read:
"A battery is typically over 1000 mAh at 4.2 volts, that's 4.2 watts. N900 is 5000 mWh, and 10% of that is 500 mWh. That's 500 mW for 3600s, that's 1800Ws, and on a 10 second powerup you'd need to eat 180W (my math may be off it's really late)."

Considering the hour, I was surprisingly accurate.

what ndi means here is that it uses 180Ws (Watt seconds) not 180W, he just forgets the s at the end (e.g. Ws)

Because as i wrote the BL-5J cannot deliver much in terms of POWER, even in a 1 second spike you wont even get 20W

Now i don't know why he insists on using Ws instead of just Wh.
This then calculates to 10% being drawn in 30 sec = 1320mAh / 10 = 132mAh
In watts thats 132mAh * 3.8V = 0.5Wh that have been draw

But the fact is here that this is equal to 12C for 30sec which i can assure you that the battery or circuit can't handle

Now i know that it does not use 10% on a reboot/boot, because if you measure the battery in REAL TIME then you will see the voltage go up when it is done with the boot, e.g. it "recovers" some %, the real drain is more like max a few %, but why keep speculating, someone here please charge their battery full up, and then do either
10 Boot
20 Wait 10sec after idle
30 Shutdown
40 GOTO 10
or
10 Reboot #after full charge
20 Wait 10sec after idle
30 GOTO 10
and then report back how many boots or reboots the 1320mAh battery can handle, or any other battery you have.