Power, Charging, Etc

[htl2001]

So I know there is a thread about charging over USB but things were a bit confusing in it

First off, everyone keeps referring to weather it will just be able to run without draining the battery on just USB, calculating the amount of mAh that the Pandora draws (estimated 400mA) vs the max (spec) USB output (500mA). However, I didn't see any mention of what voltage the Pandora's battery is running at, I know USB is at 5V.

Basically, the number I'm looking for is the total mWh, since that's a better way to look at power consumption/supply. I wanted to see how long this could power it/charge it for:
http://www.ladyada.net/make/mintyboost/
or any similar device

 

[Tom`]

So I know there is a thread about charging over USB but things were a bit confusing in it

First off, everyone keeps referring to weather it will just be able to run without draining the battery on just USB, calculating the amount of mAh that the Pandora draws (estimated 400mA) vs the max (spec) USB output (500mA). However, I didn't see any mention of what voltage the Pandora's battery is running at, I know USB is at 5V.

Basically, the number I'm looking for is the total mWh, since that's a better way to look at power consumption/supply. I wanted to see how long this could power it/charge it for:
http://www.ladyada.net/make/mintyboost/
or any similar device


The battery is 3.7V, I don't think that the Pandora could run solely off USB though - it consumes considerably more than 500 mA at full load (MWeston stated 710 mA at 900 MHz, but I don't think that included Wi-Fi or BT). Source

You could still use something like that to charge the Pandora, but it would be slower than charging it via the regular adapter. It might be more worthwhile to do a AA to 5V DC power supply.

 

[dj51d]

The Pandora battery is 4000mAh at 3.7V. If I did the math right, that would mean the charger has enough power to charge the Pandora battery approximately 50% once you take the charger's efficiency into account.

 

[htl2001]

the point is that 500mA at 5V is more power than 500mA at 3.7V (specifically, if 100% efficient conversion was possible, it would be about 675mA, at 80% like the device I mentioned, it would be 540mA)

Anyway, assuming the battery is 3.7V (I expected this, as that's the Li-ion cell voltage, its going to be some multiple of this) the device I linked can do the following:
using 2000mAh rechargeable Ni-MH batteries (4000mAh at 2.4V, 9600mWh)
80% conversion = 7680mWh after conversion
7680mWh/3.7V = ~2075mAh
which would be about 1/2 the capacity of the internal battery
somewhere in discussions about the charger it was said that it may get as low as 60% which puts it at 1556mAh (about 39% of the internal battery)
using the recommended formula of about 1000/(device mAh) = # of full charges, it would come out to 1/4 of a charge on a set of batteries

anyway I thought that was an interesting thing to look at, food for thought I guess. I used a pretty low capacity for Ni-MH batteries (I've seen 2500mAh ones commonly, and 1.2V/batt is the lowest they go to normally) to get a conservative estimate. I bet one of these could easily give you the equivalent of 1/3 to 1/2 of an extra battery

 

[Squidge]

Don't forget that rechargeable Ni-MH AA batteries, when charged fully are about 1.4V. They are about 1.2V when they they are flat and need a recharge.

 

[kattle87]

the point is that 500mA at 5V is more power than 500mA at 3.7V (specifically, if 100% efficient conversion was possible, it would be about 675mA, at 80% like the device I mentioned, it would be 540mA)

Never forget that the internal battery does NOT have any limit like 500 mA peak. Probably the limit is somewhere between 900 mA and 1300 mA. So peak power can be reached with battery but not with USB.
However, the pandora will have a DC in. I suggest you use it for recharging instead of USB. I feel like it will be more efficient.

 

[Polossatik]

This is basically similar to the solar dicsussion.
Best way would be to use a spare battery (and maybe make a charger for it).
Otherwise use the DC in ( 5v ) not the usb as input.

 

[htl2001]

Don't forget that rechargeable Ni-MH AA batteries, when charged fully are about 1.4V. They are about 1.2V when they they are flat and need a recharge.

Yea I know, I was using low-ish numbers. I wanted to see how it performed in pretty much the worst case senario, which is still pretty good. I mention at the end of my post about the voltage and mAh being low estimates.

This is basically similar to the solar dicsussion.
Best way would be to use a spare battery (and maybe make a charger for it).
Otherwise use the DC in ( 5v ) not the usb as input.



The device I mentioned would still only provide power at USB levels anyway, as that's its limit