4 ms·

a typical plug in a home can provide 10A. So to charge the phone you will need some big connector, not a tiny microUSB like connector. Current capacity f

a typical plug in a home can provide 10A. So to charge the phone you will need some big connector, not a tiny microUSB like connector. Current capacity for conductors depends on conductor length as well as conductor size. Short runs of a given size have a higher current capacity than longer lengths due to the larger losses in larger lengths. Also, current capacity depends on the length of time the current has to be carried: a given length of a given conductor size will carry a larger current for a short amount of time compared to the carrying capacity for a longer duration. While a micro-USB connector will not suffice, the needed connector would not be that much bigger: a 3.5mm jack would suffice to carry 18A for a few seconds given a good socket. It would not be hard to design a plug/socket combination with a high-enough capacity which would fit in a phone. To recharge the phone once a week, I guess you will need a 20000mAH battery I recharge my phone (a Motorola Defy+) once a week. It has a ~1500 mAh battery, the thing is 5 years old. I don't think you'd need a 20.000 mAh battery to be able to keep your phone running for a week unless you plan to use it as a hand warmer.

Turing_Machine 10y ago
The original source is dead, so I can't tell whether these points are addressed or not. Yep, you'd probably need a connector with a positive mechanical clamping action rather than just a spring connector (as in a standard 3.5 mm jack or USB port), but that seems doable. Also, there's no reason you couldn't play internal series/parallel (or transformer, or DC-DC converter) games to charge the thing up at a higher voltage (and thus lower amperage) while discharging it at normal phone operating voltage. Nothing says the charger has to operate at USB voltage levels. Heck, why not charge it at full line voltage? (it goes without saying that you'd need to engineer it so there's no way that full line voltage could possibly reach the phone internals).
- lutusp 10y ago
  > Also, there's no reason you couldn't play internal series/parallel (or transformer, or DC-DC converter) games to charge the thing up at a higher voltage (and thus lower amperage) while discharging it at normal phone operating voltage. Capacitors aren't charged by voltage, they're charged by current. One does not apply a voltage to a capacitor, one applies a current. If you doubt this, try connecting a capacitor to a high-powered constant-voltage power supply. But first, step back. A capacitor's charging rate is determined by its value in farads and the applied current. Nothing else matters. One farad, one ampere, one second produces one volt and one joule of stored energy. Nothing could be simpler. If you raise the voltage expecting to see a faster charging rate, you will see the opposite, because a higher voltage for a given power rating means a lower current, and a lower current will require more time to charge the capacitor. > Heck, why not charge it at full line voltage? Why not indeed? But don't be in the same room when this idea collides with reality.
  - jwatte 10y ago
    The voltage of the interconnect doesn't need to match the voltage at the capacitor. Buck converters for the win!
    - lutusp 10y ago
      Yes -- translation: "Capacitors are charged by current, not voltage."
  - Turing_Machine 10y ago
    Really? So C=Q/V isn't a thing any more? What other laws of physics have changed since I was in school? The important factor for charging time is how many joules/second you can stuff in, not how much current (coulombs/second) is flowing. You can pull a helluva lot more joules/second out of a wall outlet than you can a micro-USB connector.
    - lutusp 10y ago
      > What other laws of physics have changed since I was in school? Sorry for the late reply -- I'm traveling. What I said in my original post is quite uncontroversial. When charging a capacitor one must not think in terms of voltage (which is an effect, not a cause of charging) but of current. > The important factor for charging time is how many joules/second you can stuff in, not how much current (coulombs/second) is flowing. You're confusing cause and effect. Capacitor charging is accomplished with current -- voltage is an effect, not a cause. The voltage on a capacitor is the time integral of past applied currents. Want to change the charge level on a capacitor? Apply a current and let the voltage change in response. > You can pull a helluva lot more joules/second out of a wall outlet than you can a micro-USB connector. Non sequitur, the issue is how much current the capacitor can tolerate. And attaching a capacitor to a wall outlet will not work for multiple reasons. The ideal charging source for a capacitor is a constant-current supply set to provide a high, but not damaging, level of current, then when the capacitor's voltage limit is approached, shut down the current supply. The worst possible source for a capacitor is a constant voltage with substantial power available, which will destroy the device. As I originally said, "Capacitors aren't charged by voltage, they're charged by current." Concise, and quite accurate.
striking 10y ago
How I wish there were a 4G phone that could do what you describe. My OnePlus One has double that battery capacity (3100mAh) but I can really only make it last two days on that charge at maximum. Typically it needs to be recharged every day, or my usage habits need to change. So 20000mAh accurately describes my average weekly cellphone power consumption... and if I changed my habits a bit, I'd still expect to require 10000mAh. Please, if you know of a 4G phone that can last for more than two days, no matter how "dumb" it is (it could even be a flip phone, just as long as it can tether) I'd love to know about it.
- semi-extrinsic 10y ago
  FWIW, I have a Samsung A6 at work that I almost never use. It lasts 5-7 days on one charge; but this only works when display and network are off 99.5% of the time.
- cmdrfred 10y ago
  I can get 3, maybe 4 out of a OnePlus Two(Cyanogenmod 13) provided I run Greenify and only use it for text messages, emails (sync every 15 minutes), and occasional calls. Web browsing will kill it in 6-8 hours.
- monort 10y ago
  10000 mAh batteries is totally possible to buy, I use one from ZeroLemon. Of course, the phone is much bulkier, than with original battery, and phones with replaceable batteries are dying out.
jwatte 10y ago
A Deans T connector is rated at 50A long term. For shorter time, it can carry more. (It's a matter of current times resistance minus thermal dissipation.)

Turing_Machine 10y ago

The original source is dead, so I can't tell whether these points are addressed or not. Yep, you'd probably need a connector with a positive mechanical clamping action rather than just a spring connector (as in a standard 3.5 mm jack or USB port), but that seems doable. Also, there's no reason you couldn't play internal series/parallel (or transformer, or DC-DC converter) games to charge the thing up at a higher voltage (and thus lower amperage) while discharging it at normal phone operating voltage. Nothing says the charger has to operate at USB voltage levels. Heck, why not charge it at full line voltage? (it goes without saying that you'd need to engineer it so there's no way that full line voltage could possibly reach the phone internals).

lutusp 10y ago
> Also, there's no reason you couldn't play internal series/parallel (or transformer, or DC-DC converter) games to charge the thing up at a higher voltage (and thus lower amperage) while discharging it at normal phone operating voltage. Capacitors aren't charged by voltage, they're charged by current. One does not apply a voltage to a capacitor, one applies a current. If you doubt this, try connecting a capacitor to a high-powered constant-voltage power supply. But first, step back. A capacitor's charging rate is determined by its value in farads and the applied current. Nothing else matters. One farad, one ampere, one second produces one volt and one joule of stored energy. Nothing could be simpler. If you raise the voltage expecting to see a faster charging rate, you will see the opposite, because a higher voltage for a given power rating means a lower current, and a lower current will require more time to charge the capacitor. > Heck, why not charge it at full line voltage? Why not indeed? But don't be in the same room when this idea collides with reality.
- jwatte 10y ago
  The voltage of the interconnect doesn't need to match the voltage at the capacitor. Buck converters for the win!
  - lutusp 10y ago
    Yes -- translation: "Capacitors are charged by current, not voltage."
- Turing_Machine 10y ago
  Really? So C=Q/V isn't a thing any more? What other laws of physics have changed since I was in school? The important factor for charging time is how many joules/second you can stuff in, not how much current (coulombs/second) is flowing. You can pull a helluva lot more joules/second out of a wall outlet than you can a micro-USB connector.
  - lutusp 10y ago
    > What other laws of physics have changed since I was in school? Sorry for the late reply -- I'm traveling. What I said in my original post is quite uncontroversial. When charging a capacitor one must not think in terms of voltage (which is an effect, not a cause of charging) but of current. > The important factor for charging time is how many joules/second you can stuff in, not how much current (coulombs/second) is flowing. You're confusing cause and effect. Capacitor charging is accomplished with current -- voltage is an effect, not a cause. The voltage on a capacitor is the time integral of past applied currents. Want to change the charge level on a capacitor? Apply a current and let the voltage change in response. > You can pull a helluva lot more joules/second out of a wall outlet than you can a micro-USB connector. Non sequitur, the issue is how much current the capacitor can tolerate. And attaching a capacitor to a wall outlet will not work for multiple reasons. The ideal charging source for a capacitor is a constant-current supply set to provide a high, but not damaging, level of current, then when the capacitor's voltage limit is approached, shut down the current supply. The worst possible source for a capacitor is a constant voltage with substantial power available, which will destroy the device. As I originally said, "Capacitors aren't charged by voltage, they're charged by current." Concise, and quite accurate.

striking 10y ago

How I wish there were a 4G phone that could do what you describe. My OnePlus One has double that battery capacity (3100mAh) but I can really only make it last two days on that charge at maximum. Typically it needs to be recharged every day, or my usage habits need to change. So 20000mAh accurately describes my average weekly cellphone power consumption... and if I changed my habits a bit, I'd still expect to require 10000mAh. Please, if you know of a 4G phone that can last for more than two days, no matter how "dumb" it is (it could even be a flip phone, just as long as it can tether) I'd love to know about it.

semi-extrinsic 10y ago
FWIW, I have a Samsung A6 at work that I almost never use. It lasts 5-7 days on one charge; but this only works when display and network are off 99.5% of the time.
cmdrfred 10y ago
I can get 3, maybe 4 out of a OnePlus Two(Cyanogenmod 13) provided I run Greenify and only use it for text messages, emails (sync every 15 minutes), and occasional calls. Web browsing will kill it in 6-8 hours.
monort 10y ago
10000 mAh batteries is totally possible to buy, I use one from ZeroLemon. Of course, the phone is much bulkier, than with original battery, and phones with replaceable batteries are dying out.

jwatte 10y ago

A Deans T connector is rated at 50A long term. For shorter time, it can carry more. (It's a matter of current times resistance minus thermal dissipation.)