Sick of Apple chargers? Solution: SupperCharger+, the Next Generation

[IanB]

I do not find any issue with the standard Apple chargers. My devices run for days between charges meaning I don't normally have to charge them in a hurry. When the battery gets down to about 30% I just plug them in overnight and they are ready by the morning. That said, if I do charge them during the day it only takes about 2 hours to get most of a full charge. Also, the times when you do want a super fast charge (like sitting in an airport lounge waiting for a plane) you are not likely to have your super charger in your hand baggage...

[JacquesBBB]

The 12 W charger does not do much better, maxing out at (IIRC) 1.8 A at (from memory):

        1.8A

V+:  5.06V - 5.08V (at load)
D+:  2.75V (Note: 2.75 D+/2.75 D- is the 12 W charger indicator config)
D- :  2.75V

Are you sure ?

I just made an additional test, by using pots at D- and D+, and  for  D- = 2.75 V current remains below 1 A.

This is also what can be found  here
http://www.righto.com/2012/10/a-dozen-usb-chargers-in-lab-apple-is.html

with a lot of additional information on this topic.

[JacquesBBB]

I do not find any issue with the standard Apple chargers. My devices run for days between charges meaning I don't normally have to charge them in a hurry. When the battery gets down to about 30% I just plug them in overnight and they are ready by the morning. That said, if I do charge them during the day it only takes about 2 hours to get most of a full charge. Also, the times when you do want a super fast charge (like sitting in an airport lounge waiting for a plane) you are not likely to have your super charger in your hand baggage...

By the way,  I participate to this discussion because I   have interest in building USB  PSU that is adapted to  Apple devices, but I have not much problems with the USB  PSU of  Apple and  I had no failures on Apple Ipad PSU.
I do not think the  self made PSU  could be a large gain  with respect to the original from Apple.

I have solved  the problem of fast charge  (when I need to go and realize that my ipad is empty). I have a
power bank containing 4  18650 batteries that  can provide 2 A and charge a full ipad.
So if I need to go,  I just hook the ipad to the powerbank and put both in my bag.

Nevertheless, I must say  that  I just checked  the output of a  10W ipad  charger, and had 1.4 A, while I had 1.95 A on my PSU when set to 5.2V. So there is place for small improvement.

I am much more reserved on the  Apple PSU for  macbook Pro. I have experienced many failures with those. Moreover, there are not easy to repair. but I would not make a PSU for macbook pro as I have not yet understood  the way the magnetic connector works. That would be another story ...

[IanB]

I have solved  the problem of fast charge  (when I need to go and realize that my ipad is empty). I have a
power bank containing 4  18650 batteries that  can provide 2 A and charge a full ipad.
So if I need to go,  I just hook the ipad to the powerbank and put both in my bag.

There is a risk that power bank could be confiscated by airport security (due to lithium battery air transport regulations).

[amyk]

Most of the quick-charging damage is done at high state of charge (over 4.0 volts), so if you only charge your phone to approx. 70%, probably no extra damage is done, even if the current was 20-30% higher than normal. Very quick charging requires an additional current curve in addition to CC-CV, as too much damage is done near the CV voltage at high rates. For example, a cell can be fine up to 1.5C charge until 3.9V, then linearly derating to 0.5C until 4.2V, then CV.

Perhaps a simple 4.2V supply and current-limiting resistor would actually give the most "gentle" charge? The curent would be proportional to the charge voltage, being the highest at low charge and then tapering off gradually as it approaches 4.2.

It is nearly impossible to build a device that can remote sense at the Lightning connector's Vcc. This is one of the challenges I will be facing.

Hack apart a connector (carefully!) and solder a sense wire to the Vcc pin. You can also buy raw connectors and breakout boards from various Chinese shops - see this page for some more interesting details:

http://ramtin-amin.fr/tristar.html

I got tired of reading the idle speculation, so apologies if I missed this, but I went digging through teardowns and checking part numbers last year because I wondered about Apple's choice of charge controller. From that research, I learned that it looks very much like Apple iPhone and iPads made in the last 3+ years use switch-mode chargers.

Do you have any part numbers to show? The numbers posted above suggest that the current flattens off at around 5.1-5.2V (where is this measured?)

[Richard Crowley]

So you mean to say that the built-in charging circuit, rather than intelligently reducing the current load as voltage increases to maintain a constant output voltage (and current), burns it off instead (e.g., like a Zener)?

Your question is flawed.
Yes, the built-in charging circuit DOES "intelligently reduce the current load as voltage increases to maintain a constant output voltage (and current)." The reason for this is to protect the battery cells to prevent overheating, fire and explosion. And to comply with the ideal charging profile for whatever chemistry of the cells under charge.

As for HOW it regulates, yes SOME charger circuits probably use analog pass elements which dissipate the unwanted power as heat.  And likely more modern gadgets use switch-mode regulation (which, BTW, care even LESS how "regulated" their raw input power is since they literally make a "hash" of it.)  Those are the ONLY two ways of accomplishing power regulation that I am aware of. If there are others, I'm sure someone will remind us.

[SharpEars]

The 12 W charger does not do much better, maxing out at (IIRC) 1.8 A at (from memory):

        1.8A

V+:  5.06V - 5.08V (at load)
D+:  2.75V (Note: 2.75 D+/2.75 D- is the 12 W charger indicator config)
D- :  2.75V

Are you sure ?

I just made an additional test, by using pots at D- and D+, and  for  D- = 2.75 V current remains below 1 A.

This is also what can be found  here
http://www.righto.com/2012/10/a-dozen-usb-chargers-in-lab-apple-is.html

with a lot of additional information on this topic.

What device are you testing? If you it's an iPhone 5s or lower, then the most you're going to pull from 5 V is 1 A. If it's a device that is capable of charging at a higher current (e.g., an iPad that can take on 2A) then the reason lies in the D+/D- lines not staying above 2.7 V when the the (relatively small) load is placed on them by the device being charged, as part of determining the type of charger you are using. You can try using smaller resistors for your ladder (in the tens of kOhms) and see if that makes a difference.

Also, for my testing I not only set the D+/D- lines to 2.7 V, but I short them together as well. That is, I create a ladder divider using a single pot to produce the 2.7 V and then I connect both data pins to that output. This may be the difference between my test and yours. The link you provided comparing various chargers does not mention the Apple 12 W charger or its configuration, but I was able to find this on "some Chinese web site" and confirm it via a TI spec page for one of the new ICs they are producing for use in 12 W universal chargers.

[SharpEars]

So you mean to say that the built-in charging circuit, rather than intelligently reducing the current load as voltage increases to maintain a constant output voltage (and current), burns it off instead (e.g., like a Zener)?

Your question is flawed.
Yes, the built-in charging circuit DOES "intelligently reduce the current load as voltage increases to maintain a constant output voltage (and current)." The reason for this is to protect the battery cells to prevent overheating, fire and explosion. And to comply with the ideal charging profile for whatever chemistry of the cells under charge.

As for HOW it regulates, yes SOME charger circuits probably use analog pass elements which dissipate the unwanted power as heat.  And likely more modern gadgets use switch-mode regulation (which, BTW, care even LESS how "regulated" their raw input power is since they literally make a "hash" of it.)  Those are the ONLY two ways of accomplishing power regulation that I am aware of. If there are others, I'm sure someone will remind us.

So, given that the iDevices use the latter form of circuit (i.e., switch mode regulation) as several people here already pointed out, it would make sense for the current draw to decrease as the incoming voltage increases. It does the exact opposite (i.e., the current the device draws increases with voltage until the OVP kicks in).

[Richard Crowley]

So, given that the iDevices use the latter form of circuit (i.e., switch mode regulation) as several people here already pointed out, it would make sense for the current draw to decrease as the incoming voltage increases. It does the exact opposite (i.e., the current the device draws increases with voltage until the OVP kicks in).

Wrong question again.  The current doesn't increase or decrease based in the INCOMING power.  The current is regulated by the charging profile for the particular battery being charged.  It is a much more complex process than you seem to be imagining.  It is based on the beginning state of the charge, the temperature, how fast the battery is accepting the charge, etc. etc. etc.  And in many cases there is a microprocessor inside the battery pack which keep tracks of the previous discharge cycle to further optimize the latest charge cycle.

And it has almost nothing to do with the source power available. Except that it clearly can't put MORE current into the battery than is available from the source. Which is why some devices complain of "slow charge" when they can't get as much power from the source as they want.

[SharpEars]

So, given that the iDevices use the latter form of circuit (i.e., switch mode regulation) as several people here already pointed out, it would make sense for the current draw to decrease as the incoming voltage increases. It does the exact opposite (i.e., the current the device draws increases with voltage until the OVP kicks in).

Wrong question again.  The current doesn't increase or decrease based in the INCOMING power.  The current is regulated by the charging profile for the particular battery being charged.  It is a much more complex process than you seem to be imagining.  It is based on the beginning state of the charge, the temperature, how fast the battery is accepting the charge, etc. etc. etc.  And in many cases there is a microprocessor inside the battery pack which keep tracks of the previous discharge cycle to further optimize the latest charge cycle.

And it has almost nothing to do with the source power available. Except that it clearly can't put MORE current into the battery than is available from the source. Which is why some devices complain of "slow charge" when they can't get as much power from the source as they want.

Wrong answer again. There is more current going into the device as the voltage increases (up to a point). This means that there is more INCOMING power. Now this has to be transferred to the battery (1) or given off as heat by either the device's regulation circuit or the battery's own protection circuit which ultimately includes the battery's cells themselves (2). Which one of these two possibilities are you asserting, because it has to be one of them or perhaps a combination of both? Are you suggesting that the extra incoming power is 100% being converted to heat and not being used to charge the battery's cells? This is what it sounds like you are suggesting, despite concurrently stating that the entire thing is microprocessor controlled. This of course makes no sense, because if it was truly microprocessor controlled, the incoming current would be reduced as the input voltage increases. It seems to me that whatever is inside these iDevices is far more simple. My claim is that some of the incoming excess power at raised input voltages is not only going to the battery but is actually being used to charge the cells within the battery (i.e., this includes getting through any circuitry in the battery and eventually hitting the battery's cells). I am not saying that 100% of the increase in power is getting to the cells, but I am saying that a very significant percentage of it is.

Without actual objective testing, both of us are speculating (i.e., theory crafting) and either of us could be right or we can both be partially right (i.e., some power is given off as heat and some is used to charge the cells).

So, the real test will be whether or not the battery will charge faster when it is charged from 0% until the charging circuit stops charging it any further (either the device's, the battery's or as is more likely the case, a combination of both). Also of interest will be the change in charging time as the voltage is increased from 5 V to 5.3 V. Of benefit would be thermal monitoring of the phone's case during both tests, but I do not have the equipment on hand to do this. I can however keep the environmental conditions relatively constant at least, so that external temperature does not skew the test.

The above is the test that I have not yet performed, because it takes several hours and requires a fully discharged device in combination with current draw tracking over the course of those hours to see how the charging circuit behaves and when the charge actually ends. I plan on perform this test in the near future with several devices to definitively answer the criticisms once and for all (and/or be proven wrong).

[Richard Crowley]

Will a cheap commodity charging circuit output more current when you supply more voltage into it than it was designed for?  Probably, when those circuits are designed down to the tenth of a cent.  Is that good for the battery you are trying to charge?  Probably not. In any case this whole discussion seems silly at best and dangerous at worst.

[Siwastaja]

Wrong question again.  The current doesn't increase or decrease based in the INCOMING power.

What are you trying to say? The actual input current draw has been experimentally measured by the OP and some others, and it's a result you can't just go and ignore --- and it's a fair assumption that the device is not a shunt regulator as it would be totally insane (a lot more insane than linear pass type regulator). I.e., power to cell = power in minus controller losses, which will be fairly consistent percentage for a switch mode regulator. More voltage AND more current in ---> DEFINITELY more current into the cell, unless the design is totally crazy (a shunt regulator).

The current is regulated by the charging profile for the particular battery being charged.  It is a much more complex process than you seem to be imagining.  It is based on the beginning state of the charge, the temperature, how fast the battery is accepting the charge, etc. etc. etc.  And in many cases there is a microprocessor inside the battery pack which keep tracks of the previous discharge cycle to further optimize the latest charge cycle.

I, as a battery management system designer, would tend to say that it's a lot more simple than you seem to be imaging. Although, industry has shown over-complex examples, which mostly tend to fail due to wrong assumptions made.

These iDevices do not have a battery pack, they have a single cell. Previous discharge cycle has absolutely nothing to do with any "optimization" of the charge cycle; you are just throwing in fancy weasel words; please stop, it belongs to kickstarer scams, not on this forum IMO. Thank you.

Li-ion is the simplest battery type in use so far; it has practically 100% (typically >99.9%) coulombic efficiency, practically no "memory effects", no Peukert effect. All losses are ohmic, i.e., voltage drops. It is always true that charge in = charge out. Lithium plating does limit allowable charging current especially at high voltages and low temperatures, but if the freaking device actually is measured taking in more charging power, then it's most definitely going into the cell. That may have two conclusions; they don't use 100% of their "allowed" charging current normally; or they are exceeding their "allowed" budget, causing more damage than it's designed for.

It is well possible that a mere 30% increase in charging current may shorten the lifetime of the cell from 500 to 100 cycles if it happens in just the right conditions, but it's not very probable they are so close to that limit.

[SharpEars]

I have decided to do a better test:

What is the charge percent on an iPhone 5 after charging for 20 mins. The start point is a dead phone (i.e., a phone that has shut off due to running low (i.i.e, below 1%) ). This is almost exactly the case for which I am designing.

That is:

• You have a dead or nearly dead phone
• You have a very short amount of time (say 5-20 mins), because you have to leave (plane, train or automobile) and you won't be packing a battery charger
• The actual test: To what percentage level can you charge your phone within that time as a function of charging voltage?

Take aways:

• You are most likely not charging your phone to 100%
• Your phone will start off but will turn on at some point, per Apple's OS/charger policy and you don't have the time/luxury to sit there and wait for that moment to manually turn it back off again

Results for iPhone 5 after twenty minutes of charging:

Charge voltage or charger	Charge percentage	Number of tests
5.0 V	13 %/13 %	2
5.3 V	13 %	1
Apple 12W charger	13 %	1

Therefore, for the iPhone 5 at a 20 min time and with the charging voltage as specified, there appears to be no difference between a charge at 5 V and a charge at 5.3 V. An Apple 12 W charger performs as well as a well regulated supply in terms of getting the phone's percentage up to 13 in twenty minutes. I want to do additional testing with different time periods and perhaps time the discharge time as well to do a crosscheck and further verify the results.

The two non-Apple power supply tests above were made using a HP 6632A 20 V/5 GPIB lab DC power supply (calibrated and traceable to NIST) set to the voltages specified. Power supply sense cables were used to maintain a constant voltage independent of load at the point of load. The power supply power cables were connected to a taken apart USB 2.0 female connector with about five centimeters of cables dangling. A lightning cable was attached to this connector with the lightning end plugged into the DUT, an iPhone 5. The current reading on the PS was monitored to make sure that the 2.7 V D+/D- setting had taken affect (i.e., we were charging at the max current supported by the DUT).

A calibrated Fluke 8506A (6 1/2 digit) multimeter was used to monitor the voltage which stayed constant for both tests and was 3 mV above the voltage stated in the above table for the duration of the test - an insignificant deviation. The reading was crosschecked with a calibrated HP 3456A (6 1/2 digit) multimeter which differed lower by 12 counts from the Fluke - again, an insignificant deviation for the purpose of this test.

I want to run further tests for the iPad as well. I think this is where there may be a difference and at this point I definitely want to do further testing before making any additional claims.

[Monkeh]

There is a cheaper, easier, smaller, and more effective solution to the problem your scenario presents. It is called a battery.

www.amazon.com/dp/B005X1Y7I2

*wanders off*

[amyk]

Also, I found it odd in discharging the phone that it stayed at 13% longer when charged at 5.3 V than at 5.0 V, so there maybe some mysterious activity here

Forget about those percentages if you want accurate results; you really need to measure the voltage and current into the cell together to determine the overall energy put in. Current will be harder to do since you'll need to insert a shunt inline, but voltage should be relatively easy.

Don't forget to account for the slight (but measurable) capacity loss every time you cycle the cell too...

[IanB]

...and you won't be packing a battery charger

WTF? 

Why not?

[funkathustra]

Results for iPhone 5:

Charge voltage	Charge percentage	Number of tests
5.0 V	13 %/13 %	2
5.3 V	13 %	1

Therefore, for the iPhone 5 at a 20 min time and with the charging voltage as specified, there appears to be no difference between a charge at 5 V and a charge at 5.3 V.

I'm glad it only took us 90 posts to realize that Apple probably knows how to design a USB power pack that will provide good charging performance.

An iPhone 6 has an 1800 mAh battery. Assuming Apple programmed the charger to charge at 1C (to extend battery lifetime), the charger will use up to 1800 mA of current. It's a switching regulator, so it could use appreciably less current from the USB VBUS to achieve that. For example, 4V @ 1800mA = 5V @ 1.4A (assuming 100% efficiency).

I'm in the camp that thinks if you want to charge your phone faster, you'll have to swap the battery charger IC in your phone, and then pay dearly for it in 6 months when your battery is shot. That's when you'll really need your SupperFast charger (is it supper time yet? I'm hungry).

The increased current consumption you see at higher voltages is strange, but is probably due to operating near the upper limits of the battery charger. I've seen very strange behavior out of ICs before -- in deep-sleep mode, an ESP8266 will use hundreds of times more power if you're running it at 3.6V instead of 3.3V, for example. I've never tested regulators, but I could imagine protection components or other things that would end up shunting the current to ground as voltage approached a threshold.

[Monkeh]

This increased current consumption is strange, but is probably due to operating near the upper limits of the battery charger. I've seen very strange behavior out of ICs before -- in deep-sleep mode, an ESP8266 will use hundreds of times more power if you're running it at 3.6V instead of 3.3V, for example. I've never tested regulators, but I could imagine protection components or other things that would end up shunting the current to ground as voltage approached a threshold.

And before he breaks out the 'this is within USB specs' argument again: At the time this hardware was designed, VBUS max was 5.25V. Any operaton above and beyond that is unspecified.

[SharpEars]

Also, I found it odd in discharging the phone that it stayed at 13% longer when charged at 5.3 V than at 5.0 V, so there maybe some mysterious activity here

Forget about those percentages if you want accurate results; you really need to measure the voltage and current into the cell together to determine the overall energy put in. Current will be harder to do since you'll need to insert a shunt inline, but voltage should be relatively easy.

Don't forget to account for the slight (but measurable) capacity loss every time you cycle the cell too...

But this is very difficult to do with devices like iPhones and iPads, because they are not trivial to take apart. There is also the role that any circuitry built into the cell may play.

[SharpEars]

There is a cheaper, easier, smaller, and more effective solution to the problem your scenario presents. It is called a battery.

www.amazon.com/dp/B005X1Y7I2

*wanders off*

You are going off the assumption that said battery based charger will not be dead as well, due to neglect. I would not make this assumption.

[SharpEars]

I'm glad it only took us 90 posts to realize that Apple probably knows how to design a USB power pack that will provide good charging performance.

Hey, if you want short concise answers, you've picked the wrong forum .

An iPhone 6 has an 1800 mAh battery. Assuming Apple programmed the charger to charge at 1C (to extend battery lifetime), the charger will use up to 1800 mA of current. It's a switching regulator, so it could use appreciably less current from the USB VBUS to achieve that. For example, 4V @ 1800mA = 5V @ 1.4A (assuming 100% efficiency).

I'm in the camp that thinks if you want to charge your phone faster, you'll have to swap the battery charger IC in your phone, and then pay dearly for it in 6 months when your battery is shot. That's when you'll really need your SupperFast charger (is it supper time yet? I'm hungry).

The increased current consumption you see at higher voltages is strange, but is probably due to operating near the upper limits of the battery charger. I've seen very strange behavior out of ICs before -- in deep-sleep mode, an ESP8266 will use hundreds of times more power if you're running it at 3.6V instead of 3.3V, for example. I've never tested regulators, but I could imagine protection components or other things that would end up shunting the current to ground as voltage approached a threshold.

The funny thing is that if they shunted the current to ground, it would go up more drastically. It, instead, appears to rise slightly and then dip slightly after 5.3 V or so, with increasing voltage and I've tested up to 6.1 V ! I don't see it drastically spiking up or down, so it would seem that if the excess power is not going to the battery, it is being converted to heat.

[Monkeh]

There is a cheaper, easier, smaller, and more effective solution to the problem your scenario presents. It is called a battery.

www.amazon.com/dp/B005X1Y7I2

*wanders off*

You are going off the assumption that said battery based charger will not be dead as well, due to neglect. I would not make this assumption.

Then let's not assume you're sharp enough to plug a charger into an outlet either..

You'd have to neglect one of those for many, many, many months before it self-discharges, and one would imagine you'd remember to charge it if you'd used it. Plus you can just leave it permanently connected to one of those Apple chargers you hate so much.

[amyk]

Also, I found it odd in discharging the phone that it stayed at 13% longer when charged at 5.3 V than at 5.0 V, so there maybe some mysterious activity here

Forget about those percentages if you want accurate results; you really need to measure the voltage and current into the cell together to determine the overall energy put in. Current will be harder to do since you'll need to insert a shunt inline, but voltage should be relatively easy.

Don't forget to account for the slight (but measurable) capacity loss every time you cycle the cell too...

But this is very difficult to do with devices like iPhones and iPads, because they are not trivial to take apart. There is also the role that any circuitry built into the cell may play.

It's not as easy as the others with user-removable batteries but the iPhone 5 mentioned above is pretty easy to get to the battery on. No glue to fight with, at least. The battery connector doesn't look so tiny either.

Circuitry on the cell will be limited to thermistor, protection, and a fuel gauge:

https://ripitapart.wordpress.com/tag/iphone-5-battery-pinout/

The iPad is going to be rather more difficult to get into... so some good measurements on the iPhone would probably be enough to put to rest all the speculation.

[rs20]

... so some good measurements on the iPhone would probably be enough to put to rest all the speculation.

Unless the iPad attempts to draw more current than the iPhone (definitely true), which the charging circuitry dropout + supply sag can't sustain (unknown).

Also, we haven't really accounted for where all this extra current is going, have we? Perhaps only of academic interest, but interesting to consider nevertheless.

[SharpEars]

Also, I found it odd in discharging the phone that it stayed at 13% longer when charged at 5.3 V than at 5.0 V, so there maybe some mysterious activity here

Forget about those percentages if you want accurate results; you really need to measure the voltage and current into the cell together to determine the overall energy put in. Current will be harder to do since you'll need to insert a shunt inline, but voltage should be relatively easy.

Don't forget to account for the slight (but measurable) capacity loss every time you cycle the cell too...

But this is very difficult to do with devices like iPhones and iPads, because they are not trivial to take apart. There is also the role that any circuitry built into the cell may play.

It's not as easy as the others with user-removable batteries but the iPhone 5 mentioned above is pretty easy to get to the battery on. No glue to fight with, at least. The battery connector doesn't look so tiny either.

Circuitry on the cell will be limited to thermistor, protection, and a fuel gauge:

https://ripitapart.wordpress.com/tag/iphone-5-battery-pinout/

The iPad is going to be rather more difficult to get into... so some good measurements on the iPhone would probably be enough to put to rest all the speculation.

Interesting, I will have to look into that...