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34

What is the ideal way to handle data pins D+ and D- on a USB power adapter to be compatible with fast charging on devices? ... I am hoping to figure out which method provides the best compatibility with all USB devices. It is for practical purposes impossible to make a truly universal charger using any combination of unchanging shorts or resistors on the ...


30

I found this page answers your question clearly. I quote the relevant parts below. The BC1.2 outlines three distinct types of USB port and two key monikers. A "charging" port is one that delivers currents higher than 500mA. A "downstream" port signals data as per USB 2.0. The BC1.2 specification also establishes both how each port should ...


25

Although the question has provided limited details, this answer presents a somewhat different hypothesis from the standard assumption that there's an inductive coil hidden in there somewhere. The charger in question possibly uses a Piezoelectric Transformer instead of the magnetic (inductive) transformers usually seen for isolation. Does the charger looks ...


25

If you look at the charge profile of a lithium battery you'll see that at certain points it changes from constant current to constant voltage charging: - This means that some form of "in series" charge control mechanism needs to be present to act initially as a constant current source then change to a constant voltage source. This charge control circuit ...


23

While it is true that you can't actually charge and discharge a battery simultaneously, it is quite common to have a charging source, a battery, and a load, all connected in parallel so it looks like you are charging and discharging simultaneously. If the charging source can deliver more current than the load requires, then the excess current will be used ...


21

To be compatible with the original standard, USB devices should not draw more than 100mA (which is plenty to power the logic interface), until they have negotiated with the host, to find out what it can supply. After successful negotiation, they can draw up to 500mA. This is to protect the operation of a 4 port hub, should it be plugged into a PC with all ...


16

If by "use the battery" you mean "can the battery be in the circuit?" then yes. If by "use the battery" you mean "draw power from the battery while charging it?" then no. Charging is, by definition, putting power into the battery. This is the opposite of drawing power from it. You can't do both, by definition. Under normal operation, where the battery is ...


16

Historically: really old mainboards connect USB power pins to the 5V power rail, with no protection power on by keypress was added, which added a jumper or a BIOS setting that decided whether USB ports would be powered from the standby power or from the regular 5V rail. Since standby power was introduced in ATX, this does not exist on AT mainboards. USB ...


13

I was thinking "just use google", but then I saw a comment suggesting that many people wouldn't even know what to look for... A common kind of vacuum tube failure is when the gain just gradually goes to zero. This happens when the "emitter" stops emitting electrons. That happens when the "emissivity" falls. That happens when the emitter is "poisoned", that ...


12

I'm not going to try to answer all parts of your description, since part of the problem is that the charger is being used incorrectly. Only after that is resolved, might it be worth investigating the second behaviour. Those charger boards from Ebay cannot be relied upon to work (a) correctly, and (b) safely. They typically appear to use the TP4056 (...


12

The device (phone, etc.) has the special charging circuit built-in. And the battery pack typically has a special circuit inside to prevent excessive discharge (or charging). Li-ion battery cells are inherently unstable/dangerous and require special circuits somewhere (often built into the battery pack) to make them safe to use.


12

Don't. Just use one TP4056 and connect both cells in parallel after balancing them first. Don't connect batteries with more than 0.2V difference in parallel as this can risk fire and explosions (excessive charging current from one to another). This will work because Lithium cells have a wide voltage range. So when connected in parallel they will self-balance....


11

"The MCP73871 device specifically adheres to the current drawn limits governed by the USB specification." -MCP73871 You don't need to current limit anything. Your MOSFET solution would only be useful to current limit inrush due to a capacitive load but your VDD pin is not capacitive nor would that circuit limit once it's already on. All you need to do is ...


11

Another reason is standarisation - a few years ago the EU pushed for standarisation of the charging connector in phones which ended being micro USB. USB implies 5V. And as a personal note I worry about those devices which require over an amp to charge properly charging through USB 2.0.


10

simulate this circuit – Schematic created using CircuitLab Figure 1. \$ V_{OUT} = V_{IN} - 0.7 \$. A silicon diode will drop about 0.7 V when more than a few milliamps is running through it.


10

If the input power of the adapter is less than 75W, you don't need a power factor correction stage (by EU and other regulations). So they are easier to design, and they hit the largest segment of the market. Once you need a 2 stage converter you have additional components and thermal considerations and GAN probably doesn't make as big a performance and ...


9

You already have an 80V power pack and all you need is another 6V at approaching 18A. This 6V could be produced by an isolating flyback or forward converter powered from the 80V supply. Because the output will be isolated, it can be wired in series with the 80V to make 86V. So, 80V feeds an isolating forward/flyback converter that generates 6V dc. This ...


9

2017 update: There is no ideal way to handle USB data pins to provide compatibility and "fast charge". There could be many different chargers, and there are many USB devices/phones/tablets that need charging. Historically there were two approaches: The device is a "smart device". It tries to detect various signatures of the port it is connected to, and ...


9

The circuit you have shown WILL NOT WORK: - It shows a transmitter and receiver of power (top and bottom). The transmitter of power has some kind of rectifier THAT WON'T WORK - both diodes are facing the same way - it needs to be a bridge rectifier. Then it shows an oscillator - THIS WON'T WORK - there is no active device to control the oscillation and no ...


9

Feasible: yes Useful: no, because the power needed by a smartphone is orders of magnitude larger than what such a (kinetic) generator generates. A (non smart) wristwatch needs only very little power, it can run for years on a small button cell. This amount of power can easily be generated by a kinetic generator. A smartphone has a large battery and needs ...


8

The voltage on a 12V lead-acid battery will always be higher during the charge state than when it is at rest. So you can expect to see lower voltages 10 - 20 minutes after charging has stopped. This doesn't mean you're loosing power, this is just the natural chemical response of a battery after a charge cycle has stopped. As Optionparty mentioned you have ...


8

No it won't have any effect. The 220v or 110v AC at 50 or 60Hz is stepped down to a lower voltage and rectified to DC before it is used to charge anything. So since the actual charging uses the rectified DC, the original AC frequency doesn't make any difference.


8

Charging one battery from the other will result in overall drop in efficiency and run time. Consider that energy will be lost in the charging circuit and in heat losses of both the battery powering the charger and the battery being charged. The most efficient is to take power from the 'good' battery. As its voltage drops to the level of the battery with ...


8

The answer to your header question is NO! Input 19.5x3.33 = 64.935 Watts. Output 12x5.5 = 66 Watts. So your new output power is higher then your input power which is against the law of conservation of energy. But according to your question details your equipment requires only 48.5 Watts. That is do-able but only if your buck step down converter (or ...


8

First of all, the USB is not being fed directly to the phone’s battery. It’s actually being fed to the battery charger circuitry inside the phone. You cannot safely directly connect a normal regulated voltage source to a lithium ion battery without the distinct possibility of an explosion and/or fire! The charger circuitry adjusts the voltage and current fed ...


7

It's a ferrite choke, not going to retype it all but here's an article: What are the bumps at the end of computer cables?. These "bumps" are called ferrite beads or sometimes ferrite chokes. Their goal in life is to reduce EMI (electromagnetic interference) and RFI (radio-frequency interference). A ferrite bead is simply a hollow bead or cylinder ...


7

I believe the answer can only be empirical, not definitive. To examine some of the figures mentioned: there is a 20% inefficiency (which I do not know if it is true for most portable charger) A portable charger that is itself charged from USB (5 Volts) would need a boost converter to be able to supply 5 volts at its output. Boost converters commonly ...


7

You're on the right track. Lithium batteries take what is called a "Constant Current, Constant Voltage" or CC-CV charge profile. Constant Current: When the battery is discharged, you begin charging it by applying a fixed current (at the battery's voltage). Typically this current is somewhere in the neighborhood of 1C (the current required to completely ...


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