I'm trying to use a mobile phone charger for my projects, and want a high current output. I have read this, however my measurements show something else.

Here is a link to the Battery Charging Specification Rev. 1.2.

1.4.7 Dedicated Charging Port

A Dedicated Charging Port (DCP) is a downstream port on a device that outputs power through a USB connector, but is not capable of enumerating a downstream device. A DCP shall source \$I_{DCP}\$ at an average voltage of \$V_{CHG}\$. A DCP shall short the D+ line to the D- line.

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I verified on three different chargers, and all read \$R_{DCP\_DAT}\$ as ~1.5 Ohms. Now, if there is a short between the D+ and D-, there is no detection on the charging port side, and the charger should always output \$I_{DCP}\$ {0.5 - 5.0A max} on the VBUS line - is this correct?

I tested the current output of three chargers, but they are all completely different.

Charger 1 - Nokia

Rated current output: 1.3A

Measured current output: 1.34A

Charger 2 - Asus

Rated current output: 2.0A

Measured current output: 0.7A - 1.1A (unstable)

Charger 3 - HTC

Rated current output: 1A

Measurent current output: 0.1A

If all these dedicated charging ports have no current negotiation, how come only one charger is showing it's rated output?

N.B All three chargers can charge a mobile phone in a reasonable amount of time.

  • 1
    How were you measuring the current output? What load did you connect? If you just put an ammeter directly across the output, that's not a valid measurement - the supply designer did not expect someone to short its output. It might even damage the charger. – pericynthion Sep 24 '14 at 1:14
  • @pericynthion I put a few 10 ohm resistors in parallel, enough to draw their maximum load current. – tgun926 Sep 24 '14 at 1:18
  • The way it works, as I recall, is that the charging device needs to sense input voltage and back-off if it droops below some threshold. So the charger and chargee negotiate by way of their V-I curve. The details are specified. I don't think there is a requirement that the DCP supply 5A. I think the minimum is 1.5A. So devices which draw less than 1.5 don't need to back-off. – mkeith Jun 8 '16 at 18:34

there is no detection on the charging port side, and the charger should always output IDCPIDCP {0.5 - 5.0A max} on the VBUS line - is this correct?

I don't understand this completely, because the BC spec is confusing to read, but yes, dedicated chargers (DCP) short the D+ and D- together to indicate what they are. This doesn't indicate any particular current available, though, it just says that it's a DCP. Different chargers supply different amounts of current.

The charger has no brain in it; it just supplies 5 V until the current draw is too great, and then its voltage starts to droop:

DCP Required Operating Range

It's the "portable device" (PD) which has to be smart about limiting its own current draw from the DCP to stay within the dark region of the plot. So it can try to draw up to 1.5 A, but if the charger voltage drops below 2 V at 0.5 A, then you can't draw any more than 0.5 A from it.

For a dedicated charger or USB charger, the current limit is determined by loading the adapter. When the adapter’s output voltage starts to collapse, it is an indication that the current limit of the device is reached. - MAX8895 datasheet

I am not sure if this answer will solve your question, but Apple does a similar thing with their wall chargers to make sure that the devices are not being charged too quickly. To accomplish this, the chargers have a voltage divider circuit comprised of 2 resistors that gives a reference voltage to one or both of the data pins. Inside the iPhone, there is additional circuitry that reads the voltage on the data pins from the voltage divider.

Here is a diagram of an "Apple Compatible" charger featuring the voltage divider that I was talking about:

Apple Compatible Charger

I would assume that Apple has different voltage dividers for different rated chargers. This way, a phone can tell if it can safely charge itself if the reference voltage is under 2v for example.

The question is ill-posed, conceptually.

Usual wall (and BC1.1/1.2) chargers do not negotiate anything. They only "advertise" their capability by means of some signature on D+/D- wires. It is the DEVICE that decides to take maximum necessary current based on detected signature and the state of internal battery.

As other respondents noted, there is "Chinese signature" (with D+ shorted to D-), there is "Apple signature" with certain combination of voltages (using ~40k-70k resistors), there could be BC1.2 with sequential handshake. Due to total awkwardness and complexity associated with BC1.2, this signature can be hardly found ever.

In modern days things can be different with advent of new Power Delivery Specification (PD), where "provider" (charger) actively advertises its capability with "consumer" through a serial communication channel with a fairly sophisticated protocol. Originally the VBUS wire was thought for this purpose in PD1.1 (the idea is now abandoned), and now the CC pins in Type-C connector are used for this purpose.

I guess it's possible the charger is implementing data contact detection (to ensure full insertion- the power contacts mate first).. suggest you try connecting the D+ to a plausible VDP_SRC voltage.

  • It says VDP_SRC should be 0.5 - 0.7V and source 250uA. Will a resistor divider between VBUS and GND be okay, or is it more complicated than that? – tgun926 Sep 24 '14 at 4:25
  • Diode and a resistor maybe. – Spehro Pefhany Sep 24 '14 at 4:28
  • Still the same result.. I also noticed the current doesn't increase linearly. ie. 10ohm resistor = 0.43A, 10 || 10 ohm resistors = 0.75, 10 || 10 || 10 resistors = 1.0A – tgun926 Sep 24 '14 at 5:15
  • 2
    What's happening to the voltage? How much voltage is your ammeter dropping? – Spehro Pefhany Sep 24 '14 at 5:44
  • Don't have a second meter with me right now, will let you know once I do – tgun926 Sep 24 '14 at 6:32

I've seen charger using AD42011D having only 2 wires to connect to device, maybe a data loopback at the usb male. Good thing about this IC is that it is able to detect full charge and will provide trickle charge when full, so you will not overcharge your device. I couldn't find the datasheet but very simple circuit with minimal components.

  • This answer would be much better if you could track down at least a datasheet for the part, I couldn't seem to find a datasheet or even mention of the existence of a AD42011 are you sure that's the correct part number? – PeterJ Nov 3 '14 at 21:48

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