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
    \$\begingroup\$ 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. \$\endgroup\$ Commented Sep 24, 2014 at 1:14
  • \$\begingroup\$ @pericynthion I put a few 10 ohm resistors in parallel, enough to draw their maximum load current. \$\endgroup\$
    – tgun926
    Commented Sep 24, 2014 at 1:18
  • \$\begingroup\$ 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. \$\endgroup\$
    – user57037
    Commented Jun 8, 2016 at 18:34

5 Answers 5


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.

  • \$\begingroup\$ I'm quite sure that both Apple and Samsung used circuits that detected an Apple phone reliable (Apple charger) or a Samsung phone reliable (Samsung charger), so an Apple charger with a Samsung phone would be reliable and slow, and a Samsung charger with an Apple phone would be reliable and slow, and each phone with the right charger was about twice the speed. Since this was non-standard, they both tried hard not to give more power except to their own phones. \$\endgroup\$
    – gnasher729
    Commented Dec 13, 2022 at 15:18

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.

  • 1
    \$\begingroup\$ Does this mean that PD 2.0 only works over Type-C connectors? I've read elsewhere that the USB 3.0, PD and Type-C specs are completely independent, and that Type-C with electronically marked cables can only add the benefit of allowing 5 A instead of 3 A current over the cable (besides additional data wires, for alternate modes). I concluded that PD can work over Type-A, too. I only have access to a PD adapter with a fixed Type-C plug. All other chargers are QC 2.0 (Samsung) or unknown. This is so confusing. \$\endgroup\$
    – ygoe
    Commented Nov 23, 2018 at 18:32
  • \$\begingroup\$ @ygoe, if I understood correctly, PD over type-A requires a special cable, though I have never seen one. PD over type-C, at 3 A at most, works with any compliant cable. \$\endgroup\$ Commented Feb 2, 2019 at 16:22
  • \$\begingroup\$ "D+ shorted to D-" is from the USB Battery Charging specification. Not sure what you mean by "Chinese signature" \$\endgroup\$
    – endolith
    Commented Mar 3, 2021 at 21:55
  • 1
    \$\begingroup\$ @endolith, long before the BC1.2 was crafted into existence, the Chinese government has established the D+-short-D- standard. en.wikipedia.org/wiki/… \$\endgroup\$ Commented Mar 8, 2021 at 17:57
  • \$\begingroup\$ @Ale..chenski Interesting. Interface requirement of the charger side "D+ and D- should be short-circuit and does not connect to any part of the charger inside the charger." \$\endgroup\$
    – endolith
    Commented Mar 8, 2021 at 18:10

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.

  • \$\begingroup\$ 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? \$\endgroup\$
    – tgun926
    Commented Sep 24, 2014 at 4:25
  • \$\begingroup\$ Diode and a resistor maybe. \$\endgroup\$ Commented Sep 24, 2014 at 4:28
  • \$\begingroup\$ 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 \$\endgroup\$
    – tgun926
    Commented Sep 24, 2014 at 5:15
  • 2
    \$\begingroup\$ What's happening to the voltage? How much voltage is your ammeter dropping? \$\endgroup\$ Commented Sep 24, 2014 at 5:44
  • \$\begingroup\$ Don't have a second meter with me right now, will let you know once I do \$\endgroup\$
    – tgun926
    Commented Sep 24, 2014 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.

  • \$\begingroup\$ 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? \$\endgroup\$
    – PeterJ
    Commented Nov 3, 2014 at 21:48

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