USB – Purpose of Pull-Up Resistors on Data Lines

chargerusb

I recently got a car lighter to USB adapter which I planned to use to charge my mobile phone. Unfortunately, the phone won't charge (and I know that the phone is capable of charging over USB port), so I opened it to see what could be the reason. The charger itself is pretty simple. It has a rectifier, MC34063A regulator and needed components, an LED and resistor and USB port.

Of particular interest to me was the USB port itself. If I remember correctly, for USB chargers, there should be a 220Ω resistor between data lines to signal to the device that the host is a charger.

In my case there was this circuit:

Circuit

I used black rectangles to mark junctions.

So if I'm reading this correctly, the D- and D+ lines are connected via pull-up resistors to Vcc.

I'm thinking of removing these resistors and putting a 220Ω resistor between the data lines.

My questions are:

  1. Is there any obvious reason why should the data lines be connected to Vcc through this complicated system of resistors?
  2. Is there any reason why I shouldn't remove those resistors and just put a 220Ω resistor between the data lines?

Best Answer

If I am not getting confused, this system of resistors is required by iPhones and other i-Devices to charge at different current levels. Otherwise an error message will pop up. If you don't have an iPhone you can quite safely remove these and replace them.

Apple stopped being as 'lax' with the charging interface and started being very picky about having the official chargers. We still doubted that there was an enumeration chip inside each charger - too expensive and complex. So there must be something else going on in those data lines. Time to sacrifice an official Apple iPhone 3Gs charger!

Taking it apart, desoldering the 4 data line resistors and measuring them on our multimeter, we found the following as shown in the schematic: enter image description here

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