Sorry if I misunderstand, your typing uses next to no punctuation.
Currently I have Pin 1 connected to the 5v+ on the output connector, and the pin4 negative to the negative rail. As far as I know, USB spec calls for the outside case of a USB connector to be a positive ground; are you telling me to connect this to 5v?
I know the other articles are out there, and sure they may work for the iPod mini, but this is the full iPod, and if you carefully read the progression of comments, you'll see that no one has been able to successfully do this in a way compatible with the larger iPods; in fact, most people think its impossible to do (even though the iPod documents specify USB as a charging source).
It is also worthwhile to note that many people have tried charging directly through the USB connector with no power regulation. This will burn out any iPod. The iPod expects 5v potential, and will NOT regulate USB in. They figure the designer is smart enough to know the internal power potential in the device is 5v, and USB is a direct input to the power rails of the device.
I currently have the power regulator outputting 5v and everything should be dandy, heck, even some different devices will charge from it, but because of the unit identification part of the power detection spec, the data ports need to have some kind of potential on them. I don't know what it is, however, because the iPod patent doesn't go into that kind of specific detail.
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Edit ::
After a little more research, it turns out voltages for the DP and DM lines are identified by "an available power indicator 204" (204 is an identification number). Further schematics show it is detected with a resistor arrangement biasing each pin to the next. This is only in the iPod power input however.. this doesn't answer the question of how much voltage is required on DP and DM.
In english, this means that the voltage on DP and DM somehow controls the amount of watts available on my USB battery pack, depending exclusively on how much power is available on the pack, and how much voltage is present on DP and DM.
This at least answers a little of a question. The voltage difference between DP and DM must be negative to be identified as a battery pack, however the magnitude of the power detected is reliant on the voltage levels of the two data lines.
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Links of interest ::
Section [0067] ::
Detection of a USB power source on iPod - United States Patent