Can someone explain what this statment means? I don't understand where both ends of the resisters are being connected? are all resisters connected at one end to 5v source and the other leg to pin (23) and (16) There are 4 resistors and only two connection points?

33k resistor to +5v
(23) and 22k resistor to gnd(16) to obtain 2v and 33k to +5v and​

47k to gnd to obtain 2.8v

This is how far I have got. as you can see I am trying to make D+ and D- 2.0 and 2.8 volts. THe above statement says how to do it but I am not understanding it.

 

Let me give you a link to our Ebook section on voltage dividers - http://www.allaboutcircuits.com/vol_1/chpt_6/1.html

You can't make connections to the data lines (D+ & D-). To obtain those voltages, you need two resistors, one connected to the +5 line and the other to ground. The junction between the two resistors will have a lower voltage.

With 33K to +5 and the 22K to ground, the voltage at their junction is 2 volts. With 47K instead of 22K, you have 2.8 volts at the junction. Use 43K to obtain 2.8 volts.

You do understand that you get these voltages at a very low current. Any load at all will cause the divider voltage to drop.

What is your intended use of these voltages?

 

Let me give you a link to our Ebook section on voltage dividers - http://www.allaboutcircuits.com/vol_1/chpt_6/1.html

You can't make connections to the data lines (D+ & D-). To obtain those voltages, you need two resistors, one connected to the +5 line and the other to ground. The junction between the two resistors will have a lower voltage.

With 33K to +5 and the 22K to ground, the voltage at their junction is 2 volts. With 47K instead of 22K, you have 2.8 volts at the junction. Use 43K to obtain 2.8 volts.

You do understand that you get these voltages at a very low current. Any load at all will cause the divider voltage to drop.

What is your intended use of these voltages?

do you mean like this?

I am wanting to charge my iphone 3gs and need these voltages on the D+ and D- lines.

 

That is how you get the voltages, correct. I do not understand why you are attaching those voltages to the data lines.

 

That is how you get the voltages, correct. I do not understand why you are attaching those voltages to the data lines.

Without the voltage on the data lines the iphone 3gs will not accept a charge. Will this voltage change as the charging system in the car goes up from 12v to 13.8v?

 

iPhone or other products reads the voltage on these lines to decide how much current it can safely "suck" from the charger. Be it 500mA or over 1A.

 

Without the voltage on the data lines the iphone 3gs will not accept a charge. Will this voltage change as the charging system in the car goes up from 12v to 13.8v?

If you plug that into your auto electrical system, you will fry it. It's designed for 5v use from a USB port, not directly from an automotive system.

 

If you plug that into your auto electrical system, you will fry it. It's designed for 5v use from a USB port, not directly from an automotive system.

understood. It is being fed by a 12vdc cig charger circuit

 

didn't work. still no charge

 

OK, got some resistors today and hooked them up as per diagram. phone said charging not supported. tested two iphone chargers I have here. diagram shows results using the 4 usb wires as input. The chargers adapted the 30 pin phone connector. I tried to recreat the same thing only at the input end of the usb cable. didn't work.

When using a standard iphone usb cable it should work doing the same thing at the female end of the usb plug, where the iphone male 4 pin plugs in, shouldn't it?

I matched the purple charger numbers using resistors and it didn't work.

HEre is a pic of the above purple charger 2 pinout. I can't see why this can't be duplicated at the input end of the usb cable. When I did it, the phone still would not charge. That resistor or what ever it is only dropped voltage .24 volts.

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