Charge Cable equation 2

Hi, I'm trying to design a charge cable for tablets. The power supply is a 10W USB w/output 5.1V 2.1A. The cable described below shows the power conductor is 22 AWG (46 strands) and the drain is (48 strands). Will 22 AWG supply the following current and voltage over 9 foot length? The cable also has to power a LED on the tablet end.

1.0A-----  4.7V ~ 5.25V
2.1A------ 4.55V ~ 5.25V

CABLE SPECS

30AWG(19/0.05TC+250D)*1P=1 pair of 30 AWG wires, each of 19 strands of .05 diam wire (0.05 tinned copper stranded), with 250D nylon to enhance the strength of the cable (not insulation)

22AWG(46/0.10TC+250D)*1C=1 conductor of 22 AWG wire, made of 46 strands of 0.1 diam (0.1 tinned copper stranded), with 250D nylon to enhance the strength of the cable (not insulation)

AL/MY=AL+Mylar , 95% coverage, this will be used on EMI & shielding.

D.W(16/0.10TC+250D)*3C=Drain wire is 3 conductor wires, each of 16 strands of 0.10 diam wire (0.10 tinned copper stranded), with 250D nylon
 
For drain wire there are 3 conductor wires, but at the connector end they would be twisted into 1 , and then soldered onto V- , as USB connector has only 4 pins. Therefore,

30 AWG* 1P are on D+ & D-, (data lines)
22 AWG*1C is on V+
Drain Wire is on V-
 
Although this cable would be designed as non-MFi, such specifications of the cable can still meet Apple’s basic electrical requirement. If any additional information or specification is needed, we can send it to you.

The Taiwan company that designed this cable is slow to respond.

Thanks in advance
Mark S.

I will address the first question [I'm in the middle of a project so not much time ]

Based on the American Wire Gauge table on Wikipedia [http://en.wikipedia.org/wiki/American_wire_gauge]:

The fusing current for AWG 22 is a minimum of 41A, so, no problem there. As for voltage loss on the lines you specified:

From the table for 22AWG: 16.14 mOhms/ft. That's 9 * 16.14 = 145 mOhms, so using Ohm's Law [E= IR or Voltage equals current times resistance]:
* At 1A there will be a voltage drop of: 145 mOhms * 1A = 145 mVolts [or around 0.15 volts]
* At 2.1A there will be a voltage drop of: 145 mOhms * 2.1A = 305 mVolts [or around 0.3 volts]

So, there's you answer AND how to figure it out

Bottom line: The things to consider are:

1. Will the wire melt [Fusing current]
2. Will the wire get too warm [i.e. will it compromise the insulation or present a fire hazard or burn hazard or produce harmful fumes, etc.]
3. Will too much voltage be lost at the highest current level--that is determined by the Source Voltage minus the voltage drop in the cable. If the final voltage is within the specified range, then yer good. Calculated, above, the worst voltage drop is 0.3Volts. So, 5.1 - 0.3 = 4.8V and your minimum is 4.55V so no problem [unless you have to double that for the return "drain" path -- then there might be a problem].

So, sorry this is so disjointed, but I keep adding stuff as I think of it [again I'm busy trying to get another project done]: I only calculated for the path from the supply to the USB port -- there is also the return path, so do the same calculations for that -- then there is also contact resistance, but I don't know have enough real-world experience to give you any guidance about that -- try Googling "Contact Resistance"

I'm not qualified to answer question #2 BUT considering that there will only be a power loss of 145mW at 1A and 640 mW at 2.1V, that's not enough power for the cable to even get warm. [Power is computed as follows: P = I²R [that's Power equals the current squared times the resistance]

BTW: if the LED is a common indicator type that will run at 5ma to 20ma, then the additional current load is insignificant.

Thank you! for answer. ". . .4.55V so no problem [unless you have to double that for the return “drain” path — then there might be a problem]."

I think there must be a return path (drain) to complete the circuit? Which would make the voltage go below the limit. So then a larger wire is the solution. 22AWG is not large enough.

Yup

Do watch out for the drop in that return line (drain), since its resistance is probably on a par with the 22AWG wire, or maybe more. It looks like your supplier is designing it very close to the maximum spec with no margin for error or manufacturing variation.

Larger conductors of course will solve the voltage drop problem, but may make the resultant cable stiffer than you'd like. Sometimes more smaller-diameter conductors can achieve the necessary voltage drop and still have the cable be flexible, though diameter usually increases. Lots of things to balance in a simple cable!

Please see below.

Apple wants the cable to . . . "comply with DC resistance requirement (Round trip VBus with shield shorted to GND at each end must be below 200m ohm), and comply with USB 2.0 specifications"

Will this cable do this?