This is my circuit (google docs link)

It is powered by (2) 18650 lithium batteries, with protection pcbs. Do the diodes I added provide protection from reverse battery loading and such?

The LM317t should provide 5v and not burn up my iphone?

I dont think any current is drawn in the data lines so resistors work? (An official apple charger had similar design, with resistors so I should be good?) apparently 2 volts at D- and D+ means it draws 500ma

What could potentially go wrong to burn up my iphone?
[or should I buy some cheap chinese portable charger instead?]


THANKs!

 

You won't have enough "headroom" for the regulator.

You have two 3.6v batteries for 7.2v, but then you subtract roughly 0.8v*3=2.4v for the diodes and you wind up with 5v out. This means you'd need a 0v dropout regulator, and that's just not going to happen. The LM317 will drop around 1.7v from IN to OUT with a 500mA load, depending on temperature.

See the datasheet: http://www.national.com/ds/LM/LM117.pdf
Go to page 6, and see the "Dropout Voltage" plot, 2nd down on the right.

 

You won't have enough "headroom" for the regulator.

You have two 3.6v batteries for 7.2v, but then you subtract roughly 0.8v*3=2.4v for the diodes and you wind up with 5v out. This means you'd need a 0v dropout regulator, and that's just not going to happen. The LM317 will drop around 1.7v from IN to OUT with a 500mA load, depending on temperature.

See the datasheet: http://www.national.com/ds/LM/LM117.pdf
Go to page 6, and see the "Dropout Voltage" plot, 2nd down on the right.

ok so scratch the diodes, at 500ma or an [1]amp this thing will work fine?
7.2-1.7v=5.5v > 5v
would you use it on your iphone?

 

I don't have an iphone. The phone should have the battery charging circuit built-in.

You'll need a heat sink on the regulator; it's possible for the regulator to dissipate over 2W power as heat when charging a dead phone with fresh batteries at high current.

 

I am going to put a small quarter size piece of metal with fins on the regulator
It should always be drawing 500ma, so itll be dissipating 2.5 watts constantly for about an hour max
but I figured since the Volt differential is pretty small it might not heat up as much

on the datasheet http://www.national.com/ds/LM/LM117.pdf
pg 11 fig8 shows the ambient temperature graph, assuming the ambient temperature never starts more than 30C I would be okay without a heatsink?

Is it likley the iphone has any internal charging protection?

 

I am going to put a small quarter size piece of metal with fins on the regulator
It should always be drawing 500ma, so itll be dissipating 2.5 watts constantly for about an hour max
but I figured since the Volt differential is pretty small it might not heat up as much

It depends on how deeply the battery has been discharged, and how much current the phone's charger will actually draw once connected to power.

on the datasheet http://www.national.com/ds/LM/LM117.pdf
pg 11 fig8 shows the ambient temperature graph, assuming the ambient temperature never starts more than 30C I would be okay without a heatsink?

Look near the bottom of page 5 in the datasheet, at "Thermal Resistance, θ
JA Junction-to-Ambient (No Heat Sink)", and further qualify it by the T (TO220) package. Thermal resistance JA is 50°C/1 Watt.
The LM317 overheats and shuts down at 125°C.
You say your ambient temperature is 30°C
125°C - 30°C = 95°C; 95/50 = 1.9 Watts maximum power dissipation with no heat sink.

You need a heat sink.

Is it likley the iphone has any internal charging protection?

Yes.

 

Would you connect your iphone to it?

 

After I checked the output....maybe

 

The LM317t should provide 5v and not burn up my iphone?

If your voltage is truly regulated at 5v, your iPhone will be OK. It'll take what it wants, if your supply can keep up.

Are you sure about putting the same voltage on both data pins? With older iPods, I thought one got connected to minus and one to plus, thru resistors. I'm not saying it still has to be that way, just that you should be sure before you do a final build.