Sorry this isn't the correct category. Couldn't find other another form to ask for guidance and thought I'd give this a shot.
I'm Using a LM317 voltage regulator board to create a phone charging circuit. I connected a DC socket to the Vin for testing purposes. Goal is to swap it with solar panel. I have a USB type A female socket connected to output side.
Test performed using multimeter
DC socket giving 12V
USB socket receiving 5V and 1A
My phone does not charge though even though the USB is getting what it needs. I've read online the typical Iphone requires 1A to have sufficient charging time. Question I suppose is why would my phone not charge? Am I using the wrong USB socket for my phone to charge? Is there a difference between Type A USB 2.0 and USB 3.0?
When the phone is plugged into the usb there doesn't seem to be a charge symbol on the phone. Unlike when its plugged directly into wall jack. I'm assuming if the charge symbol isn't present its not charging my phone or not sufficient enough. Even when I increase the current the Vin and ground measure to be exactly the same as the Vout from the board. I'm not sure where its going or if I need some sort of USB 5V boost converter in place of the USB type A socket I'm using. Why I'm asking for guidance.
Apple devices use more than just the power and ground wires for charging. The charger has resistors connected to the D+ and D- signals that tell the phone how much current the charger can provide.
Adafruit has a video that explains what's going on.
Thanks for the tip. Ill look into it. I know the data sheet on the USB type A has 1 1K pull down resistor for D+ and D- and when I tried building those into the circuit it didn't help. Which makes sense now. The 1 K ohm resistors would produce 5mA. So ill need to look for 5-10 ohm resistors if I under stand the video's concept.
Looking at the pic above where you are measuring current with the multimeter, I can't see what load is connected to the output (is your phone connected?) or how you are connecting the multimeter. It looks like you are touching the probes to the two output terminals, short-circuiting the output of the board through the multimeter.
I also noticed that neither of the probes is connected to the COM socket on the multimeter. I would expect you would need to connect the black probe to COM and the red probe to the 10A socket. I don't know how the meter will behave if the black probe is connected to the V/Ohm/etc socket. That might even damage it.
Nice catch, this connection illustrates that one needs to know how to use a meter before one can proceed to test this setup. To measure current, the OP needs to wire from power output to red lead of meter, plugged into "10A" socket, then with black lead in "COM", wire the black lead to the USB red wire. That puts the meter in series with the phone input, which is necessary to measure the current being consumed by the phone.
Yea I don't think that's correct. When plugged into com that when my circuit actually shorts. Just tested it. Power to the board is lost. Why I have it connected the way I do.
Ok i see what I'm doing. The only thing I need to swap is the 10A port to the COM. When doing this I still get the same readings as when plugged into the 10 A. so when the LM317 is set to 5 V the output is 5V and roughly one amp. which is what I want. The load connected is a iphone. which kellygray mentioned the iphones don't charge unless you have the D+ and D- connect in a configuration. For 1 A on the iphone I need to have D+ get 2V and D- get 2.8 V.
Vout on LM317 to Ground without female socket attached, 5V roughly 1A
LM3177 Vout to Vin on female socket. Same reading
Vin to GND on female socket same reading
From your image,
config1 - using your meter as a load, no other load? NOT how you use a current meter.
config 2 - correct
config 3 - see config 1
This is not how you test current. It may work(for a moment), but it's abuse of the meter. Instead, for a 5V source that's capable of delivering 1 A, use a 5/1=5 ohm power resistor.
Seriously, some electrical knowledge is required here.
