DIY Iphone Charger  - Page 1
Forum section: electronics / beginners

author: Jbliss
author_rank: Regular Contributor
timestamp: June 15, 2015, 02:48:08 pm
content: Just checking that this circuit I am going to use is trust worthy it seem legit. very simpleanybody.....Thanks

author: Mark IV
author_rank: Contributor
timestamp: June 15, 2015, 02:52:46 pm
content: And the circuit?

author: Jbliss
author_rank: Regular Contributor
timestamp: June 15, 2015, 03:10:25 pm
content: Wow im an idiothere we go

author: ivan747
author_rank: Super Contributor
timestamp: June 15, 2015, 03:33:42 pm
content: Don't forget the heatsink.

author: suicidaleggroll
author_rank: Super Contributor
timestamp: June 15, 2015, 04:00:16 pm
content: Those are the pin voltages for a 1A charger, no?So you're planning to dissipate up to 28 W in that regulator? Good luck with that. Even with a heatsink it won't work. The max junction temperature of the 7805 is 125C. With 28W being burned off in it, even if it had a perfect 1e32 C/W heatsink, the junction to case thermal resistance is 5 C/W, which would put the junction at 142 C over ambient. So unless you're planning on running this thing at sub-zero temperatures WITH a heatsink, it's going to below the 7805 in a heartbeat.Also, your input voltage spec is way too wide. You can't take 7V on the input, because after the diode drop and 1R resistor, the voltage input to the 7805 will be ~5.3V. The 7805 has a 2V dropout, it needs at least 7V at its input, which means you need at least 9V at the circuit's input. You also don't want to use a 35V cap if you expect a 35V input. Either up the voltage rating of C1 or drop your max input voltage to <30V.

author: mtdoc
author_rank: Super Contributor
timestamp: June 16, 2015, 12:00:36 am
content: I don't think he specified his planned input Voltage, did he? That just looks like a copied circuit diagram from somewhere.Can someone enlighten me on the purpose of the wheatstone bridge in the OPs circuit?

author: Mr.B
author_rank: Unknown
timestamp: June 16, 2015, 12:06:52 am
content: Its drawn as a wheatstone, however it is really just 2 resistor dividers.Most modern phones, inc Apple, expect set voltages on the D+ and D- pins.I think it originates from being able to detect a genuine Apple charger.

author: mtdoc
author_rank: Super Contributor
timestamp: June 16, 2015, 12:22:07 am
content: Ok thanks.Right - a wheatstone bridgeisjust 2 resistor dividers ... but I suppose since it's not being used to measure a resistance one wouldn't call it that. OK got it.

author: Jbliss
author_rank: Regular Contributor
timestamp: June 16, 2015, 03:02:31 am
content: Quote from: suicidaleggroll on June 15, 2015, 04:00:16 pmHi i never intended to use 35v sorry for not being clear but that would be silly. I would use a 9 volt battery. anyone one have any better designs using easily accessible parts.

author: suicidaleggroll
author_rank: Super Contributor
timestamp: June 16, 2015, 03:19:36 am
content: A 9v alkaline battery can't supply that much current, and even if it could the voltage of a "9v" battery drops to more like 7-8v almost immediately, which is too low for that circuit to work.You could switch to a Pch mosfet for reverse voltage protection, ditch the 1R series resistor, and use an LDO instead of the 7805 to get around the voltage problem, but you'd still have issues pulling that much current from a typical 9v battery.Now if you dropped the charging current to 500mA, and moved to a switching converter to improve efficiency, you might have a chance.Here's the data sheet for an energizer alkaline 9v, FYI:http://data.energizer.com/PDFs/522.pdf

author: Jbliss
author_rank: Regular Contributor
timestamp: June 16, 2015, 04:03:41 am
content: Does any one possibly have a design

author: Jbliss
author_rank: Regular Contributor
timestamp: June 16, 2015, 04:09:39 am
content: something like this

author: Richard Crowley
author_rank: Super Contributor
timestamp: June 16, 2015, 04:18:57 am
content: You would be far better off with a pair of AA cells (or even C cells if you have enough space)Those rectangular 9V batteries are the worst possible efficiency and cost per watt.And a linear regulator is a huge waste of power.You didn't mention the raw power source so maybe none of this matters to you??

author: Jbliss
author_rank: Regular Contributor
timestamp: June 16, 2015, 04:31:06 am
content: Or this switching circut would this be more Efficient ??

author: LukeW
author_rank: Frequent Contributor
timestamp: June 16, 2015, 05:29:16 am
content: Quote from: mtdoc on June 16, 2015, 12:00:36 amIt provides a DC bias voltage on the USB D+/D- lines for "supply identification" with certain devices, eg. Apple iPhone, to tell the device what the power supply current capacity is.https://learn.adafruit.com/minty-boost/ichargingHere's a good explanation.Look at the schematic not as a Wheatstone bridge, but two separate voltage dividers connected to 5V and Gnd and D+ and D-.OP: What's your actual power supply input going to be?You can use any 5V regulator, just take your 5V circuit (buck converter or whatever), and just add the same identifier resistors.Personally, I would get rid of R4, do you really need it?Arguably you probably don't really need C2, either.R1 should be changed to 49.9k, because 49.7k is not a standard E-series value.(Also, calling both resistors R1 = bad. Even though they have the same value they should have unique names.)R3 should be changed to 75k (standard E-series, which 74.9k is not), and I would also change R2 to 75k to set the supply current identifier at 500mA.

author: Rick Law
author_rank: Super Contributor
timestamp: June 16, 2015, 05:54:16 am
content: Since you are trying different configuration anyhow, why not consider a cheap 5V buck and just work out the divider to give the D+/D- the voltage level it needs to sense.Note : EDITed - Corrections madeIf you want it to be battery driven, consider 3 or 4 serial 18650 or 14500 (AA size, more compact). With 4X18650 at say 3000mAH each =12,000mAH3000mAH@4x3.7V=14.8V@3000mAH=44WH, you can probably charge your iPhone for a week.On the other hand, if you want it really compact,1x186502x14500and with those mini RC buck, it would be smaller than a tube of travel size tooth paste (3 inches by 1 inch diameter). These RC bucks (see photo) doesn't give the cleanest (noise free) DC, but certainly good enough to use for charging. (Typical 7V to 20V in for 5V out. I would not use more than 18V in. Above 18V, the noise ripple gets very bad.)Note: I am not endorsing this seller nor am I suggesting anything about this seller. Just a link to show what I mean by the mini RC converter.http://www.ebay.com/itm/DC-DC-Converter-Mini-Step-Down-Module-BUCK-Adjustable-3V-5V-16V-for-RC-Plane-/371295065523?pt=LH_DefaultDomain_0&hash=item5672e915b3Good luckRickEDIT corrections made. (Must be asleep...)(a) In serial, 4x18650 you still get only 3000mAH but at 14.8V. Bucking it back down you get 8880mAH at 5V (at 100% efficiency). Nothing runs at 100% efficiency, so say at 80% you get around 7000mAH@5V.(b) 1x18650 doesn't work in buck mode for 5V. For just one, you need boost. So better go with 2x14500.

author: mtdoc
author_rank: Super Contributor
timestamp: June 16, 2015, 06:08:14 am
content: Quote from: LukeW on June 16, 2015, 05:29:16 amExcellent link - just the details I was looking for. Thank you!

author: Jbliss
author_rank: Regular Contributor
timestamp: June 16, 2015, 06:17:13 am
content: Quote from: LukeW on June 16, 2015, 05:29:16 amHi thanks for the clarification. so what your saying is that the first circuit i posted should be fine if i make those modifications. al i need it to do is charge my phone in case it dies in and important situation

author: Jbliss
author_rank: Regular Contributor
timestamp: June 16, 2015, 06:19:25 am
content: Input will be 12 -9v

author: Jbliss
author_rank: Regular Contributor
timestamp: June 16, 2015, 02:00:24 pm
content: After some more research i have decided to use a switching regulator such as this circutwould anyone agree

author: suicidaleggroll
author_rank: Super Contributor
timestamp: June 16, 2015, 03:16:18 pm
content: Quote from: Jbliss on June 16, 2015, 06:17:13 amNot if you still want to use a 9v battery.Quote from: Jbliss on June 16, 2015, 06:19:25 amSo you're no longer planning to use a 9v battery? What are you going to use?You can't keep posting schematic after schematic without telling us what you plan to use to power it.

author: mtdoc
author_rank: Super Contributor
timestamp: June 16, 2015, 03:26:52 pm
content: Quote from: suicidaleggroll on June 16, 2015, 03:16:18 pmI think maybe he meant 12 9V batteries in parallel..

author: Monkeh
author_rank: Super Contributor
timestamp: June 16, 2015, 04:16:36 pm
content: Quote from: Jbliss on June 16, 2015, 06:17:13 amhttp://www.amazon.com/dp/B005X1Y7I2

author: Rick Law
author_rank: Super Contributor
timestamp: June 16, 2015, 04:41:40 pm
content: Quote from: Monkeh on June 16, 2015, 04:16:36 pmGreat suggestion. This looks like a lone 18650 with a boost converter. Can't get easier than that. This production model is packaged very attractively.The boost converter is probably around $1 to $2. So one can make it cheaper - but, the packaging is going to be hard and no where near as slick.

author: suicidaleggroll
author_rank: Super Contributor
timestamp: June 16, 2015, 04:50:34 pm
content: Quote from: Monkeh on June 16, 2015, 04:16:36 pmMy wife has one of those, works great.

author: Jbliss
author_rank: Regular Contributor
timestamp: June 17, 2015, 03:56:18 am
content: Ok just to clear up i will be using a 12v lipo with 3000mah with this circutthanks.

author: Jbliss
author_rank: Regular Contributor
timestamp: June 17, 2015, 05:20:49 am
content: O0