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This question already has an answer here:

Overall goal: make a couple USB phone & tablet charging ports from a 12V deep cycle battery.

Steps taken so far: I've read that using a resistor based system to step down from 12V to 5V leads to a lot of heat inefficiency so I purchased a SMAKN 12V to 5V 3A voltage regulator. I can simply solder the + & - terminals of a female USB port to this regulator but I thought it would be more efficient to split the 5V 3A output into 5V 1A and 5V 2A to charge a phone AND tablet at their desired wattages.

Confused part: Googling this leads to surprisingly few results, most of which discuss using PSUs where the answers generally revolve around picking a more specific PSU.. [Edit]: While other answers do discuss the underlying solution, they are specific to choosing a PSU and don't get into the details of how to construct a multi-device USB charging circuit.

Using Ohm & Kirchoff's laws, if I build a parallel circuit with a 5 ohm and a 2.5 ohm resistor, this will give me my 1A (5V / 5 ohm = 1A) and my 2.5A (5V / 2.5 ohm = 2 A).

schematic

simulate this circuit – Schematic created using CircuitLab

Jump back on google to find a 5 ohm and 2.5 ohm resistor and...I see a number of results some of which say to operate at 1/4 Watt which is clearly way too low.

I called local electronic shops - they all seemed surprised that I needed such a low resistance resistor with such high wattage.

All of which adds up to me thinking I'm going about this incorrectly or am violating some basic principle? Are my resistors (R1 = 5 ohm, R2 = 2.5 ohm) correct? Are there such resistors for 5V circuits? Is this design going to lead to power loss if only of the two USB ports are in use?

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marked as duplicate by uint128_t, Sparky256, laptop2d, Chris Stratton, winny Dec 29 '17 at 11:13

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • \$\begingroup\$ You should really have a look here. \$\endgroup\$ – Harry Svensson Dec 28 '17 at 2:31
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    \$\begingroup\$ Yes, I'd read that post a couple times before posting this question, it does get to the underlying problem now that I've read the answers posted here, but when framed from the side of choosing a PSU, it still didn't provide enough clarity to explain why my approach was mistaken. \$\endgroup\$ – santeko Dec 28 '17 at 6:44
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You have a fundamental misunderstanding here, that is why there are so few results.

You do not need to split anything. No, it will not be more efficient to split it into a 5V 1A port and a 5V 2A port. Why would that be more efficient? And I don't understand what you think you're splitting. Loads (like a phone or tablet) draw varying amounts of current, and will have a maximum current draw rating. They will never draw more than this, but will draw much less (like when fully charged, for example). The device will draw however much current it needs, and beyond that, it needs 5V. You can't limit current using resistors because the loads aren't linear. Resistors limit current by reducing the voltage, but a phone or tablet will simply stop using the port if the voltage goes outside the USB spec.

When you see a USB port rated at 1A, that means it can provide UP TO 1A. You need an integrated circuit to shut off the port using a MOSFET switch in the event of over draw. You need active circuitry if you want to limit the current draw from the port, you cannot do it using resistors.

But, there is no reason to do this. You have one 5V power supply that can provide 3A. Specify the two ports as having a combined maximum current of 3A, and connect them in parallel. You're done.

Now, USB ports cannot provide more than 500mA of current unless you signal the device it can draw more. Any phone or tablet that conforms to the USB spec will refuse to draw more than 500mA from either of your ports, no matter what you do.

You have to use resistors to make voltage dividers that yield specific voltages across the D+ and D- USB pins, and depending on which you chose, will signal different devices a certain kind of charger port is available. Unfortunately, many tablets use 2.1A, not 2A, so to call a 2A port a tablet port is not really correct. It's insufficient for a great many tablets, while all smart phones expect a 1A charging port minimum. So you're 100mA shy of having the capacity for a true phone and tablet charging device. It might charge YOUR tablet, and if that is all you want, then no worries. But if you plan to have other people use this, you need a slightly beefier 5V regulator.

As for signaling the maximum capacity for each port... well, you can do that using resistors, but high resistance, low wattage ones. They create a voltage divider with the D+ and D- pins of the USB port connected in the middle. However, there is no standard voltage value you want on the D+ and D- pins, it's not part of the USB spec. So Apple generally has its own specific resistor values that are different from android devices, and Samsung has its own thing for its tablets... so its impossible to create a true universal charging port using resistors alone. You have to use a smart port IC, which will detect the type of device connected to a USB port and set the voltages on D+ and D- appropriately. The TPS2546 is one such IC.

If you only plan to use this for yourself, and only need it to work with your specific devices, then you can just use a resistor divider. There is one for 1A Apple devices, then that same divider but with the D+ and D- voltages swapped yields a port for charging 2A apple devices. There is one for sony devices, and one for samsung devices. Then European devices all need only a 200 ohm resistor shorting D+ and D-, by law.

Here is an example of what I mean: enter image description here

I don't know what device(s) you want the ports to work with, but if you simply google the voltages for a given device and current, you can find the information you need. I also highly recommend you read this app note, which is a nice quick and dirty crash course on all this.

Unfortunately, this is one of those situations where something that ought to be simple... is not. C'est la vie.

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    \$\begingroup\$ I believe that what you mean is that USB devices are forbidden to draw more without asking for it; if that's the case, I suggest rewording. (In particular, any number of poorly-behaved USB accessories dispense with enumeration entirely.) \$\endgroup\$ – chrylis Dec 28 '17 at 6:04
  • \$\begingroup\$ @chrylis I understand the individual words in "dispense with enumeration entirely" - but not the whole, can you explain? \$\endgroup\$ – Volker Siegel Dec 28 '17 at 11:46
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    \$\begingroup\$ @VolkerSiegel USB devices are required by spec to identify themselves to the hub/host they're plugged in to and to ask for power if they need much. Many "accessories" (desk fans, lights, batteries) are "dumb" and just draw power without registering properly, which can cause problems if they draw more than the host can provide. \$\endgroup\$ – chrylis Dec 28 '17 at 16:04
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If the phone and tablet are designed to be charged from a 5 volt USB source, they will draw whatever current they need from your 5 volt source. There is no need for you to try to limit or split the current as you suggest.

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  • \$\begingroup\$ Yes but I would I like to charge 2 devices using one 12V to 5V regulator given the 5V output is at 3A and thus could power two devices (say a Phone at 1A and tablet at 2A). Ideally I would like to create a power strip to power 10+ device and if I need one regulator for every USB port, that will double the cost of this project. \$\endgroup\$ – santeko Dec 28 '17 at 1:05
  • \$\begingroup\$ Unless what you're saying that all I need to do is split the 5V 3A output + wire from the regulator and connect 2 usb ports. Each port will technically have 3A to it but if I plug one 5W device in, it'll effectively leave 2A for the other port? \$\endgroup\$ – santeko Dec 28 '17 at 1:14
  • \$\begingroup\$ @skotturi sort of. What exactly happens if your devices collectively try to draw more than the supply rating depends on all sorts of internal behavior that can't be summarized with a single number per device. And BTW, most phones won't charge very fast with only the power wires connected, they expect charging indications bias resistor on the data lines, or even fancier signaling. \$\endgroup\$ – Chris Stratton Dec 28 '17 at 1:16
  • \$\begingroup\$ the 5v output from a regulator is going to be 0-3A depending on how much current is drawn - the 3A limit is simply where it gets too hot (or any internal transistors start to get damaged) it may not be a hard limit, or (if you choose right) it's where some internal circuit powers it down before damage occurs \$\endgroup\$ – Taniwha Dec 28 '17 at 1:18
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    \$\begingroup\$ @Taniwha USB power supplies normally output a constant voltage (5V) until they reach their current limit, and then output a constant current (3A in this case) while dropping the voltage. If shorted, they will shut down. Even the sketchiest chargers I've tested behave like this (although "constant" can be relative). \$\endgroup\$ – Ken Shirriff Dec 28 '17 at 4:30
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The devices regulate charging power themselves, you can just connect them to the 5V rails directly. For the devices to "quick charge", you need to connect additional circuitry to the data lines of the USB port to allow the device to detect the maximum current it is allowed to use.

If you want to make the device safe to use, you need to detect the current, compare it to the maximum allowed on this port, and turn off the port if the device draws more than that. This is a fair bit more complex and expensive to build than buying a ready-made USB charger.

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Your calculations based on Ohm's law are correct but not for the situation you imagined. If you put 5 Ω in series with your load the only way to get 1 A through it is to reduce the load resistance to zero and have no voltage available for your load.

schematic

simulate this circuit – Schematic created using CircuitLab

Figure 1. The only way to get 1 A through 5 Ω on a 5 V supply is to connect it across the supply. Similarly the 2.5 Ω can only pass 2 A when directly across the 5 V supply. There will be no voltage left for the load.

When we use regulated supplies we try to eliminate the series resistance as this would otherwise cause the voltage to drop with increasing current. Instead we regulate the voltage to maintain the 5 V supply from zero to rated current.

As others have pointed out, the current drawn is determined by the charging circuitry in the phone and tablet - not by your external supply.

Be aware that with only USB +5 V and GND that most devices will only draw 0.5 A as this was the original USB specification. To draw more current signals have to be present on the D+ and D- wires to indicate the power rating of the supply.

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You can buy three terminal DC-DC converters, (buck). Get one for each line. Maybe in parallel if you need more current for one line.

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    \$\begingroup\$ This explains very little, George. What is a three-terminal DC-DC converter (buck)? Perhaps you mean something like the 7805 pin-compatible devices. Compare the quality of your answer with the accepted one. \$\endgroup\$ – Transistor Dec 28 '17 at 11:33
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Look at the P7805 or V7805 DC to DC converters from CUI or Recom, they do 36 to 6V in and 5V out. Don't worry about the resistors, the voltage regulates, if you really built this circuit that you describe above then there would be a 5Ω*1A=5V drop across the resistor and the USB device would get little of that current. Or because it's a variable load it the resistor would get half of the voltage and the usb load would get half. Either way it would cause a voltage drop on the USB device, which is bad. Review Kirchoffs laws

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  • \$\begingroup\$ The 7805 is a linear regulator, not a DC to DC converter. A linear regulator like a 7805 is a resistor - specifically a semiconductor one controlled to wastefully achieve via heat dissipation whatever voltage drop will result in a designed output voltage. \$\endgroup\$ – Chris Stratton Dec 29 '17 at 4:53
  • \$\begingroup\$ Dude, they make DC to DC converters that have a 7805 prefix, and an equivalent form factor. They call them 7805 because they are drop in compatible,. \$\endgroup\$ – laptop2d Dec 29 '17 at 5:34
  • \$\begingroup\$ Then call them 7805 alternatives, but don't call them 7805's, because they are not - the 7805 is a wasteful linear regulator. Besides, that won't really help the proposal, unless they also offer current regulation \$\endgroup\$ – Chris Stratton Dec 29 '17 at 5:35
  • \$\begingroup\$ I'm sorry, what part of "7805 DC to DC converter" did you not understand? digikey.com/products/en/power-supplies-board-mount/… \$\endgroup\$ – laptop2d Dec 29 '17 at 5:37
  • \$\begingroup\$ The 7805 part, as that is by definition a linear regulator and not a DC to DC converter \$\endgroup\$ – Chris Stratton Dec 29 '17 at 5:42

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