# Why USB wall chargers don't reach their amperage rating? [closed]

recently I bought a power bank that requires 1,6A of charge current. I tried charging it with a 2A Samsung charger, which came with my tablet. Unfortunately it charges very slow. Then I tried another 1A HTC charger and the result was pretty much the same! I decided to test 5 chargers I had available and the results are confusing me. I loaded each charger with a rheostat in order to see how much load they can take. Practically ALL USB wall chargers failed to deliver.

Here are my results:

1: Samsung 2A : This charger drops the output voltage to 4.5V with 21 ohms, therefore achieving ~ 230mA

2: HTC 1A : This charger drops the output voltage to 4.5V with 15 ohms, therefore achieving ~ 330mA

3: Sony 850mA : This charger drops the output voltage to 4.5V with 18.5 ohms, therefore achieving ~ 270mA

4: Huawei 680mA : This charger drops the output voltage to 4.5V with 12.4 ohms, therefore achieving ~ 400mA

5: old, non-USB Samsung 700mA : This charger drops the output voltage to 4.95V with 7.4 ohms, therefore achieving ~ 670mA This one actually works!

so ... as we can see, USB wall chargers just don't achieve their ratings and the question is WHY?! I am sure they can deliver what they are rated for but they don't. They seem to expect some other signal or something in order to output their full power and I want to know what do they need?! I am sure the 2A charger can deliver this power, because it charges my tablet well! It's also interesting to see that the old charger DOES achieve the current it is rated for. What am I missing here? Do you know how I can unlock the full potential of these chargers?

## closed as off-topic by Brian Carlton, uint128_t, Voltage Spike, Daniel Grillo, brhansJan 23 '17 at 20:51

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "Questions on the use of electronic devices are off-topic as this site is intended specifically for questions on electronics design." – Brian Carlton, uint128_t, Voltage Spike, Daniel Grillo, brhans
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• Seems like a charge bank problem rather than a charger problem. Quantify "it charges very slow", also tell us what is the rated capacity of your charge bank. This will tell if it is exhibiting normal charge times or not. – Adil Malik Jan 22 '17 at 14:20
• The question doesn't relate to the power bank at this point. The power bank charges slowly because the chargers don't deliver the current. The bank is 8000mA on paper but my test show it's more like 4400mA or something like that. – Komental Jan 22 '17 at 14:26
• Asked and answered: electronics.stackexchange.com/questions/5498/… – fergbrain Jan 22 '17 at 14:30
• Smart chargers are not supposed to deliver full power to dumb load devices. Depending how the USB power demand protocol is implemented on the USB host, determines if it can obtain the full power available. USB3 standards apply and the dumb charger wins as it has no protocol limitations for safe charging. – Sunnyskyguy EE75 Jan 22 '17 at 17:18
• @user3528438 the average wall charger HAS NO NEGOTIATION. Newer fast charge ones do, but anything older than 2015 doesn't. – Passerby Jan 23 '17 at 2:44

Ok, here's a simpler answer than my other long answer. So it sounds like this is how you are testing the chargers. Rbus and Rgnd represent the resistance in your "mouse" USB cable.

simulate this circuit – Schematic created using CircuitLab

It sounds like you are measuring voltage after the high resistance mouse USB cable. So let's say you adjust Rheostat so that there is 250 mA flowing. Because of the total 2 Ohms resistance, you will get a voltage drop of 2 Ohms * 0.25 Amps = 0.5 Volts in the cable.

In other words, even though VBUS at the charger is 5V, the voltage across your rheostat (which is where the phone would be) is only 4.5 Volts because 0.5 Volts is lost in the wire: 0.25V across the VBUS wire and another 0.25V across the GND wire.

You need to measure voltage at the charger, so that's why I mentioned making your own USB cable using that DIY connector so you can measure VBUS right at the charger.

But I hope this also explains why a skinny USB cable charges your phone or power bank very slowly. It is because your phone is starved for voltage; it has to reduce its input current so it has enough voltage to work, typically 4.5V (which is actually the USB spec).

• Yes, that's my test setup and you are right that I would have voltage drop and maybe this solves the mystery. The more current I draw, the more voltage drop I get on my mouse wire, so it's much less on the rheostat. It also explains why the old charger did measure right ... I used it's own cable and I guess they compensated for the resistance of the wire, because it did deliver 4.95V/670mA at the end of the plug. :) Thanks. – Komental Jan 23 '17 at 18:01

Hooking up a simple resistor to modern chargers is not a good idea to gauge performance. Most modern chargers are "smart" chargers. They need some sort of "handshake" from the device to determine what current limit to enforce. A simple resistor will likely result in the lowest current limit being applied. This concurs with your results: the cheapest charger is likely ignoring the signaling standards and gives you the highest current limit anyways.

Here is a list lifted from apcmag. It tells you the voltage that must be present on the D+ and D- lines respectively to set a current limit:

• 2.0V/2.0V – low power (500mA)
• 2.0V/2.7V – Apple iPhone (1000mA/5-watt)
• 2.7V/2.0V – Apple iPad (2100mA/10-watt)
• 2.7V/2.7V – 12-watt (2400mA, possibly used by Blackberry) D+/D- shorted together – USB-IF BC 1.2 standard
• 1.2V/1.2V – Samsung devices

Keeping in mind this list, it might therefore be your power bank not applying the right signal to the charger.

Furthermore, what cable are you using? Cheap and long USB cables from China can have relatively high resistance (even a few ohms!). So you might be dropping voltage there instead of from the charger itself.

Furthermore, many cheap charge banks indeed charge very slowly by design. So please define "slow". What is the rated capacity of your charge bank? Seems like a charge bank problem rather than a charger problem. Battery capacity (mAh)/rated charge current(mA) = best case charge time(hours). Is this expected?

EDIT 1: The use of the word handshake (in double quotes for a reason) is perhaps misleading. The "handshake", here, simply consists of the voltages given in the table above, applied across the the D+/D- lines, if this isn't clear.

• YES, I suspect that they are smart chargers but I don't know how to unlock their current. Thanks for this table with voltages, maybe it's the key factor here. I have no voltages on D+ and D-. What I read is that some devices want D+ and D- shorted in order to demand more current but my rheostat shouldn't do that LOL. My power bank is HAMA PowerPack 8000mA and I don't think it's very sophisticated. I guess it doesn't do any handshakes with the charger ... it just expects 5 volts and enough current. – Komental Jan 22 '17 at 16:42
• Getting the right voltages on the D+ and D- lines is not that hard. You just need a few resistors to form potential dividers across the +5v and ground pin. You can easily modify an existing cable. Google the matter. – Adil Malik Jan 22 '17 at 16:49
• so if we follow the rules of your table (for Samsung), I should have 1.2V between +5V and D+ and I should have D+ and D- shorted or I should have 2 voltage dividers each going to D+ and D-? I really want to use the Samsung 2A charger, because it delivers 2A! :) – Komental Jan 22 '17 at 17:08
• Most of your question on Smart charging protocols was answered here: electronics.stackexchange.com/questions/121366/… – Jack Creasey Jan 22 '17 at 19:04
• Most modern chargers are "smart" chargers.They need some sort of "handshake" from the device to determine what current limit to enforce. patently false. Most modern chargers are still dumb chargers, and "tell" the device what it can pull with dumb resistors. No handshake of any kind. – Passerby Jan 23 '17 at 2:45

Chargers from reputable companies will put out their rated current and voltage at least at nominal temperature.

I'm going to guess that the problem you are having is because you've used a long thin USB cable and are seeing high resistance.

Most cables use 28 AWG wires and most chargers set their output a bit high at 5.1V or 5.2V to make up for cable loss since USB allows up to 5.25V. So if you use a 6 foot long typical cable, a ballpark calculation for max current you can sustain is:

I = V / R

I = (5.1V - 4.5V) / (4m*.213Ohms/m + 4*0.030Ohms/contact)

I = 617 mA

Here, .213 Ohms per meter is resistance of 28 AWG wire times 4 meters (2 meters to the phone, 2 meters ground return), and I'm assuming 30 milliOhms for each VBUS and GND contact in the USB connector (phone side and charger side), and 4.5V as the minimum the phone needs.

Some cables are even worse, using 30 AWG or even 32 AWG wire!

So no matter what rating your charger has, there is no way to get high current unless you use a shorter cable preferably with larger conductors for the VBUS and GND lines.

Let's say you get a thick USB cable, the kind using 28 AWG for data lines but 24 AWG for VBUS and GND, and let's say you use a 3 foot long (.5 meter) cable, and lets say the manufacturer is really pushing close to the spec limit and setting the open circuit voltage of your charger to 5.2V. We get:

I = (5.2V - 4.5V) / (1m*.0842Ohms/m + 4*0.030Ohms/contact)

I = 2.43 Amps

This illustrates that cable quality makes a huge impact if you want high current charging. Unfortunately, thick USB cables are unwieldy and can stress the USB connector on the phone.

There are schemes that overcome cable resistance by allowing the phone to request a higher voltage from the charger. Examples are Qualcomm's QuickCharge 2.0/3.0 and USB Power Delivery. But both phones and chargers need to support the standard, otherwise they revert to conventional USB charging.

Also, the other post is not quite right. Your simple rheostat result should be valid if performed correctly. There is no fancy handshaking that limits charger current if you simply connect a resistor from VBUS and GND to test current capability. I can explain further if needed. Hope that helps, -Vince

EDIT: If you want better test results, make your own USB cable with large short wires maybe 1 foot or shorter 24 AWG soldered directly to your meter and rheostat. Measure the voltage at the USB plug itself. Use a DIY USB connector like this one: https://www.amazon.com/Estone-10pcs-Socket-Connector-Plastic/dp/B00HN072RY

• Thanks for your comment. I actually measured the usb test wire to be 1 ohm, so 2 wires = 2ohms. However if we add these 2 ohms to the overall test load the current ratings are not going to be much better. – Komental Jan 22 '17 at 16:29
• That's my point. The 2 Ohms will cause so much voltage drop that there is no way you can get high current. You need to reduce resistance to much less than 1 Ohm. Do the math. – Vince Patron Jan 22 '17 at 17:47
• I don't really get what you mean. I charge the bank with another thick Sony cable. The test cable is from a mouse and it does have resistance but the current is low on high resistance. For example the Samsung 2A charger drops the voltage to 4.5V at 21ohms and if we add 2 more ohms = 23ohms. 5V/23ohms = 217mA. – Komental Jan 22 '17 at 18:08
• I'll post a clearer answer later today with a diagram – Vince Patron Jan 22 '17 at 18:47
• 217 mA through 2 ohms would be almost a half-volt drop, so if that is where you terminated your test, then the cable is at least a contributing reason to why you terminated it. Your admission that you used a cable from a mouse shows that you are going about this without thinking through the issues - mice are low current peripherals that need hyper flexible cords, which is the exact opposite of what you want when trying to move large amounts of power at 5v potential. – Chris Stratton Jan 22 '17 at 20:10