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Negative impedance output from 5V phone charger!?!?!?! ?!?!?!?!?!?!?!?!?!?!?!?!?!?!?!?!?!?!?!?!?!?!?!?!?!?!?!?!?!?!?!?!? Answered

I acquired a handly little USB multimeter thing that goes in-line with a phone charger to measure voltage, current, resistance (?), and voltage on the data rails as well. I was able to go through my phone and car chargers and dispose of the crappy ones that would not charge at a fast rate.

However, I noticed that my samsung (OEM) chargers appear to somehow compensate for the resistance of the USB cable! The charger would output 5.1V under no load but that would rise to 5.3V under load! It does this for any device, not just my samsung phone. The voltage rise seems to be proportional to the current. My guess is that the supply has some positive feedback within a certain current range and voltage range that allows it to roughly compensate for the drop in a long USB cable.

How would such behavior be implemented in a feedback loop in a way that avoids parasitic oscillation. It seems that it would require positive feedback, or some way to modify the voltage reference voltage in proportion to output current.. I am intrigued! Any insights?

Discussions

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Downunder35m

Best Answer 2 years ago

Good USB adapters do that on purpose.
Mine act the same way.
Comes down to the little chip used - all integrated ;)
From my understanding of the USB charging protocol a compliant charger has a no load voltage of 5.1 - 5.2V.
This is considered to be the "overshoot" to compensate for wire resistance and connector resistance so the voltage drop from them combined should result in exactly 5V at the end of the cable when measured with a 50Ohm load.
During the actual charge the voltage rises depending on the current to a specified max of 5.42V.
This is to compensate for the losses of the actual chargin electronics in the connected device.

That is why proper chargers provide the voltages on the data lines for Apple and normal products.
Only then can the phone know that it can draw to specified max rating (proper cable of course).
Modifying a cheap unregulated charger to provide let's say 2.1A for the charging could be done with a few resistors for the data lines.
But it only works if during max load the voltage won't drop under 5.3V as then the connected device should lower the draw to prevent damage to the charger.
Sadly most manufacturers go the cheap way and simply state that you have to use a certified charger, usually this means the device will try to get the max even if the provided voltage is too low.
In return the charging circuit overheats and if not properly protected fails.

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-max-Downunder35m

Answer 2 years ago

Interesting. How is just positive feedback / resistance compensation done? Do you know of the source to get the USB charging specifications, or are these propriatory to samsung. To my knowledge no other adaptor does that. Only the fast charger and the "5.3V 2A" generic samsung charger appear to do that.

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Downunder35m-max-

Answer 2 years ago

I "collected" the USB standards from Wiki when the specifications for fast charge via UBS2.0 came out.
Not much to be found apart from the specs, so no details on how to implement it.
From the (car) chargers I checked and kept I can only say their microchip does all what is required.
Unless you find some internal specs for them it will be quite hard to replicate.
But there is always the simple option of adjusting the output voltage through the feedback resistor.
Most of these chips work from around 7V up to 40V input side.
Output can be adjusted from around 3V to close to input voltage.
Some go down to 1.25V though.
So instead of re-inventing the wheel I would just modify a good car charger ;)
The mains adapters are a bit different and to be honest I have not disected a Samsung yet.
Principle is the same but of course they work directly by modifying the input side of the transformer to adjust the output load.

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-max-Downunder35m

Answer 2 years ago

im really just curious about the analog implemenration. I was trying to get that to work in a linear supply design, I suspect I can add a current sense amp to modify a voltage reference value so that as more current is drawn, the output voltage actually goes up.

Just an intresting thing. Eventally the op amp will saturate and reach a maximum value. It reminds me of how hysterisis is done in the analog world.

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-max-Downunder35m

Answer 2 years ago

I like nice basic switch mode supplies. I have been using the simple switchers and this one seems very simalar. But It does not appear to have the USB cable compensation features. I have actually build up a switcher very simalar to that using an LM339 comparator not so long ago to power an XHP70 LED. It was very problematic (really bad ripple and parasitic oscilations) but I did achieve 70--80% efficiency.

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Downunder35m-max-

Answer 2 years ago

Might be just a coincidence but I kept this one because it actually does increase the voltage with the load.
Just don't ask me why or ho it does it LOL

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-max-Downunder35m

Answer 2 years ago

OK, I managed to build up a linear supply design (using LT3080 that does EXACTLY this. It is able to compensate for 1 ohm of output impedance. at low frequencies, the output impedance is actually -1 ohm. I have (attempted) to upload the .ASC file for everyone to see. Check it out!

Screenshot from LTspice negitive_Z_power_supply.png
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-max--max-

Answer 2 years ago

The green line is voltage across the current source. As you can see, there is absolutely no feedback from the load side of the circuit back to the main circuit. The current shunt causes 100mV/A to be amplified by the op amp by a factor of 10, to 1V/A, which offsets the voltage at the set pin. The 500K resistor sets the voltage due to a 10uA current source inside the LT3080. :D

I'm sure this can be modified with any transfer function. So long as you have the mathematical model of the parasitic properties of the length of wire, (LCR I'm sure) then you should be able to compensate.

One of my fears was that the thing would oscillate, but it actually seems pretty stable so long as there is a small capacitor from the set pin to ground. This limits the bandwidth of the positive feedback since, well, positive feedback = oscillation!

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Vyger

2 years ago

I miss the good old days when things were either plugged in or not plugged in. Batteries were rationed for good behavior, I never got many of them. And batteries were never included with anything.

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-max-Vyger

Answer 2 years ago

the days when "flashlights" got their name because the zink-mercury cells of yesterday were expensive and didnt last long, so you'd turn the light on only for breif periods just long enough to see? Yeah I dont miss those days. Mostly 'cause I'm too young to remember that lol

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Vyger

2 years ago

Was your keyboard stuck or did your cat sit on the keys? That would be a reach since those keys are at the opposite ends. Front and back I suppose.

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-max-Vyger

Answer 2 years ago

Copy and paste worked well. A 555 timer connected to the !? button works well too. ;)

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iceng

2 years ago

It is possible that it ( the four wires in a USB cable ) are being used to sense delivered voltage... Also your "USB thing" adds an extra little voltage drop when measuring current...

Four_Wire_Connection_Diagram.JPG
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icengiceng

Answer 2 years ago

Oh yes, be sure to click the pic to see the whole image !

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-max-iceng

Answer 2 years ago

It'd be cool if the USB standard worked that way, as implementing a supply that does essentially a 4 wire measurement to compensate lead resistance would enable good charging.

I believe some manufacturers are working on ways so that when a phone connects and detects the correct charger, it can just straight-up connect the USB directly to the battery letting the charger push the full charging current directly into the battery, minimizing heat production in the phone.

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-max-iceng

Answer 2 years ago

I dont think remote sense is how the USB charging specification works. The USB charger seems to output random 2 to 4 V on the data lones, my understanding is that its a simple way for the phone to recognize a walladaptor from a computer USB port. Do you have any USB specification data to support this theory?

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iceng-max-

Answer 2 years ago

No published data, it just made sense when you have two extra lines to use them to do what you describe.