How to add 0.2V per Ampère that device draws? (Voltage drop compensation)

Question

I have noticed that many mobile phone chargers already do this: For each 1 A of current that the mobile phone draws, the charger's output will be increased by 0.05V or 0.1V or something similar, to compensate for the ohmic voltage loss due to the resistance of the wire (USB cable).

Example for 0.1V voltage drop compensation, assuming base voltage 5.1V:

• 0A → 5.1V
• 1A → 5.2V
• 2A → 5.3V
• 3A → 5.4V

This Powstro HKL-USB32 wall charger does exactly that:

That's because through longer cables, the charging IC of mobile phones sees a voltage drop which makes the charging controller falsely consider the charger exhausted. The charger partially compensates for that, to “fool” the mobile phone into drawing a higher current.

When connecting a female USB port to the terminals of a variable laboratory bench power supply (correct polarity regarded), increasing the voltage manually subsequently only increases the charging speed on powerbanks, which only pay attention to the momentary voltage, not the initial voltage drop.

Mobile phones however choose a current at the moment when connecting it to the charger, and then hang onto it, no matter how much the voltage increases afterwards. (No overvoltage. Many mobile phones, including the one I used to test it, support 9V and 12V for fast charging standards. All tests at my own risk.)

Example: If I select 7 volts on the bench power supply that powers the female USB port, then enable power output while the mobile phone is already connected to the USB port, the mobile phone (which supports all input voltages from 5V to 14V, therefore no overvoltage) does choses 0.83A. When I decrease the voltage to 6 volts, it is still 0.83A. If I increase to 10V afterwards, it is still precisely 0.83A. The current can not increase, only decrease under the following conditions: If source voltage goes too low, near 4.5V, or if the battery is becoming sated (full). It can also happen that the phone chooses e.g. 6V 1.6A. Increasing to 12V would result in overwattage for the battery, therefore the current would also decrease. But the phone tries to maintain it's initially decided Ampèrage.

How do I increase the voltage in conjunction to the current?

• The goal is that the mobile phone draws it's full power over any length of wire.
• Is there any module for that? (I could not find one by searching for “Voltage Drop Compensation” and “cable drop compensation”.)
• I don't have a bench power supply with a “sense voltage” feature yet.

There are already boost voltage modules, but they are just as useful as increasing the source voltage manually.
One thing I could do is placing a boost voltage module near to the tip of the USB cable. But is there a way to increase the source voltage in conjunction?

Answer 1

Here is an example solution. Schematic is from link:

If you want your power supply to compensate for cable voltage drop without sensing the actual voltage at the load, then your power supply needs an output impedance that is a negative resistance. More precisely, it needs to be the complex conjugate of the cable's impedance, but a negative resistance corresponding to the cable resistance is all you need.

Negative resistors don't exist of course, but you can synthetize one by measuring the output current with a shunt and a current sense amplifier, and tweaking the power regulator's output voltage accordingly. In the above schematic, an offset voltage is added into the regulator's feedback network to shift its output voltage in the desired direction.

You won't be able to do this with your bench supply, since it does not have an extra input for controlling its voltage, but if you build your own power supply then it isn't difficult.

The tricky bit is that synthetized negative impedances will become unstable and oscillate or misbehave in certain conditions. For example, if the load is a resistor Rl, and your power supply has output impedance -Ro, then if Ro>Rl then the output voltage will shoot to infinity. The sum of all impedances in the circuit must be positive!

If your load is constant power (like a buck DC-DC with constant output current) then it will have a negative input impedance too, increasing the voltage makes it draw less current, at this point your "negative output impedance" power supply will decrease its voltage, making the phone draw more current, and it can oscillate.

So, you must also consider the stability of your closed loop system, which is also load-dependent. It will most likely depend which charging phase the phone is doing. So it is best not to overdo it and add just the little bit of negative resistance that you need, not more.

Answer 2

I have noticed that many mobile phone chargers already do this: For each A of current that the mobile phone draws, the applied voltage will be increased by 0.05V or 0.1V or something similar.

That's because through longer cables, the charging IC of mobile phones sees a voltage drop which makes the charging controller falsely consider the charger exhausted. The charger partially compensates for that, to “fool” the mobile phone into drawing a higher current.

These statements and ideas are unsupported by any evidence and are misleading. As a matter of fact, USB chargers are designed to go into partial "constant current" mode when the load (Phone) exceeds the charging supply capability. Here is an example from Battery Charging specifications:

This "special feature" of chargers (as compared to simple voltage supplies) to drop voltage as load increases above parimary charger capability allows "smart phones" to accommodate long and skinny cables when they accidentally used for charging. The phones do have a sense of input voltage and REDUCE their intake if the voltage at input starts dropping. This is designed to continue the charge at least at some less-optimal rate other than just to fail any charge at all. Here is the spec for "mobile device" with USB port, from BC1.2:

IN CONCLUSION: USB charging specifications explicitly prohibit any voltage drop compensation at charger side to compensate for weak cables. Instead of "compensation", mobile devices employ "load correction". No one needs to "fool" any part of charging schema and set cables or device afire.

Answer 3

There is no simple solution.

This is exactly why so many residential Power Conditioners in Pakistan/India contribute to harming the unstable power grid.

Increasing the load causes a voltage drop and servo tap changers then demand more power to equalize, which then causes greater transformer and grid loss. Then they compensate the grid and at nite the voltage shoots up to over 300 V in some areas.

If was load was a constant current or even constant resistance, that would be ok, but often load current rises with voltage by some exponent >1 which may result in more error correction than your cable resistance of 200 mOhms.(est.)

Using a current sense with a 50mV drop max to correct the feedback for max current of say 3A requires a shunt of 60 mOhms and a Diff Amp. to drive the FB input using the Vfb as a reference for the conductance error correction. This can cause instabilities due low ESR loads where dI/dV = 1/ESR and nit a constant current or linear load R.

The way this is solved in universal laptop chargers is to remote-sense the plug voltage at the cable end for feedback so the source rise to compensate for the drop.

0.2V/1A = 200 mohms.

Good luck. I expect you will learn how to stabilize it after it oscillates and learn some enhanced methods of overshoot and resonance compensation.