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