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

battery-chargingmobile phonespower supplyusbvoltage-drop

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:
Powstro wall charger that does compensate for voltage losses.

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?

Best Answer

Here is an example solution. Schematic is from link:

enter image description here

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.