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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?

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    \$\begingroup\$ so, first of all, I'm not sure this is actually what's happening: How do you measure these voltages? My guess is that increased current draw increases voltage ripple, and your voltmeter misinterprets that as increased (average) voltage. \$\endgroup\$ – Marcus Müller May 25 at 12:38
  • \$\begingroup\$ @MarcusMüller I tested it both on a linear bench power supply and a switched-mode bench power supply, of which the former one is ripple-free. Same results. The bench power supply has a built-in voltage meter, but the voltage shown there at the beginnning is of course higher than at the end of a long cable. \$\endgroup\$ – neverMind9 May 25 at 12:40
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    \$\begingroup\$ I'd like to know your measurement device and where exactly you measured; the power supplies themselves are probably OK in how they work :) Say, you don't happen to have an oscilloscope that you can attach to the supply-side end of the USB cable and observe while increasingly drawing more current? And the same on the phone-side end of the cable? \$\endgroup\$ – Marcus Müller May 25 at 13:10
  • \$\begingroup\$ @MarcusMüller I don't own an oscilloscope yet, but I already have some portable multimeters. But I already know how Ohm's law and voltage drops work. More current = more voltage drop in same cable. Therefore, my goal is to compensate for the voltage drop in real time to “fool” the phone's charging controller into thinking that the cable has as low resistance as possible, in order to draw the maximum available current. \$\endgroup\$ – neverMind9 May 25 at 16:54
  • \$\begingroup\$ "The goal is that the mobile phone draws it's full power over any length of wire." - and what if the voltage increase over thin wire will cause cable melting and potentially set a fire? \$\endgroup\$ – Ale..chenski May 25 at 17:59
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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.

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  • \$\begingroup\$ The Texas Instruments is a fine company indeed, but this proposal of their young application engineer ( eetimes.com/author.asp?section_id=36&doc_id=1327857# ) contradicts USB charging specifications and should be discarded. \$\endgroup\$ – Ale..chenski May 25 at 18:04
  • \$\begingroup\$ More, if the TI proposal advises to stay within USB legal limit (5.5V per latest change in specs), there is no need to go and compensate for anything, just set the output to max of +5.5 V and it is done. \$\endgroup\$ – Ale..chenski May 25 at 18:08
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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: enter image description here

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:

enter image description here

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.

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  • \$\begingroup\$ I measured that my USB charger does actually increase output voltage by 0.1V for each 1A of current. \$\endgroup\$ – neverMind9 May 25 at 17:48
  • \$\begingroup\$ Added cable length does not mean less maximum current handling capabilities but just more resistance, which leads to less current drawing. \$\endgroup\$ – neverMind9 May 25 at 17:49
  • \$\begingroup\$ If your charger does increase output with load above the legal level of 5.5V, it is out of specifications and is dangerous to use. You should stop using it. If it stays within 5.5V but varies with load, it is likely just a cheap deficient design with a poor load regulation. \$\endgroup\$ – Ale..chenski May 25 at 18:13
  • \$\begingroup\$ Qualcomm Quick Charge and MediaTek PumpExpress don't mind applying 9V and 12V to ordinary USB cables. Never had a problem. In 2014, the Galaxy Note 4 had Qualcomm Quick Charging 2.0, which worked with 9V 1.67A. USB-B could not do 3A to reach 15W, so a higher voltage was needed. Never had an issue. \$\endgroup\$ – neverMind9 May 25 at 21:12
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    \$\begingroup\$ @neverMind9, you misunderstood the concept of current over "thin cable" and various "Fast charge" approaches. They increase voltage precisely for the purpose to avoid high currents. And if you apply high voltage to ordinary USB devices, they might blow up. Only the devices that are designed specifically for elevated voltages as for PD or QC connections will withstand these voltages, and a PD or QC charger won't output higher voltages unless some negotiation is completed successfully. Applying high voltage boldly is not a right approach. \$\endgroup\$ – Ale..chenski May 25 at 21:21

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