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Background: Hello. I am an Electronics Engineering student, and while I am familiar with basic power concepts and power circuits, I don't know how laptops are generally built and I want to cross check this before I purchase an adapter.

Specifics: I have a Lenovo Yoga 510-14ikb which uses a 4.0-1.7mm barrel jack to charge. The power adapter is rated for 20V 65W. I want to buy a USB C to 4017 barrel jack "decoy", which negotiates 20V from a suitable USB PD powersupply/charger and feeds it to the laptop.

Question: As USB PD chargers are designed to default to 5V output, it is bound to happen that the charger will give the laptop 5V. Maybe for a brief time, or that it will fail to negotiate the needed 20V. So, how would most laptops react to this? Would it damage the laptop, or would the laptop possibly overload the charger?

I am trying to rule out that this pairing will cause malfunction or something serious. Help is appreciated.

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If the laptop supports PD, then chances are it has a PD charge controller integrated circuit.

Connecting 5V to the barrel connector itself will probably do nothing. Likely it has diodes preventing flowback to the power supply, to prevent it back-powering the supply and wasting battery power when the charger is switched off. Other than that the 5V will do nothing as far as the laptop is concerned, as it is not enough to push any real current into a battery that will have a higher voltage than it does.

If that "decoy" controller is well designed then it will be able to function at 5V in order to do the power negotiation, and will then negotiate with the PD device to supply a higher voltage and current. Before that negotiation it will likely have a FET (or one on each line) that isolates the input from the battery until the PD controller enables power to go to the battery, which will mean that there is no chance of back-powering the 5V supply, and will make sure that the batteries only charge from a "good" supply.

If the power supply doesn't properly negotiate more than 5V then nothing at all will happen. There simply isn't enough power or a high enough voltage to charge the battery.

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  • After consulting with an engineer, your answer is very close so I accepted it. Only thing to add is that the voltage is actually reduced through a buck converter, so the reverse blocking is done with mosfets and not just a diode in the schematic they showed me.
    – Anas Malas
    Jun 7, 2021 at 6:09
  • A diode is simply the absolute minimum I would expect to see for protection and indeed a FET of some description would be preferable as it can be controlled at the same time as providing the basic protection required.
    – Mokubai ♦
    Jun 7, 2021 at 10:00
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Probably nothing should happen - voltage too low.

If the battery is dead (voltage too low) nothing happens.

I do not recommend long term wrong voltage, but for a short time and correct the issue, nothing should go wrong.

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