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I have a lenovo yoga 730 13-ikb without a charger, it uses TYPE C usb charger with PD

specifications for the charger in online stores are: AC Input: 100-240V ~ 1.5A 50-60Hz DC Output: 20V 3.25A (65 watt charger)

Laptop Specs sheet say: AC adapter 65W USB Type-C AC adapter

I was thinking of creating a DIY charger with 20V power pins and 65w from some other source.

Can I damage the USB C port or the laptop at all if I don't do it with the Power Delivery specs, I mean does it need it to be PD compliant or giving it the required power as the spec says be good to go without worrying that it will only work for this laptop and nothing else?

I don't care if it does not work for my phone or any other lower consumption device, I don't want it to be a universal USB type C charger, just want t to charge this specific laptop.

is Power Delivery supposed to be switching voltages once it is powering the laptop , maybe for powering different parts of it, or once negotiated it continually delivers settled voltage until disconnection?

Can I Bypass the negotiation of power by bridging some pins or somehow modify it so it just goes with single power available, is negotiation mandatory?.

I have seen more questions with answers in but apparently all talk about a charger compliant with PD, my goal is not to create a PD charger, just directly be able to charge the laptop.

Update While checking the service manual I found this Checking the AC ADAPTER so apparently it always has the 20V to start with, does anybody know if PD chargers output by default the maximum power?

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  • \$\begingroup\$ It'll be cheaper and easier to just buy any USB-C power supply. It doesn't have to be a Lenovo specific one, as long as it conforms to the proper USB PD specs. \$\endgroup\$ – Tom Carpenter Mar 30 at 22:40
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    \$\begingroup\$ "my goal is not to create a PD charger, just directly be able to charge the laptop" - the laptop requires a PD compliant charging source - you can't just force 20V into it and hope for the best, nothing good will come of that. \$\endgroup\$ – Tom Carpenter Mar 30 at 22:41
  • \$\begingroup\$ I already have everything for just creating it that way, so dont think it would be cheaper , but if negotiation has to occure then , yes it would be safer to just not try it., but the charger specs nor the product spec sheet say anything about being PD. I am assuming it is because they sell both options for the same laptop, one says PD the other does not.1 is 45 dlls the other is 20., im checking the hardware manual now \$\endgroup\$ – arana Mar 31 at 0:09
  • \$\begingroup\$ In which case I stand corrected. If there are really only two pins on the charger, they have obviously designed their system to cope with directly applying 20V without negotiation. Wire up your supply as indicated, and make sure not to plug it in to anything else. \$\endgroup\$ – Tom Carpenter Mar 31 at 1:19
  • \$\begingroup\$ It looks like the service manual you are looking at is for a slightly different laptop, 730 model 15ikb. I can't see the squarish USB-size (Type-A) port on 13ikb model. \$\endgroup\$ – Ale..chenski Mar 31 at 2:27
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This laptop model only has USB-C PD charging option. PD has controller on each side of the connection (inside the charger and inside the laptop) and those two controllers will negotiate voltage/current. This negotiation is a must in PD connections.

Initial connection happens at standard USB 5V/max 0.5A, where devices then negotiate V/I increase in specific steps. PD protocol is rather complex and if you were to try to feed it permanent 20V DC it most likely will burn PD controller inside the laptop.

Just get any USB-C PD charger capable of minimum 65W (that is what you laptop rated at), and it should work.

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It is not uncommon to modern laptops to have dual power option. Lenovo laptops use squarish plugs 20V (which should not be confused with Type-C connector), and their Type-C ports have an alternative means to power the laptop if the port is connected to a powerful hub, or the Type-C port can supply power to something like external monitor. To function as power provider or power consumer, Type-C port uses Power Delivery negotiations. These negotiations involve CC-channel pulls up-down first to determine the base role of "link partners", and then PD negotiation may revert one of roles (power role and/or data role) into necessary direction. If you just make a 20V source over Type-C connector, nothing will happen.

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  • \$\begingroup\$ this one does not have the squarish connector., I have other lenovo that does, you can see in the image it is not a squarish connector but a Type C, and also it states +20V, it does have another type C that you can use to charge other devices but only one that caan be used for charging it. if it is giving 20+ when it is unplugged from the laptop, then it is not a PD correct? or it would just give to the initial +5V? \$\endgroup\$ – arana Mar 31 at 20:04
  • \$\begingroup\$ @arana, unconnected/unplugged Type-C ports don't output any voltage initially, per specifications. Regarding the posted image, it is unclear. But Type-C connector has 24 pins, not just "1 and 2". Please post a better picture of it. \$\endgroup\$ – Ale..chenski Mar 31 at 20:32
  • \$\begingroup\$ thing is I dont have the original charger, I can give you a picture of the port in the laptop, in the laptop i.imgur.com/oGkJQ7p.png the left one is the one for charging, the right one can be used to charge a phone or something else from the powered laptop., here is from the laptop manual also i.imgur.com/bQWcqxF.png , the top one is the one I have (the one with type C), but the other illustration confuses me also, as it shows only 2 pins :S, or maybe they only refer to those 2 pins andd do not show the rest? \$\endgroup\$ – arana Mar 31 at 23:15
  • \$\begingroup\$ here is from the laptop manual also i.imgur.com/bQWcqxF.png , the top one is the one I have and this other is for the 730-15(squarish) i.imgur.com/0oSSTUD.png, maybe they just didnt check that instructions on the 730-13 type C one are totally wrong and just changed the picture and not the instructions? \$\endgroup\$ – arana Mar 31 at 23:36

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