can i charge my phone with copies

Giraffe76

Hi all,

i have a xaiomi pocophone F1 phone which i bought from abroad and came with the charging cables etc

I got a 2nd 2 pin european adaptor(exactly like others ive got),usb type c adaptor (with xiaomi on the side in small writing) and a usb type c charging cable from ebay that i thought would do the job but when i first used them it took alnight for my phone to charge to 95%.

Hope this all makes sense any help would be most appreciated

kind regards Mark

Treboeth

Yes you can charge with copies.

But they may not be
a. Genuine
B. The same  voltage as the original.

Either buy the Original from the manufacturer or a reputable brand ,Anker are usually recommended.

Giraffe76

Would anker be ok to use with a xaiomi phone?

Treboeth

Anker will be fine with any phone.
https://www.amazon.co.uk/Anker-Charger-PowerIQ-Technology-Motorola-Black/dp/B016LPMFUA/ref=sr_1_10?keywords=usb anker charger&qid=1564615252&s=telephone&sr=1-10
£18.98 with cable
Motorola
https://www.amazon.co.uk/Motorola-TurboPower-SPN5917A-Charger-Family/dp/B07L4YXMZL/ref=sr_1_6?keywords=usb motorola charger&qid=1564615149&rnid=1642204031&s=electronics&sr=1-6
£10 if you have Prime

TerryClarke

Yeah, Anker is absolutely ok, just be aware of chargers from China (Aliexpress, etc.).

Giraffe76

would this not be a better option

https://www.amazon.co.uk/Anker-Charger-Delivery-PowerPort-Indicator/dp/B07GSNZJ8R/ref=dp_ob_title_wld

Giraffe76

My applogies but i am now confused.
The charger in the first link was a 10.99 charger and i couldnt find any link to a type c cable worth 7.99 to make the total 18.98 but i could find the motorola lead worth £10.
are you saying that either option would work? if so which would do a better job?

Mark

TylerDurden

On that link scroll about half way down to the Frequently bought together section

Treboeth

Always the danger with amazon the links may update.
The Motorola for £10 is fine.

Giraffe76

Thank you all for your help and advice.
I have now bought a charger plug and lead.