My phone charges on 5v .7a and I'm afraid to buy a cheap $2 eBay charger because I've heard bad things about them. Walmart sells one that is their store brand (onn) and it's rated as 5v .6a. Will the .1a difference really matter? It's still made in China but it's $7 so hopefully it's better than the eBay cheapo.
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(Solved) Car charger question
- Thread starter jmcook79
- Start date
Rukbat
Extreme Android User
It'll make about a 16% difference in the amount of charging time - that's all the current difference changes. The phone takes whatever it needs - up to the capability of the charger. The only thing I'd be concerned about is thw quality of the chatger - will it last, will it overheat? Since it's Walmart, it won't be too much trouble to bring it back if there's a problem. (Just keep touching it every 15 minutes or so for the first hour or teo to make sure it's still cool enough to not peel the paint off the wall.)
I've used chargers from Ebay shipped from China for the past 10 years on all my phones and have NEVER had a problem. The phone will take the charge it needs. Just might be a little faster or slower with a different charger. I charge overnight on my night stand. 100% in the morning and only down to 43% now at 9PM.
Any charger is fine!
Any charger is fine!
Well I know Walmart buys from china, I just thought since it actually had a brand name (even if it's Walmart's store brand) it may be a higher grade of China 
Honestly I've never had a problem with cheap China chargers either, but this is my first phone I've ever owned that cost over $50 and I am babying it lol. I was just worried about articles that I've read about voltage differences and how cheaply made chargers sometimes go out of bounds and provide too much voltage.
Honestly I've never had a problem with cheap China chargers either, but this is my first phone I've ever owned that cost over $50 and I am babying it lol. I was just worried about articles that I've read about voltage differences and how cheaply made chargers sometimes go out of bounds and provide too much voltage.