Jabra makes a lot of wireless earbuds, and the Elite 85t have the distinction of being among the best that the brand has to offer. Yes, I’m aware that the Elite 7 Pro is the outright leader if you’re looking for the best wireless earbuds, but they’re costlier than the 85t.
With the Elite 85t now selling at $150, you don’t want to miss out on this deal. The standout here is that they have active noise cancellation, and it works incredibly well even in demanding situations with a lot of ambient noise in your vicinity.
That alone gives the Elite 85t an edge over its siblings, and another area where they’re outstanding is calls; with wind protection and three mics on each earbud, you can make crystal-clear calls when you’re on the go. The controls on each earbud are also highly customizable, and you can invoke Google Assistant and Alexa with ease.
Then there’s the sound quality. Jabra has always managed to deliver on this front, and the Elite 85t come with large audio drivers with a booming bass and soaring highs. The fit also ensures that the earbuds stay in your ears even during workouts, and on that note, you get IPX4 water resistance. You can easily adjust the sound balance to your tastes, and with Bluetooth 5.1, these earbuds connect effortlessly to your phone.
Save $80 on the Jabra Elite 85t
Jabra Elite 85t
$150 at Amazon $110 at Walmart
The Elite 85t do a fantastic job tuning out ambient noise, deliver excellent sound quality, and you get all the extras: customizable sound, water resistance, wireless charging, and digital assistant integration. If you want high-end earbuds with a great value, these are the ones to beat.
As for battery, the Elite 85t last 5.5 hours on a full charge, with the case offering another two full charges. So you’ll only need to charge the case once a week even with heavy use, and you get wireless charging as standard here.
The Elite 85t may not be the latest from Jabra, but they continue to excel in this category, and at $150, these are among the Best Black Friday headphone deals you’ll find.