Huawei Honor V8 Max with 6.6-inch Display, 3GB RAM spotted on TENAA


Honor V8 Max

Honor V8 Max a new device with EDI-DL00 model no from Huawei is certified on TENAA and reveals most of its key specs. The device will sport a 6.6-inch Display which we can call a phablet. This new device will directly compete with Xiaomi Mi MaxAsus Zenfone 3 Ultra. Let’s check out the specs of the new V8 Max.

Honor V8 Max TENAA

Specifications of Honor V8 Max

  • 6.6-inch QHD AMOLED Display
  • Octa core 2.3 GHz HiSilicon Kirin 950 SoC
  • 3GB RAM
  • 32GB internal storage
  • External storage up to 128GB via MicroSD card
  • Android 6.0 Marshmallow with EMUI 4.1 on the top
  • 13MP Rear Camera with dual tone LED flash
  • 8MP front-facing Camera
  • Dual SIM
  • Fingerprint Sensor
  • 178.8 x 90.9 x 7.2mm dimensions
  • 4G LTE with VoLTE
  • 4,400mAh battery

Honor V8 Max will sport a 6.6-inch QHD (2K) AMOLED Display, Octa core 2.3 GHz HiSilicon Kirin 950 processor, 3GB RAM, 32GB internal storage inside it. The internal storage can be extended up to 128GB using MicroSD card. The device will be powered by Android 6.0 Marshmallow with EMUI 4.1 on the top.

Also Read  Honor Pad 2 with 8-inch Display, 3GB RAM, 4800mAh battery announced in China

Honor V8 Max TENAA

The primary camera is a 13MP shooter with Dual tone LED Flash and the front-facing camera is an 8MP shooter. The device comes with Dual SIM with 4G LTE with VoLTE and a fingerprint sensor at the back. It weighs around 219g and dimensions are 178.8 x 90.9 x 7.2mm. It has a 4,400mAh battery under the hood which will probably come with fast charging support.

Still there’s no news on the pricing and the announcement date yet. But Huawei has called an event on 11th July for the honor 8 launch so it is expected that the phablet will be announced there.