Thanks for your help.
I don't have Sky Q.
When I tried to find out about 5GHz capability I got this:
802.11n 802.11g 802.11b 802.11ac 802.11n 802.11a
I obviously don't have a clue what it means!
No I haven't got an ethernet port on my laptop.
Then there is not much you can do about the wireless connection issue. It would be unusual to find a laptop without an Ethernet port. If you have a spare USB port, then you can get a Ethernet to USB adapter.
Some laptops don't have a physical RJ45 connector on them, they rely on a docking station instead to provide that functionality. What make and model is yours?
"802.11n 802.11g 802.11b 802.11ac 802.11n 802.11a"
Those are difference WiFi standards and from those listed .11ac would typically be the fastest. Do you have a frequency range choice usually 2.4GHz or 5GHz?
Have you tried a WiFi Scanner app to see if changing your wireless channel might improve throughput if the local area is congested?
I have an Asus M513I
I've had a go at changing the wireless channel and will see how my zoom connection is when I next use it tomorrow evening.
Thank you for your help and advice everyone. 🙂
Hi,
Did that help at all? I had that issue but it seems like it is something to do with the firewall specific to Zoom connection.
It was definitely better but I still got the 'unstable connection' message twice.
You really need to know which band your laptop is connected to before changing WiFi channel. For W11 go to Settings > Network & Internet & double-click WiFi. Then double-click on (YourNetworkName) Properties & scroll down to where it shows the network band & properties. If you're connected to 2.4GHz then obviously changing the 5GHz channel won't help & vice-versa.
If you have the original Smart Hub or earlier you can split the WiFi bands to give them different names (SSID) so you know for sure which you're connecting to. Generally you want 5GHz for speed, 2.4GHz for distance. But if you have a Smart Hub 2 you can't do that.
As already said, a USB to ethernet adapter will take WiFi out of the equation altogether.