cancel
Showing results for 
Search instead for 
Did you mean: 

Do we have to pay for a bluetooth license if we use the bluetooth within the STM32WB? Looking at the website it cost a lot of money , approx $5k . https://www.bluetooth.com/develop-with-bluetooth/qualification-listing/qualification-listing-fees/ Scott

SSmit.13
Senior
 
1 ACCEPTED SOLUTION

Accepted Solutions
Peter BENSCH
ST Employee

We know that this is inconvenient, especially for small system quantities. But yes, to be Bluetooth compliant requires both registration with the BT SIG and the corresponding payment of licence fees, which are incurred per new project. Details can be found on the BT SIG website you mentioned.

Does it answer your question?

Regards

/Peter

In order to give better visibility on the answered topics, please click on Accept as Solution on the reply which solved your issue or answered your question.

View solution in original post

3 REPLIES 3
Peter BENSCH
ST Employee

We know that this is inconvenient, especially for small system quantities. But yes, to be Bluetooth compliant requires both registration with the BT SIG and the corresponding payment of licence fees, which are incurred per new project. Details can be found on the BT SIG website you mentioned.

Does it answer your question?

Regards

/Peter

In order to give better visibility on the answered topics, please click on Accept as Solution on the reply which solved your issue or answered your question.
SSmit.13
Senior

Thanks Peter.

Unfortunately the approx $5k per project for just to use bluetooth is far to expensive for our projects. We will use WiFi .

Best Regards

Scott

Peter BENSCH
ST Employee

When the problem is solved resp. the question answered, even if it doesn't quite meet your expectations, please mark this thread as answered by selecting Select as best, as also explained here. This will help other users find that answer faster.

Good luck!

/Peter

In order to give better visibility on the answered topics, please click on Accept as Solution on the reply which solved your issue or answered your question.