How to set Uart Baud Rate beyond the max limit

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How to set Uart Baud Rate beyond the max limit

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davidzhou
Contributor V

Hi,

I have K60 tower. According to document, the maximum baud rate for UART can be 1/16 of clock freq:

The maximum baud rate is 1/16 of related source clock frequency

But when I change the setting in the user_config.h

#define BSPCFG_SCI5_BAUD_RATE 115200

to

#define BSPCFG_SCI5_BAUD_RATE 921600

it shows an error:

145499_145499.pngpastedImage_0.png

How to set a higher baud rate and what is the limit for this setting? And how to set beyond the current limit?

 

Thank you

David Zhou

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TICS_Fiona
NXP Employee
NXP Employee

Hello David

If the Bus Clock that you configured in CPU component is 60MHz, the UART5 of Baud Rate can be up to 3750000. So there is no problem to configure the Baud Rate into 921600

I checked in the source code, the macro definition for BSPCFG_SCI5_BAUD_RATE is set correctly and it should work well. You can ignore the message of exceeds max limit.

Best Regards

Fiona Kuang

Technical Information & Commercial Support

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Alice_Yang
NXP TechSupport
NXP TechSupport

Hello David,

What about the bus clock ?

Also the full name of your board , for example TWR-K60N100M or others.

BR

Alice

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davidzhou
Contributor V

Hi Alice,

TWR-K60F120M

Clock:      120M

It displayed in red color with tag message of exceeds max limit. But it compiles and runs. I try to find where in the IDE stores and checks this limit.

Thanks,

David Zhou

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TICS_Fiona
NXP Employee
NXP Employee

Hello David

If the Bus Clock that you configured in CPU component is 60MHz, the UART5 of Baud Rate can be up to 3750000. So there is no problem to configure the Baud Rate into 921600

I checked in the source code, the macro definition for BSPCFG_SCI5_BAUD_RATE is set correctly and it should work well. You can ignore the message of exceeds max limit.

Best Regards

Fiona Kuang

Technical Information & Commercial Support

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davidzhou
Contributor V

Hi Fiona,

I tried 921600 it worked. But above that baud rate, it has problem of Tx/Rx.

Thank you,

David Zhou

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