How to calculation UART bit transmitter and receiver time?

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How to calculation UART bit transmitter and receiver time?

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felixhsu
Contributor III

Hi All

Hardware platform: i.MX7D

As below calculation is correct or not?

Thanks.

baud rate is 115200 / Tref_CLK is 24MHz (0.0416us)

 

uart_01.JPG
UA1 Spec min= 1/115200 - 0.0416 = 8.6384 us

UA1 Spec max= 1/115200 + 0.0416 = 8.7216 us

uart_02.JPG
UA2 Spec min = 1/115200 - 1/(16*115200) = 8.68 - 0.00000054 = 8.67999

UA2 Spec max = 1/115200 + 1/(16*115200) = 8.68+ 0.00000054 = 8.68000054

 

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Carlos_Musich
NXP Employee
NXP Employee

Hi Felix,

UA1 Spec min= 1/115200 - 0.0416 = 8.6384 us
UA1 Spec max= 1/115200 + 0.0416 = 8.7216 us

I don´t know how you calculated these values, but they should be ipg_perclk/RFDIV


Regards,
Carlos
NXP Technical Support
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Carlos_Musich
NXP Employee
NXP Employee

Hi Felix,

UA1 Spec min= 1/115200 - 0.0416 = 8.6384 us
UA1 Spec max= 1/115200 + 0.0416 = 8.7216 us

I don´t know how you calculated these values, but they should be ipg_perclk/RFDIV


Regards,
Carlos
NXP Technical Support
-----------------------------------------------------------------------------------------------------------------------
Note: If this post answers your question, please click the Correct Answer button. Thank you!
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felixhsu
Contributor III

Hi Carlos

Do you have any comments for this case?

Thanks.

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Carlos_Musich
NXP Employee
NXP Employee

Hi Felix,

I am sorry for the delay, that looks corrects now.

Regards,

Carlos

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felixhsu
Contributor III

Hi Carlos

reference as below result,

uart01.JPG

uart02.JPG

T_ref_clk is 24MHz/6=4MHz.

1/T_ref_clk=1/4MHz=0.00000025

1/buad rate=0.0000086806

UA1 Spec min= 0.0000086806 - 0.00000025 = 0.0000084306 s 
UA1 Spec max= 0.0000086806 + 0.00000025 = 0.0000089306 s

Is the above is correct?

Thanks.

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