I'm little confused about the baud rate calculating equation of LPC2148
In data sheet equation is
But in every tutorial I could find
equation is
Second equation sets the baud rate as required.
Am I missing something?
Tutorials that I checked
\$\begingroup\$
\$\endgroup\$
2
-
\$\begingroup\$ I have no experience with LPC2148, but equations are same if U0 stand for oversampling which is usually 16 (or 8 with faster baud rates). \$\endgroup\$– RoktaCommented Oct 11, 2018 at 6:53
-
\$\begingroup\$ @Rokta Here U0 stands for UART0, U0DLL & U0DLM are device latch registers of UART0 \$\endgroup\$– AthulCommented Oct 11, 2018 at 6:56
Add a comment
|
1 Answer
\$\begingroup\$
\$\endgroup\$
3
The only thing that makes sense is that the equation in the datasheet has an error.
That same datasheet says:
The U0DLL and U0DLM registers together form a 16 bit divisor where U0DLL contains the lower 8 bits of the divisor and U0DLM contains the higher 8 bits of the divisor
That suggests it should be 256*DLM+DLL as written in tutorials. In example tables below, it's also treated as 16-bits altogether (upper 8 + lower 8).
Note that April 2012 version of datasheet [1] contains the corrected equation.
-
\$\begingroup\$ I gave DLM = 1, DLL = 97, MULVAL = 14 & DIVADDVAL = 1, If equation was the one with 16 in denominator, Calculated baud rate would be 7743, and Equation with 256 in denominator would give 2478. I checked the value in Keil debugger, and the answer is 2478. So equation with 256 is the right one. \$\endgroup\$– AthulCommented Oct 12, 2018 at 19:15
-
\$\begingroup\$ Can you tell me how you got the denominator as 256 by the info of U0DLL & U0DLM \$\endgroup\$– AthulCommented Oct 12, 2018 at 19:17
-
\$\begingroup\$ Upper 8 bits means
1<<8or2**8, which is 256. \$\endgroup\$– domenCommented Oct 14, 2018 at 9:03