Hi All.
The baud rate calculator at http://mspgcc.sourceforge.net/baudrate.html is
very useful. Unfortunately it seems to contain an error since it doesn't
take into account the error at the end of the first bit - the start bit -
and therefore gives non-optimal results. Additionally it reports an
incorrect % error in the summary. It also calculates the modulator using
end-bit calculations, not mid-bit which I think might be more useful (see
below).
MSP-UartCal.pl details:
1. $max=$error; needs to be moved outside the for() loop so that the first
bit error is considered.
2. The error % column is incorrect and is a factor of 10x too high by the
last row:
"$err*100/$t" in the print should be "$err*100/$desired"
Mid-bit or End-bit?
=============
This business of baud rates is more complicated than at first is apparent.
On a transmitted signal it would seem reasonable to adjust the bit
boundaries as close as possible to the nominal. This can be done by either
choosing a crystal that gives exact baud rates or by using inexact baud
rates and adjusting the error to a minimum via the modulation register. In
the former case typically a 7.3728MHz crystal would be used in place of
8MHz giving well-behaved divisors (64 for 115200 baud, 768 for 9600, baud
and so on). In the latter case choose the lowest integer value and adjust
the actual bit timing via the modulation register (8MHz @ 9600 baud
requires a divisor of 833.33, so 833 is used). For high clock rates it is
clear that the error is so small (ie. error<<2%) that it isn't really
necessary to use the modulation register correction at all.
With the availability of only a 32.768kHz crystal in low power or low cost
implementations the situation is rather different. Unless the DCO is used
in PLL mode locked to the crystal (which makes using LPM3 mode difficult
since the DCO is then off while asleep) the divisors are much more coarse.
32768kHz @ 9600 baud requires a divisor of 3.41 and the closest integer
value of 3 will give large errors. This is where the modulation register
comes in.
As the MSP430 manual states, the BITCLK can be adjusted from bit to bit
with the modulator to meet timing requirements when a non-integer divisor
is needed. Timing of each bit is expanded by one BRCLK clock cycle if the
modulator bit Mi is set. Each time a bit is received or transmitted, the
next bit in the modulation control register determines the timing for that
bit. A set modulation bit increases the division factor by one while a
cleared modulation bit maintains the division factor given by UxBR. The
timing for the start bit is determined by UxBR plus M0, the next bit is
determined by UxBR plus M1, and so on. The modulation sequence begins with
the LSB. When the character is greater than 8 bits, the modulation sequence
restarts with m0 and continues until all bits are processed. Note that
values are positive deviation from nominal - ie only 0 or 1.
However, just what are we adjusting? On a transmitted signal it would seem
reasonable to adjust the bit boundaries bit by bit as close as possible to
the nominal boundary value, thus minimising the error. But on a received
signal the bit boundary is pretty much irrelevant, since the receiver will
only ever sample at the mid-point of each bit period. For this reason, from
the point of view of the receiver we should be minimising the error at the
mid-point, and not the end point, of each bit. This is important because it
turns out that to do this a slightly different value of modulation is
required for the transmitter and receiver. This would imply that 2
modulation registers are required, both 10 bits long.
Since there is only a single modulation register, it is probably better to
optimise it for the receiver rather than the transmitter, but this is going
to be application dependent. Some systems will only ever transmit, for
example. Since all timing is referenced from the falling edge of the start
bit (which triggers the time-to-first-sampling point and affects all
subsequent sampling points) clearly the most important bit to minimise the
error on is the first (or start) bit. I use a spreadsheet to do these
calculations, which I could post if anyone is interested in using it.
Hope this is of some help.
Regards, Hugh
