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Table 32. BRP and PT0 values
BRP
PT0_min
PT0_max
Comments
0
1
464
1451
2901
4351
5801
7251
..
1450
2900
4350
5800
7250
8700
..
2
3
4
5
..
43
44
45
..
20416
20880
21344
..
63800
65250
66700
..
Possible timer overflow
63
X
X
The error coming from the measurement of the 29 bit is:
e1 = 6 / [PT0]
It is maximal for the smallest BRP value and the smallest number of ticks in PT0. Therefore:
1 Max = 1.29%
e
To improve precision, the aim is to have the smallest BRP so that the time quantum is the
smallest possible. Thus, an error on the calculation of time quanta in a bit time is minimal.
In order to do so, the value of PT0 is divided into ranges of 1450 ticks. In the algorithm, PT0
is divided by 1451 and the result is BRP.
The calculated BRP value is then used to divide PT0 in order to have the value of (1 +
Tseg1 + Tseg2). A table is made to set the values for Tseg1 and Tseg2 according to the
value of (1 + Tseg1 + Tseg2). These values of Tseg1 and Tseg2 are chosen in order to
reach a sample point between 70% and 80% of the bit time.
During the calculation of (1 + Tseg1 + Tseg2), an error e2 can be introduced by the division.
This error is of 1 time quantum maximum.
To compensate for any possible error on bit rate, the (Re)Synchronization Jump Width is
fixed to 2 time quanta.
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Doc ID 12303 Rev 3
STMICROELECTRONICS [ ST ]
