PIC18F44800TESLASHML datasheet(337/484 Pages) MICROCHIP

337 / 484 page

 2004 Microchip Technology Inc.

Preliminary

DS39637A-page 335

PIC18F2480/2580/4480/4580

23.9.2

TIME QUANTA

As already mentioned, the Time Quanta is a fixed unit

derived from the oscillator period and baud rate pre-

scaler. Its relationship to TBIT and the Nominal Bit Rate

is shown in Example 23-6.

EXAMPLE 23-6:

CALCULATING TQ,

NOMINAL BIT RATE AND

NOMINAL BIT TIME

The frequencies of the oscillators in the different nodes

must be coordinated in order to provide a system wide

specified nominal bit time. This means that all oscilla-

tors must have a TOSC that is an integral divisor of TQ.

It should also be noted that although the number of TQ

is programmable from 4 to 25, the usable minimum is

8TQ. There is no assurance that a bit time of less than

8 TQ in length will operate correctly.

23.9.3

SYNCHRONIZATION SEGMENT

This part of the bit time is used to synchronize the

various CAN nodes on the bus. The edge of the input

signal is expected to occur during the sync segment.

The duration is 1 TQ.

23.9.4

PROPAGATION SEGMENT

This part of the bit time is used to compensate for phys-

ical delay times within the network. These delay times

consist of the signal propagation time on the bus line

and the internal delay time of the nodes. The length of

the Propagation Segment can be programmed from

1TQ to 8 TQ by setting the PRSEG2:PRSEG0 bits.

23.9.5

PHASE BUFFER SEGMENTS

The phase buffer segments are used to optimally

locate the sampling point of the received bit within the

nominal bit time. The sampling point occurs between

Phase Segment 1 and Phase Segment 2. These

segments can be lengthened or shortened by the

resynchronization process. The end of Phase Segment

1 determines the sampling point within a bit time.

Phase Segment 1 is programmable from 1 TQ to 8 TQ

in duration. Phase Segment 2 provides delay before

the next transmitted data transition and is also pro-

grammable from 1 TQ to 8 TQ in duration. However, due

to IPT requirements, the actual minimum length of

Phase Segment 2 is 2 TQ, or it may be defined to be

equal to the greater of Phase Segment 1 or the

Information Processing Time (IPT). The sampling point

should be as late as possible or approximately 80% of

the bit time.

23.9.6

SAMPLE POINT

The sample point is the point of time at which the bus

level is read and the value of the received bit is deter-

mined. The sampling point occurs at the end of Phase

Segment 1. If the bit timing is slow and contains many

TQ, it is possible to specify multiple sampling of the bus

line at the sample point. The value of the received bit is

determined to be the value of the majority decision of

three values. The three samples are taken at the sam-

ple point and twice before, with a time of TQ/2 between

each sample.

23.9.7

INFORMATION PROCESSING TIME

The Information Processing Time (IPT) is the time

segment starting at the sample point that is reserved

for calculation of the subsequent bit level. The CAN

specification defines this time to be less than or equal

to 2 TQ. The PIC18F2480/2580/4480/4580 devices

define this time to be 2 TQ. Thus, Phase Segment 2

must be at least 2 TQ long.

TQ (

µs) = (2 * (BRP+1))/FOSC (MHz)

TBIT (

µs) = TQ (µs) * number of TQ per bit interval

Nominal Bit Rate (bits/s) = 1/TBIT

This frequency (FOSC) refers to the effective

frequency used. If, for example, a 10 MHz external

signal is used along with a PLL, then the effective

frequency will be 4 x 10 MHz which equals 40 MHz.

CASE 1:

For FOSC = 16 MHz, BRP<5:0> = 00h and

Nominal Bit Time = 8 TQ:

TQ = (2 * 1)/16 = 0.125

µs (125 ns)

TBIT = 8 * 0.125 = 1

µs (10-6s)

Nominal Bit Rate = 1/10-6 = 106 bits/s (1 Mb/s)

CASE 2:

For FOSC = 20 MHz, BRP<5:0> = 01h and

Nominal Bit Time = 8 TQ:

TQ = (2 * 2)/20 = 0.2

µs (200 ns)

TBIT = 8 * 0.2 = 1.6

µs (1.6 * 10-6s)

Nominal Bit Rate = 1/1.6 * 10-6s =

625,000 bits/s

(625 Kb/s)

CASE 3:

For FOSC = 25 MHz, BRP<5:0> = 3Fh and

Nominal Bit Time = 25 TQ:

TQ = (2 * 64)/25 = 5.12

µs

TBIT = 25 * 5.12 = 128

µs (1.28 * 10-4s)

Nominal Bit Rate = 1/1.28 * 10-4 =

7813 bits/s

(7.8 Kb/s)

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