 2001 Microchip Technology Inc.

DS00754A-page 9

AN754

EXAMPLE 3:

Maximum Bit Rate

The previous example showed that for a given bus

length, the maximum data rate is inversely affected,

due to oscillator tolerance (as oscillator tolerance goes

up, the data rate goes down). To achieve the maximum

bit rate for a given bus length, the emphasis is placed

on configuring the bit time for the propagation delays

(i.e., adjusting PropSeg to maximum). The oscillator

tolerance must be minimized.

Given the same delays as the previous example:

Since the oscillator tolerance is minimum, the phase

segments and SJW can be set to the minimum. Assum-

ing the bit time is 10TQ total, the PropSeg can be set to

6TQ which sets TQ = 125 ns. Figure 8 shows the bit

timing for maximum bit rate.

FIGURE 8:

BIT TIMING FOR MAXIMUM BIT RATE

REFERENCES

MCP2510 Data Sheet, DS21291, Microchip Technol-

ogy, Inc.

Lawrenz, Wolfhard, “CAN System Engineering From

Theory to Practical Applications”, Springer, 1997

“CAN Specification”, Version 2.0, Parts A and B, Robert

Bosch GmbH, 1991

“ISO11898”, International Standards Organization,

1993

Controller Area Network (CAN) Basics, DS00713

PIC18C Reference Manual, DS39500

PIC18C58 Datasheet, DS30475

CONCLUSION

Setting up CAN bit timing is not an arbitrary process.

The system designer must be aware of the compo-

nents that affect bit timing and compensate to get opti-

mal performance across the network. For example, if

the desired system uses oscillators with the maximum

tolerance, the maximum bus length is reduced. Like-

wise, if maximum bus length is desired, the oscillator

tolerances must be minimized. CAN data rates must

also be considered because the data rate is a third vari-

able that determines maximum length and maximum

oscillator tolerances.

This application note should help assist system engi-

neers design a controller area network for optimal per-

formance based on requirements of the system.

SyncSeg

PropSeg

PS1

PS2

TQ

SyncSeg = 1TQ

PS1 = 1TQ

PS2 = 2TQ