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DESIGN OF A CUSTOM ASIC INCORPORATING CAN™ AND 1 ...

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n w 1<br />

m J n<br />

j <br />

<br />

bit<br />

wm Bm Cj<br />

(4.5)<br />

j hpm Tj<br />

<br />

A value of zero for w can be used. The iteration process proceeds until convergence (i.e.<br />

n 1 n<br />

m m<br />

0<br />

m<br />

<br />

w w ). Equations 4.1 through 4.5 do not assume anything about how identifiers and hence<br />

priorities are chosen [67].<br />

From analysis of Equations 4.1 through 4.5, the minimal latency time occurs when the<br />

CAN bus is running at its maximum speed of 1 Mbps and when a frame consisting of zero<br />

data bytes is transmitted (i.e. sm 0 and bit 1μs ). Solving Equation 4.2 for a value of Cm<br />

yields:<br />

C<br />

C<br />

m<br />

m<br />

34 <br />

47 8.5 47 55.5 μsec 11-bit header<br />

4 <br />

<br />

101<br />

<br />

54 <br />

67 13.5 67 80.5 μsec 29-bit header<br />

4 <br />

<br />

<br />

The maximum latency time occurs when the bus is running at its slowest speed of 10 kbps and<br />

when a frame consisting of eight data bytes is transmitted (i.e. sm 8and<br />

bit 100μs ).<br />

Resolving Equation 4.2 for Cm yields:<br />

C<br />

C<br />

m<br />

m<br />

(4.6)<br />

34 88 98 <br />

47 8 8 10μs 111 10μs 1.355msec=1355μsec (11-bit)<br />

4 <br />

<br />

<br />

4 <br />

(4.7)<br />

54 88 118 <br />

67 88 10μs 13110μs1.605msec=1605μsec (29-bit)<br />

4 <br />

4 <br />

From this analysis, the worst-case latency value comes from the bus operating at the<br />

slowest speed of 10 kbps when each message contains the maximum amount of data at eight<br />

bytes and the values for the queuing jitter (Jm) and worst-case queuing delay (wm) are zero. The

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