
MOTOROLA
SYSTEM INTEGRATION MODULE
M68HC16 Z SERIES
5-28
USER’S MANUAL
The periodic interrupt timer modulus counter is clocked by one of two signals. When
the PLL is enabled (MODCLK = 1 during reset), fref is used with a slow reference os-
cillator; fref 128 is used with fast reference oscillator. When the PLL is disabled (MOD-
CLK = 0 during reset), fref is used. The value of the periodic timer prescaler (PTP) bit
in the periodic interrupt timer register (PITR) determines system clock prescaling for
a factor of 512, can be selected. The value of PTP is affected by the state of the MOD-
CLK pin during reset, as shown in Table 5-11. System software can change PTP val-
ue.
Either clock signal selected by the PTP is divided by four before driving the modulus
counter. The modulus counter is initialized by writing a value to the periodic interrupt
timer modulus (PITM[7:0]) field in PITR. A zero value turns off the periodic timer. When
the modulus counter value reaches zero, an interrupt is generated. The modulus
counter is then reloaded with the value in PITM[7:0] and counting repeats. If a new val-
ue is written to PITR, it is loaded into the modulus counter when the current count is
completed.
The following equation calculates the PIT period when a slow reference frequency is
used:
The following equation calculates the PIT period when a fast reference frequency is
used:
The following equation calculates the PIT period for an externally input clock frequen-
cy on both slow and fast reference frequency devices.
5.4.7 Interrupt Priority and Vectoring
Interrupt priority and vectoring are determined by the values of the periodic interrupt
request level (PIRQL[2:0]) and periodic interrupt vector (PIV) fields in the periodic in-
terrupt control register (PICR).
Table 5-11 MODCLK Pin and PTP Bit at Reset
MODCLK
PTP
0 (External Clock)
1 (
÷ 512)
1 (Internal Clock)
0 (
÷ 1)
PIT Period
PITM[7:0]
() 1 if PTP = 0, 512 if PTP = 1
() 4
()
f
ref
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=
PIT Period
128
() PITM[7:0]
() 1 if PTP = 0, 512 if PTP = 1
() 4
()
f
ref
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=
PIT Period
PITM[7:0]
() 1 if PTP = 0, 512 if PTP = 1
() 4
()
f
sys
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=