XC2000 datasheet(89/112 Pages) INFINEON

89 / 112 page

XC2230L, XC2234L

XC2000 Family / Econo Line

Electrical Parameters

Data Sheet

V1.3, 2015-02

The timing in the AC Characteristics refers to TCSs. Timing must be calculated using the

minimum TCS possible under the given circumstances.

The actual minimum value for TCS depends on the jitter of the PLL. Because the PLL is

constantly adjusting its output frequency to correspond to the input frequency (from

crystal or oscillator), the accumulated jitter is limited. This means that the relative

deviation for periods of more than one TCS is lower than for a single TCS (see formulas

and Figure 19).

This is especially important for bus cycles using waitstates and for the operation of

timers, serial interfaces, etc. For all slower operations and longer periods (e.g. pulse train

generation or measurement, lower baudrates, etc.) the deviation caused by the PLL jitter

is negligible.

The value of the accumulated PLL jitter depends on the number of consecutive VCO

output cycles within the respective timeframe. The VCO output clock is divided by the

output prescaler K2 to generate the system clock signal

SYS. The number of VCO cycles

is K2

× T, where T is the number of consecutive f

SYS cycles (TCS).

The maximum accumulated jitter (long-term jitter) D

Tmax is defined by:

Tmax [ns] = ±(220 / (K2 × f

SYS) + 4.3)

This maximum value is applicable, if either the number of clock cycles T > (

SYS / 1.2) or

the prescaler value K2 > 17.

In all other cases for a timeframe of T

× TCS the accumulated jitter D

T is determined by:

T [ns] = D

Tmax × [(1 - 0.058 × K2) × (T - 1) / (0.83 × fSYS - 1) + 0.058 × K2]

SYS in [MHz] in all formulas.

Example, for a period of 3 TCSs @ 33 MHz and K2 = 4:

max = ±(220 / (4 × 33) + 4.3) = 5.97 ns (Not applicable directly in this case!)

3 = 5.97 × [(1 - 0.058 × 4) × (3 - 1) / (0.83 × 33 - 1) + 0.058 × 4]

= 5.97

× [0.768 × 2 / 26.39 + 0.232]

= 1.7 ns

Example, for a period of 3 TCSs @ 33 MHz and K2 = 2:

max = ±(220 / (2 × 33) + 4.3) = 7.63 ns (Not applicable directly in this case!)

3 = 7.63 × [(1 - 0.058 × 2) × (3 - 1) / (0.83 × 33 - 1) + 0.058 × 2]

= 7.63

× [0.884 × 2 / 26.39 + 0.116]

= 1.4 ns

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