AD652KPZ datasheet(8/28 Pages) AD

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AD652

Rev. C | Page 8 of 28

Another way to view this is that the output is a frequency of

approximately one-quarter of the clock that has been phase

modulated. A constant frequency can be thought of as

accumulating phase linearly with time at a rate equal to 2πf

radians per second. Therefore, the average output frequency,

which is slightly in excess of a quarter of the clock, requires

phase accumulation at a certain rate. However, since the SVFC

is running at exactly one-quarter of the clock, it does not

accumulate enough phase (see Figure 7). When the difference

between the required phase (average frequency) and the actual

phase equals 2π, a step-in phase is taken where the deficit is

made up instantaneously. The output frequency is then a steady

carrier that has been phase modulated by a sawtooth signal (see

Figure 7). The period of the sawtooth phase modulation is the

time required to accumulate a 2π difference in phase between

the required average frequency and one quarter of the clock

frequency. The sawtooth phase modulation amplitude is 2π.

TIME

φMOD (t)

ACTUAL PHASE

EXPECTED

PHASE

MODULATION

AVERAGE

CARRIER FREQUENCY

VOUT (t) = COS (2π× fAVE × t + φMOD (t))

Figure 7. Phase Modulation

The result of this synchronism is that the rate at which data may

be extracted from the series bit stream produced by the SVFC is

limited. The output pulses are typically counted during a fixed

gate interval and the result is interpreted as an average

frequency. The resolution of such a measurement is determined

by the clock frequency and the gate time. For example, if the

clock frequency is 4 MHz and the gate time is 4.096 ms, a

maximum count of 8,192 is produced by a full-scale frequency

of 2 MHz. Thus, the resolution is 13 bits.

OVERRANGE

Since each reset pulse is only one clock period in length, the

full-scale output frequency is equal to one-half the clock

frequency. At full scale, the current steering switch spends half

of the time on the summing junction; thus, an input current of

0.5 mA can be balanced. In the case of an overrange, the output

of the integrator op amp drifts in the negative direction and the

output of the comparator remains high. The logic circuits

simply settle into a divide-by-two of the clock state.

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