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CPISpeed()
In this case, the
instruction could be used to change the bit rate from the
250 kbps default to 125 kbps, allowing a longer cable. Then, referring to the table:
Network
topology, bit rate, and cable length
(p. 36), the longest cable length possible is 1200 ft with
the use of a termination plug, or 1000 ft without.
c. What should the scan BufferOption be set to for this example?
The general rule is that the number of scan buffers should be enough to hold 2 seconds'
worth of data. With a scan rate of 20 ms, the BufferOption should be 2 s ÷ 0.02 s =
100. The
CRBasic Scan()
instruction would be written this way:
Scan
(20,msec,100,0)
d. If there were four other VOLT 116s on the network with this same configuration, what
restraints would there be on the system?
Because the system can run in pipeline mode, each VOLT 116 can start its measurements at
the same time, not requiring any change in the scan rate. The data generated (102.4 kbps)
would be multiplied by a total of five VOLT 116s, giving 512 kbps total. In order to keep up
with this data output, the CPI speed would need to be increased to 1000 kbps. This reduces
the total allowable cable length to 50 ft with the use of a termination plug, or 1 ft without
(see
Network topology, bit rate, and cable length
(p. 36)).
D.2 Example 2
a. How many differential measurements can a VOLT 116 or VOLT A108 make at 5 ms (200Hz) if
input reversal is used?
This example uses the measurement time equation for measurements with input reversal
from
Appendix D
(p. 51):
The equation shows that making multiple measurements by using the Repetitions
parameter within a measurement instruction is more efficient than using multiple
measurement instructions. Solving the previous equation for repetitions gives the
following:
With f
= 30000 Hz, T
= 100 µs, and measurement time = 5 ms (5000 µs), repetitions
N1
settle
is calculated to be 7.866, so 7 repetitions, or measurements, are possible at that scan rate.
VOLT 100 series
52
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