This function accepts an analog input in engineering units/ time and produces a pulsed output signal where the pulse rate is
proportional to the analog input.
NOTE: The output of this function block must be directly connected to a digital output. Use function code 79 or 83 for
the Harmony controllers.
Outputs:
Blk |
Type |
Description |
N |
R |
Repetitive pulse output having a duration proportional to the analog input |
Specifications:
Spec |
Tune |
Default |
Type |
Range |
Description |
S1 |
N |
5 |
I |
Note 1 |
Block address of input signal |
S2 |
Y |
0.000 |
R |
Full |
Scaling parameter (units / pulse) |
S3 |
Y |
0.000 |
R |
Full |
Low cutoff (no output below value) |
NOTES:
Maximum values are: 9,998 for the BRC-100, IMMFP11/12 31,998 for the HAC
5.1 Explanation
The <S1> input is an analog signal representing rate in terms of engineering units per unit of time. The S2 term sets the
number of input engineering units that produce a 50 millisecond output pulse. The number of output pulses is according to
the equation:
Suppose the input signal represents zero to 100 gallons per second. It is necessary to obtain one output pulse for every 100 gallons. To accomplish this, set S2 to 100.00. If the input signal is 100 gallons per second, the output is one pulse per 100 gallons or one pulse per second. If the input signal decreases to 50 gallons per second, the output would be one pulse every two seconds, and so on. If the input flow rate is in units per minute or units per hour, then S2 must be scaled accordingly. The application section gives the procedure for determining S2.
The output pulses are always 50 milliseconds in duration, and the minimum time between pulses is 50 milliseconds so there is a limit of ten pulses per second.
5.2 Applications
The output of this function may be used to drive a counter via a digital output. To implement this function to drive a counter,
follow these steps:
Determine the maximum flow rate for the input. Although this function always calculates the number of output pulses by units per second, flows in units per hour and units per minute may be used in the equation given in Step 5 because a factor can be inserted to adjust the scaling.
Determine the maximum input value to the pulse rate function at maximum flow.
Determine the counter capacity as follows:
Divide the results of this equation by 60 to obtain counts per minute or by 3600 to obtain counts per second. The minimum reset time should generally be more than 24 hours.
4. Determine the desired output in terms of counts (or pulses) per hour (assuming the flow rate remains at maximum). Choose the desired counts per time to be less than what was determined in Step 3. It is generally best to make the output differ from the input by a factor of some power of ten (10, 100, 1000 etc.).
5. Calculate the S2 scaling factor using the following equation:
This equation is to be used when the flow rate is units per hour. When the flow rate is in terms of units per minute, use 60 seconds per minute in place of the 3600 seconds per hour and substitute the units of minutes for units of hours in the equation. If the input flow rate is in seconds, omit the conversion factor entirely and use units of seconds for the terms.
Figure 5-1 shows one example of function code 5 used to obtain a count of total pounds of flow. In this example, the range of the flow is zero to 500,000 pounds per hour:
The maximum flow rate is 500,000 pounds per hour.
The input range is zero to 500, so the maximum input is 500.
The counter to be used has six digits and the counter should not reset in less than 24 hours. So the maximum count per hour allowable is:
4. The desired counts for maximum flow is 500. This means each count will represent 1,000 pounds. This is considerably less than the counter capacity for 24 hours determined in Step 3.
5. The Scaling factor is: