Laversab 6300 User Manual page 69

High accuracy automated pressure controller
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6300 Rev H User's Manual
If the commands are sent correctly, the 6300 will echo back '0'. When the offset is entered, the 6300
Actual value (Pitot or Static) will immediately change to reflect the correction made due to the offset
entered. The objective is to enter an offset that causes the Actual value to match the value of the Pressure
Standard.
An offset may be entered as a value with 4 digits after the decimal point. This allows offset entry of
values as small as ±0.0001 inHg. For example, if the Static Actual value is fluctuating between two values
like 3.200 and 3.201, it may be best to enter an offset of S1=0.0005.
Note: There are no offsets for the 0% and 100% points for the Pitot output. Also, there are no offsets
for the 5% (1.600 inHg) and 100% points for the Static output.
Note: When offset values are entered, they are actually added to the existing offset value already
resident in the 6300.
Verification Procedure
1.
Generate the pressures shown in the Verification points table for a specific point.
2.
Only for point 1, wait at least 5 minutes to allow temperature effects to stabilize. For other points,
wait at least 1 minute.
3.
If an offset needs to be entered for the point, enter it through the RS232 port.
4.
Note the Actual value, either for Pitot or for Static, on the calibration report. Only at point 6, you
will need to note the Actual values for both Pitot and Static.
5.
Repeat steps 1 through 4 for all the 21 points in the table.
6.
The verification process is complete.
7.
If you wish to record the offsets stored in the 6300 (recommended), then through the RS232 port,
send the 'SO=0' command to see all the Static offsets and the 'SO=1' command to see all the Pitot
offsets. The offsets will be displayed in 5% increments. Although the offsets for the 5% points
are never entered, these values are interpolated based on the 10% offset values. Please bear in
mind that the offsets displayed may not match what you entered in step 3 above. This is because
the entered value is added to the existing value already stored in the 6300.
125-9106A
Page 69

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