Differential Pressure: Module 1136_0

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jsmad
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Joined: Wed Feb 11, 2026 4:46 pm

Differential Pressure: Module 1136_0

Post by jsmad »

I am trying to read the pressure from my 1136_0 pressure sensor using the code below (the second line converts the voltage ratio into "inches of H20").

getPressureVoltageRatio = channelPressureSensor.getVoltageRatio()
convertPressureInH2O = (20.073*getPressureVoltageRatio) - 10.037

However, the output gives me a non-zero number when there is no airflow. Is this normal? Do I need to "zero" the reading in the code and is there a native method for doing this or is the best way just to subtract the "zero airflow" output from my above formula?
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burley
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Location: Calgary

Re: Differential Pressure: Module 1136_0

Post by burley »

The sensor has a ±2kPa input range and an output of 0-5Vdc. If the sensor is measuring 0 pressure differential then the output will be 2.5V since that is the middle of the range.
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jsmad
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Joined: Wed Feb 11, 2026 4:46 pm

Re: Differential Pressure: Module 1136_0

Post by jsmad »

I think I am following. I am testing several sensors, but my 1126 seems a little off zero (zero-pressure voltage ratio of ~0.54); is this within spec?Here is the test code I am running with three different pressure modules with the sensor type adjusted for each run: 1126, 1136, & 1137

from Phidget22.Devices.VoltageRatioInput import *

channelPressureSensor = VoltageRatioInput()
channelPressureSensor.setIsHubPortDevice(True)
channelPressureSensor.setDeviceSerialNumber(779045)
channelPressureSensor.setHubPort(3)

channelPressureSensor.openWaitForAttachment(5000)
print("Pressure Sensor Channel Open")

channelPressureSensor.setSensorType(VoltageRatioSensorType.SENSOR_TYPE_1136)

getPressureVoltageRatio = channelPressureSensor.getVoltageRatio()
getPressureSensorValue = channelPressureSensor.getSensorValue()
getPressureSensorUnit = channelPressureSensor.getSensorUnit()
print("getPressureVoltageRatio = " + str(getPressureVoltageRatio))
print("getPressureSensorValue = " + str(getPressureSensorValue))
print("getPressureSensorUnit = " + str(getPressureSensorUnit))

Here is the output for each.

1126:
getPressureVoltageRatio = 0.501671
getPressureSensorValue = 0.093
getPressureSensorUnit = [UnitInfo] (unit: 11, name: kilopascal, symbol: kPa)

1136:
getPressureVoltageRatio = 0.540688
getPressureSensorValue = 0.2034
getPressureSensorUnit = [UnitInfo] (unit: 11, name: kilopascal, symbol: kPa)

1137:
getPressureVoltageRatio = 0.50049
getPressureSensorValue = 0.0086
getPressureSensorUnit = [UnitInfo] (unit: 11, name: kilopascal, symbol: kPa)
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Patrick
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Location: Calgary

Re: Differential Pressure: Module 1136_0

Post by Patrick »

The 1136 is definitely out of spec, it it's reading that when both ports are open. It may be a bad board.

-Patrick