2 wire differential pressure transmitter 4-20ma for hvac
Micro differential pressure transmitter is suitable for differential pressure measurement of dry gas. It is the responsibility of the operator to verify that the equipment is suitable for the operating conditions of the application. This series of micro differential pressure transmitter adopts piezoresistive pressure sensor chip, and uses the film resistance on the substrate to carry out zero point correction, zero point temperature compensation and sensitivity compensation. The high-performance and stable silicon chip package makes it high static pressure resistance, anti-interference, stable and reliable.Therefore, the product can be applied to the micro differential pressure field combination of various gas measurement. It is an ideal micro differential pressure measuring instrument in the field of industrial automation.
Features 2 wire differential pressure transmitter 4-20ma for hvac
Measuring medium: gas (compatible with contact material, humidity <90%, no condensation)
Overall material: shell ABS engineering plastic
Diaphragm silicon chip (contact)
Silicone for impulse pipe (contact)
Φ 6 pagoda mouth H59 copper (contact)
Cable locking head nylon (locking wire diameter 5-8mm)
LCD display without backlight
Parameter 2 wire differential pressure transmitter 4-20ma for hvac

Parameter 2 wire differential pressure transmitter 4-20ma for hvac
Pressure mode: | differential pressure |
Output signal: | 4-20mA, RS485 |
Power supply voltage: | 12-36vdc |
Accuracy class: | 0.5% FS (range ≥ 1kPa) 1% FS (range < 1kPa) |
Working conditions: | medium temperature - 40-100 ℃ |
Ambient temperature | - 40-85 ℃ |
Temperature compensation: | - 10 ~ 60 ℃ |
Seismic performance: | 10g (20... 2000Hz) |
Response frequency: | ≤ 50Hz |
Stable performance: | ± 0.3% FS / year (range ≥ 1kPa) ± 0.5% FS / year (range < 1kPa) |
Temperature drift: | ± 0.03% FS / ℃ (within the temperature compensation range) |
Protection grade: | IP65 |
Load characteristics: | current type load ≤ {(us-7.5) ÷ 0.02 (US = supply voltage)} Ω |
