It is not differential, it is an analog output.
Good luck transmitting that voltage 75 meters. If you'll notice, these are identified as BOARD MOUNT pressure sensors. That should give you a clue that they are not intended for transmission over 75m.
You want to send an analog voltage over 75m not digital data. Provided the current drawn is low you should be able to use a screened cable and preferably a low-pass filter at the measuring end to remove any residual noise pick up. The response of that unit is very slow (1mS) so the frequencies down the cable will be low. There may be an issue with it driving a capacitive load (the cable inner to screen capacitance) in which case a buffer amplifier may be needed at the sensor end.
If you want to convert it to serial data format, differential is the way to go. RS485/422/432 should all work to >1Km with no problems.
Brian.
Hi,
If you really want to transmit analog signals, then I recommend "quasi symmetric" signalling over the cable.
Both signals should be transferred via twisted pair shielded cable. 2 twisted pairs: one for supply, the other for signals.
Sensor output --> R-C-R filter --> cable --> common mode filter --> RC filter --> difference amplifier
Sensor GND --> R-C-R filter --> cable --> common mode filter --> RC filter --> difference amplifier
The key to success is the symmetric signal flow with identical impedances.
Then external influence should be equal to both signals ... and thus eliminated by the difference amplifier.
Klaus
Hi,
at the source (sensor) side:
Yes, 100R / 10n / 100R is a good sart.
The output impedance of the sensor should be much smaller than the 100R.
at the reciver side:
it could be higher impedance, maybe 10k.
And if you want to limit bandwidth to about 100Hz (and thus increase S/N ratio) then you may use about 100nF
Klaus
is it a passive wheatstone bridge? I don´t think so. It has built in amplifier with single, buffered output.but for Wheatstone bridge
Hi,
is it a passive wheatstone bridge? I don´t think so. It has built in amplifier with single, buffered output.
Yes, you should use a linear regulator at the sensor side of the cable.
Select a regulator with high initial accuracy and low temperature drift.
Filtering is required and LowESR capacitors to suppress noise. Use a bulk capacitor + ceramics capacitor at the input side of the regulator and a ceramics capacitor at the output side of the regulator. Mind that the regulator error will have 100% impact on the sensor output. --> 5% in input voltage change will result in 5% sensor reading change.
Your part number is not complete
DRRN004xDAA5
Please select the correct type.
Klaus
Hi,
is it a passive wheatstone bridge? I don´t think so. It has built in amplifier with single, buffered output.
Yes, you should use a linear regulator at the sensor side of the cable.
Select a regulator with high initial accuracy and low temperature drift.
Filtering is required and LowESR capacitors to suppress noise. Use a bulk capacitor + ceramics capacitor at the input side of the regulator and a ceramics capacitor at the output side of the regulator. Mind that the regulator error will have 100% impact on the sensor output. --> 5% in input voltage change will result in 5% sensor reading change.
Your part number is not complete
DRRN004xDAA5
Please select the correct type.
Klaus
Hi,
Yes, then you need an extra pair of wires and the extra filters ... but how do you process the two analog signals at the receiver side...to get the true sensor value?
Now you need 4 channels of very symmetric characteristic...
As others mentioned above: I also recommend to go to digital communication.
Otherwise you need high effort to generate, transmit and process analog signals .... but they will never be that exact as the digital transmitted values.
I can´t see the benefit. But for sure you are free to what you want.
Klaus
But what do you think about this voltage reference at the sensor end for sensor excitation/power?
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