hart475 series resistance issue
Thread Content
Let’s take a look – are both of these methods for connecting resistors acceptable? The connection method shown below is used in all the references; is the connection method above also valid?
Series resistance occurs due to insufficient load
This post was last edited by cqdfwy on 2019-7-11 at 17:39. The resistor and the transmitter are connected in parallel; the resistor causes current to be Shunting, which results in the output signal current from the transmitter not matching the current received by the secondary meter downstream. Therefore, it is an incorrect connection.
Yesterday, one of the meters here wouldn’t connect to 475. The technician tried using the method of connecting resistors as described above, but still no communication was possible; in the end, it was determined that there was a problem with the meter. Is there really only this one way to connect a 250-ohm resistor, with no other methods available?
The resistors must be connected in series, but the operator can be connected in various parallel configurations
Yes, it can only be connected in the following way: a 250-ohm resistor is connected in series with the transmitter (it can be connected to either the positive or negative terminals of the transmitter).
The configuration shown above is parallel; to distribute the current, the internal resistance of the power supply must be taken into account. If that internal resistance is high enough to be ignored, then it is possible to connect it in that way. , otherwise it won’t work
In my opinion, more consideration should be given to the transmitter’s output resistance (usually, the input resistance of instruments is high while the output resistance is low). When the output resistance is low and is comparable to the parallel resistance, current sharing will definitely occur.