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Isolation amplifier with AWG value

Created by: aktsarabdikar at: 7/17/2022 7:40 PM (1 Replies)

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7/17/2022 7:40 PM	Rate (0)
aktsarabdikar Member Joined: 5/4/2021 Last visit: 9/5/2023 Posts: 22 Rating: (0)	Great explanation sir. It applies the same to isolation amplifier with the same AWG value right? For example, if I wire that transmitter sensor to isolation amplifier with <50 ohms input resistance? I just substract 700 with 50 right? ------------------------------------------------------------------------------------------ Split from Permitted transfer distance for current signal 4-20mA.
	Last edited by: Jen_Moderator at: 07/18/2022 05:58:55 New subject after splitting
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7/18/2022 1:25 AM	Rate (0)
danw Gold Expert Joined: 3/28/2010 Last visit: 11/8/2024 Posts: 1074 Rating: (214)	The example (above) was based on a transmitter that requires a minimum of 10Vdc to operate. That minimum 10Vdc means that that device can drive 700 Ohms of loop resistance with a 24Vdc power supply. If the minimum voltage needed by the transmitter (in your case) is different, then the 700 Ohm value changes according to the calculation in the example. Assuming your question is about a field transmitter that needs 10Vdc minimum and that transmitter needs to drive current through the input of an isolation amplifier with a 50 Ohm input resistance, then the remaining loop resistance can be up to 650 Ohms, which is a very, very long run of copper wire. 650 Ohms of copper cable (5+ Km of 22 gauge cable) would likely have so much capacitance that HART would not work, although the 4-20mA signal would work.

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