Help with Isolation amplifiers

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Discussion Overview

The discussion revolves around the need for isolation amplifiers in two specific applications: measuring current using a LEM current transducer and measuring temperature with Pt 100 probes. Participants explore the requirements for converting current and voltage outputs to suitable levels for a data acquisition system.

Discussion Character

  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant questions whether the LEM current sensor is already isolated, referencing its datasheet which indicates galvanic isolation from the current being sensed.
  • Another participant clarifies that the LEM sensor outputs current and that a voltage output is needed for the Personal DAQ 3000 card, expressing concerns about power losses when connecting a resistor to convert current to voltage.
  • A participant suggests that the term "isolation amplifier" may be misunderstood and emphasizes the need for current-to-voltage conversion rather than isolation, proposing the use of a small resistor and an op-amp to achieve the desired voltage output.
  • There is a discussion about the nature of the current being measured, with one participant asking if it is AC, DC, or a mix, noting that an offset may be required for AC signals to fit within the desired output range.
  • Another participant confirms that the current is AC and discusses the use of a low-value precision resistor to minimize power loss, prompting further inquiry about the power rating needed for the resistor and the op-amp circuit design.

Areas of Agreement / Disagreement

Participants express differing views on the necessity of an isolation amplifier for the LEM current sensor, with some asserting that the sensor's existing isolation suffices while others argue for the need for additional circuitry to convert current to voltage. The discussion remains unresolved regarding the optimal approach for the applications described.

Contextual Notes

Participants have not reached a consensus on the best method for achieving the required voltage outputs from the current and temperature measurements. There are also unresolved considerations regarding the specific circuit designs and power ratings for components used in the proposed solutions.

SEYOboy
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Hello. I need help in finding the following isolation amplifiers:
1. I'm measuring current of approximately 600 A using LEM current transducer LF 1005-S (which gives an output of 200 mA for 1000 A). LEM transducer need DC power supply, and i use bipolar 15 V dc power supply. I need isolation amplifier (200 mA - 500 mA to 10 V) that i'll connect to the DC supply and LEM transducer. Then i'll take voltage from amplifier and connect it to the acquisition card Personal DAQ 3000 USB (card has only voltage input).
2. I have 6 Pt 100 probes connected in series with current source (1 mA). I need isolation amplifiers for Pt 100 probes (measuring temperature) which converts cca 200 mV - 500 mV to 10 V.
Thanks
 
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SEYOboy said:
Hello. I need help in finding the following isolation amplifiers:
1. I'm measuring current of approximately 600 A using LEM current transducer LF 1005-S (which gives an output of 200 mA for 1000 A). LEM transducer need DC power supply, and i use bipolar 15 V dc power supply. I need isolation amplifier (200 mA - 500 mA to 10 V) that i'll connect to the DC supply and LEM transducer. Then i'll take voltage from amplifier and connect it to the acquisition card Personal DAQ 3000 USB (card has only voltage input).
2. I have 6 Pt 100 probes connected in series with current source (1 mA). I need isolation amplifiers for Pt 100 probes (measuring temperature) which converts cca 200 mV - 500 mV to 10 V.
Thanks

Welcome to the PF.

On your first question, isn't the LEM current sensor already isolated? The datasheet says that it is galvanically isolated from the wire whose DC/AC current it is sensing...

http://www.google.com/url?q=http://www.lem.com/docs/products/lf%25201005-s%2520sp12.pdf&sa=U&ei=8-J1T-eTAoewiQK3jr2nDg&ved=0CBAQFjAA&usg=AFQjCNEIUDLje6hxKyMaCO5jqPw84QG8RQ

.
 
Last edited by a moderator:
berkeman said:
Welcome to the PF.

On your first question, isn't the LEM current sensor already isolated? The datasheet says that it is galvanically isolated from the wire whose DC/AC current it is sensing...

http://www.google.com/url?q=http://www.lem.com/docs/products/lf%25201005-s%2520sp12.pdf&sa=U&ei=8-J1T-eTAoewiQK3jr2nDg&ved=0CBAQFjAA&usg=AFQjCNEIUDLje6hxKyMaCO5jqPw84QG8RQ

.
LEM sensor has current output, and I need voltage output for personal daq 3000 card. If you connect LEM sensor with resistor there'll be too huge losses (around 1.8 W for 200 mA and 46 Ω resistor).
 
Last edited by a moderator:
SEYOboy said:
LEM sensor has current output, and I need voltage output for personal daq 3000 card. If you connect LEM sensor with resistor there'll be too huge losses (around 1.8 W for 200 mA and 46 Ω resistor).

Ah, I think it is just a language issue with the term "isolation amplifier". I think you are asking about how to do the current-to-voltage conversion efficiently, which has nothing to do with isolation amplifiers.

The term Isolation Amplifier usually refers to an amplifier that gives some amount of galvanic isolation between input and output:

http://en.wikipedia.org/wiki/Isolation_amplifier

You would use such an amplifer if you had a sensing circuit connected to the AC Mains, and you wanted to get that sensed information into a low-voltage circuit (like a microcontroller). The isolation may be done with optoisolators or other means.

But in your first application with the current sensor, the LEM module already gives you the galvanic isolation. I think you are asking how to convert the 200mA max output current of the sensor (which does appear to have a current mode output) into a voltage for reading by your ADC. Since it is a current mode output, you should be able to use the minimum size resistor specified in the datasheet (5 Ohms), and then just use an opamp with the appropriate gain to give you the maximum output voltage for your ADC range (which I think you list as 10V).
 
BTW, will the current you are sensing be AC, DC or a mix? If AC, then you will have to add an offset to the output of your current-to-voltage converter opamp circuit, if you want the output to be in the 0V to 10V range.
 
SEYOboy said:
LEM current transducer is this one http://docs-europe.electrocomponents.com/webdocs/028f/0900766b8028ffae.pdf
Current is AC. Input in daq card is +-10V.

So you can use a load resistor R(M) of just a few Ohms to save on power.

Let's say you use a 1 Ohm precision (0.1%) resistor. What power rating does it need to have in your application? And what does the opamp circuit look like to turn the voltage across R(M) into your +/-10V output signal?
 

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