Help with Isolation amplifiers

In summary: You can use a load resistor R(M) of just a few Ohms to save on power. The opamp circuit looks like this: Opamp: Gain: Input: Output: R(M): 5 Ohms is the minimum resistor required for the LEM sensor current output. .In summary, The LEM current sensor is already galvanically isolated, so you don't need an isolation amplifier to convert the 200mA max output current to a voltage. You can use a load resistor of just a few Ohms to save on power.
  • #1
SEYOboy
3
0
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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  • #2
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

.
 
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  • #3
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).
 
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  • #4
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).
 
  • #5
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.
 
  • #7
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?
 

FAQ: Help with Isolation amplifiers

1. What is an isolation amplifier?

An isolation amplifier is an electronic device that is used to measure and amplify signals while providing electrical isolation between the input and output signals. This isolation helps to eliminate any unwanted noise or interference in the signal, making it more accurate and reliable.

2. How does an isolation amplifier work?

An isolation amplifier uses a combination of optical and magnetic isolation techniques to achieve electrical isolation between the input and output signals. The input signal is converted into an optical or magnetic form, transmitted across the isolation barrier, and then converted back into an electrical signal at the output. This process helps to eliminate any potential ground loops or noise from interfering with the signal.

3. What are the benefits of using an isolation amplifier?

There are several benefits of using an isolation amplifier, including improved accuracy and reliability of the signal, protection against electrical interference, and the ability to measure signals from different voltage levels without the need for additional circuitry. It also helps to protect sensitive equipment from potential damage due to electrical faults.

4. How do I choose the right isolation amplifier for my application?

The right isolation amplifier for your application will depend on several factors, including the type of input signal, the required isolation level, and the desired accuracy and frequency response. It is important to carefully consider these factors and consult with a technical expert to select the most suitable isolation amplifier for your specific needs.

5. Are there any limitations to using isolation amplifiers?

While isolation amplifiers offer many benefits, there are some limitations to consider. These include cost, size, and power consumption. Additionally, isolation amplifiers may have limited bandwidth and response time, which may not be suitable for certain applications. It is important to carefully evaluate these limitations and determine if an isolation amplifier is the best solution for your specific needs.

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