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Is is a proper way to solve the problem of high common-mode voltage?

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Would you mind to help me download it and transfer it to me? I can't open it or open a blank page (from google search)...

 

Thank you.

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Message 21 of 30
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here we go

Greetings from Germany
Henrik

LV since v3.1

“ground” is a convenient fantasy

'˙˙˙˙uıɐƃɐ lɐıp puɐ °06 ǝuoɥd ɹnoʎ uɹnʇ ǝsɐǝld 'ʎɹɐuıƃɐɯı sı pǝlɐıp ǝʌɐɥ noʎ ɹǝqɯnu ǝɥʇ'


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Message 22 of 30
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Got it! Thank you!

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Message 23 of 30
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Before spending time and money in building your own system:

Use a industrial isolation amplifier:

Here a source directly made for shunt measurements (sorry, in german, buy just call them 🙂 )

http://www.relmatic.ch/an/signalwandler-trennverst%C3%A4rker/im-industriegeh%C3%A4use-multispannungs...

 

Or (ab)use a isolation bridge amplifier  like this one:

With two additional resistors you can use it for shunt measurement. Or again ask them if they have something for shunts.

 

 

Greetings from Germany
Henrik

LV since v3.1

“ground” is a convenient fantasy

'˙˙˙˙uıɐƃɐ lɐıp puɐ °06 ǝuoɥd ɹnoʎ uɹnʇ ǝsɐǝld 'ʎɹɐuıƃɐɯı sı pǝlɐıp ǝʌɐɥ noʎ ɹǝqɯnu ǝɥʇ'


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Message 24 of 30
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I am confused.  In the other thread you indicated that your switching frequency was 30-40 kHz.  Now you are talking about 350 us rise time which implies a much lower frequency.

 

I put together a simple program to show the effect of reduced bandwidth on measurement accuracy.  I generated a square wave. I ran it through a first order Butterworth low pass filter with the cutoff frequency set to twice the signal frequency.  This filter is a good representation of the behavior of the amplifier with the same bandwidth. I measure the RMS value of the original signal and the filtered signal. For the values indicated above the error is 17%.  To get the error below 0.5% the filter (amplifier) bandwidth needs to be >35.2 times the signal frequency. If the waveform is more complex than a square wave, such as having narrow spikes, the bandwidth requirement will probably be greater.

 

Error from filter.png

 

Your 2 kHz bandwidth system can measure with <0.5% accuracy square waves up to 56.8 Hz.

 

Also, the datasheet for the IMC-A-R0001 gives no indication that it has any isolation or that it can handle any common mode voltages. With a sampling rate of 900 samples/second it will not give meaningful results for square waves above a few tens of hertz.

 

Lynn

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Message 25 of 30
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Sincerly appreciate for your help though I can't understand the German and the page of  isolation bridge amplifier can't be open. Sigh~

 

 

 

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Message 26 of 30
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The control principle is complex. In a word the switches turnr on/off so frequently that the output current can precisely follow the instruction. So even during the rise/fall, the  switches are turning on/off frequenty.

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Message 27 of 30
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@xcuresme wrote:

Sincerly appreciate for your help though I can't understand the German and the page of  isolation bridge amplifier can't be open. Sigh~

 

 

 


Well, finding the contact (Kontakt) page with a phone number shouldn't be too hard 😉

And like all small countries, their population is usually able to speak multiple languages ..

Spoiler
frankly, I have never met any swiss who wasn't able to speak (at least) german, french and english

 

 

and the second link plus comments include enough key word for any search engine 😛

Greetings from Germany
Henrik

LV since v3.1

“ground” is a convenient fantasy

'˙˙˙˙uıɐƃɐ lɐıp puɐ °06 ǝuoɥd ɹnoʎ uɹnʇ ǝsɐǝld 'ʎɹɐuıƃɐɯı sı pǝlɐıp ǝʌɐɥ noʎ ɹǝqɯnu ǝɥʇ'


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Message 28 of 30
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Just came across this IA for current shunt monitoring with +-250V input range

http://cds.linear.com/docs/en/datasheet/1990fb.pdf

 

However the layout with HV input pins next to -V and V_ref  in S8 package leaves <0.8mm isolation .... 

Greetings from Germany
Henrik

LV since v3.1

“ground” is a convenient fantasy

'˙˙˙˙uıɐƃɐ lɐıp puɐ °06 ǝuoɥd ɹnoʎ uɹnʇ ǝsɐǝld 'ʎɹɐuıƃɐɯı sı pǝlɐıp ǝʌɐɥ noʎ ɹǝqɯnu ǝɥʇ'


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Message 29 of 30
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Hi, Henrik~

 

Thank you for your efforts sincerely!

 

Now I prefer to float the reference "ground" of IA because the available IA's CMRR is not high enough especially in high frequency. For example, the common-mode voltage is 160V and the  differencial signal is 50mV. If CMRR is 80dB, the common-mode noise will be 16mV resulting the error of 32%!!!

 

Best Regards.

 

xcuresme

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Message 30 of 30
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