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I tried to simulate the effect of common-mode interference on unbalanced sources for three different inputs: single-ended, differential-ended and isolated. Below I inject a 1V common-mode voltage to three identical unbalanced sources(1k Ohm source impedance for positive lead and 0.1 Ohm for negative lead) with identical cables but coupled to three different receiver types:

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And here is the AC analysis for the common mode voltage between 1 Hz to 100 kHz:

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The above plots show that the single-ended inputs do not reject the common-mode related noise as expected.

But if we compare the differential input receiver's output(the green plot) and the isolated receiver's output(the blue plot), for example at 50Hz the isolated output is 130dB better than the differential-ended one.

Does that make sense? Can we say based on these plots that isolation actually acts as a differential input? Does that mean an isolation amplifier can be more immune to common-mode related noise than a differential amplifier?

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2 Answers 2

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All models are faulty, unless you make them as accurate as possible.

  • Include stray common mode E-field impedance or capacitance for cables
  • transformers have a coupling capacitance and they are mismatched
    • Baluns, GFCI's and Hybrid Splitters are better than normal transformers due to these balance requirements.
  • input output impedance matters for both CM and DM to get CMRR
    • this is why CM chokes with X,Y caps are used for line filters to achieve CMRR for FCC,IEC immunity and lightning suppression.
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Does that make sense?

No, because your test is fundamentally flawed due to the 1 uH inductance for primary and secondary in your galvanic isolating transformer. It will act as 0.63 ohms at 100 kHz and pretty much kill any signal up to that frequency. Think what you are doing and be more reasonable in your inductor choices.

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  • \$\begingroup\$ Thanks for the feedback but the reason they are 1uH because the diff ended inputs are ideal in my simulation so to make the galvanic isolation near perfect I set the inductance low. And if I set it to 100mH instead still it dominates the DE inputs by 30dB. So the question is still valid imao. The only reason I asked this question in practice I want to know whether an isolation amplifier can be used instead of a differential amplifier, \$\endgroup\$
    – user1245
    Jan 9, 2019 at 13:20
  • \$\begingroup\$ What I mean is Im more asking about electronic isolation amplifiers. Since I dont know how to simulate them I did it with transformer just to represent mimic the isolation. \$\endgroup\$
    – user1245
    Jan 9, 2019 at 13:25
  • \$\begingroup\$ @user164567 Sorry but you still don't understand - the inductor (1 uH or 100 mH) is still reducing the signal voltage (common mode or otherwise) being seen by the amplifier. 100 mH at 50 Hz has an impedance of 31 ohms hece the signal voltage is being reduced by a factor of approximately 31/1000 = 30 dB. You are fooling yourself into thinking that isolation brings something new to the party when it doesn't. Try putting a 100 mH inductor across the input pins to the differential amplifier and see what I mean. \$\endgroup\$
    – Andy aka
    Jan 9, 2019 at 13:32
  • \$\begingroup\$ Here it says en.wikipedia.org/wiki/Isolation_amplifier "Isolation amplifiers are a form of differential amplifier" \$\endgroup\$
    – user1245
    Jan 9, 2019 at 13:38
  • \$\begingroup\$ That is irrelevant to your faulty simulation circuit. \$\endgroup\$
    – Andy aka
    Jan 9, 2019 at 15:05

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