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As far as I understand the isolation is achieved by a transformer(and besides this sometimes also using an optoisolator for the feedback part) in a DC-DC converter. And this is a galvanic isolation.(I don't know what really "truly isolated" means btw.)

But the transformers have leakage capacitances which means a trandformer will block unwanted DC but it will partially attenuate the AC interference. I guess the higher the ferq. the higher it will pass the unwanted interference.

I have two things where I'm totally blank about this topic. Here is an example of an isolated buck converter. Take for example the model TEN 5-2411.

What type of isolation will such a buck converter achieve? Will it help to attenuate common mode interference or differential mode interference between input and the output? And how can one infer from the datasheet the frequency range and for such isolation.

For example lets say we have this buck converter where we want to step down 24V to 5V DC. And imagine at the input these is 1MegHz 1Vpkpk common-mode noise or differential mode noise. How can we estimate the amount of the noise would appear at the 5V DC output?

Will the transformer block CM noise but not DM noise? If so, how can it help preventing ground loops? A ground loop is not CM.

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  • \$\begingroup\$ Some texts say one reason for isolation is that isolation avoids disruptive ground loops where different ground potentials are involved. But would such ground loop be blocked even-though the unwanted interference like ground bounce is like 1MegHz? \$\endgroup\$ – user1999 Dec 17 '18 at 13:01
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The datasheet for your part says:

Isolation capacitance – Input/Output 380 pF typ.

Isolation resistance – Input/Output >1‘000 M Ohm (500 VDC)

Thus, as far as common mode noise is concerned, your converter can be modelled as a 380pF capacitor between input and output. The Isolation resistance would not usually be a concern as it is very high, but capacitance can be a problem.

The internal transformer uses high frequency. Since a transformer can never be perfectly symmetrical due to fabrication tolerances etc, some part of the the input chopped waveform is converted into common mode which appears at the output. This is why the manufacturer will also put a capacitor between transformer primary and secondary to provide a short return path for this stray HF common mode current. Without the capacitor, the current would find a way to close the loop anyway, via the rest of your circuit, turning it into an antenna, and that would be undesirable.

For example lets say we have this buck converter where we want to step down 24V to 5V DC. And imagine at the input these is 1MegHz 1Vpkpk common-mode noise or differential mode noise.

There can't be a 1MHz 1Vpp differential mode noise at the input because there is a decoupling capacitor at the input. Differential mode is measured between both input pins.

However, if there is a 1MHz 1Vpp common mode noise at the input this means both input pins' voltage wiggle relative to ground. In this case, the common mode current through the converter will be the same as you would have through a 380pF cap. It is impossible to know the exact amplitude of this current without knowing its complete path, as it will have to make a loop somehow, which may contain various impedances.

How can we estimate the amount of the noise would appear at the 5V DC output?

The differential noise (ripple) will be as specced by the DC-DC datasheet.

The common mode noise will most likely not prevent whatever is powered by this 5V supply from working, especially if it is a digital circuit, but it may convert the circuit into an antenna and radiate.

Since current always travels into a loop, if this is a problem you can reduce the amount of current by increasing the total loop impedance, ie adding a common mode choke at the input or output.

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  • \$\begingroup\$ So you mean the common mode voltages will pass all the way to output at 1MegHz. But will a 1MegHz ground loop pass to the output ground as well? If so what is the purposes of isolation here if it is passing CM interference and ground loops? \$\endgroup\$ – user1999 Dec 17 '18 at 14:14
  • \$\begingroup\$ @newage2000 what matters is how much current gets through. Since the insulation is a capacitance, it will be excellent at 50Hz, and rather bad at 1MHz. \$\endgroup\$ – peufeu Dec 17 '18 at 15:01
  • \$\begingroup\$ I dont understand. For 1p to 400p for instance these parasitic capacitance pass all 1MegHz without any attenuation yet people use isolated DC-DC converters or optosilotars to break grounds and block EMI effects. I really dont get it. \$\endgroup\$ – user1999 Dec 17 '18 at 15:08
  • \$\begingroup\$ I get your point that the current gets very low btw maybe thats why the ground loop voltage decreases. \$\endgroup\$ – user1999 Dec 17 '18 at 15:15
  • \$\begingroup\$ Common mode voltage is not important, common mode current is what creates EMI problems. \$\endgroup\$ – peufeu Dec 17 '18 at 15:23

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