# Does Skin Effect Increase With Voltage?

Joined Oct 26, 2012
434
Here's a simpler, better question, than my original question .

Will these two sines experience the same percentage power loss on the same wire due to skin effect (passed separately, not at the same time) .

Not sure if that's equivalent to asking if they will experience the same amount of impedance.

Vertical axis is voltage .

question as previously stated (ignore):

Assume a wire has a given skin effect depth at a particular frequency. Compare the following scenarios:

Scenario 1: One such wire. We send a bipolar AC current at that frequency, at a particular peak to peak voltage amplitude down that wire. Due to skin effect, it will have a particular impedance, and thus a particular power loss.

Scenario 2: Same such wire, identical to the above wire. Identical AC current. Except, we're sending the positive half ONLY down the wire. That is, it's half rectified .

Will the 2nd scenario suffer the same percentage power loss (due to skin effect) as the 1st scenario?

Examples using square and sine.

Last edited:

#### crutschow

Joined Mar 14, 2008
34,680
Will the 2nd scenario suffer the same power loss (due to skin effect) as the 1st scenario?
No. because now the signal has a DC component which is not affected by the skin effect.

Joined Oct 26, 2012
434
No. because now the signal has a DC component which is not affected by the skin effect.
i've been told skin effect results from changing current, not AC.

#### crutschow

Joined Mar 14, 2008
34,680
i've been told skin effect results from changing current, not AC.
Changing current is indeed AC.

Joined Oct 26, 2012
434
Changing current is indeed AC.
To me, "alternating" indicates forward, reverse, forward, reverse...

"Changing" indicates "higher, lower, higher, lower" ...

But that's just terminology. I think we're on the same page.

Thx

Last edited:

#### crutschow

Joined Mar 14, 2008
34,680
o me, "alternating" indicates forward, reverse, forward, reverse...

"Changing" indicates "higher, lower, higher, lower" ...
I know the name "alternating current" is a little confusing if you try to be too literal in its interpretation.
It refers to the changing part of the waveform after you subtract (or ignore) the DC component.
Thus a 1V sinusoidal "changing" signal riding a 5V DC offset is a 1VAC signal.

It can also apply to a changing high-frequency signal riding on a larger low frequency signal.
Thus a 1V HF sinusoid riding on a 5V LF sinusoid is called a 1VAC riding a 5VAC even though the 1V waveform does not actual "reverse" direction for much of the LF waveform duration.

Another way to determine what is AC is to look at the Fourier transform of the signal.
That separates out all the AC components and harmonics, leaving any DC component as a constant.
The AC components would be subject to the skin effect according to their frequency.

Joined Oct 26, 2012
434
It refers to the changing part of the waveform after you subtract (or ignore) the DC component.
Thus a 1V sinusoidal "changing" signal riding a 5V DC offset is a 1VAC signal.
Great explanation! That helps .

It can also apply to a changing high-frequency signal riding on a larger low frequency signal.
Not sure if you mean summing or AM, I agree that's AC, but I question whether that's part of the definition of AC.

AC components would be subject to the skin effect according to their frequency.
Great point, which I believe means a half rectified square will experience different total impedance than a half rectified sine. Cuz their harmonic content is different .

Plz see my updated question .thx .

Last edited:

Joined Oct 26, 2012
434
Here's a simpler, better question.

Will these two sines experience the same percentage power loss on the same wire (passed separately, not at the same time) .

Not sure if that's equivalent to asking if they will experience the same amount of impedance.

#### crutschow

Joined Mar 14, 2008
34,680
Will these two sines experience the same percentage power loss on the same wire (passed separately, not at the same time) .
Yes.

They will also experience the same impedance.
Impedance is independent of voltage unless it's a non-linear impedance.

#### MrAl

Joined Jun 17, 2014
11,565
Assume a wire has a given skin effect depth at a particular frequency. Compare the following scenarios:

Scenario 1: One such wire. We send a bipolar AC current at that frequency, at a particular peak to peak voltage amplitude down that wire. Due to skin effect, it will have a particular impedance, and thus a particular power loss.

Scenario 2: Same such wire, identical to the above wire. Identical AC current. Except, we're sending the positive half ONLY down the wire. That is, it's half rectified .

Will the 2nd scenario suffer the same percentage power loss (due to skin effect) as the 1st scenario?

Examples using square and sine.

Hello there,

Any circuit scenario like this is subject to Fourier analysis. That means that to determine exactly what happens you only need to examine the effect of each Fourier component acting individually.

What this means in relation to your question of 'changing' and 'AC' is that any change is AC, even if the main AC is just changing amplitude. That is because if the main AC is changing amplitude, then there's another Fourier component in there to consider, and that could be considered along with the main AC component.
Granted this could get a little more complicated then we'd like to see sometimes, but using a little trig also helps resolve the signals in to workable forcing functions that can be analyzed in the more typical way.

Now DC is a special case, because DC does not typically create skin effect directly. Indirectly it can cause a temperature rise which will change the DC resistance, and it could cause a magnetic effect that changes the way the signal penetrates the wire. Thus we could see some changes here too, although usually we dont bother with the DC effects except as to how they increase wire heating because wire heating could be a profound effect on the total AC and DC resistance.

As general rule, the faster the frequency the more skin effect we see. That means that if we have one low frequency signal it will have less skin effect than if we have a complex signal with many harmonics.
A half wave rectified sine has an infinite number of harmonics that decrease in amplitude as we observe higher and higher frequencies, but there are some that are very significant so they would have to be considered.

Now to quantify all this, you would have to analyze each case separately for all harmonics and then compare the two numbers. The pure sine has no harmonics, but is present for a longer time period, while the half wave sine is there for less time but has some significant higher harmonics. If you were interested in the loss for both cases you'd have to first calculate the Fourier components, then calculate the skin effect for each harmonic.

Another interesting case is the pure sine vs the full wave rectified sine wave. The full wave rectified sine probably shows more loss because of the harmonics which are not present in the pure sine (same amplitude of course for each).

Joined Oct 26, 2012
434
Hello there,
Hi @MrAl
Thx for your awesome answer .If possible, plz see my updated question .

Cheers

#### crutschow

Joined Mar 14, 2008
34,680
No because I^2*R. Higher currents experience higher proportional losses.
Good point.
The skin effect causes an increase in the apparent wire resistance, which is constant with signal amplitude.
But the loss from that is proportional to the square of the current.

Joined Oct 26, 2012
434
But the loss from that is proportional to the square of the current.
but those sines in my question are voltage sines, not current sines. Would we still see less percentage loss on a lower voltage-amplitude sine?

#### nsaspook

Joined Aug 27, 2009
13,418
but those sines in my question are voltage sines, not current sines. Would we still see less percentage loss on a lower voltage-amplitude sine?
I think your sine voltage question was perceived by most as a changing 3X voltage with stable load impedance question to make a lick of sense so there would be current waveforms that also changed.

#### crutschow

Joined Mar 14, 2008
34,680
but those sines in my question are voltage sines, not current sines. Would we still see less percentage loss on a lower voltage-amplitude sine?
Of course.
How is current related to voltage?

Joined Oct 26, 2012
434

#### crutschow

Joined Mar 14, 2008
34,680
Last edited:

#### MrAl

Joined Jun 17, 2014
11,565
Hi @MrAl
Thx for your awesome answer .If possible, plz see my updated question .

Cheers
Hi,

I think it is very bad form to go back to the first post and change the question entirely. That makes the posts that follow seem very out of touch and even totally incorrect when they were correct to begin with.
It would be good for you to correct that so the thread replies again make good sense.
It is more typical to just ask another question in the thread, rather than go back and modify the first post so much.

Just to note here, your first question was to compare a pure sine with a half wave rectified sine.
Your new question appears to be to compare a pure sine of some amplitude and a second sine with approximately 1/2 the amplitude of the first.

The simplest way to look at this second question is to consider first what resistance is. Resistance is a property that we often take to be unchanged by voltage or current in order to simplify the analysis. In the case of a wire however, wire heating is not often overlooked because it has a big effect on the resistance. In this kind of analysis however voltage isnt as much of a concern as current is unless you are applying that voltage directly across the wire and you know a lot about the wire itself. The more typical case here is when we apply a voltage at one end of the wire and we have the load at the other end. In that case the current becomes a more convenient parameter to look at in relation to loses in the wire, as well as the frequency.
With this current and resistance in mind, a shortcut way to handle this is to calculate the AC resistance from the DC resistance and the frequency. You then end up with a number for the AC resistance that is related to the DC resistance:
Rac=Rdc*K

and from there you can go on to calculate approximate heating effects and see how it changes with increased current levels. So it ends up being almost the same as when we have just DC current, but now we also have to take into account the AC resistance that we did not have to do with the pure DC case.

This leads us to the conclusion that if we dont consider heating effects then the loss is proportional to the square of the current, but if we do consider heating effects then the DC and AC resistances increase and so the losses increase more.

#### ian field

Joined Oct 27, 2012
6,536
i've been told skin effect results from changing current, not AC.
Its mostly an RF thing - an ultrasonics company I worked for used Litz wire for transformer windings in the range 20 - 80kHz. Skin effect isn't huge in that range, but they were desperate to reduce ANY losses. From about mid SW & up; its common to silver plate the wire for tuning inductors to improve Q.