Does Skin Effect Increase With Voltage?

[johnyradio]

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.

(Not homework)

 

[dknguyen]

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.

View attachment 115062

No because I^2*R. Higher currents experience higher proportional losses.

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

View attachment 115062

Same frequency so same skin depth so same impedance in this case. But it is not an equivalent question as proportional losses as evidenced by the different answer.

 

[johnyradio]

Fantastic answer, dknguyen!

So this is the formula for percentage power loss?
I^2*R.

 

[dknguyen]

Fantastic answer, dknguyen!

So this is the formula for percentage power loss?
I^2*R.

That's just the formula for power dissipated in an ideal resistor, not percentage power loss. For a fair comparison you have to muck with the load on the other end of the wire so that the same amount of power is delivered to the load. So I^2*R results in higher proportional losses even if you don't delve into more detail using skin depth.

 

[JimB]

Just a little wrinkle, but the trace identified as sin(3x) is actually 3Sin(x).

JimB

 

[johnyradio]

For a fair comparison you have to muck with the load on the other end of the wire so that the same amount of power is delivered to the load.

Same?

 

[johnyradio]

No because I^2*R. Higher currents experience higher proportional losses

but those sines in my question are voltage sines, not current sines. Still "no"?

thx

 

[unclejed613]

if you look at the formula for skin effect, voltage and current themselves do not appear in the equation. it's a function of frequency, resistivity, permeability, etc... in simple terms, it's a function of the wire resistance, frequency, and permeability of the conductor. if for instance the skin effect increases the loss on a piece of wire at a particular frequency by 1 ohm, that's 1 ohm no matter what voltage or current.

 

[dknguyen]

but those sines in my question are voltage sines, not current sines. Still "no"?

thx

Voltage sine through a resistor AND a resistive load makes a current sine. If your load is not resistive something else...well, then things are different.

 

[johnyradio]

I've learned something new:
1/2 voltage doesn't necessarily translate into 1/2 current. Current is whatever the load tries to pull.

dknguyen, is that what you meant by "same current"?

Cuz i think if it's the same current, then skin effect will be the same. Right?

 

[dknguyen]

I've learned something new:
1/2 voltage doesn't necessarily translate into 1/2 current. Current is whatever the load tries to pull.

dknguyen, is that what you meant by "same current"?

Cuz i think if it's the same current, then skin effect will be the same. Right?

When did I say same current? I did say same power, but I don't remember ever saying same current. Are you talking about the graphs in post #21?

If those are voltage graphs, and you muck around with the load so the resulting currents are the same, then yes the absolute value of the skin losses will be the same because the currents will have identical waveforms (same magnitudes and same frequency). But mucking around with it will probably the total power delivered to the load to be different which would mess up your question of proportional losses. That's why it tends to be difficult to answer your question. By asking for proportional losses and not specifying the load, you frame your question in strange scenarios that tend to produce unfair comparisons.

The straightforward thing to ask would just be how much actual power is dissipated in the same wire with X current waveform vs Y current waveform. By asking about proportional losses you introduce a load which affects everything and makes it difficult to have a fair comparison between two different waveforms.

 

[johnyradio]

any flaws in my question are due only to my lack of understanding.
i admit, i was concerned about load as well, but i wasn't sure what the problem was.

now i get it! (i hope)
the voltage swing doesn't matter.
only current swing matters.

1/2 current-swing = less % skin effect.
Same current-swing = same skin effect.
1/2 voltage-swing = unknown % skin effect.

 

[dknguyen]

1/2 current-swing = less % skin effect.
Same current-swing = same skin effect.
1/2 voltage-swing = unknown % skin effect.

Ehhh, again because by saying % loss you need to know what the power loss in the wire is AND what the power being delivered to the load is. Just knowing current flowing through the wire is not indicative of the power being delivered to the load.

I could have 1A flowing to a load through a wire, and have 100W being dissipated in the load and 1W being dissipated in the wire. Or I could have a ten amps flowing through a wire, and 1W being dissipated in the wire and 0.1W being dissipated in the load. The % loss has no ties to the current level because the load is a wildcard. Your asking about generalized % loss just doesn't work.

What you can say is that higher frequency in a wire causes more power to be dissipated in the wire. You can also say that higher currents cause more power to be dissipated in the wire. What you cannot do is generalize the % loss in these cases because the specific nature of the load has not been defined in either case that could throw the answer either way.

It's sort of like asking is it more efficient to transport something by truck or train without any further information. It can't be answered because it depends on how much stuff you're transporting. A truck is more efficient if it's one box, but not a million boxes. You can't make a generalization without that info.