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In my understanding having a cable that has a too large diameter can lead to power dissipation, having one that is too small can lead to the same result.

apparently the second case is much more popular in the "high frequency phenomena", the first case is probably caused by the material of the cable if i understood this correctly mainly because there will be an higher impedance.

i don't really get:

  • why this can happen? it's true what i have understood?
  • how i can calculate the correct diameter?
  • there are differences between AC and DC in this scenario?

I have X Volts and i have to provide a maximum of Y Watts, where i have to start to pick the best cable for the job?

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You start by calculating the current: Y watt/X volt. The voltage is relevant for the cable's isolation, but not for the diameter. (That's not entirely true. If you work at Really Low voltages the voltage drop due to the cables resistance and possibly high current may become significant. Usually not for mains voltages and higher, though.)

Thicker cables has less resistance, so less power dissipation. I don't know where you read otherwise. This page has a calculator for the cable's required diameter. The same site also has tables for different kinds of cables.

There's indeed a difference between AC and DC. AC has skin effect, where the current will flow more need the outside of the cable. That "skin" is thinner as frequency gets higher, but already exists to a small extent for 50/60 Hz. So an AC cable may need a somewhat larger diameter, though this skin depth calculator gives a more than 9 mm skin depth for 50 Hz in copper, so that won't be a problem for most cables.

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  • \$\begingroup\$ ok, i'm starting getting confused here, i speak to a person and he says to me that, when i need a lot of power ( in AC ) and i have to wire a circuit, it's better to have multiple "small" cables in diameter rather than having a bigger one, in his words this is true because of the impedence. This person is a physicist and a teacher in a college. I should be happy having my home electrical implant ( saying for example that is for a 200Volts current ) with few big cable or not? \$\endgroup\$ – user827992 Jul 11 '12 at 12:25
  • \$\begingroup\$ Stranded wire is better for higher frequencies, because of the skin effect. 50 thinner wires have more surface than 1 thick wire. At 50/60 Hz it won't make much difference, see the 9 mm skin depth in my answer. \$\endgroup\$ – stevenvh Jul 11 '12 at 12:27
  • \$\begingroup\$ when i started this topic i was thinking about power-saving and efficiency, but the more we discuss about that the more i see that i have to dissipate and waste power, it's a price that i have to pay in every possible scenario? What about low-frequency scenario in both AC and DC? Basically i have to consider my cable like a small heater or utilizer in both high-frequency and low-frequency, for both AC and DC, everytime ? \$\endgroup\$ – user827992 Jul 11 '12 at 12:34
  • \$\begingroup\$ Fortunately not. If you're talking about mains powered appliances you don't have to take high frequency effects like skin effect into account. Then it's basically the cable's cross-section (which determines resistance) and isolation (which determines heat losses), though that will only be important if your cable's temperature rises significantly, which means your cable is too thin or you have a large load, like hundreds of amperes (not likely). \$\endgroup\$ – stevenvh Jul 11 '12 at 12:39
  • \$\begingroup\$ So, when i care about voltages i need to be sure about cable isolation, when i care about Ampere i have to care about power heating, if i care about this 2 i can get the best "wattage profile" from my circuit? And this is true for both AC and DC with just this possible variation about high frequency phenomena in AC? What can happen if i use a tiny cable, well isolated, with an high voltage and small amperage? \$\endgroup\$ – user827992 Jul 11 '12 at 12:53
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If you need to carry current at high frequency (as is typically the case in switched-mode power supply transformers) the current will have a tendency to flow through the outermost part of the wire. This is caused by skin effect. Middle of the wire will not conduct any current and will just be wasted copper (expensive and heavy). To avoid this effect you will typically parallel many wires or use a litz wire. Increasing the radius of the wire above the skin depth will neither yield in increased losses nor will it decrease losses.

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  • \$\begingroup\$ it's possible to quantify the loss due to this skin effect? Many power company around the have this problem i think, because everytime you are near a powerstation you can hear an "electrical noise" coming from the implant and in particular from the cables; sometimes this also happens with domestic-level voltages like 100/200 Volts. \$\endgroup\$ – user827992 Jul 11 '12 at 12:28
  • \$\begingroup\$ @user827992 When a current flows a magnetic field is generated which exerts a physical force on materials which are non-magnetic. That is the principle used in ordinary loudspeakers. Cables carrying AC can make a sound that way, as can any metal parts near them. \$\endgroup\$ – Andrew Morton Jun 5 '18 at 18:09
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In my understanding having a cable that has a too large diameter can lead to power dissipation ...

I know of no physical effect which would cause this. Speaking for DC: Bigger is always better when you dont mind the extra cost and weight. The selection parameter for a DC cable is the current (the voltage does matter in terms of insulation, but not for the diameter). You think of a length of cable like a simple resistor: It will have a resistance (per meter), and also the ability to dissipate heat (per meter). The current flowing will generate a specific amount of heat: $$P = \frac{I^{2}}{R}$$ This amount of heat (P) has to be dissipated, unfortunately the cable has a thermal resistance and this leads to a rise above ambient temperature (more on that topic can be read if you google a basic tutorial about heat sink calculations, those are the same).

In reality these calculations are not needed but you can use tables for that purpose, these tables give you a maximum current for a give diameter of a copper cable. I can't point you to one, since the hardware I deal with usually does not handle any significant current, so the minimal diameter of cables which is necessary to withstand the usual mechanical forces is sufficient.

AC on the other hand is quite different: I think it is safe to say that anything under 1kHz behaves much like DC, higher frequencies will show things like the skin effect (not so much diameter, as you suggested).

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  • \$\begingroup\$ larger = more impedance, isn't true? Anyway i'm interested in both AC and DC. \$\endgroup\$ – user827992 Jul 11 '12 at 12:12
  • \$\begingroup\$ If you think about how electricity is conducted you will see how a larger diameter helps the flow of electrons. The beefy cables witch connect your cars battery are not so strong because of mechanical forces, but because they have to handle a huge amount of (DC) current. Having a larger diameter in a cable will never do any harm in terms of greater impedance, as long as you don't compare wires of different shapes (e.g. straight wire vs. coil, which alters the inductance of the wire thus the overall impedance). \$\endgroup\$ – 0x6d64 Jul 11 '12 at 12:36
  • \$\begingroup\$ thanks, where i can read something about shape variations? \$\endgroup\$ – user827992 Jul 11 '12 at 12:53
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    \$\begingroup\$ "larger = more impedance": Incorrect. Larger cable diameter gives lower impedance. As others have pointed out, multiple small wires will be better than one large wire for AC above a certain frequency. \$\endgroup\$ – The Photon Jul 11 '12 at 16:59
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To determine proper wire gauge you need to have at least one of bellow as prime measure calculation .
1-Voltage
2-Amperage(current)
3-Resistance
4-Watt (output or input power)
Formula is V=I.R _____voltage= Amp X Resistance
Formula is W=I.V______ Watt = Amp x Voltage
Then there's chart for wire gauge that indicate require Amp and resistance for each gauge number per foot or per 1000 feet.
Regardless of the length you need you can calculate what gauge will carry the voltage and amp you need.

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Normally cable size is based on current carrying capacity. An old but still valid rule of thumb is 1000 Amps/square inch (changing to metric did not invalidate the basic physics) The overload capacity before the fuse blows or breaker trips (normally 1.25 of the rated or label value) is the value here rather than the working load.

For DC or mains AC you can forget "skin effect" for all practical purposes. But volt drop can be an issue at DC, while on AC with heavy loads and long runs, it can be the factor determining cable size and type.

At KHz or MHz frequencies when feeding transmitting aerials from medium or high power transmitters; small bore hollow copper tubing is commonly used for the feed line regardless of the transmitter having single ended arrangement as in a ships radio room or balanced line intended to feed a large aerial such as a Rhombic or Curtain Array aerials as found at land based shortwave broadcast and other stations.

Switch mode power supply transformers and drive circuitry, and its design; so it performs even at the limits of supply voltage and temperature while meeting all regulatory requirements including better than minimum EMC. EMC quieting is more an art based as much on experience as any theory due the number of interacting variables. Modeling helps up to a point, but gut instinct and intuition is as important if your product is up for the type approvals (CE mark etc) required for mass production. OldBlueBear

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