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RESISTANCE of COIL 1 Temperature Effects

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RESISTANCE of COIL 1 Temperature Effects Chapter 6—Resistance of Coil 6–1 RESISTANCE OF COIL The resistance RTC in the RLC model is an effective or equivalent resistance which rep- resents all the losses in the Tesla coil. It includes 1. Ohmic or copper losses 2. Dielectric losses, coil form and conductor insulation 3. Eddy current losses in toroid, strike ring, and soil 4. Radiation losses 5. Losses in the spark It is surprising how difficult it is to calculate these various losses. Making meaningful measurements can also be challenging. If we operate at low input voltage so we are below spark breakout, then we can ignore the last term for the moment. I will ramble through some considerations for the other losses. 1 Temperature Effects Almost all coils are wound with copper wire. It is moderately priced and widely available. There might be an occasional aluminum coil, usually from some ‘bargain’ at a surplus auction. Aluminum has higher resistivity than copper, so to get a given resistance the wire must be physically larger. We saw earlier that to get a high toroid voltage we needed a coil with large L and/or small R. If we use larger wire to keep the resistance the same, the coil must be physically larger and the inductance will decrease. We would expect therefore that aluminum coils would always be inferior to copper coils. Example You are given a choice between two spools of magnet wire, each 1000 feet in length. The copper is 24 gauge, with nominal resistance 25.67 Ω, while the aluminum is 22 gauge, with nominal resistance 26.46 Ω. You have a piece of 5 inch diameter PVC pipe and are considering using the entire 1000 ft to wind a coil. You wish to compare the inductances of the two prospective coils. You assume a build (thickness of dielectric) of 1.65 mils. The diameter of the 22 gauge wire is then 23.35 + 2(1.65) = 28.65 mils and the diameter of the 24 gauge wire is 20.1+2(1.65) = 23.4 mils. The nominal number of turns would be 1000 N = = 764 π(5/12) Solid State Tesla Coil by Dr. Gary L. Johnson December 6, 2001 Chapter 6—Resistance of Coil 6–2 The winding length for the 22 gauge wire would be w = 764(28.65) = 21, 890 mils = 21.89 inches and 17.19 inches for the 24 gauge wire. By Wheeler’s formula, the inductance for the 22 gauge coil would be r2N 2 (2.5)2(764)2 L = = =15, 110 µH 9r +10w 9(2.5) + 10(21.89) The 24 gauge coil has an inductance of 18,120 µH.Forthisparticularexamplewherer and N are held fixed, w is proportional to wire diameter and since it appears in the denominator, a smaller wire diameter results in a larger inductance. If we hold the winding length fixed, say at 17.88 inches, then the 1000 ft of aluminum wire will fit on a coil of 624 turns and 6.121 inches diameter. The inductance is now 17,670 µH, larger but still less than that of the copper coil. The aluminum coil has more resistance and less inductance than the copper coil, both undesirable. As a side issue, if we hold the winding length fixed, the coil gets to be relatively short and fat. Standard wisdom is that the ratio of length to diameter needs to be on the order of four or five, so this is another negative factor. You should buy the copper wire. Aluminum forms a coating of aluminum oxide when exposed to air. This oxide is not a conductor (copper oxide is) so there is always a problem in making electrically solid connec- tions. This is not an unsolvable problem since electric utilities use aluminum wire almost exclusively, for cost and weight reasons. The trained individual with the right tools might be able to get acceptable performance from an aluminum coil. The rest of us should stick with copper. Silver is slightly more conductive than copper, but the price is much higher. Copper losses are not considered that big a factor in coil performance, so there is little incentive to go to silver wire. The dc resistance of copper wire is determined by table look-up or by measurement with an ohmmeter. The tables typically give the resistance of 1000 ft of wire at a specified temperature, say 20oCor68oF. The tables give the resistance to four significant places, but using more than three is somewhat of a joke because of the resistance variation with temperature. If the table resistance is R1 at a temperature T1, then the resistance R2 at some other temperature T2 is T2 − Ti R2 = R1 (1) T1 − Ti o where Ti is the inferred absolute zero temperature, −234.5 C for copper. If the table resistance o is given for 20 C, then the resistance of a copper wire at temperature T2 is T2 + 234.5 R2 = R1 (2) 20 + 234.5 Solid State Tesla Coil by Dr. Gary L. Johnson December 6, 2001 Chapter 6—Resistance of Coil 6–3 The resistance has dropped to 90% of the tabulated value at −5.45oC (not impossible for an unheated shop in some parts of the world) and is at 110% of the tabulated value at 45.45oC. The wire could certainly rise to this temperature during operation even on a pleasant day. The inferred absolute zero temperature does not have much to do with absolute zero (−273oC). Pure iron has an inferred absolute zero of −162oC and Manganin has an inferred absolute zero of −167, 000oC. The fact that the inferred absolute zero of copper is not dras- tically different from absolute zero makes it useful as a temperature sensor. If one is curious about how warm a Tesla coil got during operation, the best way of measuring the temperature is to measure the dc resistance and work backwards through the above equations. 2SkinEffect At dc, current flows uniformly across the entire cross section of the conductor. As frequency increases, however, a phenomenon called skin effect causes less of the total current to flow in the center of the wire. Having less conductor available causes the resistance to increase. An expression for skin depth can be derived as δ √ 1 = πfµσ (3) where f is the frequency in Hz, µ is the permeability of the conductor (4π × 10−7 for nonfer- romagnetic materials), and σ is the conductivity. The skin depth for copper at 20oCis . δ 0√066 = f m(4) Most introductory electromagnetic theory books derive the expression for ac resistance as b R = R (5) ac dc 2δ where b is the radius of the wire and δ is the skin depth, in consistent units. This equation is only valid for δ b. As might be expected, this excludes most Tesla coils, so we must find other expressions. A number of more advanced electromagnetic theory books derive an expression for the ac resistance. Typical is the treatment by Ramo, Whinnery, and Van Duzer [6]. They start with Maxwell’s Equations, and write a differential equation for the current density Jz(r)insidean isolated straight cylindrical conductor centered on the z axis. Solid State Tesla Coil by Dr. Gary L. Johnson December 6, 2001 Chapter 6—Resistance of Coil 6–4 d2J dJ z 1 z T 2J dr2 + r dr + z =0 (6) where T 2 −jωµσ −j πfµσ −j 2 = = (2 )= δ2 (7) The solution for current density is Jz = AJ0(Tr)(8) where J0(Tr) is the Bessel function of the first kind and zero order. The infinite series for this Bessel function is Tr 2m ∞ − m ∞ m 2 m 2m ( 1) 2 (−1) (T ) r J0(Tr)= = (9) m 2 2m m 2 m=0 ( !) m=0 2 ( !) When we insert the above expression for T 2, the series becomes m ∞ − m −j m 2 r2m ( 1) ( ) δ2 J0(Tr)= (10) 2m m 2 m=0 2 ( !) Note that (−1)m(−j)m =(j)m (11) 2m 1 = (12) 22m 2m r2m r 2m = (13) δ2m δ With these substitutions, we can write 2m ∞ m r (j) δ J0(Tr)= (14) m m 2 m=0 2 ( !) The series can now be split into its real and imaginary parts. Solid State Tesla Coil by Dr. Gary L. Johnson December 6, 2001 Chapter 6—Resistance of Coil 6–5 r 4 r 8 δ δ Real[J0(Tr)] = 1 − + − + ··· (15) 22(2!)2 24(4!)2 r 2 r 6 r 10 δ δ δ Imag[J0(Tr)] = − + − + ··· (16) 2(1!)2 23(3!)2 25(5!)2 These are related to the ber and bei functions as follows. √ Real[J0(Tr)] = ber( 2r/δ) (17) √ Imag[J0(Tr)] = bei( 2r/δ) (18) There are several ways to proceed to find the ac resistance. One is to use one of Maxwell’s curl equations to find the magnetic field inside the wire, a constant times the derivative of the electric field. This magnetic field is used to find the total current in the wire. The electric field times the wire length is the total voltage. The ac resistance is just the real part of the ratio of voltage to current. To my knowledge, every method of finding the ac resistance is tedious. I will proceed with a basic Circuit Theory I type approach. Consider the straight wire as formed of N concentric cylinders, as shown in Fig. 1. .................................. .................. .................. ............ ............ ......... ......... ........ ........ ...... ....... ...... ............................................... ...... ...... .............. .............. ...... ...... ......... .......... ...... ..... ........ ....... ..... ..... ...... ....... ..... ..... ...... ...... ..... .... ...... ..... .... ..... ..... .................................... ..... ..... .... ..... ............. ............. ..... .... .... ..... ........ ......... ..... .... .... ..... ....... ....... .... .... ... .... ...... ...... ..... ... ... .... ...... ..... .... ... ... .... ..... ....................................... ..... .... ... ... .... ..... .......... .......... ..... ... ... ... ... ..... ....... ....... .... ... ... ... ... .... ...... ...... ..... ... ... ... ... .... ..... ..... .... ... ... .. ... ... ..... ..... ... ... .. .. .. ... ..... ..... ... ... .. .. .. ... .... ............................. .... ... .. .. .. .. ... ... ........ ........ ... ... .. .. .. .. ... ... ...... ...... ... ... .. .. .. .. .. ... ..... ..... ... .. .. .. .. .. .. ..... ..... ... .. .. .. .. .. .. ... ... .. .
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