ExtraClassSylalbus2009jan-AD7FO

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Rev 2.02 E5B05 It will take .020 seconds (or 20 milliseconds) for an initial charge of 20 V DC to decrease to 7.36 V DC in a 0.01microfarad capacitor when a 2-megohm resistor is connected across it. To discharge to 7.36 VDC would take one time constant it is 20V –(.632 x 20V) or 7.36 Volts TC = 2 x .01 or .02 seconds or 20 milliseconds E5B06 It takes 450 seconds for an initial charge of 800 V DC to decrease to 294 V DC in a 450-microfarad capacitor when a 1-megohm resistor is connected across it. To discharge to 294 VDC would take one time constant 800V – (.632 x 800V) = 294.4V TC = 1 x 450 or 450 seconds Basic Trigonometric Functions For a number of problems associated with electronics involving series circuits of resistance and reactance and the Extra class Exam you will need a basic understanding of trigonometry. The problems center on a right triangle (that is a triangle that has one angle that is 90° and the sum of the remaining two angles is equal to 90°). Using trigonometric functions if we know two sides, or an angle (other than the 90°angle) and one side of the triangle we can calculate the remaining angles and dimensions. Examples: If X=3 and Y=4 then R=5 and θ = 53.1° The sides of the triangle are given names from the rectangular coordinate system with the horizontal side called X (also called the Adjacent side) and the vertical side is called Y (also called the opposite side) and the side connecting the X and Y sides is called the Hypotenuse called R in this example If two of the three sides are known the third side can be found using the following equation: Hypotenuse² = X² + Y² or as commonly expressed R = There are 6 trigonometric functions that can be used to calculate the angle between X and R and between Y and R. We will focus on three of these functions; Sine, Cosine and Tangent to solve for the angle between the X side and the R side ( θ ). Sine of θ = Y / R Secant = R/Y Cosine of θ = X / R Cosecant = R/X Tangent of θ = Y / X Cotangent = X/Y If X=50 and Y=100 then R=111.8 and θ = 63.4° If X=5 and Y=10 then R=8.66 and θ = 63.1° If X=153 and Y=52 then R=161.6 and θ = 18.7° X² + Y² Example 1: To find the angle, θ for a triangle with side X value of 3 and a side Y value of 6: Tangent of θ = 6 / 3 or 2.00. To find the angle enter 2 into your calculator and press the Arc Tan key which will show you the angle represented by the tangent value, in this case 2.00. The Arc Tan of 2 is 63.43°. Jack Tiley <strong>AD7FO</strong> Page 40 3/15/2009
Rev 2.02 Example 2: To find the angle, θ for a triangle with a side X value of 3 and a side Y value of 4: Tangent of θ = 4 / 3 or 1.333 To find the angle enter 1.333 into your calculator and press the Arc Tan(or Tan^-1 on some calculators) key which will show you 53.03° Example 3: To find the angle, θ for a triangle with a side X value of 12 and a side Y value of 12 Tangent of θ = 12/12 or 1 To find the angle enter 1 into your calculator and press the Arc Tan key which will show you 45.0° Reactance (AC resistance of capacitors and inductors) Capacitors and inductors exhibit a resistance to current flow much like a resistor but with values that change with the frequency of the applied circuit. The AC resistance of a capacitor is called capacitive reactance (Xc) and is calculated using the formula: Xc= 1/(2π x F x C) with C in µF and F in MHz Examples: Find the capacitive reactance of a 1 µf capacitor at 200 Hz Xc= 1/(2π x F x C) or Xc= 1/(6.28 x .0002 x 1) or Xc= 796 Ω Find the Capacitive reactance of a 10 PF capacitor at 7 MHz Xc= 1/(2π x F x C) or Xc= 1/(6.28 x 7.0 x 10 ^-6 ) or Xc= 2,275 Ω The AC resistance of an Inductor is called Inductive reactance (XL) and is calculated using the formula: XL= 2π x F x L with L in µH and F in MHz Examples: Find the inductive reactance of a 1 µH inductor at 100 MHz XL= 2π x F x L or XL= 6.28 x 100 x 1 or XL= 628 Ω Find the inductive reactance of a 100 µH inductor at 7 MHz XL= 2π x F x L or XL= 6.28 x 7 x 100 or XL= 4,396 Ω Circuit Impedance The term Impedance refers to the equivalent circuit resistance in ohms for a circuit consisting of resistance and capacitive reactance and / or inductive reactance. To solve these problems for series circuits we use a rectangular coordinate graph and basic algebra and trigonometry. When working with complex circuits containing resistance and reactance, the reactive components are shown with a lower case j prefix. Inductive reactance is shown with a +j prefix and capacitive reactance with a –j prefix. Rectangular Coordinate System When solving problems involving impedance and phase angle of AC series circuits we show the circuit element values in a rectangular format. The rectangular format consists of a horizontal line intersected at 90° by a vertical line. Values on the horizontal or X axis are positive to the right of the vertical line and negative to the left. Values on the vertical or Y axis are positive above the X axis and Negative below the X axis. Jack Tiley <strong>AD7FO</strong> Page 41 3/15/2009
Page 1 and 2: Rev 2.02 Preparing for the Extra cl
Page 3 and 4: Rev 2.02 Guide to using this syllab
Page 5 and 6: Rev 2.02 SUBELEMENT E4 -- AMATEUR R
Page 7 and 8: Rev 2.02 ELEMENT 4 - EXTRA CLASS QU
Page 9 and 10: Rev 2.02 E1A09 (Band Plan Chart) Th
Page 11 and 12: Rev 2.02 E1B13 (FCC Rules) Communic
Page 13 and 14: Rev 2.02 E1D08 (FCC Rules) The 2 me
Page 15 and 16: Rev 2.02 E1E20 (FCC Rules) You must
Page 17 and 18: Rev 2.02 SUBELEMENT E2 - OPERATING
Page 19 and 20: Rev 2.02 E2A13 Geosynchronous satel
Page 21 and 22: Rev 2.02 E2B17 Immunity from fading
Page 23 and 24: Rev 2.02 E2D02 The definition of
Page 25 and 26: Rev 2.02 SUBELEMENT E3 -- RADIO WAV
Page 27 and 28: Rev 2.02 E3C01 Auroral activity cau
Page 29 and 30: Rev 2.02 E4A06 A spectrum analyzer
Page 31 and 32: Rev 2.02 E4B09 High impedance input
Page 33 and 34: Rev 2.02 E4C06 If t he thermal nois
Page 35 and 36: Rev 2.02 E4D11 Third-order intermod
Page 37 and 38: Rev 2.02 SUBELEMENT E5 -- ELECTRICA
Page 39: Rev 2.02 E5B Time constants and pha
Page 43 and 44: Rev 2.02 j 50 -j 25 E5B12 The phase
Page 45 and 46: Rev 2.02 j 600 -j300 E5C04 In polar
Page 47 and 48: Rev 2.02 Parallel circuit solutions
Page 49 and 50: Rev 2.02 R=300 Ω XL= (2 π FL) or
Page 51 and 52: Rev 2.02 Power Factor represents th
Page 53 and 54: Rev 2.02 SUBELEMENT E6 -- CIRCUIT C
Page 55 and 56: Rev 2.02 Field-effect transistors e
Page 57 and 58: Rev 2.02 E6B12 Junction diodes are
Page 59 and 60: Rev 2.02 E6D02 Cathode ray tube (CR
Page 61 and 62: Rev 2.02 E6E04 The technique used t
Page 63 and 64: Rev 2.02 E6F07 Cadmium sulfide will
Page 65 and 66: Rev 2.02 E7A07 An AND gate produces
Page 67 and 68: Rev 2.02 When a Class C rather than
Page 69 and 70: Rev 2.02 E7C01 In a low-pass filter
Page 71 and 72: Rev 2.02 E7D02 One characteristic o
Page 73 and 74: Rev 2.02 E7E05 A pre-emphasis netwo
Page 75 and 76: Rev 2.02 E7F08 One purpose of a mar
Page 77 and 78: Rev 2.02 E7G17 The typical output i
Page 79 and 80: Rev 2.02 E7H19 A phase-locked loop
Page 81 and 82: Rev 2.02 E8A06 The approximate rati
Page 83 and 84: Rev 2.02 E8B13 In time division mul
Page 85 and 86: Rev 2.02 E8D07 An electromagnetic w
Page 87 and 88: Rev 2.02 E9A10 Antenna bandwidth is
Page 89 and 90: Rev 2.02 E9B10 The “Method of Mom
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Rev 2.02 E9C09 The elevation angle
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Rev 2.02 E9D09 An advantage of usin
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Rev 2.02 E9E08 An SWR greater than
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Rev 2.02 E9F11 A 1/8-wavelength tra
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Rev 2.02 E9H05 The main drawback of
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Rev 2.02 SUBELEMENT E0 -- SAFETY [1
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Rev 2.02 Jack Tiley AD7FO Page 103
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Rev 2.02 Capacitors in series C= __
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Rev 2.02 Impedance of complex serie
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Rev 2.02 Calculating Component valu
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Rev 2.02 AC Voltage Calculations Th
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Rev 2.02 Jack Tiley AD7FO Page 113
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