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A coil of wire, like the ones used in radio circuits, has an electrical property known as inductance. Basically, this is the ability of the coil to produce a magnetic field when current passes through it. The field so generated produces a voltage which opposes the flow of the current which produces it. Inductance depends on the diameter, the length, and number of turns in the coil. The amount of opposition to AC current which a coil offers is called inductive reactance. This opposition has an effect similar to resistance in a DC circuit; it is measured in ohms and it causes an AC voltage drop across the coil. But in some other ways, inductive reactance is different from resistance. One of the important differences is that the amount of reactance increases as the frequency of the AC current in creases. The exact relationship is expressed in the formula : XL= 2 pi fL (see example 3 in appendix) where XL is the inductive reactance in ohms, f is the frequency of the current in cycles per second, L is the inductance in henrys. It can be seen that as F is made larger in the formula, XL also becomes larger. Therefore, less current will flow in the circuit at higher frequencies, and more current at lower frequencies. A capacitor also exhibits a similar effect known as capacitive reactance. This also acts like resistance in an AC circuit, but again there are some important differences. Capacitive reactance depends on the size of the plates in the capacitor, the spacing, and the insulating material between them. The expression for reactance produced by a capacitor is: Xe= 2 pi r C (see example 6 in appendix) where Xe is the capacitive reactance in ohms, f is the frequency in cycles per second, C is the capacitance in farads. This formula shows that Xe decreases as the frequency in creases. This is the opposite effect from XL, and it means that the AC current flowing in a circuit will be greater at higher frequencies, and less at lower frequencies. The fact that the effects of Xr, and Xe are exactly opposite with changes in frequency means that when a capacitor and an inductance are both present in a circuit their reactances tend to cancel each other. For any circuit containing both Xr, and Xe, there will be one frequency where they are equal and will completely cancel, leaving no reactance. This frequency is called the resonant frequency of the circuit and represents the one condition when maximum current will flow. Because their reactance is dependent upon frequency, coils and capacitors are used to filter out certain frequencies and pass others. A capacitor passes high frequencies easily, but resists the flow of current at low frequencies. A coil blocks the highs and passes the lows. POWERS OF TEN Technicians use a convenient way of writing very large and very small numbers used in electronics calculations. To multiply a number by 10 it is only necessary to affix a zero as: 2 X 10 = 20 And to multiply by 100, add two zeros; by 1000 add three zeros, etc: 2 X 100 = 200 2 X 1000 = 2000 2 X 1,000,000 = 2,000,000 This idea can easily be extended to numbers containing a decimal point when one recalls that the number "2" used above is also "2.0" : 2.5 X 10 = 2§, 2.5 X 1000 = 2~ .25 X 10 =J_.5 .025 X 1000 = 5 In each case above, the product is obtained by moving the decimal point to the right as many places as there are zeros in the power of ten used, adding zeros as place holders when necessary. Technicians indicate the power of ten used for multiplication as an exponent written above and to the right of the number. Thus, 100 can be written as 102 1000 can be written as 103, and 1,000,000 can be written as 10^6, etc. By using the powers of ten it is no longer necessary to write all the zeros ordinarily needed to express a large number: 20 = 2 X 10^1 200 = 2 X 10^2 2000 = 2 X 10^3 2,000,000 = 2 X 10^6 It is sometimes more convenient to write numbers containing decimals by stating them as a number between 1 and 10, times the appropriate power of ten: 2500 = 2.5 X 10^3 250,000 = 2.5 X 10^5 2,500,000 = 2.5 X 10^6 Dividing by powers of ten uses the same rule for moving the point except that it is moved to the left: 20 + 100 = 20 ~ 200 ; 100 = 2.e 250 + 10,000 = .e, 3.59 + 100,000 = e59 To express a number which is divided by a power of ten, the technician uses the exponent with a minus sign : 2 can be written as 200 x 10^2 2.5 can be written as 25 x 10^1 .0000359 can be written as 3.59 x 10^5 The negative exponent indicates the number of places the decimal should be moved to the left if the number is to be written without the power of ten. The answers to the following exercises are given at the end of the appendix. Group 1 Write each of the following as a number between 1 and 10 times the appropriate power of ten. Example: 250 = 2.5 X 10^2 .0093 = 9.3 X 10^3 1. 2000 5. .16 9. 2¼ million 2. 17.9 6. .0257 10. 17 ten thousandths 3. 208 7. .001 4. 189.763 8. 675,000 Write each of the following as a number without using the power of ten. Example: 2.7 X 10^2 = 270 11. 16.7 X 10^3 15. 2.6 X 10^6 19. 100 X 10^3 12. 129 X 10^1 13. .007 X 10^2 20. 10 X 10^1 13. . 067 X 10^2 17 . 124.6 X 10^3 14. 100,000 X 10^5 18. .1009 X 10^2 Multiplication When two numbers are multiplied, the powers of ten are added: 100 X 1000 = 10^2 X 10^3 = 10^2 H = 10^5 = 100,000 650 X 8 =00 = (65 X 10^1 ) x (8 X 10^2 ) = 520 X 10^1 +2 = 520 X 10^3 = = 5205 When necessary, negative powers can be used: .039 X .08 = (39 X 10^3) X (8 X 10^2 ) = 312 X 10^5 When positive and negative powers of ten are mixed, the powers are combined algebraically, considering the signs: 490 X .006 = (4.9 X 10^2 ) X (6 X 10^3 ) = 29.4 X 10^1 = 2.94 .7 X 8000 = (7 X l01 ) X (8 X 103) = 56 X 10^2 = 5600 Division When dividing, the powers of ten are subtracted: 1000..:... 10 = 1000 = 10^3 The answers to the following exercises are given at the end of the appendix. Group 2 Write the product of the following as a number between 10 and 100 times the appropriate power of ten. Example: 1.69 X 10^2 X 3 X 10^4 = 5.07 X 10^2 = 50.7 X 10^3 1. 17.3 X 10^2 X 70 X 10^3 6. 2 X 10^3 X 5 X 10^2 2. . 069 X 1.7 X 10^6 7 . 49,000 X .0008 3. 8.49 X 106 X 116 X 10^2 8. 9.4 X 10^6 X .08 X 10^6 4. 2 X 10^3 X 1.7 X 10^4 9. 22.2 X 2.22 X 10^2 5. 12 X 10^3 X 8.73 X 10^3 10. 1.2 X 10^1 X .008 X 10^2 X .0006 Write the quotients of the following as a number between 10 and 100 times the appropriate power of ten. Example: 1.8 X 10^3 = 18 X 10^2 = 2 10^4 20 10^3 9 X 10^2 9 X 10^2 X 'or X 16.9 X 10^3 2 X 10^3 ll. 1.3 X 102 15· 150 X 10^6 12. .209 X 106 1.8 X 10^4 .0168 l3. 8 X 10^3 12,700 14 3.6 X 10^2 16. 17. 18. 27.3 X 106 3.9 X 10^3 148 X 10^3 2,500 390 X 10^6 27 X 10^6 19. 20. 5 X 10^2 5 X 10^2 1.5 mv 2.2meg Powers of ten gives a great advantage when dealing with the kind of numbers used in electronics because the units used for various quantities are usually extremely large or very small. Current is often measured in milliamperes (ma). A milliampere 1s 1000 o an ampere. 1 f Capacitance is often measured in microfarads (mfd). A microfarad is 1 00 ~ 000 of a farad. Frequency is 'sometimes stated in kilocycles (kc or khz), which is 1000 cycles. The table on the next page gives common units and their abbreviations. Units Used Abbreviation voltage volts V millivolts mv microvolts µV current amperes A milliamperes ma microamperes µ,a frequency cycles per cy or hz second kilocycles khz megacycles mhz resistance ohms n kilohms K megohms meg capacitance farads none microfarads mfd micromicro mmf or pico farads or picofarads inductance henry hy millihenry mh micro henry µ,h In making calculations it is convenient to shift from one kind of unit to another, as: .032 amp might be written 32 ma, or 32 x 103 amps. 1250 ohms is often expressed as 1.25K. 120,000 cycles could be written 120 khz or .12 mhz or 1.2 X 105 Cy. The technician must be able to change from one kind of unit to another quickly and accurately. To this end the conversion table on the next page should be memorized. Nearly all electronic formulas require that the values be used in the original units. That is, resistance must be used in ohms; current in amperes; capacity in farads, etc. When the values are given in other units, they must be converted and this is easily done with the powers of ten. The examples of typical circuit problems, starting on the next page, illustrate the convenience of the system. CONVERSION TABLE Example 1 To find the voltage drop across R1 : E = Il'R E = 2.5 ma x .1 meg E = (2.5 X 10^3 ) X (1 X 105 ) E = 2.5 x 102, or 250V Note that .1 meg was changed to 1 x 105 instead of .1 x 10^6 • This is often done to eliminate decimal points. Example 2 What is the input current through R1 ? I E1n In= R1 20 mv 20 x 10a 5 Iin = .5 meg = 5 X 10 +5 = 4 X 108 I1n = .04 X 100, or .04 p.a 20MV  + Final answers are usually stated in the most convenient units rather than with the powers of ten. Example 3 What is the value of XL when f = 455 khz and L = 2.3 mh? : 271' = 2 X 3.14 : : = 6.28 : XL = 6.28 X 455 X 10^3 X 23 X 10^4 = 65,720 X 101 = 65720 455 khz JI( 657211 L Another convenient way to use the powers of ten is to change all numbers to equivalents between 1 and 10: XL = 6.28 X 4.55 X 105 X 2.3 X 10^3 [Estimating: 6 x 5 x 2 = 60 x 102 helps to locate decimal] 65.72 X 10^2 = 6572n One important precaution should be emphasized. Numbers having different powers of ten can be multiplied or divided directly as we have seen, but they cannot be added or subtracted. We can multiply: 6K x 3n= 18K But we cannot add: 6K + 3n ¥a 9K, or 6K + 3n ¥a 63n In order to add numbers with powers of ten, the powers must be the same: 6K + 3n = 6000 + 3 = 6003n, or 6K + 3n = b X 10^3 + .003 X 10^3 = 6.003 X 10^3 Example 4 on the next page illustrates the idea. Example 4 The total resistance resulting from the parallel combination of R1 and R2 below is given by the formula: @ 6000 R1 x R2 RT= R1 + R2 R _ (9.2K) (600) T  9.2K + 600 Changing 600n to .6K facilitates adding in the denominator: RT= (9.2K) (600) 9.2 X 103 X 6 X 102 9.2K + .6K 9.8 x 103 6 x 9.2 can be done mentally: R  55· 2 X 105  5 63 102  563 T  9.8 X 103  . X  !l Example 5 When two resistors are in series across a known voltage, the drop across one of them is found by the formula: ~ 'iii"  ETR1 + e1 ~ ei  R1 + R2 @ Er•300V or 8:? .l lllll,j ET R2 _L .._______ e2 = R1 + R2 To find the voltage across R1 : 300 x 4.7K e ~~= 1  .1 meg + 4.7K 3 X 102 X 4.7 X 103 100 X 103 + 4.7 X 103 Note denominators with same power of ten to permit adding 14.1 X 105 104.7 X 103 = .134 X 102 = 13.4 volts To find the voltage across R2 : Estimating mentally: lo~ = ~ : .14 helps to locate decimal : : point : ET R2 3 X 102 X 1 X 105 3 X 107 e2 = R1 + R2 = 104.7 X 103 = 104.7 X 103 = .02866 x 104 = 286.6 volts Example 6 The capacitive reactance (Xe) of the coupling capacitor be low varies with frequency. Find Xe at 100 hz and at 20 khz. 1 Xe= 2,rfC 21r = 6.28 and l1r = 159 x 10a so a convenient way to write the formula is: 159 X l03 Xe= fC At 100 hz: r.... r.... 100 hz vv .... 1111111111~111 mnmtt mt 20 khz 159 X 103 Xe = 100 X .02 X 106 159 X 103 = 79.5 x 103 or 79.5K At 20 khz: 2 X 106 159 X 103 159 X 103 Xe= 20 X 103 X .02 X 106 = 4 X lQ4 = 39. 75 X lQl = 397.5 ohms GRAPHIC ANALYSIS OF AMPLIFICATION BASED ON THE LOAD LINE Plate voltage is determined by the amount of drop across the plate load resistor. As the plate current increases, more of the battery voltage is dropped across the resistor, and the platetocathode voltage becomes less. The control grid of a vacuum tube behaves like a valve, controlling the plate current. The amount of plate current is controlled by the gridto cathode voltage. Grid voltage therefore controls plate voltage by controlling the amount of current through the load resistor. For example, the 200 volts from the "B" battery in the figure below must be divided between the drop across the plate resistor and the drop across the tube. If there were 5 ma of plate current, the drop across the resistor would be: 20K x 5 ma = 100 volts This would leave 100 volts across the tube. 200V T 20K 100\/ If the plate current were 2 ma, the drop across the resistor would be: 20K x 2 ma = 40 volts This leaves: 200  40 = 160 volts across the tube. With a plate current of 9 ma, the drop across the resistor is 180 volts, leaving only 20 volts across the tube. Thus, it can be seen that the voltage across the tube is always the remainder after the resistor voltage is subtracted from the battery voltage. Also, as the plate current increases, the voltage across the tube decreases, and vice versa. If the voltages and currents found above are graphed, the results of the 20K load resistor can be seen to be a straight line. The horizontal axis on the graph shows plate voltages, and the vertical axis gives the corresponding plate current. The slanted line is called the 20K load line. The load line crosses the horizontal axis at 200 volts because this is the maximum possible plate voltage in our example and will occur when the current is zero. The load line crosses the vertical axis at 10 ma because this is the maximum possible plate current that can flow, and at this time the plate voltage is zero because all the battery voltage is dropped across the resistor. 20K x 10 ma= 200 volts To graph a load line for any size resistor it is only necessary to determine these two points. The maximum voltage is at one end of the line (on the horizontal axis) and the maximum current is at the other end ( on the vertical axis). The maximum current is the maximum voltage divided by the resistance. 200 volts _ 10 lOK – ma The curved lines in graph number 2 show the effect grid voltage has on the current through the tube. It can be seen that if the grid voltage were 3 for the triode in our example, the resulting plate current of 5 ma flowing in the 20K load resistor would produce 100 volts of plate voltage. Similarly, if the grid voltage were 0, the resulting plate current would be about 5.75 ma, and this would give a plate voltage of about 85 volts. If the grid voltage were 6, the plate current would be about 4.25 ma, and this would give a plate voltage of about 118 volts. These three points are indicated on graph number 2. Graph number 3 shows what happens to the plate voltage when the grid voltage is made to vary according to an input signal. Suppose the nosignal grid voltage is 3 and that the input signal causes the grid voltage to vary up to 0 volts and down to 6 volts, as indicated on the graph. The peaktopeak value of the input signal is then 6 volts. The meaning of the expression "voltage amplification" be comes clear when we notice that the plate voltage changes correspond to the changing currents through the load resistor. The waveform drawn below the horizontal axis shows that the plate signal is 33 volts peaktopeak. Comparing the input and output voltages, we see that the output is 5.5 times greater than the input, and this represents a voltage gain of 5.5. The importance of the load resistor in producing voltage gain should not be overlooked. A larger load resistor causes the load line to cross the vertical axis at a lower value, and this increases the peaktopeak swing of the plate voltage. Answers to Group I 1. 2 X 10^3 6. 2.57 X 10^2 2. 1.79 X 10^1 7. 1 X 10^3 3. 2.08 X 10^2 8. 6.75 X 10^5 4. 1.89763 X 10^2 9. 2.25 X 10^6 5. 1.6 X 10^1 10. 1.7 X 10^3 11. .0167 16. .00007 12. 12.9 17. 124,600 13. 6.7 18. 10.09 14. 1 19. .1 15. 2,600,000 20. 100 Answers to Group 2 1. 12.11 X 10^7 6. 10 X 10^1 2. 12.43 X 10^1 7. 39.2 3. 98.484 X 10^7 8. 75.2 X 10^14 4. 34 X 10^2 9. 49.284 X 10^2 5. 62.856 X 10^3 10. 96 X 10^3 11. 13 X 10^5 16. 70 X 10^10 12. 11.6 17. 59.2 X 10^6 13. 21 X 10^7 18. 14.44 14. 35.3 19. 10 X 10^3 15. 13.3 X 10^6 20. 68.18 X 10^11

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