The motion picture projectionist (Nov 1931-Jan 1933)

November, 1931 Motion Picture Projectionist 29 Theory and Fundamentals ==== By W. W. Jones ==== Mr. W. W. Jones, whose Department is a monthly feature of this magazine, has long been actively associated with the Motion Picture Industry. At the present time Mr. Jones is a member of the Engineering Department of RCA Photophone and has been closely identified with the educational activities of that organization since the time of its inception. He is a graduate of the Milwaukee College of Engineering and was at one time Instructor of Mathematics and Electrical Design at that institution. — The Editor. Matching the Loudspeaker and the Amplifier BECAUSE sound is emitted from the loudspeaker, the loudspeaker ■ usually receives an undue amount of criticism particularly when that sound is not satisfactory or pleasant to the listener. Sound may be unsatisfactory because it lacks + r LOAD RESISTANCE INTERNAL BATTERY RESISTANCE (r) 10 V0LT5 Fig. 1 volume or because there is distortion present. This lack in volume or distortion of signal does not necessarily mean that the speaker is at fault. The speaker may be entirely satisfactory and yet produce unsatisfactory results if it is not properly matched to the amplifier. For the purpose of discussion, the question of volume or power, and distortion will be considered separately. If we are to use efficiently the power available from the amplifier, it is necessary that the speaker be properly matched to the output of the amplifier. It is to be observed at this point that the question of matching the speaker to the amplifier involves the problem of distortion as well as power. Accordingly, the problem of distortion is not to be forgotten during the discussion of speaker matching for maximum power output. Maximum Power Output Maximum power output may be obtained from the amplifier when the load impedance of the amplifier power tube is equal to the tube plate impedance. This statement has been found to be true in practice and can be shown to be true by the following simple electrical analogy: Assume a variable resistance load (R) is placed across a 10 volt battery as shown in Fig. 1. Now assume the battery has an internal resistance (r) of 10 ohms. The resistance (r) has been drawn in the figure to represent internal resistance, and it corresponds in this analogy to the plate impedance of a vacuum tube. The resistance load (R) corresponds to the load impedance of a vacuum tube. The 10 volt battery corresponds to the voltage output of the tube. All of the values of volts and ohms selected are arbitrary and the results, of course, are independent of these values. GENERATOR ( 4 T0 1 RATIO, / r4= io amps y i n Zp 1 T mill 1ET 1 Zs»t0 0HMS Ws = 1000 WATTS Wp = 1000 WAITS Ip = MAMP5 Zp =(60 OHMS Fig. 2 The tabulated data below is obtained as follows: Assume a load resistance (R) equals zero. The current in the circuit then equals (from Ohm's Law) volts divided by ohms or 10 divided by 10 or 1 ampere. Then, the power across the load circuit equals the current squared times the load resistance (R) or 1 ampere times 1 ampere times zero ohms or zero watts as shown by the table under the heading Load Power. As a further example, assume a load resistance (R) equals 10 ohms. The total resistance of the circuit equals (r) plus (R) or 10 ohms plus 10 ohms or 20 ohms. The current in the circuit then equals volts divided by ohms or 10 divided by 20 or 0.5 amperes. The power across the load circuit equals 0.5 squared times 10 ohms or 0.25 times 10 or 2.5 watts. The remainder of the data can be calculated in a similar manner. Load Battery Load Power Resistance (R) Resistance (r) 0 10 0 2 10 1.39 5 10 2.22 10 10 2.50 15 10 2.45 20 10 2.22 50 10 1.39 By examination of the above tabulated data it can be seen that the power delivered to the load resistance (R) by the battery is maximum when the load resistance (R) is equal to the battery resistance (r). Accordingly, in the operation of a vacuum tube circuit the tube load impedance must be equal to the tube impedance when maximum power delivered to the load is desired. It is stated above that it is necessary to match the loudspeaker to the amplifier to prevent distortion. The above explanation on maximum power does not apply to maximum undistorted power, because it is necessary to sacrifice power in order to prevent distortion. In the case of loudspeaker and amplifier matching the load impedance is adjusted equal to two times the tube plate impedance, 'hen maximum undistorted power is desired. While this statement has been shown to be the true both in theory and in practice there is no simple explanation for it available as in the case for maximum power. Impedance Matching Transformer There are several means of coupling electrical circuits for proper impedance matching. However, it is common practice to couple the loudspeaker to the amplifier by means of a coupling transformer commonly known as an output transformer. Since this is true, only transformer coupling will be considered. In order to appreciate the meaning of impedance matching and in order to demonstrate the use of a transformer as a device for impedance matching, the following explanation is given: Fig. 2 represents an ordinary A. C. lighting power transformer which reduces the voltage from 400 volts to PLATE IMPEDANCE =1500 OHMS Si Zs=30>"o OHMS <1 L0UD5PEAKER' • VOICE COIL Fig. 3 100 volts. By examining the figure it will be noted that the secondary load impedance on the transformer is 10 ohms, and the power consumed is 1000 watts. We also note that the ratio of the transformer is 4 to 1, and that the power (Wp) in the primary circuit must be 1000 watts, the same as in the secondary circuit. Let it be required to find, first, the primary current (Ip) and, secondly, the primary load impedance (Zp) so that the effect of the transformer on the secondary load impedance (Z») can be determined. The primary current, Wp 1000 Ip = — — = = 2.5 amperes. Ep 400 and the primary load impedance, Ep 400 Zp = = = 160 ohms. Ip 2.5 The primary load impedance (Zp) connected across the generator as determined above is 160 ohms, whereas (Continued on page 40)