THE UNIVERSITY OF MICHIGAN INDUSTRY PROGRAM OF THE COLLEGE OF ENGINEERING LARGE SIGNAL ANALYSIS OF DISTRIBUTED AMPLIFIERS Phil H. Rogers This report was submitted in partial fulfillment of the requirements for the Degree of Doctor of Philosophy in the University of Michigan. It was originally distributed as Technical Report No. 52 by the Electronic Defense Group, the Department of Electrical Engineering, the University of Michigan, under Engineering Research Institute Project 2262, on Contract No. DA-36-039 sc-63203, Signal Corps, Department of the Army. August, 1955 IP-125

ACKNOWLEDGEMENT The Industry Program of the College of Engineering wishes to express its appreciation to Phil H. Rogers, Assistant Professor of Electrical Engineering, the University of Michigan, for making it possible to distribute this report under the Industry Program cover,

Errata Sheet ELECTRONIC DEFENSE GROUP Technical Report No. 52 Page ii - Add paragraph at beginning of Abstract, to read: "The primary importance of the distributed amplifier circuit is derived from its ability to operate in the frequency range above that covered by conventional amplifiers and below that covered by microwave amplifiers, The frequency range of conventional wide-band power amplifiers is generally limited to frequencies of the order of 10 me, Microwave power amplifiers of the traveling-wave-tube type have been constructed to operate with lower frequency limits of the order of hundreds of megacycles. The distributed amplifier can be designed to operate from tens of megacycles to hundreds of megacycles, thus filling the gap between conventional techniques and microwave techniques, Page 2, Paragraph 1, Line 3 - Add "line" after "transmission." Page 2, Paragraph 2, Line 5 - Change "amplifier" to "amplifiers." Page 3, Paragraph 1, Line 1 - Add hyphen between. "high" and "frequency." Page 8, Paragraph 3, Line 2 - Delete second "is independent." Page 10, Paragraph 1, Line 9 - Change "are" to "is" Page 10, Paragraph 2, Line 9 - Change " contribution" to "contributions." Page 11, Paragraph 2, Line 3 - Change impedances" to "impedance." Page 19, Paragraph 2, Line 3 - Add hyphen between-"low" and "frequency," Page 23, Equation No, 3.7 - Change minus to plus. Page 33, Paragraph 2, Line 4 - Change "logrithrim" to "logarithm." Page 36, Title of Section 4.3 - Add asterisk after "line." Also add footnote at bottom of page, to read: "The small signal case is discussed in Ref. 2." Page 51, Equation No. 5.4 - Delete superscript "2" on "n". Page 56, Between 2nd and 3rd paragraphs - Add paragraph to read: "Figure 4.1 shows the plate load impedances for each tube of a six-tube distributed amplifier. The plate supply voltages on the first three tubes can be reduced without appreciably affecting the output power if the frequency of operation for this amplifier is limited to the range 60 to 270 mc. However, the dc power input could be reduced to 77 percent of the normal value. This increases the efficiency by a factor of 1.3," Page 83, Paragraph 2, Line 3 - Change sentence beginning "It can be seen..." to read: "It can be seen from these figures that the efficiency tends to increase with screen voltage; however, the peaks in the power output reach a maximum and decrease as the screen voltage is further increased." Page 89, Paragraph 4, Line 1 - Add superscript "2" after "small-signal."

ENGINEERING RESEARCH INSTITUTE UNIVERSITY OF MICHIGAN ANN ARBOR LARGE SIGNAL ANALYSIS OF DISTRIBUTED AMPLIFIERS Technical Report No. 52 Electronic Defense Group Department of Electrical Engineering by Phil H. Rogers Approved by: )" ~. -4. W. Welch, Jr. Project 2262 TASK ORDER NO. EDG-1 CONTRACT NO. DA-36-039 sc-63203 SIGNAL CORPS, DEPARTMENT OF THE ARMY DEPARTMENT OF ARMY PROJECT NO. 3-99-04-042 SIGNAL CORPS PROJECT NO. 194B Submitted in partial fulfillment of the requirements for the Degree of Doctor of Philosophy in the University of Michigan July 1955

ABSTRACT It is the purpose of this dissertation to extend the analysis of distributed amplifiers to account for large signal effects, to determine a graphical procedure for analyzing the non-linear operation of the tubes in a distributed amplifier, and to investigate the efficiency of distributed amplifiers. General equations are derived describing the behavior of distributed amplifiers, including the operating load line for each tube in an n-tube distributed amplifier. The load lines for all but the last tube are ellipses and as a result graphical calculations are long and tedious. However, equations are derived whereby the output power can be calculated from the load line of the last tube at low frequencies, simplifying the graphical analysis considerably. In addition the frequency response is examined under large-signal conditions and the modifications necessary in the small-signal design equations for a flat frequency response are indicated. The frequency limitations in distributed amplifiers are determined on the basis of the realization of the constant-k artificial transmission line with finite grid and plate lead inductances. In practice the highest cut-off frequency is about 0.8 the series resonant frequency of the grid lead inductance and the input capacitance of the tube. An additional frequency limitation is present in large-signal distributed amplifiers in that clipping in the plate -circuit of the last few tubes will cause the large-signal response to differ from the small-signal response. The maximum theoretical efficiency of a distributed amplifier operating over its entire frequency range is 30 percent. This value occurs at a plate conduction angle of about 225 degrees. However, if the operating frequency range is limited to 0.2 to 0.9 of the design cut-off frequency, the efficiency can be increased by decreasing the plate supply voltage for the first few tubes. A graphical procedure is presented for calculating the power output, the efficiency, the second harmonic power output, and the frequency above which clipping occurs. ii

ACKNOWLEDGMENT The author wishes to express his appreciation to the members of his committee for their assistance, and especially to Professor Alan B. Macnee for his valuable criticisms and suggestions during the course of the work. Mr. Ward Getty assisted in the many graphical calculations. The research on which this dissertation is based was supported by the U. S. Army Signal Corps under Contract No. DA-36-039 sc-63203. iii

TABLE OF CONTENTS Page ABSTRACT ii ACKNOWLEDGMENT iii LIST OF TABLES vi LIST OF FIGURES vii CHAPTER I. INTRODUCTION 1 1.1 Statement of Problem 1 1.2 General Considerations of Distributed Amplifier 2 CHAPTER II. LARGE-SIGNAL ANALYSIS OF DISTRIBUTED AMPLIFIERS 4 2.1 Assumptions 7 2.2 Plate Load Impedance 9 2.3 Plate Load Impedance for Specific Cases 11 CHAPTER III. FREQUENCY LIMITATIONS IN DISTRIBUTED AMPLIFIERS 21 3.1 Cancellation of Lead Inductances 3.2 Conductive Loading of the Grid Line Due to High Frequency Effects 24 3.3 Frequency Tolerance 28 3.4 Summary 31 CHAPTER IV. GRID LOSSES 4.1 Attenuation Constant of Constant-k Filters with Losses 32 4.2 Determination of the Plate Load Impedance with Attenuation in the Grid Line 33 4.3 Modification of Gain Equation to Account for Losses in the Grid Line 36 CHAPTER V. EFFICIENCY 50 5.1 Calculation of Efficiency 50 5.2 Maximum Theoretical Efficiency of Distributed Amplifiers 52 5.3 Staggering the Plate Supply Voltages for Individual Tubes 56 CHAPTER VI. GRAPHICAL ANALYSIS 6.1 Introduction 58 6.2 Graphical Determination of Power Output and Efficiency 58 6.3 Approximate Graphical Solution 60 6.4 Frequency Tolerance 61 iv

TABLE OF CONTENTS (Cont.) Page 6.5 Second Harmonic Distortion 63 6.6 Sample Design of a Distributed Amplifier Using 4X150A Tubes 64 CHAPTER VII. CONCLUSIONS 89 7.1 Introduction 89 7.2 Summary of Results 89 7.3 Suggestions for Further Research 90 APPENDIX A. SERIES FOR PLATE-CATHODE VOLTAGE ON KTH TUBE OF AN N-TUBE DISTRIBUTED AMPLIFIER 91 APPENDIX B. LOW-FREQUENCY DISTRIBUTED AMPLIFIER USING 807 TUBES 94 APPENDIX C. PLATE LOAD IMPEDANCE FOR PAIRED-PLATE 97 APPENDIX D. INPUT RESISTANCE OF A VACUUM TUBE DUE TO CATHODE LEAD INDUCTANCE 100 APPENDIX E. DETERMINATION OF PLATE LOAD IMPEDANCE WITH ATTENUATION IN THE GRID LINE 102 APPENDIX F. GRAPHICAL CALCULATION OF THE OPERATION OF A 9-TUBE DISTRIBUTED AMPLIFIER USING 4X150A TUBES 106 BIBLIOGRAPHY 142

LIST OF TABLES Page 6.1 Instantaneous Values of Plate Current and Grid Current Eb = 800v; Ec2 = 400v; Eel = -13.3 volts; Esig = 33.3 volts peak 74 6.2 Results and Computations Ebq = 800v, Ec2 = 400v, EC1 = -13.3,Esig = 335.3 volts peak 81 B.1 Measurements on Tube Number 1 of 15 KC Distributed Amplifier 96 F.1 Results and Computations Ebq = 800v, Ec2 = 400v, Ecl = -20 v, Esig = 40 v peak 107 F.2 Results and Computations Ebq = 800v, Ec2 = 400v, Eel = -26.6v, Esig = 46.6v peak 113 F.3 Results and Computations Ebq = 800v, Ec2 = 400v, El = -40 volts, Esig = 60 v peak 119 F.4 Results and Computations Ebq = 800v, Ec2 = 400v, Ecl = -53.3v, Esig = 73.3v peak 125 F.5 Results and Computations Ebq = 800v, Ec2 = 400v, Ecl = -66.6v, Esig = 86.6 v peak 131 F.6 Results and Computations Eq = 800v, Ec2 = 400v, Ecl = -80v, Esig = 100 v peak 136 vi

LIST OF FIGURES Page Fig. 2.1 Block Diagram of an n-Tube Distributed Amplifier 5 Fig. 2.2 Zp and G as a Function of cp for Tube 1 of a 6-Tube Distributed Amplifier 12 Fig. 2.3 Zp and G as a Function of cp for Tube 2 of a 6-Tube Distributed Amplifier 13 Fig. 2.4 Zp and 0 as a Function of cp for Tube 3 of a 6-Tube Distributed Amplifier 14 Fig. 2.5 Zp and G as a Function of cp for Tube 4 of a 6-Tube Distributed Amplifier 15 Fig. 2.6 Zp and G as a Function of cp for Tube 5 of a 6-Tube Distributed Amplifier 16 Fig. 2.7 Zp and Q as a Function of cp for Tube 6 of a 6-Tube Distributed Amplifier 17 Fig. 2.8 Plate Load Impedance for a 9-Tube Distributed Amplifier 18 Fig. 2.9 Plate Load Impedance for a 6-Tube Paired-Plate Distributed Amplifier 20 Fig. 3.1 Schematic Diagram for a Low-Pass, Pi-Section, Constant-k Filter 21 Fig. 3.2 Negative Mutual Transformer and Its Equivalent "T" 23 Fig. 353 Plot of Cut-off Frequency of Constant-k Filter as a Function of the Coefficient of Coupling of the NegativeMutual Transformer Necessary to Cancel the Lead Inductance 25 Fig. 3.4 Equivalent Circuit for the Determination of Input Conductance Caused by the Feedback Resulting From Cathode Lead Inductance 26 Fig. 3.5 Variation of Mid-Shunt Image Impedance as a Function of Frequency 29 Fig. 3.6 Variation of the Plate Load Impedance of the Last Tube of a 6-Tube Distributed Amplifier as a Function of Frequency 30 Fig. 4.1 Plate Load Impedance as a Function of Frequency for a 6-Tube 4X150A Distributed Amplifier with a 90Q Plate Line 35 vii

LIST OF FIGURES (Cont.) Page Fig. 4.2 Plate Load Impedance as a Function of Frequency For Tube No. 1 and Tube No. 6 of a 6-Tube 4X150A Distributed Amplifier 37 Fig. 4.3 Normalized Voltage Amplification as a Function of Frequency for no 5 and odr 0.05 41 Fig. 4.4 Normalized Voltage Amplification as a Function of Frequency for na = 0.5 and odr = 0.10 42 2 co Fig. 4.5 Normalized Voltage Amplification as a Function of Frequency for n 0o and Codr = 0153 Fig. 4.6 Normalized Voltage Amplification as a Function of Frequency for n = 0.45 and odr = 0.20 44 Fig. 4.7 Normalized Voltage Amplification as a Function of Frequency for n% 049 and %odr 02045 =~- o.49 cr, 0.20 Fig. 4.8 Normalized Voltage Amplification as a Function of Frequency for no- 0.5 and %odr = 0.20 46 2 Go Fig. 4.9 Normalized Voltage Amplification as a Function of Frequency for n2o = 0.45 and od = 0.25 47 2 a0O Fig. 4.10 Normalized Voltage Amplification as a Function of Frequency for no 4and 0odr5 48 4.48 a0n25 a8 a0 Fig. 4.11 Normalized Voltage Amplification as a Function of Frequency for n = 0.5 and odr 0.50~= 0.25 49 Fig. 5.1. Idealized Constant-Current Tube Plate Characteristics With Load Line Drawn for Class A Operation 53 Fig. 5.2 Idealized Constant-Current Tube Plate Characteristics With Load Line Drawn for Class B Operation 53 Fig. 5.3 Instantaneous Value of Plate Voltage and Plate Current 54 Fig. 5.4 Variation of Maximum Theoretical Efficiency of a Distributed Amplifier as a Function of the Conduction Angle of Plate Current 57 Fig. 6.1 Frequency Variation of the Plate Load Impedance of the 6th Tube in a Distributed Amplifier Using 6, 4X150A Tubes with a Nominal Grid Line Impedance of 50 Ohms and a Nominal Plate Line Impedance of 90 Ohms 62 viii

LIST OF FIGURES (Cont.) Page Fig. 6.2 Second Harmonic Output Current Factor, Fn o) for n = 6 and n = 9 Plotted as a Function C of Frequency 65 Fig. 6.5 Variation of Equivalent Shunt Resistance of a 4X150A Caused by Lead Inductances and Transit Time 66 Fig. 6.4 Attenuation Constant Per Section of Grid Line Due to Small Signal Effects of a Distributed Amplifier Using 4X150A Tubes. Rg = 50 S 68 Fig. 6.5 Variation of Grid Voltage as a Function of Frequency for a 300 MC, 4X150A Distributed Amplifier for Peak Grid Voltages Iess Than 0 Volts 69 Fig. 6.6 Plate Characteristics for a 4X150A with Ec2 = 400 Volts Load Lines Drawn for a 9-Tube Distributed Amplifier at Low Frequencies for Various Q-Points Considered in Graphical Analysis 71 Fig. 6.7 Dynamic Transfer Characteristic for a 9-Tube Distributed Amplifier with Ebb = 800V, Ec2 = 400V. 72 Fig. 6.8 Grid Current as a Function of Grid Voltage for a 4X150OA 73 Fig. 6.9 ib VS ct for Ebq = 800V, Ec2 = 400V, Ecl = -13.3V Esig = 33.3 V peak 75 Fig. 6.10 ib COS wt VS. act for Ebq = 800V, Ec2 = 400V, E1 = -1353V Esig = 33.3V Peak 76 Fig. 6.11 ib COS 2at VS. wt for Ebq = 800V, Ec2 = 400V, EC1 = -13.3V Esig = 33.3V Peak 77 Fig. 6.12 Instantaneous Value of Grid Current as a Function of cot for Ebq = 800V, Ec2 = 400V, E1 = -13.3V Peak Signal Voltage = 33.3 Volts 78 Fig. 6.13 icl COS cot VS. ct for Ebq = 800V, Ec2 = 400V, Ecl = -133V Peak Signal Voltage = 33.3 Volts 80 Fig. 6.14 Graphical Summary of Results Calculated for a 9-Tube Distributed Amplifier Using 4X150A Tubes for Ebq = 800V, Ec2 = 400V, and a Maximum Instantaneous Grid Voltage of +20V 82 Fig. 6.15 Graphical Summary of Results Calculated for a 9-Tube Distributed Amplifier Using 4X150A Tubes for EbO = 800V, Ec2 = 500V, and a Maximum Instantaneous Grid Voltage of +25V 84 ix

LIST OF FIGURES (Cont.) Page Fig. 6.16 Graphical Summary of Results Calculated for a 9-Tube Distributed Amplifier Using 4X150A Tubes for Eb = 80oV, Ec2 = 600V, and a Maximum Instantaneous Grid Votage of +10V 85 Fig. 6.17 Graphical Summary of Results Calculated for a 9-Tube Distributed Amplifier Using 4X150A Tubes for Ebq = 1000 Volts, Ec2 = 400 Volts, and a Maximum Instantaneous Grid Voltage of +20 Volts 86 Fig. 6.18 Graphical Summary of Results Calculated for a 9-Tube Distributed Amplifier Using 4X150A Tubes for Eb lOOO1V, Ec2 = 500V, and a Maximum Instantaneous Grid Voltage of +25V 87 Fig. 6.19 Graphical Summary of Results Calculated for a 9-Tube Distributed Amplifier Using 4X150A Tubes for Eb = l1000V, Ec2 = 600V, and a Maximum Instantaneous Grid Voltage of +10V 88 Fig. B.1 Schematic Diagram for 15kc Distributed Amplifier 95 Fig. C.1 Block Diagram of Paired-Plate Distributed Amplifier 97 Fig. D.1 Equivalent Circuit of a Tube Used for the Determination of Input Conductance 100 Fig. F.l ib VS. wt for Ebq = 800V, Ec2 = 400V, E1 = -20V, Esig = 40V Peak 108 Fig. F.2 ib COS cot VS. wt for Ebq = 800V, Ec2 = 400V, Ecl= -20V, Esig = 40V Peak 109 Fig. F.3 ib COS 2 wt VS wt for Ebq = 800V, Ec2 = 400V, E = -20V, Esig = 40V Peak 110 Fig. F.4 Instantaneous Value of Grid Current as a Function of ct for Ebq = 800V, Ec2 = 400V, EC1 = -20V, Peak Signal Voltage = 40.0 Volts 111 Fig. F.5 i COS wt VS wt for Eb= 800V, Ec2 = 400V, Ec1 = -20.OV Peak Signal Volts = 40.0 Volts 112 Fig. F.6 ib VS ct for Ebq = 800V, Ec2 = 4V, E1 = -26.6V, Esig = 46.6V Peak 114 Fig. F.7 ib COS wt VS wt for Ebq = 800V, Ec2 = 400V, Ec1 =-26.6V, Esig = 46.6V Peak 1x15

LIST OF FIGURES (Cont.) Page Fig. F.8 ib COS 2 wt VS. wt for Eb = 800V, Ec2 = 400V, Ecl = -26.6V, ESig = 46.6 Peak 116 Fig. F.9 Instantaneous Value of Grid Current as a Function of ut for Ebq = 800V, Ec2 = 400V, Ecl = -26.6V, Peak Signal Voltage = 46.6 Volts 117 Fig. F.10 icl COS ot VS. wDt for Eh = 800V, Ec2 = 400V, Ecl =26.6V, Peak Signal Voltage = 46.6 Volts 118 Fig. F.11 ib VS. cnt for Ebq = 800V, Ec2 = 400V, Ec1 = -40Vj Esig = 60V Peak 120 Fig. F.12 ib COS cot VS. ot for Ebq = 800V, Ec2 = 400V, E1 = -40V, Esig = 60V Peak 121 Fig. F.13 ib COS 2cot VS. wct for Ebq = 800V, Ec2 = 400V, Ecl - 40V, Esig = 60V Peak 122 Fig. F.14 Instantaneous Value of Grid Current as a Function of ct, for Eb = 800V, Ec2 = 400V, E = 40V, Peak Signal Voltage = 60.0 Vo~is 123 Fig. F.15 icl COS wt VS. cwt for Eb = 800V, EC2 = 400V, E1 = O40V, Peak Signal Voltage = 60qVolts 124 Fig. F.16 ib VS. wct for Eb = 800V, Ec2 = 400V, Ee1 = -53.3V, Esig = 73.3V Peak 126 Fig. F.17 ib COS c)t VS. cot for Ebq = 800V, Ec2 = 400V, Ecl = -53.3V, Esig = 73.3V Peak 127 Fig. F.18 ib COS 2wot VS. wct for Ebq = 800V, Ec2 = 400V, E = -53.3V, Esig = 73.3V Peak 128 Fig. F.19 Instantaneous Value of Grid Current as a Function of ct for Eba = 800V, Ec2 = 400V, El1 = -53.3V, Peik Signal. Voltage = 73.3 Volts 129 Fig. F.20 icl COS ct VS. ct for Ebq = 800V, E2 = 400V, Ec = -53.3V, Peak Signal Voltage = 73.3 Volts 130 Fig. F.21 ib and i COS ct VS. ct for Eb = 800V, Ec2 = 400V, El = -6.6V, Esig = 86.6V Pea 132 Fig. F.22 ib COS 2ot VS. ct for Ebq = 800V, Ec2 = 400V, Ec1 = -66.6V, Esig = 86.6V Peak 133 xi

LIST OF FIGURES (Cont.) Page Fig. F.23 Instantaneous Value of Grid Current as a Function of wt for Ebq = 800V, Ec2 = 400V, Ecl = -.66.6 Volts, Peak Signal Voltage = 86.6 Volts 134 Fig. F.24 icl COS ct VS wt for Ebq = 800V, Ec2 - 400V, EC1 66.6V, Peak Signal Voltage = 86.6 Volts 135 Fig. F.25 ib VS wt for Ebq = 800V, Ec2 = 400V, Ecl = -80V, Esig = lOOV Peak 137 Fig. F.26 ib COS wt VS wt for Ebq = 800V, Ec2 = 400V, Ecl = -80V, Esig = lOOV Peak 138 Fig. F.27 ib COS 2wt VS wt for Ebq = 800V, Ec2 = 400V, Ecl = -80V, Esig = lOOV Peak 139 Fig. F.28 Instantaneous Value of Grid Current as a Function of ct for Ebq = 800V, Ec2 = 400V, Ecl = -80V, Peak Signal Voltage = 100 Volts 140 Fig, F.29 ib COS wt VS ct for Ebq = 800V, Ec2 = 400V, Ecl = -80.0V, Peak Signal Voltage = 100.OV 141 xii

LARGE SIGNAL ANALYSIS OF DISTRIBUTED AMPLIFIERS CHAPTER I. INTRODUCTION The primary importance of the distributed amplifier circuit is derived from its ability to operate in the frequency range above that covered by conventional amplifiers and below that covered by microwave amplifiers. The frequency range of conventional wide-band power amplifiers is generally limited to frequencies of the order of 10 mc. Microwave power amplifiers of the travelingwave-tube type have been constructed to operate with lower frequency limits of the order of hundreds of megacycles. The distributed amplifier can be designed to operate from tens of megacycles to hundreds of megacycles thus filling the gap between conventional techniques and microwave techniques. 1.1 Statement of Problem Equations describing the behavior of distributed amplifiers when transmitting small signals have been derived and studied by Ginzton, Hewlett, 1 2-9 Jasberg, Noe- and others. They have described various methods of obtaining a specified frequency response, and have also described special design procedures necessary at high frequencies. It is the purpose of this dissertation to extend the analysis of distributed amplifiers to account for large-signal effects, and to determine a graphical procedure for analyzing the non-linear operation of the tubes in a Distributed amplifier. The efficiency of devices designed to deliver power is of major importance in comparing power amplifiers; therefore, the efficiency of the distributed amplifier is discussed in detail in this paper. 1

2 1.2 General Considerations of Distributed Amplifiers Basically, distributed amplification is based upon a physical distributic of vacuum tubes between two transmission lines. The grids of the tubes are connected to the input transmission and the plates are connected to the output transmission line. Adjacent grids and adjacent plates are connected to the transmission lines in such a manner that the phase shift of the input voltage in traversing the line between adjacent grids is equal to the phase shift in the output signal in traversing the plate line between the corresponding plates. All of the individual tube currents add in phase at the output. The significant contribution of distributed amplifiers to wide-band amplification lies in the fact that the input and output capacities of the tubes are used as the shunt elements of artificial transmission lines in which the gain of the individual tubes add rather than multiply as in conventional cascaded amplifier. Hence, the gain of the individual tubes may be less than unity and yet the gain of the amplifier may be adjusted to any desired value by the number of tubes. This circuit essentially removes the gain-bandwidth limitation of conventional amplifiers. The voltage gain of distributed amplifiersl may be written as ngm 2 g P where n = the number of tubes gm = the transconductance of the tubes Zgg = the mid-shunt image impedance of the artificial grid line and Zap = the mid-shunt image impedance of the artificial plate line. Equation (1.1) neglects the attenuation on the grid and plate lines. The voltage amplification has a very objectionable peak as the cut-off frequency is approached due to the nature of Zir. If the attenuation on the grid line is

caused by the transit time and other high frequency effects, it is possible to adjust the product of the number of tubes and the cut-off angular frequency2 for a relatively flat amplification over the design frequency range. nwc 2- X(1.2) Gin where Wc = the cut-off angular frequency Gin = the input conductance of the tube due to high-frequency effects measured at the angular frequency w and Cg = the input capacity of the tube. The large-signal analysis of any active network in general involves the characteristic curves for the active element. The normal procedure in a largesignal analysis is to plot the voltage-current locus of the passive network on the characteristics of the active element. This locus determines the operation of the active element and allows its operation to be calculated. A distributed amplifier differs from most active networks in that each of the tubes in the amplifier contributes to the operating locus of any given tube in the amplifier. Hence the operating locus of plate voltage versus plate current for each tube in a distributed amplifier is different from any other tube. Chaper II is devoted to the determination of this operating locus.

CHAPTER II. LARGE-SIGNAL ANALYSIS OF DISTRIBUTED AMPLIFIERS The analyses of distributed,nplifiers presented in the literature have concerned themselves with the small-signal theory. From the small-signal standpoint the voltage gain and the frequency response are of prime concern; however, in distributed power amplifiers, saturated power output level and efficiency are also factors of prime interest to the designer. In fact, in many large-signal applications more emphasis is placed on the saturated power output level than is placed on the gain. The saturated power output level can only be determined from the tube characteristics. Therefore, in large-signal analysis, the tube characteristic curves will normally be used instead of the s;mall-signal concept of transconductance. In order to be able to utilize the tube characteristic curves for calculations involving distributed amplifiers it is necessary to know the operating locus of plate voltage versus plate current for each tube. This locus is generally referred to as the operating load line or as defining the plate load impedance. In the low-pass distributed amplifier circuit shown schematically in Figure 2.1 it is easily ascertained that all of the tubes are essentially in parallel at frequencies sufficiently low so that the phase shift between sections (i.e., between adjacent plates or grids) may be neglected. If all the tubes are identical, and if the grid voltage applied to each tube does not vary with the position of the tube in the amplifier, then each tube will deliver identical currents to the plate line. The voltage across the output resistance will be greater by a factor n (the number of tubes in the distributed amplifier) than the output voltage would be if only one tube were operating. This voltage also appears across each tube in the distributed amplifier. Hence the impedance seen by each tube is n times the impedance that tube would see if it were the

PLATE LINE OUTPUT REVERE TERMINATION TUBE ix TIt* IUB3T TUTIE t TUBEEE + TuE' n-I TUSE nt ~ eg2 eg~3 e ~-3 egk eg(k+i ed n ekINPUT.0+>-'. GRID LINE INTPUT j TERMINATION, FIG. 2.1. BLOCK DIAGRAM OF AN n-TUBE DISTRIBUTED AMPLIFIER.

6 only one operating, or Z z = (2.1) 2 This equation is valid for operating frequencies sufficiently close to zero so that all the grids and all the plates are essentially in parallel. As the operating frequency increases, the phase angle between the voltages on adjacent grids becomes more important since the grid voltages determine the phase angle of the currents that are fed into the plate line. In a conventional distributed amplifier the phase shift per section of the plate line is the same as the phase shift per section of the grid line, so that each of the paths a signal may pursue through a distributed amplifier produces the same amount of phase shift on the signal. The voltages appearing between plate and cathode of an individual tube will vary with the operating frequency, the position of the tube in the distributed amplifier, and the characteristic impedance of the plate line. Tetrodes or pentodes are generally used in distributed Amplifiers because of their inherently low capacitance between plate and grid. This low inter-electrode capacitance minimizes coupling between the plate line and the grid line and increases the stability of the distributed amplifier. This restriction of the choice of tube types to tetrodes or pentodes means that the tubes used in distributed amplifiers approximate constant-durrent devices in which the output current is independent of the plate voltage over a large region of operation. The output current is dependent only on the grid voltage. The plate load impedance of each tube will be calculated as the ratio of the voltage appearing between the plate and cathode of the particular tube in question to the plate current of the same tube. This plate load impedance can then be used in conjunction with the plate characteristics of the tube to determine the operating path of the tube, the point at which the tube will begin to saturate, and the amount of distortion generated in the tube. The efficiency of the

7 distributed amplifier can be calculated, given the operating locus of each tube and the plate characteristics. 2.1 Assumptions Four assumptions were made in order to simplify the analysis. These assumptions are discussed in some detail in this section. The first assumption is that the grid line has no loss; this means that the signal voltage is the same on every tube in the distributed amplifier. The errors incurred in using this assumption will be analyzed later; however, when the distributed amplifier is designed to operate at higher frequencies, it is obvious that many factors will influence the loss in the grid line. The input impedance of the tubes used in the distributed amplifier is one of the major factors involved in the grid-line dissipation. The'input impedance of the tube can be considered as a capacity in parallel with a resistance connected between grid and cathode. The capacity remains fairly constant with frequency,but the resistive component decreases as the square of frequency. This resistive component arises from the fact that the cathode lead inductance, in conjunction with the capacity between the grid and the cathode, gives rise to grid current that has an in-phase component. In addition, the inductance in series with the screen, and the capacity between the screen and the grid, give rise to an in-phase current. However, the latter in-phase component of current is 1800 out-of-phase with the applied voltage,which leads to a negative resistance. Also there will be some radiation from the grid line which will have the same effect as increased dissipation; however, this radiation can be reduced to a point where it is negligible by proper shielding. The series coils in the grid line give rise to some loss, but this loss can be kept small in comparison with that caused by the grid loading. Most of the effects discussed here are highfrequency effects that are common to any wide-band high-frequency amplifier.

8 The second assumption is that the plate line has no dissipation,and also that it has a constant transfer characteristic with frequency up to the cut-off frequency. The characteristic impedance of the plate line is assumed to be Zo. However, the magnitude of Zo may vary with frequency, but it isassumed to be resistive over the operating frequency range. The plate line is assumed to be matched at both ends. These assumptions are met fairly well in practice with the constant-k line with m-derived terminations. Radiation from the plate line can be a serious problem, but again,by adequate shielding,it is possible to reduce this loss so that it is negligible. The loss due to skin effect can be reduced by increasing the size of the wire used to wind the plate line. The third assumption made in this analysis is that the tubes are identical. This implies that all of the tubes have the same transconductance and the same input and the same output capacitances. Variation of transconductance can change the results of the calculation of plate load impedance since a cancellation takes place at some frequencies which reduces the voltage between the plate and cathode of the first few tubes. The variation of input and output capacities from tube to tube will affect; the frequency response of the amplifier and the characteristic impedance of the artificial transmission lines. However, the major effect of variation between tubes in input and output capacitances is to introduce ripple in the frequency response of the amplifier. The final assumption is that the tube characteristics are ideal constant-current curves in that the output current is independent is independent of the plate voltage as long as the plate voltage is positive. This still allows the idea of clipping to be introduced since the point where the operating load line intersects the Eb = 0 axis can be determined. This assumption allows the use of the transconductance of the tube as a measure of the output current, and also assures that the plate current is a function of the grid voltage alone.

2.2 Plate Load Impedance An expression for the plate load impedance (the ratio of the platecathode voltage to the plate current) of any tube in an n-tube distributed amplifier is derived in the following analysis. It should be noted that the four assumptions discussed in Section 2.1 apply to this analysis. From these plate load impedances the efficiency of a given amplifier can be determined from the characteristics of the tubes involved. Once the operating path is known, it is a simple matter to determine also the point at which the amplifier begins to saturate. As the frequency of operation increases the phase shift along the grid and plate lines cannot be neglected, and the plate load impedance becomes markedly different from that given in Equation 2.1. Since it has been assumed (Section 2.1) that there is no dissipation in the grid line, the individual grid voltages differ only in phase. If an equivalent large-signal transconductance (Gm) is defined as the ratio of the fundamental component of plate current to the fundamental component of grid voltage, the plate current of each tube can be written as il = Gm Egm sin ot i2 = Gm Egm sin (cut-p) ii = G Egm sin [at -(k-l)p] (2.2) i = GEgm Egm sin [t - (n-l)] where Eg sin cot is the instantaneous value of the grid voltage on the first tube of the distributed amplifier and p is the phase shift per section of the grid and plate lines at the operating angular frequency a. The plate voltage on the kth tube as a result of the application of a signal voltage to the input of the grid line can be calculatedby the Superposition Principleas the sum of the voltages that would appear there as the tubes are turned on one at a time.

10 n ek = ekj (2.3) j=1 where j = the number of the tube that is operating. If no loss is assumed in the plate line, the voltages appearing on each tube will have the same magnitude; however, the phase of the voltage appearing on the kth tube plate will be dependent on the phase shift per section of the plate line and the number of sections the current traverses in arriving at the plate of the kth tube. From tube number one, up to and including the kth tube, the total number of sections of both grid line and plate line that are traversed by an input signal in arriving at the plate of the kth tube (as these tubes are turned on one at a time) are equal. Hence the voltage on the kth tube that results from any tube from one to k is equal to the plate current of the particular tube turned on multiplied by half the characteristic impedance of the plate line. The phase angle with respect to the grid voltage on the first tube will be (k-l) times the phase shift per section of the filter networks that constitute the grid and plate lines. This results from the fact that the input signal must traverse (k-l) filter sections in arriving at the plate of the kth tube. kl = Gm Egm ~ sin [wt -(k-l)' * = (2.4) ekk = GGm Eg Z~ sin [ot -(k-l)p] The contribution of the plate currents of tubes k+l to n to the plate voltage across the kth tube are all different in phase since the number of sections of filter traversed in going from the input to the plate of the kth tube is different for each path. The voltages due to tubes (k+l) through n operating alone are ZO sin ~t-(k-1)~-2 ek(k+l) =Gm Egm sin t-(k-l)2-2 2k GE sn Ut(- l(2.5) ekn' = G Eg 2 sin [cot-(k-l)p-2(n-nk)q]

ll Substituting Equations 2.4 and 2.5 into Equation 2.3, and summing the resulting finite series as shown in Appendix A, yields ek = GmEgm 2.{k sin [ut-(k-l)P (2.6) sin cp The plate load impedance that the kth tube sees is given by the ratio of the plate voltage to the plate current (see Equation 2.2). ek Zo [ sin(n-k) -(n-k+l) (2.7) Equatn 2 Zk2 l inc Equation 2.7 shows that the plate load. impedtances of all except the nth tube possesses a phase angle which is different from zero except at discrete frequencies. The plate-voltage plate-current locus will in general be an ellipse, or over the majority of the frequency range all the tubes with the exception of the last tube operate with elliptical load lines. 2.3 Plate Load Impedance for Specific Cases Equation 2.7 is plotted for each tube of a six-tube distributed amplifier in Figures 2.2 through 2.7. It can be seen in Figure 2.2 that the phase angle of the plate load impedance is greater than 90~ in some regions. This indicates that in these regions the first tube is absorbing power from the plate line rather than delivering power to it. This power is dissipated in the form of heat at- the plate of the first tube (in the case of a six-tube distributed amplifier). This increase in plate dissipation can give rise to serious tube burnout problems in practical distributed amplifiers where the amplifier is operated under full dc input conditions. Figure 2.8 shows the magnitude of the plate load impedances for each tube of a 9-tube distributed amplifier. The regions in which the angle of the plate load impedance is greater than 90~ is indicated by the regions of increased width of the curves. It will be noted

6 FIG. 2.2. Zp AND 8 AS A FUNCTION - - OF 4 FOR TUBE I OF A 6-TUBE DISTRIBUTED AMPLIFIER.,NOTE: CIRCLED POINTS WERE EXPERIMENTALLY DETERMINED'C)...... - - - - ~~~~~~~~~~~~~~~~~~~~~~~~~~~1800 5~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ N PLATE LOAD IMPEDANCE 2.j~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~G dark~ ~~~~~~~~~~~~~~r o 900 N"' PA S PHAsE ANGLE \\ \ - o x, \ Z \1 m ~ ~ ~ ~ ~ m m'om.. m- of 0 0 0. 0~ I0~ 0~ 3, 0~40~ 50~ 60~ 70~ 80~ 90~ I00 I10~ I120~ 1:300 140~ 150~ 16CP 170~' 184: 4~), PHASE SHIFT PER SECTION OF LINE

0 I ~~~~~~~~~~~~~~~IlJ / ~~~~~~~~I,, 0 ~~~~~~~FIG. 2.3. Zp AND 9 ASA FUNCTION OF 0 FOR TUBE 2 8o0 OF A 6S-TUBE DISTRIBUTED u,PLATE LOAD IMPEDANCE O AMPLIFIER.' NOTECIRCLED POINTS WERE ___ EXPERIMENTALLY DETERMINED 04 ~~~~___ 400,, 0~~~~~~~ 4 PHASE ANGLE / 0 r o G A Ile,, ~ ~ ~ ~~~~~~~~~~ \4 /'-D B~~~~~~~ ~ ~ 2~ \__ -20 S~~~~~~~~~~~~~~~~~~~oo rr',' I I 4 0 P S o~~~ 0 I ~~~~ 0 12K 0 0* ~~~~~~~~~~~~800I 0~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~r 00~ ~ ~~~~~' 20 - 0o(080 0010 10 6010 rAPAESIFPESETOOFLNH

-0~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~IFIG. 2.4. Zp AND G AS A 6 ____ FUNCTION OF p FOR TUBE 3 600.0 OF A 6-TUBE DISTRIBUTED AMPLIFIER. NOTE:CIRCLED POINTS WERE 5o, ~PLATE LOAD IMPEDANCE EXPERIMENTALLY DETERMINED.,,_____ 400 ~ ( 5~~~~~~~~~~~~~~~~~~~~~~~~~~~~ ____ ________ ____/,oo_ \~~~~~~~~~~~~~~~~~~~~~~~~~ CM PHASE ANGLE I o,,/ j1 0,4 -- J- 20 N 0 w~ ~ 0 0 _ _0 z~~~~~~~~~~ / 0~~E.w~ I0l oo a —2 / 00 200 40, e60 800 lo00 1200 14CP 1600 PHASE SHIFT PER SECTION OF LINE

FIG. 2.5. Zp AND 8 AS A 6 ___ FUNCTION OF 4 FOR TUBE 4'_ OF A 6-TUBE DISTRIBUTED AMPLIFIER..-PLATE LOAD IMPEQANCE 0 -s- 50 5~~~~~~~~~~ 5 _____ _____ -II -- _____ _____ _____ _____ 400 4 00 ( z~~~~~~~~~ O~~~~~~~~~~~~~~~~~ r 4 ___ ___ ___..-S 200 ZG 100'0 ~~~~s 00~~~~~~~~~~.0 o IN N ~~~~~~~~~~~~~~~~~~~~~~~m 4 1 0 - U0 00 _00 400 60so0 1 120P 1400 16_ _ -100 PHASE SHIFT PER SECTION OF LINE

7~~~~~~~~~~~~~~~~~~~~~~~ 0-D_ 0 __ _____aDISTRIBUTED AMPLIFIER. O~~~~~~~~~~~~~~~~~~ I' I 1 I I I I:1~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~' 0~~~~~~~~~~~~~~~~~~~0'*% 4 0~~~~~~~~ WON~~~~~~~~~ I~~~~~~~~~~~~~ ~ ~~~~~~~~~~~~~~~~~~~~~~~~~~ I iII Ii i i i I 1_ i i i 0~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~00 N oo 0 o -t,00 __ __ 0 j~~~~~~~~~~~~~~~~~~12 d Nu~~~~I PHASEN A A NGLE CTION CIRCLED POINTS WERE i 2 ___ __ LRLMEr4IMLLY DETER- m —~O~~~~~~~~~~~~~~~~~~~~~~~-) MINED. ~ ~ ~ ~ ~ __ _ _~.1'- 10 20 40 60 SO 100' 120 140 60'~~~~.~,~' ~~ ~,PHASE SHIFT PER SECTIlON OFC UNE

"~~~ 0 -G~~~o 6 -,~~~~~~~~~~~~~~~~~~~~~~~~~ PLATE LOAD IMPEDANCE 0~~~~~ 5 ~ ~ ~' F IG. 2.7. Z~ AND8 AS A FUNCTION OF 4 FOR TUBE 6 OF A 6-TilDE DISTRIBUTED z ~~~~~~~~AMPLIFIER. w 4 ____ __ __DETERMINED. __ ____ __IO 2p 30.40 5 e 7e s 90 IO e.6 1__0___0 1__ 160 PHASE SHIFT PER SECTION OFLINE2d PLATE LOAD IMPPAEDANGLE ~. ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~~~~~~~~~0 ~~~~~~~4, PHS SHFTG.7 PE SE S UCTION OF LIN.....o

10 N I I tS N / I N I I k I V I'! I I I It9 I t9 o____ ___ I I I ANGLE Of PLATE LOAD IMPEDANCE IS GREATER nz9 THAN 90. k. 4. 2, 3......78, & 9 MAGNITUDE [ 4L1 vs..

19 here that the plate load impedance has a negative resistance component for certain regions of the second tube as well as for the first tube. Hence some special precautions must be taken in order to prevent overheating of the first few tubes in long-string distributed amplifiers. A low-frequency model of a distributed amplifier was constructed in order to verify experimentally this analysis of plate load impedance as determined by Equation 2.7. The design of this low frequency distributed amplifier,as well as the method of data reduction,are shown in Appendix B. The points on Figures 2.2 through 2.7 were experimentally obtained. The experimentally determined points are, in general, higher than the theoretical curve. The disagreement of the curves is caused by three factors. First, it is impossible to match the transconductance of the tubes exactly. Second, the dissipation in the grid line was neglected in the theoretical curves. And third, the low-frequency model has more dissipation in the plate line than a wide-band distributed amplifierand it affects the plate load impedance. However, the experimental points are in general agreement with the theoretical curves. Figure 2.9 shows the plate load impedance for a six-tube paired-plate distributed amplifier. The derivation of the equations for this figure are shown in Appendix C. In addition the schematic diagram for a paired-plate distributed amplifier is shown in Appendix C (Figure C.1). Figure 2.9 is included to show the effects of one variation (the paired-plate) to the standard distributed amplifier circuit. Many other variations are possible and for each one a plate load impedance plot such as shown in Figure 2.9 could be calculated by the techniques described above.

6 60*-S —ei em i.'UBESS Il6 5 i I.I I r I I I I /! I1 4~~~~, _ 1 /I N.. ~! ~'. ~r 4!I ~~ I rY I1\ / f.. r / f j~~~~~~~~~~ N! ~~~~I').' ____ ~~~~~~~~ N o N ~~N II R I r TU~~OBESI2 2,4 0 o I w I-L a. 0 - r 00 30 6cf 90 l20 1500 ISO PHASE SHIFT PER SECTION OF LINE - o' o FIG. 2.9. PLATE LOAD IMPEDANCE FOR A 6-TUBE PAIRED-PLATE 4DISTRIBUTED AMPLIFIER..J G.30 ~ 090120Ic"10 FIC/) -..!Lf iOA iM~AC O -UEPIE-L Dt~STR!BUTED AMPLIIERi

CHAPTER III. FREQUENCY LIMITATIONS IN DISTRIBUTED AMPLIFIERS Standard distributed amplifiers use constant-k filter sections between kdjacent grids and also between adjacent plates. The impedance levels of the plate and grid lines are adjusted so that the phase shift per section of the rid line and the plate line are the same at any frequency in the operating ange of the amplifier. The following formulas for constant-k low-pass filters an be found in many text books. The mid-shunt image impedance Z = R 1 (3.1) f 1- - 2' The phase shift per section = 2 sin-1(tD) (3.2) Ithere c = 2 VLC R C jIultiplying the equations for R and cc the following useful equation is obtained. _ = 2 (3.3) & and C are defined by the schematic of a constant-k filter shown in Figure 5.1. T 2 FIG. 3.1. SCHEMATIC DIAGRAM FOR A LOW-PASS, PI - SECTION, CONSTANT- k FILTER. 21

22 Since the phase shift per section of the filter sections in the plate line and the grid line must be equal, it can be seen from Equation 3.2 that the cut-off frequency of the grid section must be the same as the cut-off frequency for the corresponding plate line section. In the usual distributed amplifier all the sections of the plate line are identical and all the sections of the grid line are identical. If Equation 3.3 for the plate line is divided by a similar equation for the grid line, it can be seen that the impedance level of the plate and grid lines are directly proportional to the input and output capacities of the tubes, or R=~p f (3.4) Rg Cp The above equations would seem to indicate that there is no upper limit for the cut-off frequency for which a distributed amplifier can be designed. Equation 3.3 leads one to the conclusion that with a given tube capacitance the cut-off frequency can be raised as far as is desired if the impedance of the line is reduced so that the product of the angular cut-off frequency and the impedance level of the line remain equal to twice the reciprocal of the tube capacitance. However, the realization of the constant-k structure in a practical wide-band amplifier is, in general, dependent upon the cancellation of the inductance of the grid lead and the inductance of the plate lead. 3.1 Cancellation of Lead Inductances The standard procedure in cancelling lead inductances is to utilize air core transformers with negative mutual inductance as shown in Figure 3.3. Here the lead inductance is represented by Li and the capacitance between tube elements by C. The mutual inductance is adjusted so that it is equal in magnitude to the lead inductance. Figure 3.2 shows the negative mutual transformer and its "T" equivalent.

23 L*+ M L+M L 8L -M FIG.3.2. NEGATIVE MUTUAL TRANSFORIER AND ITS EQUIVALENT "T ". It can be seen from Figure 3.2 that half the constant-k inductance is given by L *( L L + M (3.5) where: L = primary or secondary inductance of transformer M = mutual inductance between primary and secondary of transformer. The coefficient of coupling is defined as -i MI I~MI k = IMI I(3.6) LLf Lt(1)i If Equation 3.6 is solved for M and the result substituted in Equation 5.5, 2- - (3 7) The magnitude of the mutual inductance is made equal to the lead inductance in order to cancel the effect of the lead inductance. When this equality is substituted into Equation 3.6 and the result applied to Equation 3.7, the constant-k inductance becomes L = 2L [ l+kl (3.8) If this value for the inductance of the constant-k section is substituted into

24 Equation 3.2 for the cut-off angular frequency, then 2 2k Wc = = /' (3.9) 2L2 kl+k If the series resonant angular frequency of the tube element being considered is designated by 0o2 — c = / (3.-10) Wo 1 +k Equation 3.10 is plotted in Figure 3.3. It can be seen from this figure that it is theoretically possible to operate an amplifier up to the series resonant frequency of the controlling element of the tubes; however, this requires a coupling coefficient of unity. In air-core coils it is possible to realize a coupling coefficient of about 0.5. Hence the upper cut-off frequency of the amplifier is limited to about 0.8 of the series resonant frequency of the controlling element of the tube. In most tetrodes and pentodes the input capacity is larger than the output capacity, and the le-ad inductance of the grid is of the same order of magnitude as that of the plate (in most power tubes the lead.inductance of the grid is larger than that of the plate lead). Therefore, in general, the grid circuit is the controlling element as far as frequency is concerned. As a rule of thumb, the highest cut-off frequency can be taken as f.2 Conductive Loadin 0.8 fDue to High Frequency Effects 3.2 Conductive Loading of the Grid Line Due to High Frequency Effects There are three major frequency-dependent factors that contribute to attenuation in the grid line. Each of these factors have been considered in 12,13 threliteraure and are repeated here only for the sake of completeness. In general, the effect of cathode lead inductance within the tube is to provide a common element through which both the output current and the input current flow.

;-z-; 3f, ~t-i9-e-V z9ZZ 25 a80 FIG. 3.3 - I. PLOT OF CUT-OFF FREQUENCY OF CONSTANT-k FILTER AS A FUNCTION OF THE COEFFICIENT OF COUPLING OF THE NEGATIVE-MUTUAL TRANSFORMER 0.6 NECESSARY TO CANCEL THE LEAD w INDUCTANCE. 0.2 o 0.2 0.4 0.6 0.8 1.0 RATIO OF THE CUT-OFF fo) FREQUENCY TO SERIES NEGATIVE MUTUALS RESONANT FREQUENCY OF La ANO C.

26 This feedback within the tube affects the input conductance of the tube (Equation 3.12). G. = g:D2 (C W2CpLk) Gin gnu2g~ gm a2LX2 + [o2IX(Cg g d)CS 1 (3.12) The constants in this equation are defined by Figure 3.4. gm is the transconductance of the tube and o is the operating angular frequency. This equation is derived in Appendix D and involves the assumption that the impedance in the 0 _ CP 3Lk.M. 3.4. EQUIVALENT CIRCUITFMR THE DETERMINATION OF INPUT CONDUCTANCE CAUSED BY THE FEEDBACK RESULTING FROM CATHODE LEAD INDUCTANCE. plate circuit is so small that the Miller Effect is negligible. If the corrective term is neglected, Equation 5.12 simplifies to Equation 3.13, which is the one most quoted in the literature. Gin = gM(2 CgLk (3.15) In order to neglect the corrective term the resonant frequency of the cathode lead inductance and the capacitance from grid to cathode must be high in comparison to the operating frequency. If this term is not negligible, Equation 3.12 must be used. Inductance in the screen grid lead results in a negative conductive component in the input to the tube. In some amplifiers this fact has been used to

27 )ffset, to a certain extent, the effect of cathode lead inductance. Equation 3.14 shows the manner in which the input conductance varies if the screen lead inductance alone is considered Gin gm2CD2Lg2Cgs (3.14) where gm2 is the transconductance from the control grid to the screen grid, Lg2 Ls the inductance of the screen lead, and Cgs is the capacity between the control grid and the screen grid. Equation 3.14 neglects all the impedances associated with the tube with the exception of the screen lead inductance and the capacitance between the control grid and the screen grid. At frequencies above 30 me transit time affects the operation of tubes. As the transit time becomes a measurable fraction of the period of the input frequency, the current induced in the grid due to the passage of the electrons will contain a fundamental component that no longer leads the grid voltage by 90. As a result the input admittance of the tube contains a conductive component. The input conductance is given approximately by5 Gin K- gmc2T2 (3 15) where T is the electron transit time. The three effects mentioned above can be represented in Equation 3.16. Gin - [gmCgLk + - gm2Lg2Cgs] (3.16) This equation contains the most important loading effects present in a small-signal distributed amplifier. These loading effects are present in a large-signal amplifier also,but Equation 3.16 must be modified by the substitution of the equivalent large-signal transconductance (Gm) for the transconductance (gm)). However, in power distributed amplifiers there is an additional effect in that the tubes are generally driven into the positive grid region where a considerable grid current flows. This gives rise to an additional loading effect on the grid line; however, this loading is independent of frequency being dependent on the

28 magnitude of grid voltage to a first approximation. This effect will be considered in a later chapter. 3.3 Frequency Tolerance Equation 3.1 shows the variation of the mid-shunt image impedance of a constant-k filter as a function of frequency. This equation is plotted in Figure 3.5. It was shown in Chapter II that the plate load impedance of the various tubes in a distributed amplifier is dependent upon the value of the characteristic impedance of the plate line. For a constant-k plate line, this value is the mid-shunt image impedance of the constant-k filter sections. This gives rise to a plate load impedance which increases rapidly as the cut-off frequency is approached. This is particularly true of the last few tubes in a distributed amplifier, where the plate load impedance does not vary as radically with frequency as it does for the first few tubes. This rising plate load impedance can result in plate clipping in the last few tubes in a distributed amplifier, and cause the large-signal response to be different from the smallsignal response. This is shown in Figure 3.6 where the plate load line for the last tube in a six-tube distributed amplifier is drawn for four different frequencies. It will be noted from Figure 3.6 that for the particular operating point chosen, if the grid is driven 20 volts positive, clipping will begin to occur at a frequency of o.685 of the cut-off frequency. However, if the grid drive is such that the grid maximum positive excursion is only 15 volts, then clipping will occur above about 0.8 the cut-off frequency. Since the frequency at which plate clipping occurs is a function of both the operating point and the grid drive, the proper determination of the operating point and the grid drive is not a simple procedure. In order to facilitate the location of the operating point, an additional parameter, called

29 6.0.....V5.0 _ N 4.0 I.O I 3.0.. i w FIG.3.5. VARIATION OF MID —SHUNT IMAGE IMPEDANCE AS A FUNCTION OF FREQUENCY

30 g;-;-g Rr gi-IO9*V ZtZ 2.8 0 1.8. l l. I I I l I - I....1-. O lc__ Y___ +96 +00 PL A' I TE GRID VOLTME -V __._.... __. m..._ — m-. m m -'15 0.2 — 30 0 0 200 400 600SW tooo l000 PLATE VOLTAGE-VOLTS FIG. 3.6. VARIATION OF THE PLATE LOAD IsMPEDANCE OF THE LAST TUBE OF A 6 TUBE DISTRIBUTED AMPLIFIER AS A FUNCTION OF FREOUENCY.

31 the frequency tolerance, is defined. Frequency tolerance is defined as that frequency above which clipping will be present in the last tube of a low-pass Distributed amplifier. This concept will be explored further in a later chapter. 3.4 Summary The high-frequency effects discussed in this chapter can be classified into two main categories. One of these categories deals with an upper limit to the cut-off frequency for the constant-k filter sections used as the plate line and the grid line of distributed amplifiers. This limit is imposed by the finite grid and plate lead inductances. However, this frequency limit could be increased considerably if tubes possessing satisfactory high-frequency characteristics were constructed with two grid leads and two plate leads. These lkads should preferably come out opposite sides of the tube. The second category includes the high-frequency effects which tend to distort the normal frequency response curve of the amplifier. The loading on the grid line due to the input conductance of the tubes can serve the useful function of flattening the normally rising frequency response of distributed amplifiers without losses. However, for wide-band distributed amplifiers, every effort should be made to reduce the input conductance of the tubes as much as possible. This implies tubes with as small an inductance in the cathode lead as possible and with small transit time. Sonie inductance in the screen lead is a definite advantage; however, care must be taken to insure stable operation of the tube. Also included in this category is the frequency coverage of power distributed. amplifiers, which is determined by the onset of clipping in the final tube of the distributed amplifier. The frequency tolerance is determined by the location of the Q-point and the amplitude of the grid drive. For a given frequency tolerance and grid drive (this is the same as requiring a certain power output) the operating po.int for the tubes is determined.

CHAPTER IV. GRID LOSSES 4.'1 Attenuation Constant of Constant'-k Filters with Losses All of the analytical assumptions discussed in Section 2.1 are close approximations to the real situation with the exception of the lossless grid line assumption. The losses associated with the input impedance of the tubes in a distributed amplifier give rise to attenuation in the grid line. At the higher frequencies this attenuation is serious enough so that it cannot be neglected. In distributed amplifiers designed to deliver power, the grids will, in general, be driven positive giving rise to an additional loading effect on the grid line. Consequently, the attenuation on the grid line will be a function of the grid driving voltage and of frequency. The constant-k filter sections of the grid can be represented as shown in Figure 3.2 with the addition of a resistance across the capacities representing the input admittance of the tube. If it is desired to take the losses in the coils into account, a resistance in series with the coils can represent this loss. R is defined as the resistance in series with the inductances comprising the grid line, and G is defined as the conductance shunting the input capacity of the tubes in the distributed amplifier. The attenuation constant per section of the grid line, a, may be written asl6 t ( +2C dmu 2L 2C ) 2 29,,+' (4.1) providing <<1. 32

33 or a constant-k filter Equation 4.1 becomes /R G 2 1 c\2L 2C ) 2 2 +( - + — ) c c +.. (4.2) Jo a high degree of approximation in wide-band power distributed amplifiers R 2L.s negligibly small in comparison to G. Therefore, a may be written approxi2C ately as 2 g G ion of the frequency region the second term in Equation 4tu may be neglected in;omparison to the first term. However, this assumption is not valid if the amplifier is operating near cut-off frequency. i.2 Determination of the Plate Load Impedance with Attenuation in the Grid Line Less case. However, their amplitudes will differ by a, where e is the natural Logrithrim base and a is the attenuation constant described in the preceding 3ection. The plate load impedance for any tube, k, in an n-tube distributed

34 Chapter II. In this derivation it is assumed that there is no attenuation in the plate line, that the attenuation constant for each section of the grid line is the same, that each artificial transmission line is matched at the ends, and that each of the tubes have identical large-signal transconductances. This analysis is carried out in Appendix E, and only the results are given here. Z sinh sinh[(n-k) + j(n-k) O -2- e(k-1)H. sioh~n-k >z. j[~-k~m] -(n-k+l)6 -j (n-k+l) Zk e sinh k2 sien + (4.4) Figure 4.1 is a plot of Equation 4.4 for a six-tube distributed amplifier using type 4X150A tubes. In order to determine these curves it was necessary to know the input impedance of the 4X150A tube in the circuit in which it is finally to be used, i.e., with the same lead lengths and the same by-pass condensers. This measurement indicated that the input circuit of the 4X150A may be approximated by 2260 a resistance of ohms (where f is the frequency of measurement expressed in _2 hundreds of megacycles) in parallel with a capacity of about 20 micro-microfarads. These curves represent the plate load impedance with no dc grid current. The curves will vary to some extent when the grids are driven positive with respect to the cathode. The magnitude of the plate load impedance will increase somewhat for small grid currents. The curves of Figure 4.1 represent the plate load impedanc of the six-tube amplifier to a reasonable degree of approximation even under large~ signal conditions. Note that the plate load impedance on the first few tubes is quite low over the major portion of the frequency range; hence, it would seem feasible to utilize a lower supply voltage for the first few tubes without appreciab affecting the power output. This would increase the overall efficiency of the amplifier. Calculations indicate that it should be possible to increase the efficiency of a standard distributed amplifier by about 30l by staggering the plate supply voltages.

gg-o-si N3r -9*1,-1-V zs9Z 35 900 FIG. 4.1. PLATE LOAD IMPEDANCE AS FUNCTION OF.FIQUENCY FOR A 6-TUBE 4XI 50A DISTRIUTE0' AMPLIFIER WITH A 90A PLATE LINE. 8oo es~~~~~~oo ~~f t f FREQUENCY IN HUNDREDS OF MC TOO 600 200 100 - O 5 100 1/0 - 200 2R MC

36 The plate load impedance for the first and sixth tube are plotted in Figure 4.2. The solid curves are reproduced from Figure 4.1 and the dashed curves indicate the plate load impedance calculated for no attenuation in the grid line. It can be seen from Figure 4.2 that the plate load impedance of the first tube is not affected appreciably by attenuation in the grid line; howeverthe plate load impedance for the sixth tube is seen to rise faster as frequency increases. 4.3 Modification of Gain Equation to Account for Losses in the Grid Line Without loss in the grid line the voltages on the various tubes of a distributed amplifier are the same. The grid voltages are related to the driving power as shown in Equation 4.5. Eg = /Pdr Z (4.5) This expression holds if matched conditions are assumed on the input and output of the grid line. Here, Pdr is the driving power. However, if the grid line has attenuation, the voltages on the individual tubes will be given by Equation 4.6. gl =Vr.g - e1 Eg Pdr ~z e(l+2) drEg2 dr 4. g e (a1+0(4.6) -(Cvl+al2+ -"+an) Egn JPr ~/ zg e where Egk is the grid voltage on the kth tube, al is the attenuation constant for the first half section filter, and ak is the attenuation constant for the filter section terminating on the kth tube. The current flowing into the output halfsection filter of the plate line can be written as shown in Equation 4.7. GmI = / * [.e.. + e'( +... + e( 22'n)] (4.7) 2

99-91 -. 3V 90Z-19-V Z9Z3 37 900 FIG. 4.2. PLATE LOAD IMPEDANCE AS A FUNCTION OF FREQUENCY FOR TUBE NO. I AND TUBE NO. 6 OFA 6-TUBE 4X150A DISTRIBUTED AMPLIFIER. I 800 NOTE: THE DASHED CURVES ARE FOR A LOSSLESS GRID LINE. 700 - I 600 __ 400 TUBEIII I# I U, / o / /:a~~ ~ ~~.3 ~ ~ ~~ / I/ 0 _ _ _ I FR~O VTUBE #U 1 I 000 FREQUENCY~ MC j

38 The output power is given by the square of the current flowing into the last half-section of the plate line multiplied by the mid-shunt image impedance of the plate line. P I 2 z 9P (4.8) Pout = 2 (4.8) The output power can be calculated for any given set of operating conditions on the tubes in a distributed amplifier from Equation 4.8; however, the procedure is rather long and tedious. The main difficulty arises from the fact that the attenu4 tion constants are different for each section of the grid line. From Equation 4.3, a can be expressed approximately as one-half the product of the input conductance of the tube and the mid-shunt image impedance of the grid line..GgZ... (4.9) 2 The input conductance of the tube contains two components. One varies as the square of frequency and the other depends on the magnitude of the driving voltage. Gg = Goaw + Gdr (4.10) Equation 4.9 can be written in terms of Go and Gdr as shown in Equation 4.11 2 Goc)Rg Gi) drc 2Let 0 2 2 Cg GdrRg alCode r = (4.11)

39 in is the small-signal input conductance of the tube at the frequency w, and dr is the effective conductance of the tubes due to the flow of grid current in he positive grid region. Equation 4.11 can be written as 2 1 a = o WC Cl'to _ ( C ) + aodr 1 (1w/ c)2 (c )2 a Moj/M Cl aod~r l -o _ _ _ _ _ _ + (4.14) 0A 1 - (O/coC)2 (o /21 [f the variation of aodr for the various tubes in a distributed amplifier is Lssumed small and, in addition, the variation of %odr with frequency is assumed imall, the same value of a can be used in Equation 4.7 for c..., and an. o-1 would be half this value since it is for the input half-section. With this assumption, Cquation 4.7 can be written as Gmo~ gg -= [1+ e i + e 2x +. + e.. (n-l)a (4.15) 2.he exponential series shown in brackets can be summed in a manner similar to the,ne shown in Appendix A, and the result is shown in Equation 4.16. P GmV'V e __ _ (4.16) 2out 4,i~~~ (~22 2 sinh (2) Gm Pdr ZIg ZIp 2 2X G d ZPge (4.17) e~~~~

40 Substituting Equation 3.1 in Equation 4.17 the power output becomes 2,._ /na' Pout = -- g s ( inh( ] (4.18) Th factor outsin he power output from one tube operating alone in the distributed amnrplifier at low frequency and with no attenuation in the grid line. This factor is easily calculable from the characteristics for the particular tube as the square of one-half the rms value of the fundamental component of plate current times the nominal plate line impedance. The normalized value of the factor in the Brackets in Equation 4.18 is plotted in Figures 4.3 through 4.11 as a function of frequency. From these curves it can be seen that a value of ~ of about 0.5 should be used for a flat-gain characteris1 2 However, as the value of odr increases, the value of n_~ for a flat response - fo 2 decreases. This change is slight for odr e 0.25. nO' 1 ~o If a value of 2 2= is chosen as an approximate design parameter, the product of the number of tubes and the cut-off frequency for any given tube type is automatically determined. 2 ~c = G-. cg (4.19) in where: n = the number of tubes (c = the angular cut-off frequency Gin = the input conductance to the tube at the frequency w Cg = the input capacity of the tube. If aodr is determined from the calculated value of the grid voltage on the first tube, the results calculated from Equation 4.18 will tend to be too low, since the value of a will actually be decreasing with k, where k is the number of the tube under consideration.

gs-is9 r3Pr Lt1-19-V Z9Zz 41 10, 0.9 0.8. 0 O 0.4 0.6 0.8 1.0 W/wc FIG. 4.3. NORMALIZED VOLTAGE AMPLIFICATION AS A FUNCTION OF FREQUENCY FOR nClo Codr 0.Oo5 AND 0.05. 2 ao

gG-OZ-9 R3r 8lt1-19-V;92Z l,.. 0.90 0, 80 0 0. E 0.4 0.6 0.8.0 IX, i,,, FIG. 4.4. NORMALIZED VOLTAGE APPUUFICATION AS A FUNCTION OF FREQUENCY FOR n2 0.5 AND ~ d.O.lO.

9g-OZ-9 I3P 6*t1'-19-Vt 89Z 43 1.0 F'. -L 1 1 1. II nti 00 0.9 0.8 O 0.2 0.4 0.6 0.8 1.0 W/ WC FIG. 4.5. NORMALIZED VOLTAGE AMPLIFICATION AS A FUNCTION OF FREQUENCY FOR 0.5 AND i..

44 gg-IZ-9 N3P OGI —Y. Z9;Z 1.0 0 -'.... I 9 0 0.2 0.4 0.6 0.8 1.0 W/Wc FIG. 4.6. NORMALIZED VO.LTAGE AMPLIFICATION AS A FUNCTION OF FREQUENCY FOR 2..45AND 0aor 0. 20. 2 a0

45 Gg-Z-9 fip I9I-I9-V Z9ZZ 1.I. 1.0 a2 0.4~~~~nl2 -n - W/WC., at 0 aod 2, o..7.. V A A 0.8 O'0 0,2 f.4 l.6 0. 1.0 OF FREQUENCY FOR n9 O 0.49AND O0.20. 2

46 I L 0.. 0 0.2 0.4 0.6 0.8 1.0 C/WC FIG. -4.8. NORMALIZED VOLTAGE AMPLIFICATION AS A FUNCTION no0 L X odr OF FREQUENCY FOR-.S AND o.2O. Ga 9a...

g-zz-9 3Pr t9-19-v zezz 47 1.1'_.. -1- - - - - 1A tO m~ un u~ ____ i _A_, 0.S --. -o..2 0.4 0/ 0.8 1.0 - S.1 4.. NORMALIZED VOLTAGE AMPLIFICATION AS A FUNCTION OF FREQUENCY FOR.10.4,5 AND edr0 *0.25. 2 ~ l,,

go —9Y W3P ginl-g 1 489Z 1.0 1'' A.9 0.9 0.8'0 0.2 0.4 0.6 o0. 1.0:/ WC FIG. 4.1Q NORMALIZED VOLTAGE AMPLIFICATION AS A FUNCTiON OF Fodr. OF FREOUENCY FOR - 0.48 AND 0.25. ~/~~~~~cl

49 Ig-ZZ-9 I43Pr ff1 —V Z9ZZ nx1 1.0,,,.. -.... 0.9 0.6 0 0.2 0.4 0.6 0.8 L.O D/ WC FIG. 4A 14 NORMALIZED VOLTAGE AMPLIFICATION AS A FUNCTION l 00 aodr OF FREQUENCY R 0.5 ND.. 22 0.4 0.6

CHAPTER V. E-FFICIENCY It the attenuation in the grid line is neglected in order to obtain a general analysis of distributed amplifiers operating as large-signal devices, rather elementary reasoning will suffice to show that the power output varies as the square of the number of tubes. This comes from the fact that all of the tube currents add in-phase as they progress toward the output end of the amplifier; consequently, the output current is just n times the current that one tube produces in the output resistance. Hence, the output power varies as the square of the number of tubes. This is in contrast to the dc power input which varies directly with the number of. tubes. The above reasoning leads to the conclusion that the efficiency increases directly with the number of tubes. Therefore, if the attenuation in the grid line is neglected, it would seem desirable to use as large a number of tubes as possible; the number being limited only by the plate load impedance and the plate characteristics of the tube under consideration. 5-.1 Calculation of Efficiency If attenuation in the grid line is taken into consideration in a distributed amplifier which is designed to have a relatively flat gain over the frequency region, the output power no longer increases as the square of the number of tubes. From Equation 4.18 the output power can be written as I~X / nhC 2 [e 2 2sinh(2) Pout = Poli J (5.1) (03/c2..... (5. a1 where Po1 is the output power for one tube operating in the distributed amplifier without attenuation in the grid line. At low frequencies Equation 5.1 can be writt as shown in Equation 5.2. 5o

51 Pout: _ e sinh (.. Gdr og) (5.2) sinh (Gdr Rg ) nGdr Rg f < 0.2 4 nG dr Rg) 2 sinh:4. he error involved here is less than 1.25 percent. The output power can be written pproximately as nGdrRg Pout Polne 2 (5.3) ro0 If the value of - is chosen in a manner described in Chapter IV for constant power gain over the frequency region, the efficiency can be calculated y Equation 5.4. nGd.Rg l = 1 n2e- 2 (5.4) here q1 = the efficiency of one tube operating alone in the distributed amplifier. The efficiency as calculated from Equation 5.4 is valid for low frequencies.'ince the magnitude of the grid voltages varies with frequency, the average value of late current will also vary with frequency if harmonic distortion is present. In L distributed amplifier designed for a relatively flat gain over the operating freLuency range, the output power will be constant if the driving power is constant. Eowever, the average value of the dc current drawn from the plate supply will de-:rease with frequency. Hence, the distributed amplifier will tend to become more!fficient as the operating frequency increases. The change in efficiency is related;o the change in the average plate current from no-signal conditions to full-signal ~onditions.

52 5.2 Maximum t'heoretical Efficiency of Distributed AmplifiersFigures 5.1 and 5.2 show idealized characteristics that are assumed for the calculations of maximum theoretical efficiency. The load lines that are drawn on these characteristics are for a pure resistance load on the tubes. At low frequencie's the tubes in a distributed amplifier are operating in parallel as discussed in Chapter II. In a distributed amplifier designed for a flat frequency response the output power is constant with frequency as long as the driving power remains constant. Hence the output power calculated for low frequencies should apply for any frequency in the operating range. The instantaneous values of plate voltage and plate current are shown in Figure 5.3.. The instantaneous value of the plate voltage may be written as eb = Em (1 - cos ct) <- < (5.5) = Em [1 - cos (2 )zt< 22 - 2 2 The instantaneous value of the plate current may be written as ib = Im [cos Mt - cos()] - <ct < - - = 0< Wt < 2 - 2The dc value of the plate voltage must be equal to the plate supply voltage, Ebb. Eb m + [1 - cos at] dot + [2r o] [ c 1 os] =2-bcos( 2 )cos()-i 2s() (5.7) n e 2 c',,, ( Zo)] i( ~

gg-8-9 I3i' LI-Is - Z9gZZ SLOPE - IbMAX =) b Ebb EbMAX PLATE VOLTS FIG. 5.1. IDEALIZED CONSTANT-CURRENT TUBE PLATE CHARACTERISTICS WITH LOAD LINE DRAWN FOR CLASS A OPERATION. SLOPE A t bMAX I..L. Ib Ebb Eb MAX PLATE VOLTS FIG. 5.2 IDEALIZED CONSTANT-CURRENT TUBE PLATE CHARACTERISTICS WITH LOAD LINE DRAWN FOR CLASS B OPERATION.

54 GG-ZZ-9 fW3r9g-19-V 29ZZ eb I I i E I m I Ln" I COS(e/2) II -/2-72 ir/2 e/2/2 IAND PLATE CURRENT. - e/2 -'ff'/2 0 77'/2 /2t FIG. 5.3. INSTANTANEOUS VALUE OF PLATE V.OLTAGE AND PLATE CURRENT.

55 The dc plate current is the average value of the instantaneous plate current. Ib 2 f 9 [cos ct - cos ( d)] tt'2 t [sin( 2-) 2 cos] (5.8) she peak value of the fundamental component of plate voltage is -given by Equation 5.9. E1 = 2Jf (1 - cos Ot) cos wt dt + Q/ [1 - cos cos oft dwtt = 2 [sin - g] (5.9) Lhe peak value of the fundamental component of plate current is given by Equation 6.10 Il = fh [cos ot - cos ()] cos ct dot =2- [ sin] (5.10) Lhe efficiency can be calculated as the ratio of the output power to the input dc Dower. At low frequencies the plate circuit of distributed amplifiers is only;0 percent efficient since half the output power from the tubes is dissipated in;he reverse termination. The output power is one-half the product of the fundamental current and fundamental voltage.

56 ElI1 Ebb Ib (5.-1) + os( sin( [2 sin @ cos)] Equation 5.11 is plotted in Figure 5.4 as a function of conduction angle of the pla current. The maximum theoretical efficiency is seen to be about 30 percent and occurs with a plate current conduction angle of about 225 degrees. 5.3 Staggering the Plate Supply Voltages for Individual Tubes The curves of plate load impedance as a function of frequency show that the impedance for the first few tubes of a distributed amplifier remain low over an appreciable portion of the frequency range. If the actual operating frequency range is limited to a value somewhat below cut-off, and also at the low-frequency end of the nominal region of operation, the dc voltage supplying the first few tube can be decreased from that used for the last tubes with very little effect on the output of the distributed amplifier, but a corresponding increase in efficiency will be realized. The efficiency of distributed amplifiers is inherently low due mainly to the inefficiency of the plate line. However when the distributed amplifier circuit is compared with other wide-band amplifiers, it is seen that the distributed amplifier, the wide-band video power amplifier, and the traveling wave amplifier have comparable efficiencies.

57 gG —g9 043r 1 gl-o-lt Z9n 50-. 40 _ so 120 IO 240 300. 360 CTION ANGLE EES OF A DISTRIBUTED AMPLI0FIER AS A FUNCTION OF THE CONDUCTION ANGLE OF PLATE CURRENT.

CHAPTER VI. GRAPHICAL ANALYSIS 6.1 Introduction In any large-signal device utilizing vacuum tubes a graphical analysis is required to account for the non-linearity of the tubes. In this chapter a graphical method for the determination of the output power and efficiency of distributed amplifiers is outlined, and the frequency tolerance and second harmonic distortion problems encountered in distributed amplifiers are considered. It is further shown that under the proper operating conditions, the graphical analysis procedure may be simplified into one of reasonable length. A sample design of a distributed amplifier is given in order to illustrate the graphical analysis procedure. 6.2- Graphical Determination of Power Output and Efficiency The design of a distributed amplifier having essentially constant power gain over the design frequency range was discussed in Chapter IV. If a distributed amplifier is designed in this manner, the output power calculated at low frequencie will be valid for any frequency in the operating range as long as clipping does not occur. Clipping at the plate of the last few tubes is considered in Section 6.4. Equation 4.6 shows the method of calculation of the grid voltages on any tube in the distributed amplifier. The value of the various a's in this equation can be determined from Equation 4.3. The value of Gg becomes Gdr and can be calculated for any given operating condition of the tubes in a distributed amplifier. However, in actual calculations it is easier to start with the last tube in the distributed amplifier. For an assumed voltage applied to the last tube, and an assumed bias condition, the fundamental component of grid current is calculable by 58

59 standard graphical Fourier Analysis techniques. The ratio of the fundamental component of grid current to the grid voltage yields the equivalent grid conductance. This determines the value of An. The voltage on the grid of tube (n-l) is ean times the voltage assumed on the last tube. This procedure is repeated for each tube in the distributed amplifier. If the equivalent large-signal transconductance (the ratio of the fundamental component of plate current to grid voltage) of each tube is assumed to be equal, the plate load impedance of the kth tube at low frequencies (see Equations 4.6 and 4.7) may be written as Rk =- [e(a2+a35+' +ak) + e(a3+a4+ k)+ + e +.. + + e e (k+l+~] k ~ Le (6.1) These load lines may be drawn through the operating Q-point and the fundamental component of plate current determined from a graphical Fourier analysis. The currents so determined may now be used to determine to a closer degree of approximation the plate load impedance on each tube. The above procedure is continued until the change in the fundamental component of plate current between successive approximations is negligible. The power output can be determined as the square of one-half the sum of the fundamental components of plate current found above multiplied by the nominal impedance of the plate line, Rp. The efficiency at low frequencies may be determined as the ratio of the output power calculated above and the dc power input. The dc power input is the product of the sum of the average values of the plate current and the plate supply voltage. In general, the efficiency of large-signal distributed amplifiers will not be constant with frequency since the grid voltage on each tube varies with frequency, and hence the average value of the plate current will be a function of frequency. In a distributed amplifier designed for a flat frequency response the

60 efficiency will increase to some extent with frequency since the magnitude of the majority of grid voltages decreases with frequency which results in a lower dc supply current. 6.3 Approximate Graphical SolutionIf the variation in the a's at low frequencies between the various tubes in a distributed amplifier is small, a single value for the a's may be used with the exception of al which would be one-half this value since it is for a halfsection of the grid-line filter networks. a may be determined by a method discussed in Section 6.2. If the equivalent large-signal transconductances are assumed to be the same for all the tubes, the plate load impedance (See Equation 4.4) on the kth tube is given by Rk [e (2k-nel)J sinh ( (6.2) If a is small, Rp (6.3) The approximate load line may be drawn through the operating Q-point and the fundamental component of plate current determined from a graphical Fourier analysis. The output power can be determined from Equation 4.18, or as rewritten in Equation 6.4 for low frequencies. [ —~ sinh() 2 out = o1' L sinh( ) (6.4) where Pol is the output power of one tube operating alone in the distributed amplifier without any attenuation in the grid line. However, if nGdrRg -v — < 0.2,' 2,., 2

The output power can be written approximately as Pout = Pol n e (6.5) 6.'4 Frequencry Tolerance Frequency tolerance as defined in Chapter III is the frequency above which clipping occurs in the last tube of the distributed amplifier. This frequency may be found with the aid of the plate characteristics of the tube and two equations. If n is substituted for k in Equation 4.4 the plate load impedance for the last tube is given by na Z~Tp e(n-l) s sin2h.' Zn = ih (6.6) sinh 2 The grid voltage on the last tube of the distributed amplifier can be calculated from Equation 4.6. However, if all the a's are assumed to be equal, the voltage on the last tube can be written as En =/P~'/Z/ e2n (6.7) The plate load impedance for several different frequencies can be drawn on a set of plate characteristics using Equation 6.6. The grid voltage on these load lines is determined from Equation 6.7. As an example of these calculations, Figure 6.1 shows the final tube load line for a six-tube 4X150A distributed amplifier with a nominal grid line impedance of 50 ohms and a plate line impedance of 90 ohms. The dotted line crossing the load lines indicates the extent of the grid excursion. It will be noticed that above a frequency of about 250 me, clipping begins to occur in the plate circuit of the final tubejand as a result the power output begins to decrease. It is interesting to note that the same frequency is obtained as the frequency tolerance if the normals increase in Zt is used without attenuation in the grid line. This is indicated by the dashed curve in Figure 6.1.

62 g9-0194 R3: i —19-v Z92 z3i 11' w4 m +': + 5 =i'. O 0'.4 -20 398 0 o00 400 0o0 00o 000 PLATE VOLTAGE, VOLTS FIG. 6.1. FREQUENCY VARIATION OF THE PLATE LOAD IMPEDANCE OF THE 6th TUBE IN A DISTRIBUTED AMPLIFIER USING 6, 4X 150A TUBES WITH A. NOMINAL GRID LINE IMPEDANCE OF 60 OHMS AND A NOMINAL PLATE LINE IMPEDANCE OF 90 OHMS.

63. 5 Second tarmonic Distortion In a distributed amplifier operating under specified conditions it is possible to determine the shape of the plate current waveform, and, by means of i graphical Fourier analysis, to determine the value of the second harmonic compolent of plate current. In order to determine the second harmonic output power Lt is necessary to take into account the phase difference of the individual curL'ents as they appear at the output tube. Once the magnitude of the current which inters the filter section between the last tube and the output termination has )een determined, the second harmonic power can be calculated by multiplying the 3quare of this current by Zir of the plate line. If the attenuation in the grid line is neglected over the frequency range there the second harmonic would be transmitted by the plate line, the second harnonic component of the individual tubes can be written as i! = I2 sin 2wt 2 = 12 sin 2((t - p) in = 12 sin 2(wt - [n-l] cp) (6.8) {here cp is the phase shift per section of the grid line at an angular frequency )f a, and 1 is the peak value of the second harmonic component of plate current. If the phase shift per section of the plate line at the second harmonic is designated at P2' the current entering the last filter section is i1 in iout = 21 (n-l) 2) +'... + 2 I2 2s i(, - 2 22 2 2ct [e J(nl )2 + elI29r+ (n-2)c2] + ej e + e +(~.9

64 This finite exponential series may be summed in a manner similar to those of Appendix A, and Equation (6.9) can be written in closed form as shown in Equation 6.10 I2 ej [?ot —.(n-1) -2 out 2 2 E;1`C 11 sin(T ) 2 u =- 2 Fn(' ) ei[ 2() (6.10) where cp= 2 sin'l { — " ~where c = 2 sin'(.) and cp2 = 2 sin 1() (6.11) Figure 6.2 shows Fn( - )for a six-tube and a nine-tube distributed amplifier. It can be seen from Figure 6.2 that the variation of Fn (2) as a function of frequency is not great over the major portion of the frequency range where the second harmonic distortion is of interest. However, at the higher frequencies the phase angles of the second harmonic component of plate current of the individual tubes which appears in the output cause a cancellation effect and Fn ( ) decreases. The second harmonic output power can be calculated from Equation 6.12. P2out =) 2..) Z: (2a) (6.12) where Z,(20) is the mid-shunt image impedance of the plate line evaluated at the second harmonic frequency. 6.6 samle Desini of a Distributed Amplifier Using- 4X150A.Tubes After the tube type has been picked., the first step in the design procedure is to measure the small-signal effects. Figure 6.3 shows the input resistance for a 4X150A as a function of frequency. The input capacity of a 4X150A was measured as 21.2 micro-microfarads. If the experimentally determined values above are substituted in Equation 4.19, the product of the angular cut-off frequency and.

65 FJG. 6.2. SEGOND HARMONIC OUTPUT CURRENTf B~oz-~"Xt o | -lot loz P AS A FUNCTION \F- FREQUENCY. 4.. fJG. 6.2. SECOND HARONIC OUTPUT CURRENTl |FACTOR, Fn(-), FOR nuSAND n39ll PLQTTED AS A FUNCTION OF FREQUENCY, 0 0.1 0.2 0.3 0.4. CA/Wc

66 gg-t*Z-9 V43P 09l-19-V Z9ZZ 5000 - 4000 IN 2000 f FREQUENCY IN HUNDREDS OF MEACYCLES O A "..........'600 1 00 20 30 40 0o 6o 100 200 300 400 UEEUWENCY, Mc FIG. 6.3. VARIATION OF EQUIVALENT SHUNT RESISTANCE OF A 4X 1O5A CAUSEDo. BY LEAD:INDUCTANCES AND TRANSIT TIME.

67;he number of tubes for a 4X150A is ic = 189 x 108 (6.13) %he inductance of the grid lead of the 4X150A is about 0.007 microhenrys. The;eries resonant frequency of the grid lead inductance and the input capacity of the tube is 410 mc. Equation 3.11 indicates a maximum cut-off frequency of about 530 mc. The solution of Equation 6.13 with 330 mc for the cut-off frequency yields,% value of nine tubes. The impedance level of the grid and plate lines can be Determined from Equation 3.3. R = cg =45Q = 81.5 OcCp ror Cp= 11.8 pLif. In order to simplify the impedance matching problem the upper cut-off frequency is modified so that the grid line has a nominal impedance of 50 Dhms. This lowers the cut-off frequency to 300 me and raises the nominal impedance ~f the plate line to 90 ohms. The values of the inductances for the grid line and the plate line can be letermined from Equation 3.2 and the coefficient of coupling of the negative mutual soils from Equation 3.10. The attenuation of the grid line can be calculated for small signals from Cquation 4.3, and the results of Figure 6.3. Figure 6.4 is a plot of the attenuation per section of the grid line under small-signal conditions. If the values of x from Figure 6.4 are substituted into Equation,4.6, the value of the various grid voltages can be found as a function of frequency. The results of this calculation Jre plotted in Figure 6.5. The next step in the design of a distributed aplitfier is to determine the Dperating voltages. Unfortunately there is no quick and easy method for graphically

9gg-~Z-g g3i sI-19-vt;9Zz 68 0.7 0.3 aos I I I I I / I I I I 1 0.4 o CIO 0.3 0.0 i /_ I i00 040 180 2_ 2_ z 0.0...4.....REQUENC I00 I 140 ISo 20600 /I~REUNCM

69 Wb~ r?'iS l-r IO-V I9 VARIATION.'OF GRID VOLTAGE AS A FUNCTION OF FREQUENCY FOR A 300M C, 4 X 150A DISTRIBUTED AMPLIFIER FOR -A.4 PEAK GRID VOLTAGES LESS THAN VOLTS. k ~ NUMBER OF THE TUBE k:I 2 / 04 hi~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ FREQUENCY, MC..

70 determining the proper operating conditions for the distributed amplifier. The final operating conditions will in general represent a compromise between efficienc: power output, frequency tolerance, and driving power. The weighting of each of these parameters depends upon the application. One method of attack is illustrated in the graphical analysis that follows. A plate voltage at the Q-point, Ebq of 800 volts and a screen voltage of 400 volts were assumed. Figure 6.6 shows the plate characteristics for a 4X150A extrapolated to a screen voltage of 400 volts. This extrapolation process is described in detail in the Eimac tube handbook. Several possible values for the grid bias were assumed and the plate load line drawn. For this first approxi1 mation the load line is drawn with a slope of - mhos (attenuation in the grid line is neglected). The dynamic transfer characteristic is obtained from these lines and is plotted in Figure 6.7. It will be noticed from Figure 6.7 that the 4X150A is a good approximation to a constant-current device in that all the curves lie very close to one another. Figure 6.8 shows the variation of grid current with grid voltage for a 4X150A operating with a screen voltage of 400 volts. This curve will be used to determine the equivalent loading resistance due to grid current. Table 6.1 shows the instantaneous values of plate current and grid current for a bias voltage of -13.3 volts and a peak signal voltage of 33.3 volts. Figure 6.9 is a plot of the instantaneous value of plate current as a function of mt. From this plot the dc plate current is found to be 0.86 amperes. Figure 6.10 is a plot of the instantaneous value of the plate current multiplied by cos mt. Twice the average value of this curve is the peak value of the fundamental componen of plate current. From this curve the fundamental value of plate current is found to be 0.75 amps. Figure 6.11 is a plot of the instantaneous value of plate current multiplied by cos 2at. Twice the average value of this curve is the peak value of the second harmonic component of plate current. Figure 6.12 is a plot of the

2.0 ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~~~~~~~~~~~~~~~~2. 1.8 Ut) 1.6 (41.4 q).. cc~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~h hi.2 z~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ z 14 1.2 Z hi Z 10 20 So 40 w00 9.0 co-.0 10 lU o.a 1.6 0.2I DRAWN FOR. A 9-TUBE DISTRIBUTED AMPLIFI-ER AT LOW FREQUENCIES FOR VARIOUS 0- POINTS CONSIDERED IN GRAPHICAL ANALYSIS.

tg-9-zl W3r t91-19-V zgz 72 -26.6 &-20 1.8 FIG. 6.7. DYNAMIC TRANSFER CHARACTERISTIC FOR A 9-TUBE DISTRIBUTED AMPLIFIER / WITH Ebb 800 V, EC2 400V. 1.6 // 1.24..oL<<mS. II ECI 2E 1.0 0.6 0.4 0.2 -I00 -80 -60 -40 - 20 0 20 ECI

N-.S1.9 m3r 991-19-v Z9EZ 73 120 Ec2 400V 100 sO SO _5 _ 1S 20 25 30 RID VOLTAGE, VOLTS FIG. 6.8. G-RID CURRENT AS A FUNCTION OF GRID VOLTAGE FOR A 4 X 150 A. VOLTAGE FOR A 4XI50A.

74 TABLE 6.1 Instantaneous Values of Plate Current and Grid Current Ebq = 800 v; Ec2 = 400v; Ec1 = -13.3 volts; Esig = 33;.3 volts peak 33.3 cOSW t 1t -13.3 ib ib Cs)t ib COS2Wt ic (ma) ic cosWt 0 20 1.68 1.68 1.68 64.7 64.7 15 18.9 1.64 1.58 1.42 57.5 55.6 30 15.5 1.56 1.35 0.78 40.5 35.1 45 10.2 1.42 1.0 0.0 20 14.1 60 3.4 1.25 o.63 -0.63 4 2 75 -4.7 1.06 0.27 -0.92 0 0 90o -133.82 0 -0.82 0, 0 105 -21.9.62 -0.16 -0.54 0 0 120 -30.0.45 -0.23 -0.23 o 0 135 -36.8.34 -0.24 0.0 0 0 150 -42.1.26 -0.23 0.13 0 0 165 -45.5.2.2 -0.21 0.19 0 0 180 -46.6.21 -0.21 0.21 0 0

75 Gg-6-g W3r t91S9-v9 z;zz.8 - - - - - -- 1.6 FIG. 6.9 Lb VS. wt FOR Ebq 800V, Ec2 400OV, ECI -13.3V 1.4 I r Esig ~ 33.3V PEAK 1,2 1.0 8DC 0.88 AMPS Q6 0.4 0.2-_, 0 30~ 60~ 90~ 120~ 150~ 1800 ct

GS -LI -ZI M3vr 3P 91-19-V Z97Z 1.6 FIG. 6.10 Lb COS wt VS. wt FOR Ebq BOOV, EcZ 400V, E00V, E -13.3V Esig ~ 33.3V PEAK 1.2 1.0 __ _.I. Cl) 0.8 I. - -- 5 AMPS 0.6 0.4 Q2 -0.2 -Q3 00 30~ 60~ 90 o 120~ 150~ 180o Cot

77 g -L -Zl I3r 991-10-V Z9ZZ 22,0~~~~~~~~~~~~~~~.0 1.8 1.6,,, FIG. 6.11 1.4 ib COS 2wt VS. wt FOR Ebq = 800V, Eca 400V, Ec~ -13.3 V 1.2 Esi'g33.3V PEAK 0.8 0.68 0.2 120.06AMPS 0. - - - - - - - - -- -04 -0.6 -0.8 i. 1.0 00 300 600 900oo 1200 1500 1800, t, DEGREES

e1-t~1-g rr e1-1g-v zgz; 78 70 FIG. l. __ _.. 6.12 INSTANTANEOUS VALUE OF'GtID-, CURRENT AS A FUNCTION OF Wt FOR Ebq 800V, EC2O 400V, E c -13.3V PEAK SIGNAL VOLTAGE ~ 33.3 VOLTS 60 XX_ _ 50 __ 4-0' 20. -\ tC|~ 12..9 MA 10 - -. -' - 0 IS 30 45 60 75 WCt,.gSREES

79 nstantaneous value of grid current. The dc grid current is the average value of his curve. Figure 6.13 is a plot of the instantaneous value of grid current multi)lied by the cos mt. Twice the average value of this curve is the peak value of;he fundamental component of grid current. Due to the non-linearity of the dynamic transfer characteristic of the LX150A, the load line will shift from the small-signal operating point. Each of the load lines drawn on Figure 6.6 is drawn for a voltage at the Q-point of 800 rolts. The actual dc supply voltage, Ebb, will in general be less than the plate roltage at the Q-point. Ebb is easily determined from the load line as the voltage Lcross the tube when the plate current has its average value. This is shown in Figure 6.6. Table 6.2 summarizes the calculations and results for the 9-tube di - tributed amplifier Qperating under the conditions stated above. The graphical calculations for the other operating points shown in Figure 6.6 are given in Appendix F. Figure 6.14 is a compilation of the calculated results for a 9-tube distributed amplifier operating with a plate voltage at the Q-point of 800 volts, a screen voltage of 400 volts, and a peak signal voltage adjusted so that the grids are driven 20 volts positive. It will be noted from these curves that the dc input power is in excess of the plate dissipation rating of the tubes. In order to realize these operating conditions in a distributed amplifier it is necessary to operate the amplifier intermittently so that the tubes remain in a safe operating region as far as heat dissipation is concerned. The maximum duty cycle for safe operation of the tubes is indicated in Figure 6.14 The compromises which are involved in the final selection of the operating Joint may be seen from Figure 6.14. For the particular plate voltage at the Q-point

80 ci,COS (t VS wt FOR Eb-q 8CXV, Ec ~ 4OOV, Ec, x-133V PEAK SIGNAL VOLTAGE 33.3 VOLTS FIG. 6.13 60 4 30 80 ____ _____ ____ _~, 23.3 MA O 15 30 45 60 7_ (at, DEGREES

TABLE 6.2RESULTS AND COMPUTATIONS Ebq = 8O, Ec2 400v, Eci = -13.3, Esig = 33.3 volts peak Ib = 0.88 Amps. I1= 0.75 Amps. I2 = o.06 Amps. Ic = 12.9 Ma. IC, = 23.3 Ma. R = 1.43 IK geq aodr = 0.0175 nepers/section aodr-... = 0.175 ao Ebb = 770 volts Power Output = 437 watts Power Input = 6090 = 7.18% Driving Power = 11.1 watts Frequency Tolerance = 0.9 Power Out 2nd Harmonic = 3.3 watts

Gg-91-9 i3Pr 00Z-19-V Z9zz O2 14.0 100 16.0 o 1s Q 1''20 / 5.0 4I 14.0 s60 4_ w t I B 12.0 c4, 4_,0 ____ w' / IL 5 bb 0 0 FIG. 6.14 Tow - 0 0.9 SC GRAPHICAL SUMMARY OF RESULTS CALCULATED FOR A 9-TUBIE DISTRIBUTED AMPLIFIER USING 4X150A TUBES FOR EbqUsOOV, Ec34OOv, V AND 6000 GOO - "6A MAXIMUM INSTANTANEOUS GRID VOLTAGE OF 4' WV. 0.s 7000 SW. E CfC 0 0~ -10 -20 -30 -40 -50 -60 -70 -80

83 and screen voltage assumed, the output power peaks at a bias voltage of about -35 volts; however, the efficiency reaches a maximum value in the neighborhood of -60 rolts bias. The cost of this increase in efficiency comes from a larger driving )ower requirement and a decreased frequency tolerance. Figures 6.15 through 6.19 show a graphical summary of the results of calculations for a 9-tube distributed amplifier using 4X150A tubes for several ralues of plate voltage at the Q-point and screen voltage. It can be seen from these figures that the peaks in the power output curves tend to increase with;creen voltage; however, the efficiency reaches a maximum and decreases as the;creen voltage is further increased. If curves were drawn for a constant input )ower, one would expect to find an optimum value of screen voltage from the stand)oint of power output and efficiency. In general, the frequency tolerance decreases as the operating bias increases. The driving power increases as the bias.ncreases. The general trends mentioned in the preceding paragraph are useful in Letermining the proper direction in which to move in assuming new operating voltages Dn distributed amnplifier tubes after an unsatisfactory design has been conpleted.

gg-91-9 m3P 10-19-V 92Z9 8. 7O 2 0.0 0.6 4t 60 18.0. 00 O I1 a~~~~~~~~~~~~~.~,2 ia 1. 0.4. s o'6.... ~ m CD Ii. I-~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~l w z / +~] I ii I (.3 30 12.0 _00o 0.2 r t 30. 1.0o0 FI G. 6.15 GRAPHICAL SUMMARY OF RESULTS CALCULATED FOR 20o IQo,...A 9-TUBE DISTRIBUTEDo -20 laIO 6.0 AMPLIFIER USING 4XI50A TUBES FOR Ebq 8 800'Vs, EcOt a 5OOV AND A MAXIMIUM INSTANTANEOUS GRID VOLTAGE OF +25V 7000 1100 -5.0 ~- __ ~I 9L5ooo ILooo & O. 4 lad O I ~ l0 z'6000' 9000 a~~~~~~~.2 40.00 O0....._,2.0 -20 -30 -40 -5o -60 ECi BIAS VOLTAGE

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PEAK POWER INPUT, WATTS PEAK DRIVING POWER, WATTS PEAK POWER OUTPUT, WATTS PER CENT EFFICIENCY )( 4 P -'Ol - -__-x O CrO~ o'-'.... III m~. OD C; CO) o rn C.. I) C'1,-U'VI~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~. _,_ < r 0'. C -r ~ O r ms2;DQ 65 G) 0 CI +0 C b PI < S4 -~~~~~~v4 rn FI~~~~~~~~~~~~~~~',, / -~~ ~ ~ ~ ~~~1A.Ll/ ~''. 3~V31J A/~l3L r- V) t~~~~~~~~~~~~~~~~~~~~~~~~~~~~~.Jlr.GI) C qq 0CnmI k~ k, ~~o- N P" /.rn c: 0obh CD~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~c 00 +~~.~,~o.~ (1 ~ J` ~ 3-iuo;3 Aina/ 1i A3NnaO.~~Q, a.<8..5.., I',~ / 1~.. ~~~~~~~~~~~~~~~~~~~~~~S-1A

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CHiAPTER VII. CONCLUSIONS 7.1 Introduction In the preceding chapters the small-signal distributed amplifier Equations have been modified to account for the additional complications Encountered in large-signal operation. A graphical procedure for analysis of power distributed amplifiers operating in a non-linear region was also presented. 7.2 Summary of Results 1. It was shown in Chapter VI that an exact graphical analysis is possible. However, this method of analysis involves a cut-and-try procedure,;hich is long and tedious. If certain assumptions are made, it is possible to ise an approximate graphical analysis for distributed amplifier operation. 2. It was shown in Chapter V that the maximum theoretical efficiency )f a distributed amplifier operating at low frequencies is 30 percent. It was lso pointed out in Chapter IV that it should be possible to increase the actual Efficiency of a distributed amplifier by about 30 percent if the low-frequency peration is limited so that the plate load impedance remains low over the )perating frequency range. This increase in efficiency is accomplished by )perating the first few tubes at a reduced plate supply voltage. 3. The equations for a flat frequency response for a small-signal listributed amplifier are modified to account for the loss due to the positive )peration of the grids. Curves showing the frequency response of a distributed %odr Lmplifier for several values of are presented in Chapter IV. CIO

90 4. The frequency limitations are pointed out in Chapter III. The inductance of the grid lead is found to be the most serious limiting factor on the upper cut-off frequency as far as the small-signal response is concerned. A serious limitation on the large-signal frequency response is incurred by the on-set of clipping in the plate circuit of the last few tubes in a distributed amplifier. 7.3 Suggestions for Further Research During the course of this investigation several problems were encountered which seem worthy of investigation. 1. Power tubes designed to operate in the VHF-UHF region are not designed for grounded cathode operation. The frequency limitation imposed by the grid lead inductance and the plate lead inductance could be essentially removed if tubes were built with two grid leads and two plate leads. One of the important parameters in the determination of the proper number of tubes to be used in a distributed amplifier is the input conductance. If the transit time and the cathode lead inductance were decreased, it would be possible to use more tubes in a distributed amplifier and, therefore, present the possibility of achieving greater output power and increased efficiency. 2. The plate circuit of a distributed amplifier is only 50 percent efficient. If a more efficient plate circuit for a distributed amplifier could be synthesized without incurring other serious defects, a large increase in the efficiency should be possible. 3.'Triodes operating in a grounded-grid distributed amplifier circuit in which the impedance of the cathode line is changed at each tube in order to match all the tubes to the input seems worthy of consideration when bandwidths of the order of 500 me are contemplated.

APPENDIX A SERIES FOR PLATE-CATHODE VOLTAGE ON KTH TUBE OF AN N-TUBE DISTRIBUTED MPLIFIER From Equation 2.3 Chapter II the voltage on the plate of the kth tube.s given by n ek Z 2 ekj (A. 1) j=l Z here: ekl = GmEgm sin [ut- (k-l)p] m 2 * Zo = Gm......gm sin[t - (k-l)- (n-l-k)] ek(kJ*) = GmEgm 2 sin[c1t - (k-l)cp _ 2cP] z ek GmEgm {k sin[wt - (k-l)p] + sin[t - (k-l)cp - 2c] + sin[cut - (k-l)qc - 4c + + sin[t - (k-l1)p - 2(n-l-k)cp] + sin[wt - (k-l)cp - 2(n-k)cp] }. (A.2) Consider the series S = sin(A - 2cp) + sin(A - 4q) +.... + sin[A- 2(m-l)p] + sin[A - 2mp] (A.3) 91

92 ejA. -e e 2.p _ eA e2 r'p eJ e-4-JA...A eJ4 S = -- - - +... 2j 2j e A e-j2(m-1) ) ejA e j2(m-1K) +..... - 2j eiA e-j2mrp- ejAA J2mp + e * (A.4) 2j 2[eje ej2CP + e +.. + e + (m) e - ejA{ej2c + ej4? +... + ej2(m-l)qp + eij22mP} (A.5) Let S1 =e-j2cP + e'j4 +,,, + e'j2(ml)CP + e'j2mP (A.6) le j e- j4j + -e-6JP +,,, + e-j2mcp+ e-j2(m+1)cp (A.7) S=e e- +,...+e +e Subtracting Equation A.7 from A.6 Sl(i - e2) = e j2 -_ e'j2(m+1)e (A.8) or S m= -2c 1,1,, j2-, i= e 1 - e-j2cP (A.9) Similarly if S2 = ej2 + eJ49 + +... j+ ej2(ml)cp + e2m(A10) then j2m ~1S2 = e = l-e~2~ (A.11) Equation A.5 can be written as S j = j 1 [ —e- e j'-2m - e'JA e2j2cP 1_ e2m P 1 (A.12) 1 - ej2' J

93 1 - ej2m P can be written as 1 - eJ2cp em (ejmp - e -m) -j(m-1)p sin m (A.15) e j'P (eiJP - eP) jPfsin. 3imilarly _1 -_ e-' ~ e PJmp -e"J'' e'Jmp') (A.14) 1 - e j2c ei (e'jp - eij) - eJ(m'-l );in mq sin cp So that Equation A.12 can be written as s = 1 [ej[A - 2 - (m-l)] sinm e-j[A - 2- (m-l) sin (A.15) 2j [ sin cp sin p Now S can be written as — S _ s sin[A 2p - (m-1)c] sin cp sin mrM' sin[A - (m+l)p] (A.16) sin cp Identifying the terms of this series with those in Equation A.2, A.2 can be written as ek = GmEgm 2 k sin - (k-l) + sinQ (n-kh ) s wt-(k-l)p - (n-k+l) (A.17) sin cpJ

APPENDIX B LOW-FREQUENCY DISTRIBUTED AMPLIFIER USING 807 TUBES The following values were picked for convenience. n = 6 fc = 15 kc C =.04 gf P Co =.044 gf 2 2 - = OcCp 2nXl5Xo103X.044X106 _ = 2 - = 2 x 530 = 11.25 mh R -2 2 - 482 S g -cCg 2 1 5X103X.044X0lo -p=~,~ 482 Lg 2 2 x = 10.25 mh. Dc 212X15X103 The schematic diagram is shown in Figure B.1. The 10 ohm resistors in the plate of each tube are used in order to be able to measure the plate current on an oscilloscope. Table B.1 shows the experimental data obtained for the plate load impedance for the first tube in the amplifier. The method of data reduction is also shown in Table B.1. The absolute value of the plate load impedance of the first tube is normalized to one-half the mid-shunt impedance of the constant-k plate line. In order to plot the experimental points on Figure 2.2, the phase shift per section of the artificial transmission lines was calculated from the ratio of f/fc' 94

rDWO 0.04O0 0.041Lf 0.04If t O 530An n 11T25mhT IL25rnh T 112 0T 126h 563mh [ P emh 11.25- m 5 5.63MA &3P00:I OP6-~-~-~t -- 00 -09~ Pmh ~ K~~n >lon OstJoi. > ix >.ka soSTSl807 ~ 80T 0t8077Ot~~. Kl C015111 BIOJ1~f ) 101EjLf gIO ~a~tr en la~E: ^~~ izs~.fU~~~~~o -. I - 1I -. - 1 1 f. 10.04~ z.04/ 04..04pf o-O4X.044/.,Jk ~~~0.~~0,~.-.,3pf~.6f +o.v - FIG. -.l. SCHIEMATIC DIAGRAM FOR Ikc DISTRIBUED AMPflER.

TABLE B.1 MEASUREMENTS ON TUBE NUMBER 1 OF 15 KC DISTRIBUTED AMPLIFIER freq.(kc) E(VOLTS) I(mo) Z f/fc 0DEGREES Z 7T/R Z/R/2 Z/ R/2 X R/Z7T 1 13 8.5 1530.0667 7.5 1.002 5.74 5.73 2 11 9. 1220.1335 13.5 1.004 4.57 4.54 3 5.5 9.5 580.2 23 1.02 2.17 2.13 3.9 0.5 9.5 52.26 30 1.028.195.19 4 1 10 100.266 30.5 1.03 ~375.364 5 5.5 10 550.333 38.5 1.06 2.06 1.945 5.5 5.5 9-5 580.367 43 1.075 2.17 2.02 6 5.5 10 550.4 47 1.09 2.06 1.89 7 2.7 9.5 284.467 55.5 1.125 1.065.947 7.5 1 9.5 105.5 60 1.15.394.343 8 2.5 10 250 ~533 64 1.175.938.798 8.5 4 10 400.566 69 1.21 1.5 1.24 9 5 10.7 467.6 74 1.25 1.75 1.4 10 4.5 10.5 428.667 83.5 1.35 1.605 1.19 10.7 0 10.5 0 0. O O 11 2.5 11.5 217.734 94 1.49.815.547 11.5 5.5 12 458.767 100 1.515 1.72 1.08 12 8 12 666.8 106 1.67 2.5 1.5 12.5 8 11.5 695.834 113.5 1.82 2.6 1.43 13 4 11.5 348.867 121 1.99 1.305.657 13.2 1.8 11.5 156.88 123 2.07.585.282 13.5 3 12.5 240.9 128.5 2.3.9.39 14 12 13 924.934 134 2.77 3.46 1.25 14.5 5 13 385.967 152 3.75 1.445.386

APPENDIX C PLATE LOAD IMPEDANCE FOR PAIRED-PLATE The circuit for paired-plate connection is shown in Figure C.1. _ _, components which are eual in magnitude but differ in phase by (n/2) P 2-2 FIG. C.I. BLOCK DIAGRAM Of PAIRED-PLATE DISTRIBUTED AMPLIFIER. components which are equal in magnitude but differ in phase by y. I- = 21I 1 cos I -' 2 Ie = 2IPI COS - - (2k-2)p 2 2 2 97

~~th ~98 Voltage on k Pair Plates Ek1 = ILpl Cos - -(2k-2)cp Ekk = IIpl cos T Z~ - 2 -(2k-2)c 2 2 Ek(k+l) = IIpI cos 2 Z 2 -(2k)cp-2cp Ek(k+2) = Iplcos Zo - (2 k+2) pE6 | Ip plCos Zo - I -(n-2)p- (z - k)2c n/2 Now Ek = Eki i=l kII I cos Zo - [ -(2k-2)p + CPl os ZO {1 - -(2k+2)cp + 1 - — ((2k+6)cp + 1- 2 -(n-2)p -(n-2k)cp Consider the series in brackets. 1 2i'' - (2k-2)cp -4p 2 +1 L - 2 (2k-2)cp- 8qp +i |- - (2k-2)cp - 2(n-2k)Cq e e + e8 +... + e-j2(n-2k)] Let -j4cp -j8cp -j2(n-2k)(p S = e +e +.... +e

99 hen S e'j4c = -j8p + + -j2(n-2k)cp+ e-j [2(n-2k)cp+49] e'j4 - ej [2(n-2k)p+4p] e -j e - 1- e-4 e-j4cp e-j(n-2k)cp[n (n-2k)] e-j2cP[sin 2p] e-j(n-2k+2)q~ sin [(n-2k) c] sin 2cp [S e i [ (2k-2)]_ -j (n + s) in[(n-2k)j S e 2 + = e sin 2cp Ek = Ipl cos 2 ZO [k e + (2k-2)] sin [(n-2k)p] e-j(n + 2)] + - eJ sin 2cp Ik 21Ipi os 2 - I - (2k-2)qcp EkIk = + sin [(n-2k)qp] e-j [n-2k+2]p ] Zk I 2 sin 2cp

APPENDIX D INPUT RESISTANCE OF A VACUUM TUBE DUE TO CATHODE LEAD INDUCTANCE Figure D.1 shows the equ:ivalent circuit used for the determination of input resistance. IG. Cp Eg E k FIG. D. I. EQUIVALENT CIRCUIT OF A TUSE USED FOR THE DETERMINATION OF INPUT CONDUCTANCE. The node equations for this circuit are shown in Equation D.1 and D.2 I = Eg[j(Co + Cg)] Ek[ji3Cg] (D.1) gmEgk = -Eg[ g] + Ek[(Cg + Cp)+. (D.2) But Egk = Eg - Ek. Equation D.2 can be written as shown in Equation D.3. o = - Eg[g + jaCg] + Ek[m + iD(Cg + CP)+ ] (D-3) Equations D.1 and D.3 can be solved for Eg as shown in Equation D.4 Eg [ (Cg +cp ), ] (D.. E9 L (D.4) j[((Co + Cg)] i W(Cg + Co) + j [i + m + + p) + m+ Jg] 100

101 rhe input admittance is given by the ratio of Ig to Eg or [jw(Co + Cg)][gm + ji(CgC + Cp) + ] 4[ji [gm + jCg] y =........ -,. -. -. in gm + jC(Cg + c,) +'he real part of Equation D.5 can be written as shown in Equation D.6 2 1. --,,C p Gin = gm 2TkCg (D.6) 2gm"2 + Lk( + [2C Cp)L11-]

APPENDIX E DETERMINATION OF PLATE LOAD IMPEDANCE WITH ATTENUATION TN THE GRID LINE The grid voltages on the various tubes are assumed to be of the. form E1 = Eg LO E2 = Ege I-c (E.1) E Ege -|2p 3 g E E e-(n-l)a Ege These grid voltages will result in the following fundamental component of plate current 12 = Ege GM -cp (E.2) I = Ege (nl)aGm (n-l) The voltage on the plate of the kth tube can be written as n Ek = E Ek (E.3) j=l i where Ekj = the voltage that appears on the plate of the kth tube with only the jth tube operating. These voltages may be written as shown in Equation E. 4. 102

103 zo Ekl = G 2m Eg -(k-l)cp Ek2 = Gm - Ege | -(k-1)cp Zo -(k-1) -(k-)cp (.) Ekk = Gm Eg(k-l)cp (E.4) z~g m 2g - (-k)q-a Ek(k+l) = Gm Ege )-(k-l)cp-2cp 2 Zo - (n-lw1) Ekn = Gm Ege (k-)cp2(nkp Substituting Equation E.4 into E.3 the voltage on the kth tube is Zo -j -(k-1)a Ek = Gm 2 Ege'j(k)ep [l+ea +... + e + e-k ej2P + e -(k+l)a e-j4 +... (E.5) + e (n-l)Cx e-j2(n-k)P ] et S1 indicate the series -a - (k-1)a S1=l+e +...+e (E.6) Sle = e a +... + e(k-l)o + e-ka (E.7) f BFuation E.7 is subtracted from Equation E.6 the result is Sl(1-e'G) = 1-eka (E.8) a kc ka -e'ka e -k + 2-e,-e e = e 2 e 2 i-e -a -ea a (B -(k-l) sinh kc( sinh et S2 indicate the series S2 = e'kc i-2c[ + e e' J +... + e (nkl)a.-J2(n-k-l)p] (E.10)

104 S2 e e2-j2 =ek e-k 2 [ea e-22 [ e2 + *** + e'(n-k-l)U e-J2(n-k-1)p + e (n-k)a e-j2(n-k)cp] (E. 11) Subtracting Equation E.11 from Equation E.10 we have S2(1-ea ea-j2) = e'J) = ej2 [l-e(n-k)o e.'j2(n-k)cp] (E.12) or S = -ke.kj2~P l-e-e(n-k)a-j2(n-k) (E13) = 1 (E. 13) e-aj2cp e-ka-j2cp e-(n-k)a-j(z-k)cp sinh [(n-k)k + j(n-k)cp] eZ t sinh i[a CJ )- -( j(n +) sinh [(n-k)a + j(n-k)cp I....(E.14) sinh[2+ J+ ] Substituting Equations E.14 and E.9 into Equation E.5 the plate voltage on the th kth tube becomes zo -j (k-1)cp Ek =G Eg e x a kc | e-(k-1)2 sinh(E.15)......... (E.- ) sinh e- e(n+k~-l -i (n-k+l )qT sinh [(n-k)1 i -(n-k) p] sinh [2+Jq)]

105 The plate load impedance is given by the ratio of the plate voltage to;he plate current or Ek Z kcZk = 2k (k1)a e-(k-l)2 sinh Zk - e elm..... k 2a sinh a _ +e -(n+k-l) -j(n-k+l)cp sinh [(n-k) +j(n-k)cp] sinh[ a +jp],r Zk o s (kl)- (n-k+l) -2 (n-kLJ +l)q sinh (n-k) +j,(n-k)cp Zk e 4] 2 +

APPENDIX F GRAPHICAL CALCULATION OF THE OPERATION OF A 9-TUBE DISTRIBUTED AMPLIFIER USING 4X150A TUBES Figure F. 1 shows the instantaneous value of plate current for a ninetube distributed amplifier operating at a plate voltage at the Q point of 800 volts, a screen grid voltage of 400 volts, a control grid bias of -20 volts, and with a peak signal voltage of 40 volts. Figures F.2 through F.5 are for the same operating condition. Table F.1 summarizes the results of the distributed amplifier operating under the above conditions. The rest of the figures in this appendix repeat the calculations mentioned above for several different operating conditions on the amplifier. 106

107 TABLE F.)1 RESULTS AND COMPUTATIONS Ebq = 800v, Ec2 = 400v, Ecl = -20 v, Esig = 40 v peak Ib = 0.76 amps. I1 = 0.77 amps. I2 = 0.10 amps. Ic = 11.7 a. Icl = 21.2 a. Rgeq 1.89 K a Codr = 0.01325 nepers/section ~odr = 0.133 Ebb = 740 volts Power Output = 480 watts Power Input = 5060 watts Tl = 9.5% Driving Power = 16 watts Frequency Tolerance = 0.85 Power Out 2nd Harmonic = 9.1 watts

o108 S-9SI-2 W3P ILI-19 -V Z9ZZ 18 1.6 FIG. F, I Lb VS cot FOR EbqzBOOV, EC2:400V, ECi:2OV 1.4 1.4r~ rv ~~~Esig 40V PEAK 1.2 I.0 0.8 * DC 0.76 AMPS 0.6 0.4 a2 00 300 60~ 9Q0 1200 1500 1800 Cot

;g-gl-R I3P U-19-V Z93Z 9,.0~ gg-.g-,fl6r-I9v ~109 1.4. FIG. F. 2 Lb COS (t VS. wt FOR 1.2 _ Ebq8BOOVv Ec2400 V, E400V, Ec-20V Esig= 40V PEAK 1.0 0.8 o I - - T= Q0.77 AMPS C) 0' 0. r_____ 0.4 0.2 ___X o~ 30~ 9 o\ 1200~ 15 Io -0.2

110 Sgg-g1 - 3r r~w-19-v Z9ZZ 2.0 I,. 8 1.6 FIG. F.3 1.4 ib COS 2 Ct VS. oWt FOR Ebq =80OV, EC2 400V, Ecl- 2aV ESig 40V PEAK 0.8..... 10 0.6..... 0 0 0.6 0.8 1.0

gg-gl-e IlI H9i-l'O-V Zg ll INSTANTANEOUS VALUE OF GRID CURRENT AS A FUNCTION OF wt FOR Ebqz800V, EC2 40OV, Ec i-20V PEAK SIGNAL VOLTAGE = 40.0VOLTS. FIG. F.4 40. 4 30 I I I 1 - - - - -.- 11.7 -MA Wt, DEGREES

G.;l-9 IV3 2gLI-19-V ZZZ 112 TO70 LC, cOSwt VS. ct FOR Ebq 800V, Ec ~400V, Eci,-2E.OV PEAK SIGNAL VOLTS ~ 40.0 VOLTS FIG. F.5 60 50 40 30 tIc 21.2 M 0 15 30 45 60. 5 CWt, DEGREES

113 TABLE F.2 RESULTS AND COMPUTATIONS Ebq = 800v, Ec2 = 400v, Ec = -26.6v, Esig = 46.6v peak = 0.685 mnps I, = O.806 amps 2 = 0.14 eamps Ic = 10.7 na. ICl = 21.7 na. Rgeq = 2.15 K 2 ckodr =.01165 nepers/section aoCdr Qa0 = 0.117 Ebb = 730 v. Power Output = 532 watts Power Input = 4500 watts rl = 11.8%o Driving Power = 21.7 watts Frequency Tolerance - 0.78 Power Out 2nd Harmonic - 17.9 watts

gg-Gl-~ 1w3r SLI-19-V 9Z 114 1.6 1.4 FIG. F.6 ib VS. Wt FOR 1.2 _ Ebq 800V, EC2 400V, EcI-26.6V Esig 46.6 V. PEAK — C 0.685 AMPS 0,6.... 0.4 0.2..... 00 30~' 600 90~ 1200 1500 1800 ct

G-gI-C 3rP LLI-19-V Z9ZZ 115 1.6 1.4 FIG. F.7 Lb COS wt VS. ct FOR EbqI 800V, Ecs, 400 V, ECIl,-26V 1.2 ESlgs46.6V PEAK 1.0 4J O,.e L0 0.i - - I 0.806 AMPS 0.6 0.4 0.2 8 30~ so" 3606 120~ 15 led0o * t

116 g-1-2i W63P ILI-19-V Z9ZZ 1.2 1.0 FIG. F.8 LbCOS 2wit VS. (t FOR E0.8 bq 800V, Ecz 400V, E, -26.6 V 0.8 Eig:46.6V PEAK 0.6 u) 0 ~- 0.4 0.2 12 0..I4 AMPS -0.2 o0 ~ 300 60~ 900 g120~ 1500 180~

117 SG-gl-9 N3Pr 61-19-V Z9;$ 70 INSTANTANEOUS VALUE OF GRID CURRENT AS A FUNCTION OF uot FOR Ebq=800V, Ec 400V, EC i-26.6V PEAK SIGNAL VOLTAGE 46.6 VOLTS 6i10 ____~.__. FIG. F. 9 50. 50., 40 1 30 4560 r5 9t, DEGREES

118 29-C1-9'..r...-1i-.gZz 70 - COS ot VS. ct FOR Ebq 800V, Ec2i40XV, Ec, =-26.6V PEAK SIGNAL VOLTAGE z' 48.6 VOLTS FIG. F.IO 60. 40..3,.21.7 MA -O 27 0 15 30 45 6a r,.t, DEGREES

119 TABLE F.3 RESULTS AND COMPUTATIONS Ebq = 80Ov, Ec2 = 400v, Ecl -40 volts, Esig = 60 v peak Ib =.56amps. I1 = 0.79 amps. 12 = 0.24 amps. Ic = 9.6 ma. Icl = 18.7 ma. Rgeq 3.2 K Q %odr = 0.0078 nepers/section aodr.0785 Ebb = 680 volts Power Output = 530 watts Power Input = 3425 watts rl 15.5% Driving Power = 36 watts Frequency Tolerance = 0.63 Power Out 2nd Harmonic = 52.5 watts

120 9G;GI -~ M D 11-ED-V Z9Z 1.6 FIG. F.II 1.4 lb VS. wt FOR Ebq 800, Ec a400, E -40V Ei g60V PEAK 1.2 1.0.~ 0.8' 0.4 0.2 - 0~ 30~ 60~ 90 120 o 150~ wt

121 9g-g1-~ M9 Ztl-1-v Zg9;Z 1.6 FIG. F.12 1.4 i b COS Wt VS. wt FOR Ebq -800V, EC2 -400,EC-40V Esig 60V PEAK 1.2 1.0 C) 0 0.8 lI =0.79AMPS. 0.. 00~ 30~ 60~ 90 ~ 1200 150~ wt

gG-gj-; Me tgl-I.I-V ZO9Z 122 FIG. F. 13 1.25 LbC06 2Wt. VS. (t FOR _Eb ~s OOV, cts 400v, EC, -40V EsgS6OV PEAK 1.0 0.. 0.25 0.25 - I IOI 0.24 AMPS 0.25 X _ 00 300 60~ 9Q@ 12da 1500 3 t

*g01-9g "or'-A o-v;uos: 123 70 - INSTANTANEOUS' VALUE OF GRID CURRENT AS A FUNCTION OF cwt, FOR Ebqt800V, Ec2"400V, Ec,-40V. PEAK SIGNAL VOLTAGE z 60.0 VOLTS 60. _____ ____ FIG. F. 14 60 3 40 20 90o~~~~~ L~~I; L L a I:9.6 MA. O lE 30 45 s60 7 W t, DEGREES

gg-lI-9 V3P SS1-I9-V Z9;Z 124 70 Lci COS ct VS. Wt FOR Eb 80OV, E q:40OV, EC -40V PEAK SIGNAL VOLTAGES' 60VOLTS FIG. F.15 60, 50 _'_. -I 40 X 3 20 - - - - - I'E \ a fIcl'18.7 MA 0 I... 0 15 30 45 60 75.t, DEGREES

125 TABLE F.4 RESULTS AND COMPUTATIONS Ebq = 800v, E2 = 400v, Ec = -53.3v, Esig = 735.v peak Ib = 0.5 amps. I = 0.76 amps. 12 = 0.325 amps. I = 8.8 ma. IC = 16.6 ma. Rgeq = 4.4 K a aodr = 0.00567 nepers/section odar = 0.057 Ebb = 663 volts Power Output = 501 watts Power Input = 2980 watts TI = 16.8% Driving Power = 53.7 watts Frequency Tolerance = 0.55 Power Out 2nd Harmonic = 96.3 watts

9-.11-G I33'*91-19-V Z9ZZ 126 1,6 1.4 FIG. F.16 Lb VS. 6ot FOR Ebq "800V, EC2,400V, EC i-53.3V Esig 2 73.3 V PEAK 1.0 0,. 0.6 AVE IDC 0.5AMPS. - 0.4 0.2 0~ 300 600 900 1200 1500 CAt

gg-1I-g 43P(LIII-9-V Z9ZZ 127 1.64 L I a FIG. F.17 Lb COS ut YS. Ct FOR __ Ebq8100V, Ec2z400 V, EC -53.3V 1.2 Esig= 73.3V PEAK 1.0 0.8 I.I - 0.76AMPS. 0.6 0.4 0.2 00o 300 60s 90o 1200 1500'ot

128 gg-II-g N3P g1-J-V Z9ZZ 1.50 - 1.25 I1 FIG. F. 18 ib COS 2wt VS. at FOR Ebq= 800V, EC2= 400V, EC, -53.3V 1.00 Esaig 73.3V PEAK 4.a 0.75.8 0.50... wt oJt

;-Gl-9 Y3r i1ieel-I-V a9zs 129 INSTANTANEOUS VALUE OF GRID CURRENT AS A FUNCTION OF wt FOR Ebqs 800V, Ec2= 400V, Ec - 3.3 V. PEAK SIGNAL VOLTAGE a 73.3 VOLTS. FIG. F. 19 60 50 __ 40 wt, DEGREES,.~~~~~~~~t DEGREES,

el-O W le W-19-v @ zZ-VZ 130 Lc, cos t VS. wt FOR -,; — - Ebq=800V, EC2 400V, EC, -53.3 PEAK SIGNAL VOLTAGE z 73.3 VOLTS FIG. F. 20 60 40'' 20 - -5.I 16.6 MA O10 0 IS 30 45 60 5 Wt, DEGREES

151 TABLE F.5 RESULTS AND COMPUTATIONS Eb = 800v, E~2 = 400v, E = -66.6v, Esig = 86.6 v peak Ib = 0.44 amps. I1 = 0.701 amps. 12 = 0.36 amps. IC= 8.3ma. Icl = 15.4 ma. Rgeq = 5.6 K g 4odr =.00445 nepers/section aodr = o.0447 ao Ebb = 655 volts Power Output = 430 watts Power Input = 2685 watts 1 = 16.0% Driving Power = 75 watts Frequency Tolerance = 0.46 Power Out 2nd Harmonic = 118.1 watts

132 gg-I1*- W3P 161-1g-V z9zz.46 r ~' \ FIG. F. 21 1~~ [ T iLb AND Lb COS Ct VS. wt FOR.2 L \ \ |Ebq 80V,' Ec2 400V Ecl- 66.6V I E,'86.6V PEAK 1.0t Z 1' S II-Lb -Lb COS cot 0. 8 I = 0.701 AMPS. 0.6 oe -... mm~......,, 0.44AMPS 0.4 0.2_ oo 30~30 60 90 120 1500 t

13355 G-I-s W3rP Zl61-19-V Z9gZZ 1.50 1.25 FIG. F.22 Lb COS 2Zt VS. Wt FOR Ebq~8OOV, Ec2. 400V, E c -666V 1.00 Elg~ s 86.6V PEAK 0.75 08S 0 0.50 ______ I0.36 AMPS. ~25 -0.25 00 30~ 090 120 150i 60o wt

gG-I1-9 w3r 21 -I 9-9 ZSZZ 134 70 INSTANTANEOUS VALUE OF GRID CURRENT AS A FUNCTION OF Wt FOR Ebq =800V, EC2= 40OV, Ec,=-66.6VOLTS PEAK SIGNAL VOLTAGE = 86.6VOLTS. 60,___ ___ FIG. F.23 60 50 40.._ 3 0 0 - -. \ - c = 8.3 MA 0 15 30 45 60 Ct, DEGREES

gg-I-9 n3r tl-I19o-V z9z 135 FIG. F.24 ic _COS ot VS. Ut FOR Ebq 80OV, E2 z 400V, Ec, 66.6 V PEAK SIGNAL VOLTAGE 866.6 VOLTS 50.. 40 3 0. I \... ~ 20 _ - -T. -C i15.4 MA 1O 0 15 30 45' 60 75 Wt, DEGREES

156 TABLE F.6 RESULTS AND COMPUTATIONS Ebq = 800v, EC2 = 400v, El= -80v, E = 100 v peak Ib = 0.41 amps. I1 = 0.68 amps. I2 = 0.375 amps. Ic = 7.7 na. Icl = 14.6 ma. Rgeq = 6.85 K s'Oodr = 0.00365 nepers/section todr = 0.0367 Ebb = 650 volts Power Output = 407 watts Power Output = 2400 watts = 17% Driving Power = 100 watts Frequency Tolerance = 0.42 Power Out 2nd Harmonic = 128 watts

9; -6 - W3Pr 9SI-I9-V Z9ZZ 137 1.6 FIG. F.25 Lb VS..tt FOR Ebq =80OV, EC2 400V, Ecis -80V 1.2 I I I I EiS XIOOV PEAK 1.0 LI 0.4 C - - IDC= 0.41AMPS 0.2 0 30o 600 90 1200.1500 wt

g;-6-g V3r 9H1-tt- 9Z;z 138 1.4 FIG. F. 26 Lb COS Wt VS. (At FOR "1.2 _ _ | l l | EbqS Vi ECa 400V, ECL, -80V Esig 100 V PEAK LO -X - - I~ zI A 10.68 AMPS 0.6 C... 0.2 IIII ~~ 300 600 900 1200 1500 wt

g;-S-g rn3r z~-1o- zaez 139 1.50 1.25. FIG. F.27 Lb COS 2ct VS wt. FOR _ _ _ I_ \ I I IEbq = 0OOV, Eca 400V, E, -80V Esi9 IOOV PEAK 3 o.7 Od a.. -I= 0.375 AMPS 0.25 0 -0.25 - 00 300 600 90 120 1500 oit

g-G1-9 lt3r ts1-19-V z9ZZ 140 70. INSTANTANEOUS VALUE OF GRID CURRENT AS A FUNCTION OF cWt FOR EbqBO80OV, E2.=400V, Eci:-80V PEAK SIGNAL VOLTAGE 10 IOOVOLTS. ~60,_ FIG. F. 2.8, 50 40:' 30....... 20 10,, I, -Ic 7.7 MA 0 15 30 45 60 75wt, DEGREES

9gG-ti9 3r'661-19-IV 9ZZ 141 70 Lb COS Ct VS. ct FOR Ebq 800V, Ec2=400V, Ec,-80.OV PEAK SIGNAL VOLTAGE - 100.0 V FIG. F.29 60. 40.. 30. 20 ___ lc_= 14.6 MA 10 _____ __ _ 0 - 0 15 30 45 60 75 Wt; DEGREES

BIBLIOGRAPHY 1. Ginzton, Hewlett, Jasberg, Noe, "Distributed Amplification", Proc. IRE, Vol. 36, pp. 956-969, August 1948. 2. Horton, Jasberg, Noe, "Distributed Amplifiers: Practical Considerations and Experimental Results", Proc. IRE, Vol. 38, pp. 748-753, July 1950. 3. D. 0. Pederson, "The Analysis and Synthesis of Distributed Amplifiers with Ladder Networks", Technical Report No. 34 (N6onr251, Task 7) Stanford University (Electronics Research Laboratory), May 15, 1951. 4. A. D. Moore, "Synthesis of Distributed Amplifiers for Prescribed Amplitude Response", Technical Report No. 53 (N6onr Task 7), Stanford University (ERL) September 1, 1952. 5. D. 0. Pederson, "The Distributed Pair", Technical Report No. 70 (N6onr251, Task 7) Stanford University (ERL) October 5, 1953. 6. H. Spett, "Design Notes on Distributed Amplifiers", Technical Memorandum No. M-1480, Signal Corps Engineering Laboratories, March 18, 1953. 7. H. B. Demuth, "An Investigation of the Iterative Synthesis of Distributed Amplifiers", Technical Report No. 77 (N6onr251, Task 7) Stanford University (ERL) August 5, 1954. 8. L. Kings and W. M Furlow, Jr., "A Theoretical and Experimental Study of the Use of Triodes in a Distributed Amplifier Circuit", Technical Report issued by Melpar, Inc., Alexandria, Va. 9. D. V. Payne, "Distributed Amplifier Theory", Proc. IRE. Vol. 41, June 1953, PP. 759-762. 10. LePage and Seely, "General Network Analysis", McGraw-Hill 1952. 11. Johnson, "Transmission Lines and Networks", McGraw-Hill 1950. 12. M. J. O. Strutt and A. van der Ziel, "The Causes for the Admittances of Modern High-Frequency Amplifier Tubes for Short Waves", Proc. IRE, Vol. 26, pp. 1011-1032, August 1938. 13. Sarbacher and Edson, "Hyper and Ultra-High Frequency Engineering", pp 431436, Wiley, New York, 1943. 14. K. R. Sturley, "Radio Receiver Design", Part I, pp. 53, Wiley, New York, 1943. 15. K. R. Spangenberg, "Vacuum Tubes", pp. 493, McGraw-Hill, New York, 1948. 16. E. A. g2uillemin, "Communication Networks", Vol. II, pp. 447, Wiley, New York, 1935. 142

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