QEX 2001 MayJun

Upload
nestorescala 
Category
Documents

view
224 
download
1
Embed Size (px)
Transcript of QEX 2001 MayJun
7/28/2019 QEX 2001 MayJun
http://slidepdf.com/reader/full/qex2001mayjun 1/13
32 May/June 2001
A popular author wants to set the record straight about conjugate matching.
By Walter Maxwell, W2DUARRL Technical Advisor
243 Cranor AveDeLand, FL [email protected]
On the Nature of the Source of Power in ClassB
and C RF A mplifiers
this article, we will discussthe nature of the source of power in classB and C RF
amplifiers. As were my 197 0s QST articles, “Another Look At Reflections,”and my book Reflect ions—Transmis
sion Lines and Antennas , that clarifiedmisconceptions concerning SWR andreflected power, this article is concerned with clarifying misconceptionsprevalent among amateurs and professional electrical engineers concerning the operation of RF power amplifiers.In attempting to resolve the unfortunate and protracted controversyconcerning the conjugate matching theorem in relation to these amplifiers,
discussions with many people revealedan alarming number of misconceptionsconcerning the complex relationshipsof voltages and currents that occur inthe development of the source of powerin these amplifiers, especially in relation to the coupling to their loads. At thecore of the controversy are amateursand engineers alike, who assert thatsome of the teachings in “Reflections”are fundamentally incorrect. Therefore, it is important that the focus of this article is to highlight and clarifythose misconceptions.
Before discussing amplifier operation, however, two synonymous termsthat play a vital role in amplifier operation need clarification, becausethey are widely misinterpreted in discussions relating to the source of power delivered to a load. These terms
are maximum available power and allavailable power.
Maximum available power, or allavailable power, is simply the poweravailable for delivery from the sourceto the load whenever the source ismatched to the load. In classB and Camplifiers it is the power deliveredwhen the loading is adjusted for peakoutput at any given level of drive de
sired . It is not the absolute maximumpower that can be obtained by overdriving, or using excess plate voltageor plate current, as many amateursand engineers alike have been misledto believe.
Turning now to the discussion of amplifier operation, one misconception is that classC amplifiers cannotsupport circuit analysis using generalnetwork theorems because of the
In
7/28/2019 QEX 2001 MayJun
http://slidepdf.com/reader/full/qex2001mayjun 2/13
May/June 2001 33
nonlinearity 1 of the amplifier operation. In clarifying this misconceptionwe will show that, although the inputcircuit of the π network tank circuit inclassC amplifiers is nonlinear, theoutput circuit to the load is indeed linear, due to energy storage in the tank.Consequently, the linear relationshipbetween voltage and current appearing at the output of the tank circuitdoes indeed support the application of theorems that require circuits to belinear for their application to be valid.
Another misconcept ion concerns therelationship between the output andload resistances of these amplifiers.Because of wild speculations withoutverification by valid measurements,many people believe incorrectly thatthe output resistance is much greaterthan the load resistance, and thus proclaim that a conjugate match cannot beobtained between the amplifier and theload. However, when a linear source of power is delivering all of its availablepower to the load, there is a conjugatematch by axiomatic definition, as explained in the following paragraph. Anexample from Terman is used in clarifying the misconception concerning therelationship between output and loadresistances, shown below in “Analysisof the ClassC Amplifier.” In addition,data obtained from my own measurements, shown below in “ Measuring theOutput Resistance of the RF Power
Ampl ifier ,” prove that after the amplifier has been adjusted to deliver all of its available power at any given drivelevel, the output and load impedancesof the amplifier are equal and thus areconjugates of each other. 2 My measurements have been confirmed by TomRauch, W8JI, using an identical measurement procedure. Tom is an RFpoweramplifier engineer with
Ameritron.Now to explain two axioms of the con
jugate matching theorem that are commonly overlooked, which has resultedin widespread confusion concerning itsuse. We know that when a load impedance differs from its source impedance,a matching device is required to allowdelivery of all the available power fromthe source to the load. In this condition,we say the load is matched to thesource. The term “matched” has beenused universally for many decades, andin those earlier days, the term was usedalone. However, when all the poweravailable from the source is deliveredto the load, the matching occurs because the source and load impedances
are conjugates of each other. Consequently, during the last 5 0 years, theterm conjugate match gradually cameinto use synonymously with “match” todescribe the term more accurately. Inother words, “match” (used in this context) and “conjugate match” are oftenused interchangeably with no difference in meaning. Unfortunately, misinterpretation and misunderstanding of conjugate in the newer term has created confusion for many people when aroutine impedance match is referred toas a conjugate match. To clarify theconfusion, the following two axioms,which follow from the maximum powertransfer theorem, accurately define aconjugate match:
Axiom 1: There is a conjugate matchwhenever all of the available powerfrom a source or network is being delivered to the load.
Axiom 2: There is a conjugate match if the delivery of power decreaseswhenever the impedance of eitherthe source or load is changed in either direction.We now return to clarify the miscon
ception concerning output and load resistances. The term source resistance,
R S , of an RF power amplifier, as is oftenmisused (and confused with R P ) inreferring to the source of RF power delivered by classB and C amplifiers,reveals still another prevalent misconception. This misconception is that the
entire source of power in these classes
of vacuumtube amplifiers is a dissipative resistance. In clarifying this misconception, we will use the example byTerman to demonstrate that the sourceof RF output power in a classC amplifier is the combination of two resistances; a nondissipative resistance(related to the characteristics of the ef
fective load line) and a dissipative plateresistance R PD . R PD is not plate resistance R P , as determined from the wellknown expression R P = ∆ E P / ∆ I P . Fromthis expression it is evident that R P isthe result of a small change in plate cur
rent due only to a change in plate voltage, which is not the source of power inRF power amplifiers as is claimed bymany who have misinterpreted the expression. The source of power is actually derived by a large change in platecurrent resulting from a change in gridvoltage . This phenomenon will be discussed in more detail later.
One portion of the nondissipative resistance is the reciprocal of the totalconductance from both plate and powersupply to the input of the π networktank circuit. At that point in the typical
amateur,π
net classB and C ampli
fier, the load is the tank input. Thesource is the combination of two parallel conductive paths to the tank: (1) theblocking capacitor in series with theactive device, the tube(s) 3 and (2) thesame blocking capacitor in series withthe RF choke and the voltage of thepower supply. These two conductancepaths are paralleled at the input of thetank, operating at different, but overlapping times throughout the cycle. Theother portion of the nondissipative resistance is related to the operating loadline, which will be discussed below.
Plate resistance R PD is dissipative,whose value is determined by the power
P D dissipated as heat by the plate divided by the square of the average dcplate current I dc , the current measuredby the dc plate ammeter. Notice inTerman’s statement #3 below, that dissipated power P D is the product of theinstantaneous platetocathode voltageand the instantaneous plate current.We know that energy is transferredfrom the plate circuit of the amplifier tothe π network by periodic pulses of plate current that flow during the conducting portion of the RF cycle. Knowledge of the nondissipative portion of thesource resistance will allow you to understand why classB and C amplifierscan deliver all of their available powerinto a conjugately matched load withefficiencies greater than 5 0%. This concept is important, because the ability of these amplifiers to be conjugatelymatched has been incorrectly disputeddue to three erroneous assumptionsthat have caused many amateurs andengineers to be misled.
1Notes appear on page 44 .
Erroneous AssumptionsThe principal reason that many
people have been misled is that theyhave incorrectly estimated the amountof the source resistance in the amplifierthat is dissipative. This incorrect assumption led them to believe that half the power is dissipated in the sourceresistance, and thus, as in the classical
generator, a conjugate match wouldlimit the efficiency to 5 0%. However,this is not true, because, as noted above,the source of the power delivered to theπ network tank circuit is nondissipative, except for the dissipative plateresistance R PD. Because dissipativeplate resistance R PD is generally lessthan the load resistance R L , morepower is delivered to the load resistancethan that dissipated in the dissipativeplate resistance, thus allowing efficiencies greater than 5 0%. The lower dissipative plate resistance occurs because
plate current is allowed to flow only
7/28/2019 QEX 2001 MayJun
http://slidepdf.com/reader/full/qex2001mayjun 3/13
34 May/June 2001
when the plate voltage is at the minimum of its sinusoidalswing, as explained in Terman’s statement #7 below. Repeating Terman’s statement #3 for emphasis, dissipated
power P D is the product of instantaneous platetocathodevoltage and instantaneous plate current . (Keep in mind thatplate current is zero except during the short conductiontime, considerably less than 18 0 ° .)
The second erroneous assumption is that the conjugatematching and maximum powertransfer theorems don’t apply to classB and C RF amplifiers, because the operation of these amplifiers is nonlinear (see Note 1 ). This assumptionis also incorrect because they have failed to appreciate theisolating action of the vitally important π network tankcircuit. The π network tank is not simply an impedancetransformer, as many believe, but is also an energystoragedevice. The energystorage capacity of the tank isolates thepulsed nonlinear mode at the input from the smoothed linear mode at the output that delivers the nearly perfect sinewaves. This widely overlooked and misunderstood conceptwill be discussed in depth later.
A third erroneous assumption concerns the misuse of therole source resistance R S plays in the delivery of power to thetank circuit. Because some say the value of R S is as much asfive times greater than load resistance R
L(a condition that
violates the conjugate matching theorem), some people assert that no conjugate match is possible in systems wherethe source is an RF power amplifier. However, to obtain R Sthey erroneously used the expression R P = ∆ E P / ∆ I P, where
R P is greater than R L . The reason the expression was usederroneously is that—in this expression— I P varies only withvariation of plate voltage, not grid voltage, as explainedearlier. Because the change in plate current due to a changein plate voltage is small compared to the change in platecurrent due to a change in grid voltage, R P and the erroneous‘ R S’ are much greater than R L. . The crucial point here is thatthe source of power is derived from the much larger changein plate current due to the change in grid voltage, while theeffect of the change in plate current due to the change in platevoltage is insignificant in relation to the output impedanceof the amplifier. Consequently, as we proceed we will learnthat both R P and ‘ R S’, as perceived by some, are totally irrelevant to conjugate matching the output impedance of theamplifier to the impedance of its load, and thus impose noimpediment to the conjugate match.
Analysis of the ClassC AmplifierThe following discussion of the classC amplifier, which
reveals why the portion of the source resistance related tothe characteristics of the load line is nondissipative, isbased on statements appearing in Terman’s Radio Engineers Handbook , 1943 edition, p 445, and on Terman’sexample of classC amplifier design data appearing onp 449. Because the arguments presented in Terman’sstatements are vital to understanding the concept underdiscussion, I quote them here for convenience (parentheses and emphasis mine):1. The average of the pulses of current flowing to an elec
trode represents the direct current drawn by that electrode.
2. The power input to the plate electrode of the tube at anyinstant is the product of platesupply voltage and instantaneous plate current.
3.The corresponding power ( P D) lost at the plate is theproduct of instantaneous platecathode voltage and instantaneous plate current.
4. The difference between the two quantities obtained fromitems 2 and 3 represents the useful output, at the moment.
5. The average input, output and loss are obtained by averaging the instantaneous powers.
6. The efficiency is the ratio of average output to averageinput and is commonly of the order of 6080%.
7. The efficiency is high in a classC amplifier becausecurrent is permitted to flow only when most of the plate
supply voltage is used as voltage drop across the tunedload circuit R L , and only a smal l frac tion is wasted asvoltage drop (across R PD ) at the plate elect rode of the tube.Based on these statements, the discussion and the data
in Terman’s example that follow explain why the amplifiercan deliver power with efficiencies greater than 5 0% whileconjugately matched to its load, a condition that is widelydisputed because of the incorrect assumptions concerning classB and C amplifier operation, as noted above. Theterminology and data in the example are Terman’s,but I have added one calculation to Terman’s data to emphasize a parameter that is vital to understanding how aconjugate match can exist when the efficiency is greaterthan 5 0%. That parameter is dissipative plate resistance
R PD . (As stated earlier, dissipative resistance R PD shouldnot be confused with plate resistance R P of amplifiers operating in class A, derived from the expression R P = ∆ E P / ∆ I P .)
It is evident from Terman that the power supplied to theamplifier by the dc power supply goes to only two places, theRF power delivered to load resistance R L at the input of theπ network, and the power dissipated as heat in dissipativeplate resistance R PD. ( Again, this is not plate resistance R P ,which is totally irrelevant to obtaining a conjugate match atthe output of classB and C amplifiers.) In other words, theoutput power equals the dcinput power minus the powerdissipated in resistance R PD . We will now show why thistwoway division of power occurs. First, we calculate thevalue of R PD from Terman’s data, as seen in Eq 9 of the
example below. It is evident that when the dcinput powerminus the power dissipated in R PD equals the power delivered to resistance R L at the input of the π network, therecan be no significant dissipative resistance in the amplifierother than R PD . The antenna effect from the tank circuit isso insignificant that dissipation due to radiation can be disregarded. If there were any significant dissipative resistance in addition to R PD , the power delivered to the load plusthe power dissipated in R PD would be less than the dcinputpower, due to the power that would be dissipated in theadditional resistance. This is an impossibility, confirmed bythe data in Terman’s example, which is in accordance withthe law of conservation of energy. Therefore, we shall observe that the example confirms the total power taken from
the power supply goes only to (1) the RF power delivered tothe load R L and (2) to the power dissipated as heat in R PD ,thus proving there is no significant dissipative resistance inthe classC amplifier other than R PD .
Data from Terman’s example on p 449 of Radio Engineers Handbook :
E dc Source Voltageb = = 1000 V (Eq 1)
E E Emin b L 1000 850 150 V = − = − = (Eq 2)See Terman, Figs 76A and 76B.
I dc Plate Currentdc 75.1 mA 0.0751 A = = = (Eq 3)
E E E Peak Fundamental ac Plate VoltageL b min 1000 150 850 V = − = − = =
(Eq 4) I Peak Fundamental ac Plate Current1 132.7 mA 0.1327 A = = =
(Eq 5)
7/28/2019 QEX 2001 MayJun
http://slidepdf.com/reader/full/qex2001mayjun 4/13
May/June 2001 35
P E I dc Input PowerIN b dc 1000 0.0751 75.1 W= × = = × = (Eq 6)
P E E I E I
Output Power Delivered to ROUTb min 1 L 1
L2 2
1000 150 0.1327
256.4 W
=−( )
= =
=−( )
=
(Eq 7)
P P P
Power Dissipated in Dissipative Plate Resistance RD IN OUT
PD 18.7 W
= − =
=
(Eq 8)
R
P
I Dissipative Plate Resistance RPD
D
dc2 2 PD
18.7 W0.0751 A
3315.6= = = Ω
(Eq 9)
R
E E I
E I
Load ResistanceLb min
1
L
1
8500.1327
6405=−
= = = = Ω (Eq 10)
(6400 Ω in Terman)
Plate Efficiency P
P= × = × =OUT
IN
10056.4
100
75.175.1% (Eq 11)
Notice that in Eq 1 0, R L is determined simply by the ratioof the fundamental RF ac voltage E L divided by the fundamental RF ac current I
1, and therefore does not involve dis
sipation of any power. Thus R L is a nondissipative resistance.Referring to the data in the example, observe again from
Eq 1 0 that load resistance R L at the input of the π networktank circuit is determined by the ratio E L / I 1. This is theTerman equation which, prior to the moreprecise ChaffeeFourier analysis, was used universally to determine the approximate value of the optimum load resistance R L . (Whenthe Chaffee analysis is used to determine R L from a selectedload line the value of plate current I 1 is more precise thanthat obtained when using Terman’s equation, consequentlyrequiring fewer empirical adjustments of the amplifier’sparameters to obtain the optimum value of R L.) Load resistance R L is proportional to the slope of the operating loadline that allows all of the available integrated energy contained in the platecurrent pulses to be transferred into theπ network tank circuit. (For additional information concerning the load line, see below.) Therefore, the π network mustbe designed to provide the equivalent optimum resistance
R L looking into the input for whatever load terminates theoutput. The current pulses flowing into the network deliverbursts of electrical energy to the network periodically, in thesame manner as the springloaded escapement mechanismin the pendulum clock delivers mechanical energy periodically to the swing of the pendulum. In a similar manner, aftereach platecurrent pulse enters the π network tank circuit,the flywheel effect of the resonant tank circuit stores the electromagnetic energy delivered by the current pulse, and thusmaintains a continuous sinusoidal flow of current throughout the tank, in the same manner as the pendulum swingscontinuously and periodically after each thrust from theescapement mechanism. The continuous swing of the pendulum results from the inertia of the weight at the end of thependulum, due to the energy stored in the weight. The pathinscribed by the motion of the pendulum is a sine wave, thesame as at the output of the amplifier. We will continue thediscussion of the flywheel effect in the tank circuit with amore indepth examination later.
Let us now consider the dissipative plate resistance R PD ,which provides the evidence that the dc input power to theclassC amplifier goes only to the load R L and to dissipationas heat in R PD . With this evidence, we will show how a con
jugate match can exis t at the output of the π network with
efficiencies greater than 5 0%. In accordance with the conjugate matching theorem and the maximum powertransfertheorem, it is well understood that a conjugate match existswhenever all available power from a linear source is being delivered to the load. Further, by definition, R L is the loadresistance at the tank input determined by the characteristics of the load line that permits delivery of all the availablepower from the source into the tank. This is why R L is calledthe optimum load resistance. Thus, from the data inTerman’s example, which shows that after accounting for thepower dissipated in R PD , all the power remaining is theavailable power, which is delivered to R L and thence to theload at the output of the π network. Therefore, because allavailable power is being delivered to the load, we have a con
jugate match by definit ion. In a following section we willshow how efficiencies greater than 5 0% are achieved in classC amplifiers operating into the conjugate match.
Examining the Operating Load LineThe details of the somewhat trialanderror method of
establishing the operating load line are beyond the scope of this article. However, once established, the load line represents the nondissipative load resistance R L appearing at theinput of the π network tank circuit. The slope of the load lineis proportional to the ratio of the continuous fundamentalRF voltage and current. When the network is terminatedwith the correct output load resistance (a resistance equal tothe network output resistance as explained below), the network transforms the output load resistance up to resistance
R L at the network input. Once established (and proven bymeasurements of network output impedance), the slope of the operating load remains constant with changes in outputpower resulting from changes in drive levels. Consequently,because R L represents the slope of the load line, both thefundamental RF voltagecurrent ratio appearing along theload line and the network output impedance remain constantwhatever the power level of the integrated current pulsesenter the network. It should be clearly understood that,because the operating load line, and the optimum resistance
R L it represents, are established solely by the ratio of the RFvoltage and current, the load line and R L are nondissipative.
As explained earl ier, the entire diss ipation to heat occursonly in the dissipative plate resistance R PD .
When using the Terman equation to determine load resistance R L , an approximate load line and average platecurrent are first estimated from the tube characteristiccurves. The corresponding value of R L is used as a trialvalue and the output power and efficiency are determinedin a trial run. However, several trial runs with differentload adjustments are necessary to converge toward theoptimum value of R L that will yield the desired conditionsfor operation, simply because the first estimation of average plate current is rarely the optimum value.
When the Chaffee analysis is used to determine R L inestablishing the load line, the average value of plate current I 1 during the conduction period is obtained by firstplotting the load line on a graph of constant plate currentcharacteristics of the tube. The load line is then marked off in several increments corresponding to successive anglesof conduction of plate current. The plate current at eachconduction angle is then found at the intersection of theload line and the constantcurrent curve. The plate voltageat each conduction angle is also found on the plate voltageline directly below the above stated intersection. Theaverages of plate current and voltage are then determinedusing the trapezoidal rule. Load resistance R
Lis then de
7/28/2019 QEX 2001 MayJun
http://slidepdf.com/reader/full/qex2001mayjun 5/13
36 May/June 2001
termined by dividing the average fundamental RF plate voltage by the average fundamental RF plate current,the Terman equation. Thus, theChaffee method saves time comparedto using Terman’s equation alone, because the initial value of average platecurrent is closer to the optimum valuethan that estimated for use in theTerman equation.
Calculation of EfficiencyGreater than 50%
To show how efficiencies greater than50% are obtained while the amplifier isconjugately matched, we will dissectthe data in the Terman example to discover that load resistance R L is greaterthan dissipative plate resistance R PD ,thus allowing more power to be delivered to the load than that dissipated in
R PD . Referring again to Terman’s example in ( Eq 1 0), his calculation of loadresistance R
Lis 64 00 Ω . From ( Eq 9 ) we
find R PD is 3315.6 Ω by dividing 18.7 Wdissipated in R PD by the square of 75.1 mA dc plate current I dc flowing through R PD. Correspondingly, ( Eq 7 )shows the power delivered to R L is 56.4W, and from ( Eq 8 ), power P D dissipatedin R PD is 18.7 W. With 56.4 W deliveredto R L and 18.7 W dissipated in R PD wehave accounted for the total inputpower, 71.5 W, shown in Eq 6 . The relative power distribution is 75.1% delivered to R L , and 24.9% dissipated in
R PD . Earlier we showed that after accounting for the power dissipated in
R PD , all the remaining available poweris delivered to the load R L . Thus, thisdistribution of power clearly demonstrates why a classC amplifier can deliver more than 5 0% of its input powerto the load, because its load resistance
R L (64 00 Ω ) is greater than its dissipative plate resistance R PD (3315.6 Ω ).These calculations are in accord withwith Terman’s statement #7 that “efficiency is high in the classC amplifier,because current is permitted to flowonly when most of the platesupply voltage is used as voltage drop across thetuned load circuit R L , and only a smallfraction is wasted across R PD at theplate electrode of the tube.” None isdissipated in the nondissipative resistance related to the characteristics of the load line. As stated earlier, thenondissipative portion of the sourceresistance is the reciprocal of the totalconductance from both the plate of thetube and the power supply to the inputof the π network tank circuit. It shouldbe noticed however, that we are considering only the power delivered to thetank; we are not concerned here with
inherent loss in the tank that results insome decrease in the power delivered atthe output of the tank.
Evidence of Conjugate MatchThe example has proven that a con
jugate match exists , because all theavailable power has been delivered tothe resistive load R L , and thence to theload terminating the π network, in accordance with conjugate matching axioms 1 and 2 recited above. The example has also shown that more powerhas been delivered to the load thanwas dissipated, because 54.6 W weredelivered and only 18.7 W were dissipated. Thus, contrary to the opinion of many who fail to understand this concept, we have shown that conjugatematching to a classC RF amplifierdoes not limit its efficiency to 50%. Thesame reasoning applies to amplifiersoperating in class B (see Note 2 ).
So now you ask, “Do we have a conjugate match during SSB operation?”The answer is yes, but it begs anadditional question: Does the outputimpedance of the amplifier remain constant with SSB modulation, or does itchange during the variations of driveand output power corresponding to thevoice modulation? My measurements,described below, show that the outputimpedance does not change significantly with voice modulation. This isbecause, for a given load resistance
R LOAD , the operating load line relatedto the load resistance R
Lappearing at
the input of the tank circuit, and theoutput resistance R OUT , are established during the tuning and loading procedure when the loading is adjustedto deliver maximum available power.During this procedure, maximumavailable power is that power deliveredto the load with the drive level set toobtain the desired output power at thefull modulation level. After the loadline has been established in this manner, it remains constant for all valuesof drive. I have made extensive measurements, which show that once theoperating load line is established during this routine procedure, it remainsconstant during swings of grid voltageduring SSB modulation, as long as theplate supply voltage remains constant.
So now we ask, “Is the conjugatematch of such importance that weshould be concerned about it?” Yes it is,if we are to understand why antennatuners perform their intended task of matching the complex impedance appearing at the input of a transmissionline that is Z 0 mismatched to an antenna, while also establishing a conju
gate match that overrides the Z 0 mismatch at the antenna. The principles of conjugate matching are fundamental tothe matching function performed by theantenna tuner, and are indeed fundamental to all impedance matching obtained with any impedance matching device that allows delivery of all available power from its source!
The Vital Role of Energy Storagein the Tank Circuit: Providing Linear Operation at the Output
We now return to conduct a close examination of the vitally important flywheel effect of the tank circuit. Theenergy storage ( Q) in the tank producesthe flywheel effect that isolates thenonlinear pulsed energy entering thetank at the input from the smoothedenergy delivered at the output. Becauseof this isolation, the energy delivered atthe output is a smooth sine wave, withlinear voltage/current characteristicsthat support the theorems generallyrestricted to linear operation. We knowthat the widely varying voltage/currentrelationship at the tank input resultsin widely varying impedances, whichprecludes the possibility of a conjugatematch at the input of the tank circuit.However, the energy stored in the tankprovides constant impedance at theoutput that supports both the conjugatematching and the maximum powertransfer theorems.
The acceptance by many engineersand amateurs of the notion that theoutput of the RF tank is nonlinear is areason some readers will have difficultyin appreciating that the output of theRF tank circuit is linear and can thussupport the conjugate match. Validanalogies between different disciplinesare often helpful in clarifying difficulties in appreciating certain aspects of aparticular discipline. Fortunately,energy storage in the mechanical discipline has a valid and rigorous analogous relationship with energy storagein LC circuitry, which makes it appropriate to draw upon a mechanical example to clarify the effect of energystorage in the RF tank circuit. (A further convincing analogy involving water appears later, in which the origin of the term tank circuit is revealed.)
The smoothing action of the RF energy stored in the tank circuit is rigorously analogous to the smoothing action of the energy stored in the flywheelin the automobile engine. In the automobile engine, the flywheel smoothesthe pulses of energy delivered to thecrankshaft by the thrust of the pistons.
As in the tank circuit of the amplifier,
7/28/2019 QEX 2001 MayJun
http://slidepdf.com/reader/full/qex2001mayjun 6/13
May/June 2001 37
the automobile flywheel is an energystorage device, and the smoothing of the energy pulses from the pistons isachieved by the energy stored in theflywheel. In effect, the flywheel delivers the energy to the transmission. Theenergy storage capacity required of theflywheel to deliver smooth energy to thetransmission is determined by thenumber of piston pulses per revolutionof the crankshaft. With more pistons,less storage capacity is required toachieve a specified level of smoothnessin the energy delivered by the flywheel.The storage capacity of the flywheel isdetermined by its moment of inertia,and the storage capacity of the tank circuit in the RF amplifier is determinedby its Q.
As stated earlier , the tank circuit inthe RF amplifier receives two overlapping pulses of energy per cycle. If theeffect of the overlapping pulses wereconsidered a single pulse, we wouldhave a condition that is somewhatanalogous to an engine having only onecylinder. If we were to assume that thepiston in the onecylinder engine delivers one thrust of energy per revolution,it is evident that a large amount of energy storage is required to enable thecrankshaft to deliver a smooth outputduring the entire rotation of the crankshaft. In this case, a very heavy flywheel is required to deliver a smoothoutput. This is also the case with the RFtank circuit, which requires a Q of 10 to12 to yield a smooth sinewave outputwith an acceptable minimum of harmonic ripple. Because the tank receivesonly one pulse of energy per cycle, itmust store many times the amount of energy it passes through, to provide acontinuous sinewave output when supplied only with pulses of energy at theinput. Thus the energystorage capacity of the tank provides for thesmoothed linear output to the load circuit, despite the nonlinear pulsed input, which, for the purpose of analysis,allows the pulsed source and tank to bereplaced with an equivalent Théveningenerator whose output impedanceequals R LOAD . Although no conjugatematch exists at the input of theπ network tank (because of the largevariation of impedance in the currentpulses) the isolation derived from theflywheel action of the tank thus allowsa conjugate match to exist between theoutput of the tank and its load, a concept which will become clear as we continue.
To further clarify the action and theeffect of the energy storage in the tankthat achieves a linear voltage/current
relationship at the output of the tank,parts of the following discussion areparaphrased from correspondencewith Dr. John Fakan, KB8MU. Itshould be emphasized here that a con
jugate match can exist between theoutput of the RF power amplifier andits load because of the linear voltage/ current relationship at the output of the tank resulting from the energystorage in the tank.
The tank circuit stores the energy bypassing it back and forth cyclicallybetween the L and C components, andpasses only a fraction of that energy tothe tank’s load on each cycle. Becausethe amount passed to the load is sucha small fraction of the total stored inthe tank, and because even that smallamount is restored during the cycle,the tank can be considered an activesource. Because it can be considered asan active source, we have no need forinterest in what is going on ahead of itin the overall system (as far as whatthe downstream devices see).
Consider that when designing to getenergy from a power supply our concernis only with the characteristics seen atthe power supply terminals. Our designdoes not depend on whether the linefeed to the supply is singlephase orthreephase, 6 0 Hz or 4 00 Hz, or even if the power factor is unity or some othervalue. These things don’t matter onceyou know what shows up at the outputconnections of the supply. For our purposes, the actual source of energy is theconnection at the output of the supply,and the characteristics at that pointwill be determined by the componentsin the filter circuitry.
As a source of sinusoidal energy, ourRF amplifiers are no different. Thesource of this energy that will be passedon to our antenna system is the tankcircuit. The load connected to the output port of the amplifier can only see thevoltage swings and the impedance presented by the tank components. A properly designed tank (of any type) will notpass so much energy on each cycle thatthe relationship between its terminalvoltage and current is affected enoughto cause nonlinearity. Sometime during the cycle even that small amount of energy will be replaced, thus maintaining its operating levels.
Because this “new” source happens topresent a linear impedance to its load(the first connection in our antennasystem) we need have no concern aboutnonlinear processes occurring at pointsupstream of the tank circuit. Once wehave a linear active source in the cascade and we do nothing downstream to
subsequently cause nonlinearities, wecan take advantage of those theoremsand ideas that depend on the linearityof the network.
My teachings in “Reflections” depend on the linear nature of the energytransfer from the amplifier’s outputport right on through to the last antenna element. Because the energy tothis network is supplied by a linearsource (the tank circuit) everything inmy teachings can be assured of soundscientific basis. Objections by others,based on nonlinearities ahead of thetank are simply not applicable.
The energy pulses supplied to thetank must be sufficient to “refill” thetank’s energy store on each cycle. Theconnection where that energy transferoccurs is at the input to the tank. Asstated earlier, at that point in the typical amateur π network classC amplifier, the load is the tank input. Thesource is the combination of two processes: (1) the blocking capacitor in series with the active device (tube) and(2) the same blocking capacitor in series with the RF choke and the voltageof the power supply. These two sourcesare in parallel at the connection, butoperate at different, overlapping times through the cycle.
The load resistance R L appearing atthe input to the tank is determined bythe value required to accommodate theenergy transfer required per cycle tomake up for that being transferred bythe tank to its load. Because of the lackof linear or even simple squarewavecharacteristics of the active device, thedesigns in this region have always beenvery empirical. Actual experience anda good seatofthepants feel for the significance of activedevice data sheetshave been the main tools for the designof tank circuits. The amount of energydelivered via the action of the activedevice (the tank) is dependent on thingslike drive, feedback, supply voltagesand so on. They all can play a role in providing for the right amount of energytransfer to allow the tank to function asa linear active source.
If the tank does not receive enoughenergy to sustain the power level ithas established with its load, its output will decrease accordingly. Malfunctioning of the upstream energy“bucket brigade” can result in linearoperation at a lower level or in nonlinear operation, depending on how wellthe tank design can handle thechanges. The important point is thatonce conditions allow the tank to operate as a linear active source, everything else downstream of the tank is
7/28/2019 QEX 2001 MayJun
http://slidepdf.com/reader/full/qex2001mayjun 7/13
38 May/June 2001
linear and follows the conjugatematch theorem and other linearsystem theories. Changing the conditionsat the input to the π network (for example: changing drive, feedback andso on) affect the performance of thetank as a linear source of RF energy. If the tank is supplied with energypulses having a different integratedaverage energy than that being supplied to the tank’s load, the tank’soutput characteristics will change tofit the available energy. It will do thisby changing its output impedance towhatever value is required so that, atthe new conjugate matching point, thevoltagecurrent product will equal theenergy rate available. It has no otherchoice because the conjugatematching theorem requires a change in output impedance if it is to remain a linear source. Consequently, the loadimpedance at the output must changeaccordingly to retain the conjugatematch. If the changes are extreme, itmay not be able to accommodate therequired impedance change in a linearmanner, so waveshape distortioncould occur, for example, flat topping.
The important point is that thedesign and operation of the circuitryproviding energy to the tank circuit involves a number of issues having to dowith protection of the active device,stability, efficiency and such, as wellas the amount of energy transferred tothe tank during each cycle. It doesn’tmatter that the wave shape of the energy pulse is ugly and would be difficult to characterize mathematically,because the tank circuit doesn’t care.
It is positively uncanny how easy it isfor some people to simply ignore experimental evidence staring them right inthe eye when one of their pet understandings is in jeopardy. Many peopleconcede that amateur classC amplifiers typically operate at greater than50% efficiency. They will also agreethat it is common to tune for a powerpeak. They will then wiggle and squirmto avoid agreeing that the tuning process is simply matching the output andload impedances to a common conjugate. Their reason is that the internal“resistance” precludes the higher than50% efficiency. The fact that there aretwo independent definitions for theword resistance doesn’t seem to matter.They are completely ignoring definition#2 of the IEEE definitions of resistance,the nondissipative resistance, that is,the real part of the impedance.
RF power amplifiers are necessarilydesigned to match load impedances ator near the characteristic impedance
of common coaxial transmission lines.No other design value would makesense. The conjugatematch theoremis simple and absolute: When the energy being transferred across any linearly behaving connection cannot befurther increased simply by changing the impedance of either source or load,a conjugate match exists. That is commonly the operating condition foramateur classB and C RF amplifiers.From the tank circuit forward, thebehavior is linear, because the voltageand current at the output of the tankare continuous and sinusoidal due tothe energy storing (flywheel), smoothing action of the tank. There really isno wiggle room for debate.
Origin of the Term “Tank Circuit” Let me digress for a moment to say
that it is customary for an author of anarticle such as this to have his writing peer reviewed to uncover possible errors that may have escaped him. Because of the protracted and unfortunatecontroversy brought about by those whoclaim that a conjugate match cannotexist in an RF system powered by an RFpower amplifier, my engineering credibility as an author has been questioned. Therefore, because this articleis primarily concerned with presenting a convincing argument that a conjugatematch does indeed exist in RF poweramplifiers, I have attempted to makesure it contains no conceptual or substantive errors, or invalid statements.Consequently, I requested several professional RF engineers with unquestionable credentials and expertise toreview and critique it. All reviewers butone found my presentation accurate.This dissenting reviewer flatly rejectedthe concept that a π network tank circuit isolates its pulsed input from theoutput, and therefore he maintains thatthe output circuit of the π network cannot support or sustain linear operation,and no conjugate match. Unfortunately, during the nine years of the controversy I discovered that opposition tothe application of linear theorems toany aspect of RF power amplifier operation is prevalent in the confusion of many otherwise intelligent and capableengineers. It therefore occurred to methat others also might have similardifficulty in accepting the concept of energy storage in the tank circuit providing isolation between the input andoutput of the tank that allows lineartheorems to be valid at the output. Ihave already presented two examples of the storage of mechanical energy thatillustrate the smoothing function of
energy storage, which are preciselyanalogous to energy storage in the tankcircuit of the RF power amplifier. Inaddition, a valid water analogy wherethe operative word is “tank” in theliteral sense might further clarify theissue. I also believe you’ll find it interesting to learn how the term tank originated as an active description of the LCcircuit used in the output coupling of alldiscontinuous RF power amplifiers.
Legend has it that in the early days of RF amplifier development the watertank analogy was applied for the verypurpose of explaining the energystorage function of the LC output circuit. Itgoes like this. A water tank is filled toa specific depth that causes a corresponding pressure applied on the bottom. A hole is made in the bottom witha size that allows one gallon per minuteto flow with the specific applied pressure. Water is added at the top of thetank at the rate of one gallon perminute, thus maintaining the originallevel in the tank as the water flowssmoothly out from the bottom. Let’snow consider how the water is added atthe top. It can be added in spurts, butthe water flowing out from the bottomwill still flow smoothly without everknowing the nature of the spurts addedat the top. The spurts can be added at arate of one gallon dumped in everyminute, a half gallon twice during theminute, onethirtieth of a gallon thirtytimes per minute and so on, you get thepicture. As long as enough water isadded to maintain the level, thus maintaining the same constant pressure atthe bottom, the water will continue toflow smoothly from the bottom at therate of one gallon per minute, regardless of how the water is added. It is the
energy stored in the tank that isolatesthe intermittent additions of water atthe tank top from the continuous flowat the bottom. If the tank is filled to agreater depth, pressure at the bottomis increased, resulting in an increasedrate of flow of water at the bottom, indirect proportion to the increase inpressure.
It should be appreciated that the fluid impedance at the output of thetank is the ratio of the pressure to theflow rate. This sets the rate at which theenergy contained in the water flowsfrom the bottom of the water tank. Thetank output is established solely by thesize of the hole and the height of thewater above the hole. The same energyrate can exist with a tall tank anda small outlet hole (high output impedance), or shorter tanks withappropriate larger holes (low output
7/28/2019 QEX 2001 MayJun
http://slidepdf.com/reader/full/qex2001mayjun 8/13
May/June 2001 39
impedance). However, the impedanceand linearity of the input to the tank isirrelevant as long as it results in maintaining a constant water level. Thus,the action at the bottom is linear although the action at the top is not.
The same is true in the tank circuitof the RF amplifier. The impedance atthe output of the RF tank is the ratioof the voltage to current at whichpower is being delivered to the tank’sload. The voltage and current appearing at the output of the π network tankcircuit are analogous to the waterpressure at the bottom of the tank(voltage), and the rate of flow of thewater (current) out of the tank. As inthe water tank, the shape of the current pulses entering the π networktank has no effect on the smooth sinusoidal voltage and current appearing at the output. If the average integrated energy of the current pulsesentering the tank increases, the voltage and current at the output will increase in a linear relationship. Thus,it is shown that the output of a properly designed RF tank circuit is linear,and the theorems associated with linear circuits are applicable.
Measuring the Output Resistanceof the RF Power Amplifier
I have developed a test setup andprocedure based on the standard IEEEloadvariation method for measuring the source, or output resistance R OUTof networks, which are described below.Measurements made with this setupand procedure show that output resistance R OUT equals the load resistance
R LOAD when the amplifier is initiallyadjusted to deliver all of its availablepower to that load, thus proving theexistence of a conjugate match. However, before proceeding further it willbe helpful to obtain an appropriate perspective by reviewing some backgroundconcerning the issue.
There has never been a problem indetermining the correct value of loadresistance R L appearing at the input of the π network tank circuit of the RFamplifier. Resistance R L is routinelycalculated using either the Termanequation or the more precise Chaffeeanalysis to determine the slope andother characteristics of the operating load line, as mentioned earlier. Afterthe network has been adjusted to deliver its intended power into its terminating load, resistance R L appearing atthe input of the network is easily androutinely measured with appropriateimpedancemeasuring equipment withthe amplifier powered down.
However, determining the outputresistance R OUT appearing at theoutput of the π network has beendaunting. Wild speculations (sansmeasurements) concerning the outputresistance abound because of themisunderstandings and incorrect assumptions concerning the actions of the tank circuit as described above.The misconception that a conjugatematch cannot exist at the output of RFpower amplifiers has precluded logicalreasoning, that when the amplifier isdelivering all its available power thereis a conjugate match by definition.Consequently, it has been consideredunthinkable that the output sourceresistance could possibly be equal tothe load resistance.
I am not aware of any writings in theprofessional literature that discuss themeasurement of R OUT . A probable reason for this lack of discussion is thatknowledgeable people know that R
OUTmust equal the load resistance when allthe available power is being delivered,and it would therefore be redundant tostate it. Because of the controversysurrounding conjugate matching andamplifiers, it is now appropriate to describe the test setup and procedure thatdoes yield the correct value of source
resistance R OUT , the value that equalsthe load resistance. Consequently, using the standard IEEE loadvariationprocedure described below, it will beseen that the data resulting from mymeasurements (also shown below)prove two things: (1) source resistanceis not what some previous authors weremeasuring, and (2) my measurementsprove the existence of the conjugatematch at the output of RF power amplifiers. The data obtained from my measurements have been verified by TomRauch.
The test setup I developed for measuring the output resistance R OUT of RF power amplifiers is arranged to usethe loadvariation method of measurement, based on the IEEE expressionfor measuring the output resistance of networks. The IEEE expression is
R OUT = ∆ E / ∆ I , where ∆ E and ∆ I represent the corresponding changes inload voltage and load current, respectively, with a small change in loadresistance R LOAD terminating the network. In the measurements describedbelow, all values of R LOAD , ( R1 and R2 )are pure resistances, R + j0 . Inthese measurements, the output loadresistance R OUT ( R1 ) terminating theπ network is selected and the param
Table 1—Using Standard IEEE SmallLoadVariation Method to MeasureNetwork Output Source Resistance
Load Load Load Output Measured
Resistance Voltage Current Resistance Power Out ( Ω ) (V) (A) ( Ω ) (W)
51.2 75.9 1.482 52.7 112.544.6 70.6 1.583 111.6
∆ =5.3 ∆ = 0.101
51.2 76.9 1.502 51.2 115.544.6 71.6 1.605 114.9
∆ = 5.3 ∆ = 0.1034
51.2 69.75 1.36 49.4 94.946.4 66.29 1.43 94.8
∆ = 3.46 ∆ = 0.70
51.2 62.5 1.22 51.7 76.2546.4 59.4 1.28 76.0
∆ = 3.1 ∆ = 0.60
51.2 77.8 1.519 47.8 118.246.4 74.1 1.597 118.3
∆ = 3.7 ∆ = 0.078
51.2 77.5 1.514 47.4 117.347.75 74.9 1.569 117.5
∆ = 2.6 ∆ = 0.0549
Average 50.3 Ω
7/28/2019 QEX 2001 MayJun
http://slidepdf.com/reader/full/qex2001mayjun 9/13
40 May/June 2001
eters of the amplifier are then adjustedto obtain delivery of all the availablepower into that load at a given drivelevel. Then by varying the load resistance a small amount (to R2 ), around a10% change from R1 , and then measuring the difference in load voltage andcurrent, the output resistance is obtained by substituting the differentialvoltage and current values in the IEEEexpression for R OUT shown above.
The equipment used in the measurements consisted of two vacuumtubetransceivers using two parallel 6146tubes and a π network tank circuit inthe RF power amplifier. They are aHeathkit HW100 and a Kenwood TS830S. A HewlettPackard HP4815A RF Vector Impedance Meter modifiedfor digital readout, along with ESI250DA universal impedance bridge,for measuring RF and dc resistancesof noninductive load resistors R1 and
R2 . An HP8405A Vector Voltmetermodified for digital readout for measuring voltages appearing across loadresistors R1 and R2 , and an HP410BRF Voltmeter with an HP455A Coaxial Adapter, also modified for digital readout to indicate load voltage.The RF vector impedance meter wasused to confirm that the load resistorscontained zero reactance. The experiments were conducted at 4.0 MHz.
ProcedureThe π network output of the ampli
fier is initially terminated with R1 ,then tuned and loaded to deliver a specific maximum available output powerwith a given level of grid drive. The loadvoltage E 1 is measured with load R1 ,then the load is changed to R2 and loadvoltage E 2 is measured. Load currents,
I 1 and I 2, are then determined by calculation of I = E / R , using the measuredvalues of R and E . Finally, as statedabove, R OUT = ∆ E / ∆ I , as shown in thedata resulting from the measurementsshown in Table 1 .
The amplifier was tuned and loadedwith its drive level set to deliver maximum available power of ≈ 110 W. Alladjustments remain undisturbedthereafter. The data in Table 1 wereobtained using the Heathkit HW100.Result: Load resistance R LOAD whenadjusted for maximum power out R1 =51.2 Ω . Average measured source resistance R OUT = 50.3 Ω (See Fig 1 andTable 2).
The reason for the variation, or scatter in measured output resistance andoutput power seen in the data abovewas found to be a shortterm sag inoutput power between the measure
Table 2—Measured Network Output Resistance versus Output Power,
HW100 TransceiverAlso see Table 3 and Fig 1
Network Output Output Resistance Plate Voltage Plate Power (W) R
OUT ( Ω ) E
P (V) Current I
P , (mA)
100.0 48.4 800 27075.0 58.3 810 24050.0 57.3 820 19025.0 74.0 840 14012.5 80.0 860 110
0.0 NA 890 70Notice the increase in network output resistance with increase in plate voltage, due to poor
power supply regulation as plate current decreases with decrease in output power.
ment of R1 and R2 . This problem wassolved by changing the switching between R1 and R2 from manual to coaxial relay, thus reducing the timedelay, and by waiting until the sag inpower output bottomed out. However,the worstcase difference between R1load of 51.2 Ω and the R2 value thatyielded R OUT of 47.35 Ω is a mismatchof only 1.081:1, with a reflection coefficient ρ of 0.039, for a conjugate mismatch loss of only 0.0066 dB.
After many more measurementssimilar to those above, except for ad
just ing the π network for delivery of maximum available power prior toeach measurement, I found that withload R1 now at 51.0 Ω , the measuredvalues of R OUT varied randomlywithin 11 Ω on either side of the51.0 Ω load with each measurement,that is, from 40 Ω to 62 Ω . However, I
discovered the variation results fromthe very small slope of the power output curve near the peak. Using onlythe analog outputpower meter to observe the point at which the power wasmaximum, I found it impossible to findthe true peak in output power where
R OUT equals R1 of 51 Ω , because thecharacteristics of the matching curvenear its peak appear to be more like aroundtop hill than a peak. Evidently,the adjustment for maximum outputrequires a method of indicating thatprovides better resolution than thatobtained with the analog outputpower meter alone. On the other hand,the mismatch between 51 and 40 Ω ,and between 51 and 62 Ω , is only1.28:1, for a voltage reflection coefficient ρ of 0.12, which results in a con
jugate mismatch loss of only 0.066 dBat the maximum 11 Ω difference from
Fig 1—Measured network output resistance of a Heathkit HW1 00 and a Kenwood TS83 0Sversus output power at 4. 0 MHz, power level set by drive level (see Tables 2 and 3).
7/28/2019 QEX 2001 MayJun
http://slidepdf.com/reader/full/qex2001mayjun 10/13
May/June 2001 41
51 Ω . Thus, it is evident that during routine tuning and loading adjustments using analog meters to indicatepeak power output, the actual outputresistance of the network during themeasurements will seldom be exactlythe value of the load resistance, butthe consequence of the small difference is insignificant.
The next step in the procedureyielded the increase in resolution of thedata required to find the exact point onthe output curve where the output ismaximum and R OUT equals the load
R1 . After the maximum output thatcould be obtained by observing the indication on the analog outputpowermeter, R OUT was measured at thatpoint. The π network was then read
justed in very small i ncrements, measuring R OUT after each readjustment,until R OUT became equal to 51. 0 Ω . Theincrements were so small that although the difference in output powerat each increment was indiscernible onthe analog power meter, it was clearlyindicated by the digital voltmeter.Thus, it is shown that the output orsource resistance R OUT of an RF poweramplifier is equal to the resistance of the load when the maximum availablepower of the source is being deliveredto the load. It is therefore also evidentthat a conjugate match exists when theconditions just stated are present.
In addition to measuring R OUT withthe load resistance of 51. 0 Ω , R OUT wasalso measured with load resistances of 25 and 16.7 Ω . Using the same technique as described above, R OUT measured 25 Ω when the load resistancewas 25 Ω , and—consistent with themeasurement pattern already developed— R OUT measured 16 Ω when theload resistance was 16.7 Ω . These measurements indicate that, when theloading is initially adjusted to delivermaximum available power to any valueof load resistance within the matching capability of the π network, R OUTequals the load resistance.
There is more. So far, we have onlyconsidered the condition in which theamplifier is delivering its maximumavailable power in the CW mode. However, we would also like to know whathappens to output resistance R OUTduring SSB modulation after the amplifier is first tuned and loaded to delivermaximum available power with a specific drive level. The measurement dataappearing in Table 2 was obtained using the HW100 transceiver. These datashow that, except for the two caveatsstated below, once the operating loadline and resistance R OUT are estab
Table 3—Measuring Output Resistance of RF AmpIifier of TS830S Transceiver at 4.0 MHz at Various Power Levels Determined by Drive Level withAll Other Adjustments Undisturbed
R OUT
= ∆ E / ∆ I See Fig 1 and Table 2
Nominal Measured Measured Power Out R
LOADE
LOADI LOAD
R OUT
Power Out E P
I P
Power In
(W) ( Ω ) (V) (A) ( Ω ) (W) (V) (mA) (W)
100 51.2 72.381 1.414 50.2 102.3 790 265 20943.5 66.548 1.530 101.8
∆ E = 5.833 ∆ l = 0.161
75 51.2 62.738 1.225 49.0 76.9 790 240 19045.3 57.738 1.327 76.6
∆ E = 5.000 ∆ I = 0.102
50 51.2 50.238 0.981 50.2 49.3 810 195 15843.5 46.190 1.062 49.1
∆ E = 4.048 ∆ l = 0.081
25 51.2 35.357 0.691 50.5 24.4 810 165 13443.5 32.500 0.747 24.3
∆ E = 2.857 ∆ l = 0.057
20 51.2 32.857 0.64243.5 30.119 0.692 54.1 21.1 820 135 111
∆ E = 2.738 ∆ I = 0.051
10 51.2 23.453 0.458 58.6 10.7 825 110 9143.5 21.429 0.493 10.6
∆ E = 2.024 ∆ I = 0.035
lished at tuning and loading, R OUT remains substantially constant over theentire range of drive and output powerencountered during SSB modulation.
After sett ing the drive level for the π network to deliver maximum availablepower of 1 00 W, and leaving all adjustments except for the level of drive undisturbed thereafter, R OUT was measured at power levels decreasing from100 W to 12.5 W. (The changes in output power were obtained by changing the drive level.) This range of powerlevels, as shown in the Table 2 , corresponds to approximately the samerange of output power prevailing during SSB modulation.
However, the two caveats are necessary to explain the changes in R OUTthat appear to conflict with the statement above that R OUT is substantiallyconstant. First, and most important,due to imperfect regulation of platevoltage E
P, the increase in E
Pas the
plate current and output power decrease, causes a small change in theslope of the load line, resulting in anincrease in R OUT that would not occurwith perfect regulation of E P . Second,the scatter in the R OUT data resultsfrom the measurements being takenprior to the improved setup andmethod of taking later measurementsthat yields the more precise data.
As shown in Table 2 , the maximum
value of R OUT is 80 Ω , which appearsat the minimum outputpower level.The conjugate mismatch between the80 Ω output source resistance and the51 Ω load resistance is 1.569, establishing a voltagereflection coefficientρ = 0.2214, a powerreflection coefficient ρ 2 = 0.0490, and thus a transmission coefficient (1 – ρ 2) = 0.951, whichtranslates to a conjugate mismatchloss of 0.218 dB. This small amount of loss is insignificant when considering that the purpose of the experimentwas to establish proof that a conjugatematch exists throughout the range of output power during SSB modulation.
However, the picture is even moreoptimistic when using the same measurement procedure with the KenwoodTS83 0S transceiver. It can be seenfrom Table 3 and Fig 1 below that thevariation in output resistance with thistransceiver is much less than with theHW1 00 over the entire range of driveand output power. The reason for thenearly constant output resistance withthe TS83 0S is better platevoltageregulation of the power supply. Especially notice that, in the region of constant output resistance of the networkwith changes in power level from 1 02.3to approximately 25 W, the constantoutput resistance indicates a constantslope of the load line with changes indrive and power level. This point has
7/28/2019 QEX 2001 MayJun
http://slidepdf.com/reader/full/qex2001mayjun 11/13
42 May/June 2001
been in dispute in the absence of appropriate measurements. It is also important to observe that at the 1 0W outputlevel, where the output resistance hasincreased somewhat due to imperfectvoltage regulation (plate voltage increase with decrease in power), themismatch between the 51.2 Ω load resistance and the 58.6 Ω output resistance is 58.6 / 51.2 = 1.145:1. This smallmismatch results in a voltagereflection coefficient ρ = 0 .0674, a powerreflection coefficient ρ 2 = 0 .00 454 andthus a transmission coefficient (1 – ρ 2)= 0 .9955 (99.55%), which translates toan insignificant conjugate mismatchloss of only 0.0198 dB with minimumspeech level referenced to zero loss atmaximum speech level.
Except for the lowest levels of speech,which would result in output power lessthan 1 0 W during SSB modulation, thedata in Tables 2 and 3 and Fig 1 showthat the output resistance of the RFamplifier remains sufficiently constantover the normal range of speech levels,ensuring a realistic conjugate matchduring SSB modulation. Additionally,extrapolation of the data extending therange of output power below 1 0 Wclearly indicates that further increasein output resistance during the lowestpractical levels of speech transmissionis insignificant relative to load mismatch. Consequently, a realistic conjugate match exists over the entire rangeof speech levels.
Before concluding this subject, another way of explaining the action occurring in the amplifier during SSBmodulation is that as the average integrated energy of the current pulsesentering the tank changes with speechmodulation, both the RF voltage andcurrent at the output of the tankchange linearly in proportion to themodulation level. Consequently, theratio of output voltage to output current remains constant during speechmodulation. Since the output resistance R OUT is determined by the ratioof current to voltage, the output resistance also remains constant and equalto the load resistance during modulation. Thus all the available power fromthe source is delivered to its load at alllevels of speech, satisfying the condition required for a conjugate match atthe junction of the source and loadduring all levels of SSB modulation.
Justifying the IEEE Method forDetermining Network OutputResistance
Some authors have claimed that theload variation method cannot deter
mine the correct output resistancewhen the network is the output loadcircuit of an RF power amplifier. Theirreasoning is that the nonsymmetricalπ network circuit used in the amplifierdoes not behave like a perfect impedance transformer (perfect integralmathematical input/output ratio).Therefore, they claim even the smallchange in load resistance used in theIEEE method does not transform linearly through the network, and thusthe input impedance of the networkdoes not change directly with changesin load resistance. Their claim is thatthe imperfect transformer action of the network corrupts the results of themeasurements of network output resistance. However, as I will explainusing the data from my measurementsappearing in Tables 4 and 5 , and inFig 2 , it will become evident that thatis unfounded.
Recall that in the measurements described above to determine the networkoutput resistance, the change in loadresistance is small, around ± 10% fromthe matched load resistance. Thechange in load must be small to avoid asignificant change in the normal operation of the amplifier that would distortthe results if the change in load resistance were relatively large. For example, a change in load that wouldresult in a significant change in platecurrent would change the slope of theload line and the output resistance of the network. On the other hand, for
Table 4— π network Output Impedance Magnitude versus Load ResistanceKenwood TS830S at Approximately 100 W Output, 4.0 MHz
R LOAD
Load E LOAD
I LOAD
Z OUT
Sqr Root Power
( Ω ) Mismatch (V) (A) ( Ω ) R LOAD
× Z 0
Delivered (W)
53.1 71.0 1.34 76.7 94.927.7 1.92:1 46.0 1.66 46.1 76.453.1 73.0 1.37 67.6 100.437.3 1.42 59.0 1.58 50.2 93.353.1 72.0 1.36 56.1 97.643.9 1.21 65.0 1.48 49.6 96.253.1 75.0 1.41 52.7 105.949.0 1.08 72.0 1.47 50.8 105.853.1 74.0 1.39 41.35 103.161.1 1.15 78.5 1.28 50.25 100.953 1 73.0 1.37 39.0 100.463.0 1.19 78.2 1.24 49.57 97.153.1 74.0 1.39 31.95 103.166.6 1.25 80.1 1.20 46.13 96.353.1 74.0 1.39 29.1 103.173.3 1.38 82.0 1.12 46.13 91.753.1 74.5 1.40 22.5 104.585.7 1.61 84.0 0.980 43.95 82.3
Average 48.08
examination and comparison, the dataappearing in Table 4 and Fig 2 show thechange in network output resistance(and magnitude of impedance) that results from somewhat larger changes inload resistance. Notice that the magnitude of the output impedance changeslinearly, but in indirect proportion tothe load resistance. However, as seen inFig 2, the output magnitude remainsclose to the load resistance when theload resistances are close to the valueat which the network was adjusted fordelivery of all available power. So thequestions are: “Why are the changes inoutput impedance indirectly proportional to the load resistance?” and “Whydoes the measured output resistanceequal the load resistance when thechange in load resistance is small?” Theanswer is in the resulting change inplate current with change in load,which directly affects the slope of theload line and network output resistance. As we will see, when the changein load is sufficiently small, the changein plate current is also so small that theresulting change in amplifier operationis insignificant in relation to causing error in the measurement of the network output resistance. So let’s examine the pertinent changes.
With the matched load (53.1 Ω ) theplate current was 26 0 mA; 28 0 mA withthe 27.7 Ω minimum load resistance,(network output impedance Z = 76.7 Ω )and 21 0 mA with the 85.7 Ω maximumload resistance (network output imped
7/28/2019 QEX 2001 MayJun
http://slidepdf.com/reader/full/qex2001mayjun 12/13
May/June 2001 43
ance Z = 22.5 Ω ). However, when theload was changed from 53.1 to 49. 0 Ω toobtain Z OUT = 52.7 Ω (see Table 4 ), anychange in the 26 0 mA of plate currentresulting from this small change inload was too small to be discernible onthe 0 – 3 00 mA meter.
Observing from Fig 2, it is interesting to note that in the range from theload resistance of 37.3 Ω (mismatch1.42) to 63 Ω (mismatch 1.19), thesquare root of the product of the network output impedance and load resistance remains close to the value of theload resistance that allows delivery of all the available power ( ≈ 50 Ω ), butbegins to decrease with increased loadmismatch on either side of thematched condition. It will be shownlater that the reactance introduced bythe nonsymmetrical network whiletransforming the nonreactive loadresistances to the network input areresponsible for the decrease as themismatch increases in either directionaway from the resonant condition.
Let’s turn now to Table 5, where mymeasured data is shown to agree withthe statement that impedance transformation through a nonsymmetrical network is not the same as that through aperfect transformer having a linearratio of transformation. Although thestatement on this point is shown true,we will see that the reasoning that someclaim renders the IEEE procedure invalid is not correct. Notice that exceptfor the 5 0 Ω load resistance thatmatches the network output, all otherpurely resistive load resistances transform through the network to obtainreactive impedances at the input. Whenplotted on a Smith chart normalized to135 0 Ω , the loci of these input impedances form a straight line at an angle of 64 ° clockwise from the resistive axis, intersecting the axis at the chart centerwhere the impedance is 135 0 + j0 . Stillin Table 5, notice that when the loadmismatch is approximately 2:1 on either side of the matched point, theangles of the impedances are numerically equal with respect to that at resonance: 32 ° with the high load resistanceand –32 ° with the low load resistance.However, also notice that with the highand low load resistances, the magnitudes of the input impedances are 1 060and 178 0 Ω , respectively. The differences between the matched input impedance (nonreactive) of 135 0 Ω arethus 29 0 and 43 0 Ω , respectively. Let’snow examine the significance of thesemeasured input impedances.
First, the 1780 Ω input impedanceobtained with the lowresistance load
Table 5— π Network Input Impedance versus Nonreactive Load Resistance
HW100 Transceiver Initially Resonated with 52.0Ω
Load at 100 W OutputNetwork Mismatch Network Network Load Re 52.0 lnput Z Input Z ( Ω ) ( Ω ) Polar Rectangular
240.0 4.6:1 950@58 ° 503.4 + j 805.6160.0 3.17 980@48 ° 655.7 + j 728.3100.0 1.92 1060@32 ° 898.9 + j 561.7
83.0 1.6 1150@20 ° 1080 + j 393.352.0 1.0 1350@0 ° 1350 + j 041.0 1.22 1580@–14 ° 1533 – j 382.234.2 1.52 1630@–18 ° 1550 – j 503.726.0 2.0 1780@–32 ° 1509 – j 943.320.6 2.52 1900@–41 ° 1433 – j 124517.5 2.97 2000@–48 ° 1338 – j 1486
is 140 Ω further off the resonant valuethan the 1060 Ω with the highresistance load, resulting in the higheroffresonance plate current with thelowresistance load than for the highresistance load with same degree of mismatch (1.92:1 and 2.0:1, respectively).
Second, the measured input impedances resulting from two load resistances of ± 5 Ω relative to 52 Ω wereinterpolated on the straightline locusof all the impedances in Table 5. At theresulting two equal input mismatchesof 1.12:1, the resulting input impedances normalized to 135 0 Ω are0.95 + j0 .1 0 and 1.1 0 – j0 .1 0 , respectively, for real values of 1282.5 + j135and 1485 – j135 Ω , respectively. Withthese mismatches of only 1.12:1 on ei
ther side of resonance, the angle of themismatched impedances is only ± 6 ° andthe change from normal amplifiersystem performance of 1. 0 only to 0.9968,(99.68%) amounts to a change of only0.014 dB.
Consequently, the small change inamplifier performance under these conditions is insignificant with respect tocontributing to error while using thestandard IEEE loadvariation methodin the measurement of network outputimpedance. Ergo, criticism of the IEEEmethod of measuring the output impedance of π networks in RF power amplifier operation is unfounded and themethod is proved valid.
In conclusion, my measurementsand discussion prove that the outputof a properly designed RF tank circuit
Fig 2—Kenwood TS830S RF amplifier π network output impedance, Z , versus loadresistance. All available power of approximately 100 W is delivered initially to 53.1 Ω load.All adjustments left undisturbed during measurement of other load values (see Table 4 ).
7/28/2019 QEX 2001 MayJun
http://slidepdf.com/reader/full/qex2001mayjun 13/13
44 May/June 2001
performs with a linear relationship between output voltage and current.Consequently, the theorems associated with linear circuits, such as theConjugateMatching Theorem and theMaximumPower Transfer Theorem,are valid and applicable in both theanalysis and practical operation of RFpower amplifiers performing in bothCW and SSB modes.
Several professional RF engineersand university EE professors have reviewed the material in this article foraccuracy, all having pronounced thematerial to be an accurate and truepresentation of the subject. Theyare Warren Amfahr, W0WL; JanHornbeck, N0CS; and others. In addition are John C. Fakan, PhD(KB8MU), private consultant, formerly a consultant to NASA; John S.Belrose, PhD in Radio Science, Cantab(Cambridge, UK) (VE2CV), SeniorRadio Scientist, Communications Research Center, Canada; ForrestGehrke, BSEE (K2BT), microwaveengineer with the former RCA; AlHelfrick, PhD in Applied Sciences,(K2BLA) Professor of Avionics atEmbryRiddle Aeronautical University; Robert P. Haviland, EE, (W4MB)retired General Electric RF engineer.
This article is an excerpt from Chapter 19 of Reflections II , publ ished byWorldradio Books. Thanks to Armond
Noble, N6WR, for permitt ing us to present it in QEX.
Notes1The use of “linear” and “nonlinear” relates
to the voltage/current relationship at theoutput terminals of the amplifier tank circuit or at the terminals of a network. Thisusage does not relate to nonlinearity between the input and output of an amplifierthat results in generation of distortionproducts in the output signal.
2In addition to the data in Terman’s example,I have made measurements that determine the output impedance of RF poweramplifiers, which prove the existence of aconjugate match. The data show thatwhen the amplifier is loaded to deliver allof its available power, the output impedance of the amplifier equals the load impedance, thus signifying a linear voltage/ current relationship at the output of thetank circuit and thus a conjugate match.The description of my measurement procedure and the resulting data showing theproofs are included here in following sections.
3The material discussed in this article pertains only to RF amplifiers used in theAmateur Service with vacuum tubes andπ networks in the output circuit. The mate
rial does not necessarily pertain to amplifiers used in various commercial services,or any amplifier using solidstate components.