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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 Class-B

and -C RF A mplifiers

this article, we will discussthe nature of the source of power in class-B and -C RF

amplifiers. As were my 197 0s QST ar-ticles, “Another Look At Reflections,”and my book Reflect ions—Transmis-

sion Lines and Antennas , that clarifiedmisconceptions concerning SWR andreflected power, this article is con-cerned with clarifying misconceptionsprevalent among amateurs and profes-sional electrical engineers concerning the operation of RF power amplifiers.In attempting to resolve the unfort-unate 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 rela-tion 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. There-fore, it is important that the focus of this article is to highlight and clarifythose misconceptions.

Before discussing amplifier opera-tion, however, two synonymous termsthat play a vital role in amplifier op-eration need clarification, becausethey are widely misinterpreted in dis-cussions 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 class-B 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 misconcep-tion is that class-C amplifiers cannotsupport circuit analysis using generalnetwork theorems because of the

In

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nonlinearity 1 of the amplifier opera-tion. In clarifying this misconceptionwe will show that, although the inputcircuit of the π -network tank circuit inclass-C amplifiers is nonlinear, theoutput circuit to the load is indeed lin-ear, due to energy storage in the tank.Consequently, the linear relationshipbetween voltage and current appear-ing 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 pro-claim 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 ex-plained in the following paragraph. Anexample from Terman is used in clari-fying the misconception concerning therelationship between output and loadresistances, shown below in “Analysisof the Class-C Amplifier.” In addition,data obtained from my own measure-ments, shown below in “ Measuring theOutput Resistance of the RF Power

Ampl ifier ,” prove that after the ampli-fier 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 measure-ments have been confirmed by TomRauch, W8JI, using an identical mea-surement procedure. Tom is an RF-power-amplifier engineer with

Ameritron.Now to explain two axioms of the con-

jugate matching theorem that are com-monly overlooked, which has resultedin widespread confusion concerning itsuse. We know that when a load imped-ance 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 be-cause the source and load impedances

are conjugates of each other. Conse-quently, 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 con-text) and “conjugate match” are oftenused interchangeably with no differ-ence in meaning. Unfortunately, misin-terpretation and misunderstanding of conjugate in the newer term has cre-ated 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 power-transfer theorem, accurately define aconjugate match:

Axiom 1: There is a conjugate matchwhenever all of the available powerfrom a source or network is being de-livered 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 ei-ther direction.We now return to clarify the miscon-

ception concerning output and load re-sistances. 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 de-livered by class-B and -C amplifiers,reveals still another prevalent miscon-ception. This misconception is that the

entire source of power in these classes

of vacuum-tube amplifiers is a dissipa-tive resistance. In clarifying this mis-conception, we will use the example byTerman to demonstrate that the sourceof RF output power in a class-C ampli-fier is the combination of two resis-tances; a nondissipative resistance(related to the characteristics of the ef-

fective load line) and a dissipative plateresistance R PD . R PD is not plate resis-tance R P , as determined from the well-known 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 volt-age, which is not the source of power inRF power amplifiers as is claimed bymany who have misinterpreted the ex-pression. The source of power is actu-ally derived by a large change in platecurrent resulting from a change in gridvoltage . This phenomenon will be dis-cussed in more detail later.

One portion of the nondissipative re-sistance 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 class-B and -C ampli-

fier, the load is the tank input. Thesource is the combination of two paral-lel 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 over-lapping times throughout the cycle. Theother portion of the nondissipative re-sistance 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 di-vided by the square of the average dcplate current I dc , the current measuredby the dc plate ammeter. Notice inTerman’s statement #3 below, that dis-sipated power P D is the product of theinstantaneous plate-to-cathode 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 con-ducting portion of the RF cycle. Knowl-edge of the nondissipative portion of thesource resistance will allow you to un-derstand why class-B and -C amplifierscan deliver all of their available powerinto a conjugately matched load withefficiencies greater than 5 0%. This con-cept 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 as-sumption 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 nondissi-pative, 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 efficien-cies greater than 5 0%. The lower dissi-pative plate resistance occurs because

plate current is allowed to flow only

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when the plate voltage is at the minimum of its sinusoidalswing, as explained in Terman’s statement #7 below. Re-peating Terman’s statement #3 for emphasis, dissipated

power P D is the product of instantaneous plate-to-cathodevoltage 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 power-transfer theorems don’t ap-ply to class-B 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 energy-storagedevice. The energy-storage capacity of the tank isolates thepulsed nonlinear mode at the input from the smoothed lin-ear 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 as-sert 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 irrel-evant 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 Class-C AmplifierThe following discussion of the class-C 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 Engi-neers Handbook , 1943 edition, p 445, and on Terman’sexample of class-C 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 (parenthe-ses and emphasis mine):1. The average of the pulses of current flowing to an elec-

trode represents the direct current drawn by that elec-trode.

2. The power input to the plate electrode of the tube at anyinstant is the product of plate-supply voltage and instan-taneous plate current.

3.The corresponding power ( P D) lost at the plate is theproduct of instantaneous plate-cathode voltage and in-stantaneous plate current.

4. The difference between the two quantities obtained fromitems 2 and 3 represents the useful output, at the mo-ment.

5. The average input, output and loss are obtained by av-eraging the instantaneous powers.

6. The efficiency is the ratio of average output to averageinput and is commonly of the order of 60-80%.

7. The efficiency is high in a class-C 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 class-B 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 em-phasize 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 oper-ating 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 class-B and -C amplifiers.) In other words, theoutput power equals the dc-input power minus the powerdissipated in resistance R PD . We will now show why thistwo-way 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 dc-input powerminus the power dissipated in R PD equals the power deliv-ered 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 dis-regarded. If there were any significant dissipative resis-tance in addition to R PD , the power delivered to the load plusthe power dissipated in R PD would be less than the dc-inputpower, 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 ob-serve 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 class-C 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)

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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 funda-mental 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 more-precise ChaffeeFourier analysis, was used universally to determine the ap-proximate 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 resis-tance R L is proportional to the slope of the operating loadline that allows all of the available integrated energy con-tained in the plate-current pulses to be transferred into theπ -network tank circuit. (For additional information concern-ing 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 spring-loaded escapement mechanismin the pendulum clock delivers mechanical energy periodi-cally to the swing of the pendulum. In a similar manner, aftereach plate-current pulse enters the π -network tank circuit,the flywheel effect of the resonant tank circuit stores the elec-tromagnetic energy delivered by the current pulse, and thusmaintains a continuous sinusoidal flow of current through-out the tank, in the same manner as the pendulum swingscontinuously and periodically after each thrust from theescapement mechanism. The continuous swing of the pendu-lum 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 in-depth examination later.

Let us now consider the dissipative plate resistance R PD ,which provides the evidence that the dc input power to theclass-C 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 conju-gate matching theorem and the maximum power-transfertheorem, 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 characteris-tics 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 class-C amplifiers operating into the conjugate match.

Examining the Operating Load LineThe details of the somewhat trial-and-error method of

establishing the operating load line are beyond the scope of this article. However, once established, the load line repre-sents 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 net-work 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 voltage-current 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 re-sistance 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 aver-age 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 cur-rent 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 constant-current 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-

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termined by dividing the average fun-damental RF plate voltage by the av-erage fundamental RF plate current,the Terman equation. Thus, theChaffee method saves time comparedto using Terman’s equation alone, be-cause 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 dis-cover that load resistance R L is greaterthan dissipative plate resistance R PD ,thus allowing more power to be deliv-ered to the load than that dissipated in

R PD . Referring again to Terman’s ex-ample 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 rela-tive power distribution is 75.1% deliv-ered to R L , and 24.9% dissipated in

R PD . Earlier we showed that after ac-counting for the power dissipated in

R PD , all the remaining available poweris delivered to the load R L . Thus, thisdistribution of power clearly demon-strates why a class-C amplifier can de-liver more than 5 0% of its input powerto the load, because its load resistance

R L (64 00 Ω ) is greater than its dissipa-tive plate resistance R PD (3315.6 Ω ).These calculations are in accord withwith Terman’s statement #7 that “effi-ciency is high in the class-C amplifier,because current is permitted to flowonly when most of the plate-supply volt-age 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 non-dissipative resis-tance 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 consid-ering 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 ac-cordance with conjugate matching axioms 1 and 2 recited above. The ex-ample has also shown that more powerhas been delivered to the load thanwas dissipated, because 54.6 W weredelivered and only 18.7 W were dissi-pated. Thus, contrary to the opinion of many who fail to understand this con-cept, we have shown that conjugatematching to a class-C 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 conju-gate match during SSB operation?”The answer is yes, but it begs anadditional question: Does the outputimpedance of the amplifier remain con-stant 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 signifi-cantly 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 estab-lished 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 man-ner, it remains constant for all valuesof drive. I have made extensive mea-surements, which show that once theoperating load line is established dur-ing 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 ap-pearing at the input of a transmissionline that is Z 0 mismatched to an an-tenna, while also establishing a conju-

gate match that overrides the Z 0 mis-match at the antenna. The principles of conjugate matching are fundamental tothe matching function performed by theantenna tuner, and are indeed funda-mental to all impedance matching ob-tained with any impedance matching device that allows delivery of all avail-able 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 ex-amination of the vitally important fly-wheel 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 power-transfer 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 difficul-ties in appreciating certain aspects of aparticular discipline. Fortunately,energy storage in the mechanical disci-pline has a valid and rigorous analo-gous relationship with energy storagein LC circuitry, which makes it appro-priate to draw upon a mechanical ex-ample to clarify the effect of energystorage in the RF tank circuit. (A fur-ther convincing analogy involving wa-ter appears later, in which the origin of the term tank circuit is revealed.)

The smoothing action of the RF en-ergy stored in the tank circuit is rigor-ously analogous to the smoothing ac-tion of the energy stored in the flywheelin the automobile engine. In the auto-mobile engine, the flywheel smoothesthe pulses of energy delivered to thecrankshaft by the thrust of the pistons.

As in the tank circuit of the amplifier,

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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 deliv-ers 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 cir-cuit in the RF amplifier is determinedby its Q.

As stated earlier , the tank circuit inthe RF amplifier receives two overlap-ping 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 one-cylinder engine deliv-ers one thrust of energy per revolution,it is evident that a large amount of en-ergy storage is required to enable thecrankshaft to deliver a smooth outputduring the entire rotation of the crank-shaft. In this case, a very heavy fly-wheel 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 sine-wave outputwith an acceptable minimum of har-monic 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 sine-wave output when sup-plied only with pulses of energy at theinput. Thus the energy-storage ca-pacity of the tank provides for thesmoothed linear output to the load cir-cuit, despite the nonlinear pulsed in-put, 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 con-cept which will become clear as we con-tinue.

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 line-feed to the supply is single-phase orthree-phase, 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 pur-poses, 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 out-put port of the amplifier can only see thevoltage swings and the impedance pre-sented by the tank components. A prop-erly 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 en-ergy 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 cas-cade 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” de-pend on the linear nature of the energytransfer from the amplifier’s outputport right on through to the last an-tenna 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 typi-cal amateur π -network class-C ampli-fier, the load is the tank input. Thesource is the combination of two pro-cesses: (1) the blocking capacitor in se-ries with the active device (tube) and(2) the same blocking capacitor in se-ries 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 square-wavecharacteristics of the active device, thedesigns in this region have always beenvery empirical. Actual experience anda good seat-of-the-pants feel for the sig-nificance of active-device 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 pro-viding 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 out-put will decrease accordingly. Mal-functioning of the upstream energy“bucket brigade” can result in linearoperation at a lower level or in nonlin-ear operation, depending on how wellthe tank design can handle thechanges. The important point is thatonce conditions allow the tank to oper-ate as a linear active source, every-thing else downstream of the tank is

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linear and follows the conjugate-match theorem and other linear-sys-tem theories. Changing the conditionsat the input to the π -network (for ex-ample: 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 sup-plied 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, thevoltage-current product will equal theenergy rate available. It has no otherchoice because the conjugate-match-ing theorem requires a change in out-put impedance if it is to remain a lin-ear 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 wave-shape distortioncould occur, for example, flat topping.

The important point is that thedesign and operation of the circuitryproviding energy to the tank circuit in-volves 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 en-ergy pulse is ugly and would be diffi-cult to characterize mathematically,because the tank circuit doesn’t care.

It is positively uncanny how easy it isfor some people to simply ignore experi-mental evidence staring them right inthe eye when one of their pet under-standings is in jeopardy. Many peopleconcede that amateur class-C amplifi-ers 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 pro-cess is simply matching the output andload impedances to a common conju-gate. 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 conjugate-match theoremis simple and absolute: When the en-ergy being transferred across any lin-early behaving connection cannot befurther increased simply by changing the impedance of either source or load,a conjugate match exists. That is com-monly the operating condition foramateur class-B 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), smooth-ing 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 er-rors that may have escaped him. Be-cause 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 cred-ibility as an author has been ques-tioned. 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 sub-stantive errors, or invalid statements.Consequently, I requested several pro-fessional RF engineers with unques-tionable 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 cir-cuit isolates its pulsed input from theoutput, and therefore he maintains thatthe output circuit of the π -network can-not support or sustain linear operation,and no conjugate match. Unfortu-nately, during the nine years of the con-troversy I discovered that opposition tothe application of linear theorems toany aspect of RF power amplifier opera-tion 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 en-ergy storage in the tank circuit provid-ing 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 inter-esting to learn how the term tank origi-nated 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 water-tank analogy was applied for the verypurpose of explaining the energy-stor-age function of the LC output circuit. Itgoes like this. A water tank is filled toa specific depth that causes a corre-sponding pressure applied on the bot-tom. A hole is made in the bottom witha size that allows one gallon per minuteto flow with the specific applied pres-sure. 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, one-thirtieth of a gallon thirtytimes per minute and so on, you get thepicture. As long as enough water isadded to maintain the level, thus main-taining the same constant pressure atthe bottom, the water will continue toflow smoothly from the bottom at therate of one gallon per minute, regard-less 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 im-pedance), or shorter tanks withappropriate larger holes (low output

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impedance). However, the impedanceand linearity of the input to the tank isirrelevant as long as it results in main-taining a constant water level. Thus,the action at the bottom is linear al-though 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 appear-ing 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 cur-rent pulses entering the π -networktank has no effect on the smooth sinu-soidal voltage and current appearing at the output. If the average inte-grated energy of the current pulsesentering the tank increases, the volt-age and current at the output will in-crease in a linear relationship. Thus,it is shown that the output of a prop-erly designed RF tank circuit is linear,and the theorems associated with lin-ear circuits are applicable.

Measuring the Output Resistanceof the RF Power Amplifier

I have developed a test setup andprocedure based on the standard IEEEload-variation method for measuring the source, or output resistance R OUTof networks, which are described below.Measurements made with this setupand procedure show that output resis-tance 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. How-ever, before proceeding further it willbe helpful to obtain an appropriate per-spective 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 de-liver its intended power into its termi-nating load, resistance R L appearing atthe input of the network is easily androutinely measured with appropriateimpedance-measuring 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 as-sumptions 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 rea-son 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 de-scribe the test setup and procedure thatdoes yield the correct value of source

resistance R OUT , the value that equalsthe load resistance. Consequently, us-ing the standard IEEE load-variationprocedure 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 ampli-fiers. The data obtained from my mea-surements have been verified by TomRauch.

The test setup I developed for mea-suring the output resistance R OUT of RF power amplifiers is arranged to usethe load-variation method of measure-ment, based on the IEEE expressionfor measuring the output resistance of networks. The IEEE expression is

R OUT = ∆ E / ∆ I , where ∆ E and ∆ I rep-resent the corresponding changes inload voltage and load current, respec-tively, with a small change in loadresistance R LOAD terminating the net-work. 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 Small-Load-Variation 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

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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 resis-tance a small amount (to R2 ), around a10% change from R1 , and then measur-ing the difference in load voltage andcurrent, the output resistance is ob-tained by substituting the differentialvoltage and current values in the IEEEexpression for R OUT shown above.

The equipment used in the measure-ments consisted of two vacuum-tubetransceivers using two parallel 6146tubes and a π -network tank circuit inthe RF power amplifier. They are aHeathkit HW-100 and a Kenwood TS-830S. A Hewlett-Packard HP-4815A RF Vector Impedance Meter modifiedfor digital readout, along with ESI250-DA universal impedance bridge,for measuring RF and dc resistancesof noninductive load resistors R1 and

R2 . An HP-8405A Vector Voltmetermodified for digital readout for mea-suring voltages appearing across loadresistors R1 and R2 , and an HP-410BRF Voltmeter with an HP-455A Co-axial Adapter, also modified for digi-tal readout to indicate load voltage.The RF vector impedance meter wasused to confirm that the load resistorscontained zero reactance. The experi-ments 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 spe-cific 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 calcu-lation 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 maxi-mum available power of ≈ 110 W. Alladjustments remain undisturbedthereafter. The data in Table 1 wereobtained using the Heathkit HW-100.Result: Load resistance R LOAD whenadjusted for maximum power out R1 =51.2 Ω . Average measured source re-sistance R OUT = 50.3 Ω (See Fig 1 andTable 2).

The reason for the variation, or scat-ter in measured output resistance andoutput power seen in the data abovewas found to be a short-term sag inoutput power between the measure-

Table 2—Measured Network Output Resistance versus Output Power,

HW-100 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 be-tween R1 and R2 from manual to co-axial relay, thus reducing the timedelay, and by waiting until the sag inpower output bottomed out. However,the worst-case 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 coef-ficient ρ of 0.039, for a conjugate mis-match 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 out-put curve near the peak. Using onlythe analog output-power meter to ob-serve 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 around-top hill than a peak. Evidently,the adjustment for maximum outputrequires a method of indicating thatprovides better resolution than thatobtained with the analog output-power 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 coeffi-cient 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 HW-1 00 and a Kenwood TS-83 0Sversus output power at 4. 0 MHz, power level set by drive level (see Tables 2 and 3).

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51 Ω . Thus, it is evident that during routine tuning and loading adjust-ments 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 differ-ence 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 in-dication on the analog output-powermeter, R OUT was measured at thatpoint. The π -network was then re-ad-

justed in very small i ncrements, mea-suring R OUT after each re-adjustment,until R OUT became equal to 51. 0 Ω . Theincrements were so small that al-though 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 tech-nique as described above, R OUT mea-sured 25 Ω when the load resistancewas 25 Ω , and—consistent with themeasurement pattern already devel-oped— R OUT measured 16 Ω when theload resistance was 16.7 Ω . These mea-surements 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. How-ever, we would also like to know whathappens to output resistance R OUTduring SSB modulation after the ampli-fier is first tuned and loaded to delivermaximum available power with a spe-cific drive level. The measurement dataappearing in Table 2 was obtained us-ing the HW-100 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 TS-830S Trans-ceiver 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 re-mains 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 adjust-ments except for the level of drive un-disturbed thereafter, R OUT was mea-sured at power levels decreasing from100 W to 12.5 W. (The changes in out-put power were obtained by changing the drive level.) This range of powerlevels, as shown in the Table 2 , corre-sponds to approximately the samerange of output power prevailing dur-ing SSB modulation.

However, the two caveats are neces-sary to explain the changes in R OUTthat appear to conflict with the state-ment 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 de-crease, 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 output-power level.The conjugate mismatch between the80 Ω output source resistance and the51- Ω load resistance is 1.569, estab-lishing a voltage-reflection coefficientρ = 0.2214, a power-reflection coeffi-cient ρ 2 = 0.0490, and thus a transmis-sion 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 mea-surement procedure with the KenwoodTS-83 0S transceiver. It can be seenfrom Table 3 and Fig 1 below that thevariation in output resistance with thistransceiver is much less than with theHW-1 00 over the entire range of driveand output power. The reason for thenearly constant output resistance withthe TS-83 0S is better plate-voltageregulation of the power supply. Espe-cially notice that, in the region of con-stant 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

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been in dispute in the absence of appro-priate measurements. It is also impor-tant to observe that at the 1 0-W outputlevel, where the output resistance hasincreased somewhat due to imperfectvoltage regulation (plate voltage in-crease with decrease in power), themismatch between the 51.2- Ω load re-sistance and the 58.6- Ω output resis-tance is 58.6 / 51.2 = 1.145:1. This smallmismatch results in a voltage-reflec-tion coefficient ρ = 0 .0674, a power-reflection 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 mis-match. Consequently, a realistic conju-gate match exists over the entire rangeof speech levels.

Before concluding this subject, an-other way of explaining the action oc-curring in the amplifier during SSBmodulation is that as the average in-tegrated 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 cur-rent remains constant during speechmodulation. Since the output resis-tance R OUT is determined by the ratioof current to voltage, the output resis-tance also remains constant and equalto the load resistance during modula-tion. Thus all the available power fromthe source is delivered to its load at alllevels of speech, satisfying the condi-tion 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 imped-ance transformer (perfect integralmathematical input/output ratio).Therefore, they claim even the smallchange in load resistance used in theIEEE method does not transform lin-early 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 re-sistance. 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 de-scribed 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 opera-tion of the amplifier that would distortthe results if the change in load resis-tance were relatively large. For ex-ample, 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 TS-830S 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 re-sults from somewhat larger changes inload resistance. Notice that the magni-tude 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 propor-tional 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 resis-tance. 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 net-work output resistance. So let’s exam-ine 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-

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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 interest-ing 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 net-work output impedance and load resis-tance 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 transfor-mation through a nonsymmetrical net-work 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 in-valid is not correct. Notice that exceptfor the 5 0 -Ω load resistance thatmatches the network output, all otherpurely resistive load resistances trans-form through the network to obtainreactive impedances at the input. Whenplotted on a Smith chart normalized to135 0 Ω , the loci of these input imped-ances form a straight line at an angle of 64 ° clockwise from the resistive axis, in-tersecting 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 ei-ther side of the matched point, theangles of the impedances are numeri-cally equal with respect to that at reso-nance: 32 ° with the high load resistanceand –32 ° with the low load resistance.However, also notice that with the highand low load resistances, the magni-tudes of the input impedances are 1 060and 178 0 Ω , respectively. The differ-ences between the matched input im-pedance (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 low-resistance load

Table 5— π -Network Input Impedance versus Nonreactive Load Resistance

HW-100 Transceiver Initially Resonated with 52.0-Ω

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 high-resis-tance load, resulting in the higheroff-resonance plate current with thelow-resistance load than for the high-resistance load with same degree of mismatch (1.92:1 and 2.0:1, respec-tively).

Second, the measured input imped-ances resulting from two load resis-tances of ± 5 Ω relative to 52 Ω wereinterpolated on the straight-line locusof all the impedances in Table 5. At theresulting two equal input mismatchesof 1.12:1, the resulting input imped-ances normalized to 135 0 Ω are0.95 + j0 .1 0 and 1.1 0 – j0 .1 0 , respec-tively, 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 amplifier-sys-tem 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 con-ditions is insignificant with respect tocontributing to error while using thestandard IEEE load-variation methodin the measurement of network outputimpedance. Ergo, criticism of the IEEEmethod of measuring the output imped-ance of π -networks in RF power ampli-fier 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 TS-830S 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 ).

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44 May/June 2001

performs with a linear relationship be-tween output voltage and current.Consequently, the theorems associ-ated with linear circuits, such as theConjugate-Matching Theorem and theMaximum-Power 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 re-viewed 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 addi-tion are John C. Fakan, PhD(KB8MU), private consultant, for-merly a consultant to NASA; John S.Belrose, PhD in Radio Science, Cantab(Cambridge, UK) (VE2CV), SeniorRadio Scientist, Communications Re-search Center, Canada; ForrestGehrke, BSEE (K2BT), microwaveengineer with the former RCA; AlHelfrick, PhD in Applied Sciences,(K2BLA) Professor of Avionics atEmbry-Riddle Aeronautical Univer-sity; Robert P. Haviland, EE, (W4MB)retired General Electric RF engineer.

This article is an excerpt from Chap-ter 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 cir-cuit or at the terminals of a network. Thisusage does not relate to nonlinearity be-tween 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 deter-mine 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 imped-ance of the amplifier equals the load im-pedance, thus signifying a linear voltage/ current relationship at the output of thetank circuit and thus a conjugate match.The description of my measurement pro-cedure and the resulting data showing theproofs are included here in following sec-tions.

3The material discussed in this article per-tains 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 ampli-fiers used in various commercial services,or any amplifier using solid-state compo-nents.

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