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Calculating power dissipation in parasitic-suppressor resistors

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Sure, the resistors in VHF-parasitic-suppression networks dissipate some HF energy, especially on 10 meters. But how much? The answer may surprise you!

How much MF/HF power is dissipated in the resistors used in LR (inductor and resistor) VHFparasitic-suppression networks in tube-based MF/HF amplifiers? Apparently, most amplifier builders choose the power rating of these resistors empirically: "If it burns up, try something different." A more scientific engineering method might be useful!

The calculation of suppressor-resistor dissipation involves three major steps and a number of minor steps. The first major step is to find the maximum RF current that flows through the anode parasitic-suppression network (assumed to be aparallel LR circuit). The second major step involves finding the voltage drop across this parallel LR circuit. The third major step is to calculate the power dissipated in the resistor on the basis of the MF/HF voltage across it.(1)

Fig 1 - In a grounded-grid amplifier, the anode-to-grid capacitance (C_AG) of the amplifier tube(s) carries part of the output-network RF circulating current, as shown at A. Assuming that the amplifier is operated at the same power level throughout its frequency range, the RF current through C_AG is greatest at the highest frequency in that range because the reactance of C_AG decreases with frequency. Whatever circulating current flows through C_AG also flows through the VHF-parasitic-suppression network (L_SR_S) because this network is in series with C_AG. Redrawing the output network as a parallel tuned circuit (B) makes this series relationship more apparent.

Power dissipation in the suppressor resistor is greatest during operation on the 10-meter band. This is because more of the amplifier-output-network circulating current flows through the amplifier-tube output capacitance (anode-to-grid capacitance [C_AG] in a grounded-grid amplifier) on 10meters than on lower-frequency bands. See Fig 1A.

The reason for this is that capacitive reactance decreases as frequency increases. At the upper end of the 10-m band (29.7 MHz), the output capacitance of the amplifier tube(s) exhibits less reactance than at all frequencies lower than 29.7 MHz. Because the ac anode voltage - which is impressed across C_AG - is constant from band to band (assuming that the amplifier is operated at the same output power on all bands), the alternating current that flows through C_AG is maximum at 29.7 MHz. Once the ac anode voltage is known, the current through C_AG can be calculated by using a variation of Ohm's Law.

As Fig 1B illustrates, the anode parasitic suppressor (L_SR_S) is in series with C_AG. Because of this, whatever RF current flows through C_AG also flows through L_SR_S. Once we know the magnitude of the RF current through C_AG, we'll know how much current flows through the L_SR_S network. This information, in conjunction with the impedance of L_SR_S, will allow us to calculate the RF voltage drop across L_SR_S, and, therefore, the RF voltage drop across R_S.

Calculating the AC Anode Voltage

The peak ac anode voltage depends on several factors: the dc anode voltage, the grid voltage with respect to ground, and whether or not the amplifier is properly tuned. (For more on how amplifier tuning affects peak ac anode voltage, see the sidebar, "Why Proper Amplifier Tuning is Important.") Assuming that the amplifier is properly tuned, the peak change in anode voltage is equal to the dc supply voltage minus about 200 to 300 V. (Using the lower number results in a more conservative suppressor-resistor-dissipation estimate.)(2) This peak voltage can be converted to its root-mean-square (RMS) value by dividing it by (approximately 1.414).(3)

At this point, we can begin working a sample problem. A hypothetical MF/HF amplifier contains a 3-500Z triode (C_AG = 4.7 pF) operating in grounded grid at a dc anode supply of 4 kV. Its anode parasitic-suppression network consists of a 0.05 µH inductor in parallel with a 25-Ω resistor. From the foregoing discussion, we calculate that the peak change in anode voltage is 4000 V - 200 V, or 3800 V. To find the RMS equivalent of this voltage, we divide 3800 by 1.414. Answer: The 3-500Z in the hypothetical amplifier operates at an RMS anode potential of 2.687 kV.

Calculating the Reactance of C_AG

We can find the reactance of C_AG with the capacitive reactance formula (Xc = 1 ÷ 2ΠfC). Applying this to our sample problem: The 3-500Z's anode-to-grid capacitance is 4.7 pF; at 29.7 MHz, X_CAG is -j1140 Ω.

Calculating the RF Current Through X_CAG

Ohm's Law serves here. Substituting X_C for R, the RF current through C_AG can 'be found with the formula I = E X_CAG, where I is the RF current in amperes and E is the RMS anode voltage.

Based on an anode potential of 2.687 kV RMS, the RF current through C_AG of a 3-500Z (4.7 pF) is 2.36 A RMS at 29.7 MHz. Because the parasitic-suppression network (L_SR_S) and C_AG are in series, the RF current through L_SR_S is also 2.36 A RMS at 29.7 MHz.

Calculating the Suppressor-Network Impedance

Because the suppressor-network components are connected in parallel, the RF voltage drop across R_s is equal to that across L_SR_S. To calculate this voltage drop, we must first calculate the network impedance.

L_SR_S exhibits a complex impedance because it consists of reactance (X_LS) and resistance (R). The presence of reactance makes finding the network impedance a bit more complicated than solving a simple parallel-resistance problem.

Two approaches can be used to find the impedance of LsRs; which approach you use depends on whether you prefer to think in terms of impedance or its reciprocal, admittance. The sidebar, "Finding Impedance by Solving for Admittance," explains the admittance approach.(4) For those who prefer to work with impedance, the impedance of L_SR_S can be found with the equation

where
Z = network impedance in ohms
R_s = resistance of Rs in ohms
X_LS = reactance of L_S in ohms

Returning to our sample problem: The suppressor network in the anode of the 3-500Z consists of a 0.05-µH inductor in parallel with a 25 Ω resistor. X_LS can be found with the inductive reactance formula (X_L = 2ΠfL); in this case, X_LS is equal to j9.3305 Ω at 29.7 MHz. Solved with R_S = 25 Ω and XL = 9.3305 Ω, Eq 1 reveals a network Z of 8.74 Ω at 29.7 MHz.

Calculating the RMS RF Voltage Across the Network

Finding the voltage across the network requires only the application of Ohm's Law (in this case, E = IZ, where I is the network current (2.36 A) and Z is the network impedance (8.74 Ω). Answer: At 29.7 MHz, the RMS voltage across our hypothetical LR parasitic-suppressor network is 20.6.

Calculating the Power Dissipated in R_S

Now that we know the voltage drop across the network (and, hence, across R_S), we can calculate the power dissipated in R_S with the equation P = E² ÷ R_S, where P = power in watts, E = the RMS voltage drop across R_S, and R_S = resistance of R_S in ohms.

Concluding our sample problem: A drop of 20.6 V RMS across Rs (25 Ω) indicates that R_S dissipates 17 W at 29.7 MHz.

Finer Points

R_S dissipates 17 W as a result of the sinusoidal RF circulating current that passes through the 3-500Z anode lead and the parasitic-suppression network. An additional, lesser current flows through the anode lead and suppressor during the 180 ° (or so) of downward anode-voltage swing that occurs when the tube conducts. These pulses occur at the amplifier operating frequency - 29.7 MHz in the sample problem. This additional current causes a half-wave, RF-pulse voltage drop across R_S that adds, on the bottom half of the cycle, to the R_S voltage drop due to sine-wave circulating current. This increases the total power dissipated in R_S.

The fact that this waveform is a halfwave RF pulse instead of a sinusoid complicates the math necessary to find the additional power dissipation caused by this current. An accurate mathematical analysis would require the generation of a formula that accurately describes a waveform consisting of the half-wave RF pulse plus the sinusoidal circulating-current waveform. Calculus could then be employed to more closely calculate the RMS power actually dissipated in R_S. Such analysis is beyond the scope of this article.

It is somewhat easier to use this approximation: In a typical amplifier, the half-wave anode-current pulse increases the power dissipated in R_S by about 15% over that due to RF circulating current. Thus, in our sample problem, the total power dissipated in R_S is 1.15 × 17 W, or 20 W.

Practical Notes

Twenty watts represents, to me at least, a surprising amount of HF power dissipation in a fairly low-inductance LR parasitic suppressor. (Although the network inductance in the sample problem was 0.05 µH, most of the parasitic suppressors I've seen contain more inductance than this. The resistors in such networks probably dissipate more power than what we've calculated here.)

Of course, the sample problem represents a worst-case scenario of 29.7-MHz "lock-to-talk" RTTY or FM operation at full power. Many radio amateurs use their amplifiers at reduced anode voltage during RTTY and FM operation, however; the parasitic-suppression resistors in these amplifiers are probably not subjected to more than half of the 20-W maximum dissipation calculated in the sample problem. SSB operation, during which amplifier tubes operate at a duty cycle of about 15% with normal human voices, subjects Rs to about 3 watts of average dissipation at 29.7 MHz. This partially explains why the typical parasitic-suppressor assembly survives even though it contains only two or three 2-W resistors.

Another factor in the survival of seemingly underengineered suppressor networks is that modern metal-film and metal-oxidefilm "flameproof" resistors are very rugged: They are not as easily turned into "crispy critters" as were the heat-intolerant, phenolic-cased, carbon-composition resistors of yesteryear. (I have tested modern, 2-W [rated at 70 °C ambient], metal-and metal oxide-film resistors that can dissipate 5 watts each in free air for 1 hour with no apparent changes. These amazing resistors can dissipate 20 watts each for at least 6 seconds with no permanent change in appearance or electrical characteristics.)

An amplifier tube that exhibits more than the 4.7 pF of C_AG exhibited by the 3-500Z in our sample problem will subject a parasitic-suppression network to substantially more RF circulating current than the 2.36 A in the sample problem. The resultant higher RMS voltage drop across Rs increases dissipation exponentially (remember, the voltage term is squared in P = E2 ÷ R!). For example, an 8877 has 10 pF of C_AG - more than double the capacitance of a 3-500Z. This means that, for the same anode voltage and LR values used in the example, the 8877's anode circulating current more than doubles (to about 5 A RMS)-more than doubling the voltage drop across R_S. Because voltage is squared in the power equation, the power dissipated by R_S would increase by a factor of greater than 4 to more than 80 W at 29.7 MHz!

Finding the optimum inductance for L_S is a tricky balancing act.(5) An inductance small enough to keep R_S from roasting on 10 meters may not necessarily be enough to prevent VHF parasitic oscillation, especially if the suppressor inductor is made of (high-Q) copper (or, worse yet, silver-plated copper) instead of low-Q nichrome or similar resistance wire.

One solution to the suppressor-dissipation problem is to use a large, noninductive, Globar resistor for R_S.(6) Unfortunately, Globar resistors are expensive. A more cost-effective solution is to use two (or more) small LR parasitic suppressors, built around inexpensive 2-W metal-film resistors, in series. For an amplifier tube that has a C_AG of 20 pF or more, it's easier to build a long-lived LR parasitic suppressor using heat-tolerant nichrome wire or ribbon for L_S and R_S than it is to build a suppressor based on a conventional resistor.

Summary

The resistors used in anode LR VHF parasitic-suppression networks dissipate considerable quantities of HF power, especially at the upper end of the HF range. This power dissipation can be calculated with sufficient accuracy to allow specification of Rs wattage by design rather thanby guess. If you would like to discuss any part of this article with me, call me at 805-482-3034.

Notes

1. Power dissipation in the suppressor resistor due to do anode current is negligible because the resistor is bypassed for dc by the suppressor inductor.
2. The minimum-anode-voltage requirement of amplifier tubes is generally related to power-handling capability: Higher-power tubes generally have higher minimum anode-voltage requirements than lower-power, tower-voltage tubes.
3. This relationship is true only for pure sine waves; the relationship between peak and RMS voltages is more complex for nonsinusoidal waveforms. For the purposes of this discussion, it's acceptable to assume that the RF waveform at the anode(s) of a vacuum-tube, linear RF amplifier is sinusoidal. The concept of RMS voltage is valuable to us because it allows direct comparison of the heating effects of dc and ac. The heating caused by a drop of 1 V RMS across a given resistor is equivalent to the heating caused by a drop of 1 V dc across the same resistor. -Ed.
4. It can be argued that the relationship between current and voltage in a parallel-LR network is better analyzed in terms of admittance because an admittance diagram (Fig A in the sidebar) can be drawn for combinations of parallel L, R and C, but an impedance diagram cannot. Despite this, however, either concept-impedance or admittance-may be successfully (and simply) applied to finding the power dissipated by R_S. Thus, we present the application of both concepts for the reader's convenience. The simple, reciprocal impedance/admittance conversions discussed in the sidebar are unrelated to finding the series equivalent of a parallel circuit or the parallel equivalent of a series circuit; such transformations offer no shortcuts in finding the power dissipated in Rs. "Finding Impedance by Solving for Admittance" discusses conversion of the parallel reactance and resistance of L R to parallel susceptance and conductance (BsGs) so that an admittance diagram can be drawn. -Ed.
5. "Richard Measures" technique for measuring the inductance of the suppressor inductor will appear in a forthcoming Hints and Kinks column. -Ed.
6. A friend of mine uses a 50-W Globar resistor for Rs in his large amplifier; the resistor gets very hot during key-down testing at 29 MHz. This resistor has never burned out, but it has glowed dull red a few times during a parasitic oscillation near 68 MHz!

 

AG6K, Richard L. Measures.