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# How Vacuum Tubes in Linear Circuits Work

By: w8ji.com

## How the PA Tube Converts DC anode voltage to Radio Frequency Power

We know the tank circuit stores energy. We now understand the conversion process where DC is converted to AC (or RF) power. We also must understand energy must be transferred out of the tank at a rate equal to or exceeding the rate at which it is supplied by the downward “pull” of the tube. If we do not remove energy at a sufficient rate, voltages and currents increase until a new point of equilibrium is reached.  Voltage at point “A” can actually swing well beyond twice +HV on upward excursions, and below zero volts (becoming negative) on downward excursions.

The maximum voltage with a load or drive fault can be tremendously higher under the fault condition than typical properly operating tank working voltages (when energy is being removed at the proper rate). If we do not remove energy from the tank at the same rate the power supply is supplying energy, the voltage in the tank will increase until something absorbs that energy!

We have the same basic tank system as discussed earlier, but with light loading compared to drive level the tank is pulled down very hard by the tube. Minimum Rp is reached early in the cycle, before the tank voltage reaches it’s minimum swing.  This is easy to do, since the tube only pulls the system down and the tank stores the energy of that downward tug.As the plate voltage swings below zero (negative) from the tank energy, the tube is already cutting off. Nothing clamps or prevents point A from going negative. D1 effectively takes the tube out of the circuit. By the time the tank reaches it’s minimum, the tube’s grid-cathode voltage is already on its way positive. The tank free wheels positive, and can overshoot the +HV supply by several times the supply voltage. If loading is light enough and Q is high enough, this continues until the energy stored in the tank reaches equilibrium with energy transferred to the load, or a component fails and the arc dampens the tanks gyrations. See the practical demonstration page for a real working model with waveforms!

## Proper Plate Load Impedance or Operating Impedance of an RF Amplifier

Since the anode-to-cathode resistance of the tube varies over the RF cycle, how do we establish the plate resistance of the tube? It all goes back to energy transfer. We must extract the same energy available from the power supply power drawn by the time-varying tube resistance, less any dissipation in the tube or loss in the tank components. Otherwise the anode voltage swing will not be in equilibrium, voltage swing will continue to build until something eventually extracts the energy at the same rate as it is applied. If we try to extract too much power to the load compared to the optimum value, the output device current will increase. The tube path looks like this: The tank circuit transforms the load impedance of the antenna system or dummy load up to a new value. This resistance appears at C1, and is coupled through C1 to the tube and power supply junction. This is the resistance value we are actually calculating when we plan the plate resistance, Rp.

This resistance has to match the optimum E/I of the desired RF frequency appearing at the anode of the tube. Let's consider the fundamental frequency only in a typical class AB amplifier operating near the class B end of the operating point. Let's assume we have 3000 volts of high voltage, and the tube performs best swinging down from 3000 volts to 500 volts. This is 2500 volts of peak anode voltage swing, and the flyback effect causes the anode to reach 3,000 (B+ value) + 2,500 (downward swing) = 5,500 volts on peak. Our vacuum tube has a peak dc anode voltage of 5,500 volts in NORMAL linear operation. If you develop an antenna system problem or have a bad lightning arrestor or relay, the peak voltage could be several times that amount.

 If we think about this, all tubes should be tested to withstand at least twice the maximum dc anode voltage, and preferably more. This is a noteworthy point because test data from various Chinese manufacturers has indicated they test tubes at less than the rated dc supply voltage. They should actually be testing and gettering tubes to withstand significantly more than twice the highest expected dc supply voltage. The old Eimac 3-500Z's, rated at 4,000 volts supply voltage, used to test to over 12,000 volts of anode-to-grid breakdown voltage. They virtually never arced when new. I've seen test data from Chinese 572B's where the tubes were tested for arcing and failure at 1,500 volts dc anode voltage. Even an 811A should not be tested that low. To be reliable an 811A tube should hold off nearly 5 kV with minimal leakage and no sign of arcing. A 572B should hold off over 8,000 volts.

With the 3-500Z example, we might have a useful RF output power of 800 watts with the anode swinging 2500 volts peak (5000 volts p-p), or about 1700 volts RMS at the fundamental. There are harmonics involved, so to do this right we would have to use a Fourier (harmonic waveform) analysis of the waveform, and there is just such an analysis that was developed for vacuum tubes called a Chaffee analysis, but for casual use we can use approximations. The typical rule of thumb for tubes with a 15% minimum plate voltage is we multiply indicated anode current  by 1.6, and use the supply voltage. Ep/(Ip*1.6) = Rp. Using this example we have 3,000/(.4*1.6) = 4,688 ohms.Now let's look at another rough method. We know the swing is about 2500 volts, so RMS is about .7 times that. 1750^2 / 750 = 4,083 ohms. Using a Chaffee analysis the optimum load resistance comes out to 4500 ohms. The error in approximations is in not allowing for harmonic content.

## Tank Q

We often get far too worried about tank circuit Q. There really is very little change in efficiency as Q moves from a minimum of (SQRT Rp/Rl)+1. If we had a 4500 ohm optimum anode load resistance (Rp) and a minimum load resistance of 25 ohms, we would want a tank Q of at least SQRT 4500/25= 13.4 + 1 = 14.4. A  typical HF tank circuit, even with modest sized components and such a large impedance ratio, only has approximately 4 % power loss. If we doubled the Q, tank loss would only be 8 %. Lower impedance ratio tanks have even less loss and are less worrisome.Most of the power loss is in the conduction angle of the tube, not tank Q. It is certainly not in a blocking capacitor, would would overheat and disintegrate if it had more than a tiny fraction of just one percent loss.

Don't get overly concerned about Q. Most of the time you will be off 20-30% anyway because a handbook formulas for Q and Rp are both just rough approximations.