4.1 Output stage, or power stage, of a vacuum tube amplifier

The task of the output stage (or power stage) is to amplify the signal, produced by the preceding stage, to have the necessary power to obtain sound from a speaker. We will consider a power amplifier in Push-Pull configuration, where a pair of vacuum tubes are used simultaneously to amplify phase inverted copies of the same signal. However, before discussing it, we first need to introduce the Single Ended configuration, where a single vacuum tube is used to amplify the signal. The push-pull configuration is then obtained by appropriately pairing two single ended power amplifiers.

4.1.1    Single Ended configuration

Figure 16 depicts the basic schema of Single Ended power output stage built around a tetrode (or a pentode, here the suppressor terminal is not shown) in ultra-linear configuration (see Section 2.2.4) and an output transformer.

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Figure 16: Basic schema of a Single Ended power output stage.
The load of the vacuum tube consists of a transformer that accommodates the amplified signal so that it can be applied to the speaker. The transformer basically transforms the impedance of the speaker into an impedance that can be effectively applied to the anode of the vacuum tube.

The load of the vacuum tube is an output transformer. It is used to couple the vacuum tube and the speaker. In fact, the required load impedance of a power vacuum tube is generally much higher than that provided by commercial speakers. The output transformer transforms the speaker impedance into the load impedance needed by the vacuum tube.

As an example, Figure 17 depicts the average anode characteristics of the EL34 power vacuum tube. Red lines represents various loadlines corresponding to various loads applied to the anode of the vacuum tube. The maximum power that can be dissipated by the anode of an EL34 is 25W, corresponding to the dashed line marked as “Wa=25W”. All loadlines lay below this dashed line. In the figure, each loadline is labelled with the corresponding load impedance.

We can see that the various possible load impedances vary from 1.1K Ohm to 5.4K Ohm. These values are all far from reasonable values of impedance of commercial speakers, where impedance generally ranges between 4 Ohm and 16 Ohm. Accordingly, the main purpose of the output transformer is to adapt the impedance of the speakers to the impedance required by the power vacuum tubes, as discussed more in details in Section 4.1.2.

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Figure 17: Examples of various possible loadlines, for a power amplifier.
Red lines shows various options of loadlines possible with this vacuum tube. In this example possible values range from 1.1K Ohm, to 5.4K Ohm.

The DC voltage V+ is applied to the anode of the power vacuum tube through the centre tap of the output transformer primary. The output transformer also eliminates the DC applied to the anode, so that speakers just see the AC corresponding to the amplified signal.

The screen terminal of the power vacuum tube is connected to a tap from the primary of the output transformer, corresponding to the wanted percentage of the anode output signal, to obtain the ultra-linear configuration (see Section 2.2.4). Many power amplifier designs use a percentage around 43% of the signal to be given to the screen.

The resistor Rscreen, called the screen stopper, which connects the screen to the transformer screen tap, is mainly used to limit the current from the screen, and to avoid parasitic oscillations of the circuit. Screen current becomes dangerous when the power vacuum tube is overdriven. Overdrive should be avoided in Hi-Fi amplifiers, even if it is generally used in guitar amplifiers. Typically, values around 1K Ohm 2W are used. Smaller values or no resistance at all, is sometimes used in Hi-Fi amplifiers.

The resistor Rg, called the grid stopper, is used to block high frequency parasitic oscillations and reduce radio interference. The internal components of a vacuum tube produce some parasitic capacitances, generally referred as the Miller effect. The grid stopper forms a low-pass filter with these capacitances. The value of the grid stopper mainly depends on the capacitances. For example, values around 4.7K Ohm are generally used for EL34 power vacuum tubes.

Rl is the grid leak resistor. The vacuum tube receives the grid bias voltage through this resistor. Rl offers a high impedance path to ground to the AC signal coming from the previous stage and, as we will see in section 4.2.1, contributes to determine the loadline of the previous stage. Finally, Rl has also the important function of discharging the positive charge that might accumulate on the grid because of the gas ions forming in the vacuum tubes. This effect is very dangerous since it can cause thermal runaway and destroy the vacuum tube. When positive charges accumulate on the grid, the grid becomes less negative and the vacuum tube conducts more. More current in the vacuum tube produces more gas ions, which accumulate on the grid, and more current that will damage the vacuum tube itself. The correct value for the grid leak is a matter of compromise. Large values are preferable to provide previous stage output signal with a high impedance to ground. Small values are preferable to better help the grid to discharge the accumulated positive gas ions. Datasheets generally provide maximum allowed values for the grid leak, in terms of maximum impedance from the grid to the cathode. For instance, Philips EL34 datasheet specifies maximum 0.5M Ohm in class B, or 0.7M Ohm in class A or AB.

When fixed biasing is used, as in our example, a high value decoupling capacitor Cd is also generally connected, from the –Vg side of Rl, to ground. This capacitor, discussed in Section 3.6.1, prevents the input signal from reaching the grid of other vacuum tubes, by providing these residual signals with a very low impedance path to ground.

The capacitor Cc, is the coupling capacitor. It blocks DC in the signal arriving from previous stage. AC continues to the grid through the grid resistor, and to ground through the grid leak resistor. The coupling capacitor Cc and the grid leak Rl form together a high-pass filter. The value of the capacitor has to be chosen according to the desired low cut-off frequency. Note that no current goes in the vacuum tube, since the grid has a very high impedance and receives just a voltage signal.

Example 6: Inter-stage coupling capacitor of the power stageSuppose the grid leak is 200K Ohm and we want a cut-off frequency at 7 Hz. We can use the low-pass filter equation to obtain the capacitance of the coupling capacitor:image049

4.1.2    Impedance of an output transformer

In an output transformer, with one primary end connected to the anode of a vacuum tube, the ratio between the number of turns, of the primary and the secondary, determines the AC impedance seen by the anode, when a load (a speaker in our case) is connected to the secondary.

Let np and ns be respectively the number of turns of the primary and secondary. Let Zp, and Zs, be respectively the impedance seen at the primary and the impedance applied to the secondary (that is the impedance of the speaker). We have that the ratio between the primary and secondary impedances is equal to the square of the ratio between the primary and secondary turns. More formally:

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It is worth noting that, according to this equation, we can change the impedance, seen by the anode of the vacuum tube, which is the impedance of the primary of the transformer, by changing the impedance of the speaker. The impedance of the speaker is reflected to the primary of the transformer, according to the square of the ratio between primary and secondary turns.

Example 7: Impedance of an output transformerSuppose that we use a speaker with an impedance of 8 Ohms and we want to have a load of 3.8K Ohms to be seen by the anode. Then, we need a transformer where the square of the ratio between the primary and secondary turns is 3.8k/8=475. If we connect a speaker of 4 Ohm to the secondary of the same transformer, then the impedance seen by the anode will be 475∙4 Ohm=1.9K Ohm. If we connect a speaker of 16 Ohm, we will have 475∙16 Ohm=7600 Ohm.

It is also worth mentioning that speakers, generally do not have a constant impedance. Rather, the speaker’s impedance varies with the frequency of the signal being reproduced. Figure 18 shows the real impedance of speakers rated for 11 Ohm.

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Figure 18: Impedance of real speakers.
In a real speaker the impedance varies with the frequency of the sound being reproduced. The graph above refers to real impedance measured on speakers rated for 11 Ohm. Real impedance varies from 4 Ohms at frequencies around 20K Hz, to 11 Ohm around 3K Hz. There is also small difference between right and left speaker.

In this example, the impedance is very high around 100 Hz. It goes to 6 Ohm at 400 Hz, then jumps to 11 Ohm at 3K Hz, and then back at 4 Ohm to 20K Hz. This means that the slope of the loadline might vary significantly depending on the reproduced frequencies. The loadline, which is typically plotted as single line, is in fact a blurred area around the thin line.

4.1.3    Reactive load and loadline computation

A transformer is a reactive load that offers an impedance just when an AC signal goes through its primary. The transformer primary, practically, does not offer impedance when just DC is applied to it. When no signal is applied to the grid, the vacuum tube is in a quiescent state and no AC signal is produced at its anode. In this case, only the DC current goes through the transformer primary and no impedance is seen by the anode. In addition, no signal is transferred from the primary to the secondary of the transformer.

Since there is not impedance and there is no voltage drop, the anode receives the full V+ voltage, at the quiescent reactive operating point. Accordingly, the quiescent current is the one associated with V+ along the plot corresponding to the chosen grid bias voltage.

When the anode produces an AC signal, then the transformer offers a resistance and the anode voltage and current start oscillating around the operating point along the reactive loadline. The reactive loadline is parallel to the resistive loadline, which can be computed as described in Section 3.2, and shifted so that it goes through the reactive operating point. Figure 19 shows the resistive loadline (green line) and the reactive loadline (red line) in correspondence of a load of 3.8K Ohm, a voltage V+=400V, and a bias current of 40 mA. The red line is parallel to the green line, and shifted higher so that it passes through the reactive operating point (red spot).

Example 8 clarifies this.

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Figure 19: Reactive loadline.
In case of a resistive load, with an anode voltage of 400V, the loadline would be along the green line, and the operating point with a bias current of 40mA would be at the green spot. However, an output transformer has a reactive behaviour. In this case, there is a resistance just when an AC signal goes through it. There is no resistance when the vacuum tube is quiescent, so the quiescent voltage remains 400V independently of the bias current. In case of a bias current of 40mA the operating point is depicted by the red spot. When the vacuum tube amplifies an AC signal, an AC current goes from anode to cathode and trough the transformer, which now offers a resistance. The loadline, in this case, as depicted by the red line, is parallel to the resistive loadline, and shifted so that it passes through the reactive operating point. Note that the voltage reached by the anode, in case of a reactive load, might be higher than the voltage V+ applied to the transformer primary. This is due to the transformer reacting to current variations, virtually accumulating and releasing energy accordingly.
Example 8: Determining the reactive loadline

If we had a resistive load, we could have computed the loadline as we described in Section 3.2. For instance, suppose we had a load of 3.8K Ohm and a voltage V+ of 400V. At no conduction, the anode voltage would have been 400V. At full conduction, the anode voltage would have been 0 and the current 400V/3.8K Ohm=105mA. In this case, the resistive loadline would have been represented by the green line in Figure 19. A bias current of 40mA would have set the quiescent operating point at the green spot in the figure.

However, in our case the load is an output transformer, which has a reactive load. The transformer offers resistance just to AC signals that go through it. There is almost no resistance to DC. More specifically, when the vacuum tube is in a quiescent state, just DC goes from the anode to the cathode, corresponding to the bias current. In this case, the transformer does not offer resistance. This means that the voltage applied to the anode is the same than the V+ voltage applied to the transformer primary, 400V in our example. Therefore, a bias current of 40mA sets the reactive operating point as depicted by the red spot in Figure 19. When an AC signal goes through the vacuum tube, the transformer offers its 3.8K Ohm of impedance, so the anode voltage and current start oscillating along the reactive loadline represented by the red line in the figure. Note that reactive loadline is parallel to the resistive loadline, and shifted higher so that it goes through the reactive operating point. It might seem strange that the anode voltage can now reach values higher than the 400V applied to the transformer primary. However, this is due to the transformer reacting to current variations, virtually accumulating and releasing energy in accordance to these variations.

 

4.1.4    Push-Pull configuration

In a push-pull amplifier, two power vacuum tubes are used to amplify phase-inverted copies of the same signal. A phase splitter stage, discussed later in Section 4.2, before the push-pull power stage, creates the two phase-inverted copies of the same input signal. The two phase-inverted signals, amplified by the two power tubes, are combined together using an output transformer appositively conceived for push-pull configuration.

The basic schema of a push-pull amplifier stage is depicted in Figure 20. It is composed of two vacuum tubes having an identical configuration and circuitry. The primary of the push-pull output transformer has a centre tap that receives the V+ voltage. Anodes of the two vacuum tubes are connected to the two ends of the output transformer primary. Similarly, the two taps for ultra-linear configuration are connected to the two screens of the vacuum tubes. If we consider just half transformer and one vacuum tube, the schema is practically identical to the single ended configuration. A push-pull stage can be seen as two single ended stages connected together through the push-pull output transformer. When no signal is applied to phase splitter, the two power vacuum tubes receive no signal and just the bias current goes from the anodes to the cathodes. The current enters from the transformer primary centre tap and flows in opposite direction through the transformer toward the first and the second vacuum tube.

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Figure 20: Basic schema of a push-pull amplifier.
The phase splitter, which will be discussed later, takes the input signal and returns two copies of the same signal each with phase inverted with respect to the other. The two signals are given to two vacuum tubes that amplify them separately. The primary of the output transformer, used for a push-pull stage, has a centre tap that takes the high voltage. The two ends of the primary go to the anodes of the two vacuum tubes. Given that the two vacuum tubes amplify signals that are 180° one from the other, in the two ends of the transformer current varies symmetrically. That is, when in one end current increases, in the other end current decreases.

When a signal is received by the two vacuum tubes, the two amplified signals are identical, with inverted phases, and make the current to vary symmetrically, with respect to the bias current, in the two half of the push-pull transformer. When in one half of the transformer current increases, with respect to the bias current, in the other half current decreases. The same effect happens to the voltage measured at the two anodes of the two vacuum tubes. On the load line, this effect is seen as movements in opposite directions, as depicted in Figure 21. While one tube moves following the green arrow, the other one moves following the yellow arrow, and vice versa.

The power vacuum tubes used in a push-pull power stage should be perfectly matched to guarantee a symmetrical behaviour. In order to fine tune the two power vacuum tubes, the two bias voltages -Vg1 and -Vg2 should be adjustable so that the quiescent bias current is exactly the same in the two tubes.

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Figure 21: Dynamic behaviour of a push-pull amplifier.
When quiescent, the two power vacuum tubes are at the same operating point, depicted by the red spot. When a signal is amplified, since the vacuum tubes receive signals with inverted phases, they move in opposite directions. When one conducts more, as depicted by the green arrow, the other conducts less, as shown by the yellow arrow.

4.1.5    Push-pull advantages

Push pull amplifiers offer some advantages with respect to the single ended ones. Push-pull transformers core can be made smaller than single ended ones and without air gaps. In fact, the current from the centre tap flows in opposite directions through the two half of the transformer and one current cancel the electro-magnetic effect of the other. Two additional advantages are the reduction of the harmonic distortions, and the increased output power, as discussed below.

When the vacuum tube’s operating range is situated in an asymmetrically nonlinear area, that is when the grid voltage lines intersect more densely the loadline on one side rather than the other, second order harmonics are generated. For instance, in Figure 21, grid voltage lines intersect the loadline more densely with an anode voltage higher than 400V, than with an anode voltage lower than 200V. Therefore, half signal is amplified very differently than the other half signal. Remember that the input signal corresponds to different grid voltages, that is different points, identified by the intersection of the grid voltage lines with the loadline. Figure 22 shows the effect of this nonlinearity. The upper plot shows the anode current of two vacuum tubes in push-pull. Since each vacuum tube amplifies a phase inverted signal, with respect to the other, the currents at the anodes of the two vacuum tubes are phase inverted as well. Note that the current signal is somehow flattened in the lower parts of the plots. This is because, as can be seen in Figure 21, there is less current variation on the right side of the loadline, compared to the left side of the loadline, in correspondence of the same grid voltage variation. This asymmetric behaviour introduces even order (mainly second order) harmonic distortions. However, given that the two signals are phase inverted, signal flattening occurs in one vacuum tube at time. When one vacuum tube is operating in a more dense area, the other is operating in a less dense area. When one vacuum tube flattens the signal, the other vacuum tube does not. The two phase inverted signals are combined in the output transformer and even order harmonics, generated by the power vacuum tubes, are significantly attenuated, or cancelled, as can be seen in the lower plot of the figure. Note, however, that harmonic distortions introduced by previous stages or third order harmonics, introduced by symmetric nonlinearity, are amplified and not attenuated. Harmonic distortions produced by stages preceding the push-pull stage can be attenuated using global negative feedback, as will be discussed later in Section 4.4.

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Figure 22: Even order harmonic distortion in a push-pull amplifier.
The upper plot shows the anode current of two vacuum tubes configured as push-pull when they amplify a signal. Note that when the amplified signal produced by one vacuum tube flattens in the lower part, due to the introduction of even order harmonics, the other vacuum tube compensates. The result is that the signal coming out from the output transformer, shown in the lower plot, is again an almost perfect sinusoidal signal, where second order harmonics are significantly reduced.

The push-pull configuration also offers the possibility of increasing the headroom, that is the operating range along the loadline. Larger headroom implies higher output power. When the amplitude of the input signal is too large, either the vacuum tube saturates or cuts-off (see Section 3.3). In a Single ended configuration, this is avoided by setting the bias so that it works in Class A condition (see Section 3.5) and taking care that the input signal is not too large, limiting therefore the amplifier headroom. However, in a push-pull amplifier we can set the bias so that vacuum tubes operate in class AB, with very low harmonic distortions. In class AB, one vacuum tube amplifies most of the signal and cuts-off when the input signal is below a certain threshold. Cut-off happens when the signal reaches the point where the loadline intersects the horizontal axis. When operating in class AB, given that the two vacuum tubes amplify phase inverted signals, if one vacuum tube cuts-off, the other vacuum tube still amplifies the inverted signal. This is depicted in the upper plot of Figure 23. The output transformer combines the two signals and an almost perfect signal is sent to the speaker, as shown in the lower plot of Figure 23. Operating in class AB allows setting the bias so that the vacuum tubes cuts-off alternatively, increasing the overall headroom.

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Figure 23: Push-pull amplifier operating in class AB.
When operating in class AB, one vacuum tube amplifies just part of the whole signal, as shown in the upper plot. However, when one vacuum tube cuts-off the other still amplifies the signal, and vice versa. When the two signals are combined, in the output transformer, the original input signal is practically restored as shown in the lower plot.

4.1.6    Push-pull loadline in class AB

In a push pull output transformer, the impedance rating is referred to the whole impedance seen from one end to the other end of the primary. However, two vacuum tubes simultaneously use the transformer and each tube sees half impedance of the transformer. Therefore, when using a push pull transformer with an impedance of 7.6K Ohm, each anode sees an impedance of 3.8K Ohm. The loadline, for the power vacuum tubes of a push-pull amplifier, has to be drawn considering half the transformer impedance. The method for drawing the loadline, along with the relationships between the actual transformer impedance and the speaker impedance, has already been discussed in Section 4.1.2.

However, at a closer look, in case of a push-pull power stage, we have an additional consideration to make. We just said that the impedance seen by each anode is half that of the end-to-end impedance of the transformer primary. This is true only when both vacuum tubes are conducting simultaneously, for instance, when both vacuum tubes operate in class A. However, class AB operation is generally used In a push-pull power stage. In a class AB amplifier, as discussed in Sections 3.5 and 4.1.5, one vacuum tube might quit conducting while the other is fully active, and vice versa. When one of the two vacuum tubes does not conduct, just half transformer is used. In this case, the impedance seen by the anode is not half than that of the entire transformer. In fact, we already said that the impedance of the transformer goes with the square of the ratio between the number of turns, of the primary and the secondary of the transformer. When we use half transformer, we use half turns of the transformer primary and the impedance seen at the anode is one fourth of the entire transformer impedance. This can be verified using the equations discussed in Section 4.1.2.

Figure 24 clarifies these aspects. When the two vacuum tubes amplify each, a signal inverted with respect to the other, their operating points move in opposite directions along the loadline, starting from the quiescent operating point. The green and the yellow arrow represent the simultaneous position of the two vacuum tubes along the loadline. Suppose the signal amplified by the yellow vacuum tube is such that it reaches the bottom of the graph and quits conducting. The green vacuum tube, at this point, sees a different impedance at its anode and the loadline slope changes accordingly. From this point on, the green arrow follows the solid red line, rather than the dashed line corresponding to the continuation of the previous loadline. When the signal being amplified is such that the yellow vacuum tube starts conducting again, the green arrow will be back at the position where the impedance first changed. At this point, the green vacuum tube sees the original impedance again, and continues along the solid red line. Note, of course, that the impedance change is not abrupt as depicted in the figure. When the yellow vacuum tube is in the process of quitting conduction, the impedance seen by the other tube changes gradually.

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Figure 24: Loadline in a class AB push-pull amplifier.
In a push-pull transformer, operating in class AB, the loadline is not a straight line. While both vacuum tubes are conducting, the impedance seen at the anode is half than that of the entire transformer. When one of the two vacuum tubes quits conducting, just half transformer is used, and given that the impedance goes with the square of the turn ratio, the impedance will be one fourth of the entire transformer impedance. In the picture, when the yellow arrow, representing one power vacuum tube, reaches the bottom, the green arrow, representing the other vacuum tube, continues following the solid red line, rather than the dashed line.

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