Part 1:Control Supply |
Part 2:Meter Circuit |
Part 3:HT Supply |
Part 4:Power Amp |
[ Part 5: ]PA Driver |
Part 6:Tone Preamp |
Part 7:Effects Section |
Part 8:Line Preamp |
Part 9:Input Stage |

The driver circuit consists of a garden-variety preamplifier stage, followed by a "cathodyne" phase splitter (also known as a "concertina" phase splitter, because the plate and cathode voltages travel in opposite directions, much like the ends of the little accordion for which it's named).

There's not much to be said about this stage, which simply provides voltage gain, giving full output with an input of about 3 volts RMS.

The cathodyne phase splitter is perhaps the most common method for supplying equal-but-opposite signals to push-pull output stages. Its simplicity makes it very popular, but that same simplicity can also be deceptive. For instance, if the loads to the two outputs are identical, it can easily be proven that the output voltage will be the same. (Hint: since the grid draws no current, the currents through the plate and cathode resistors will be equal. Since the resistances are equal, it follows that the voltage drops will be equal. This is true both from a DC perspective, as well as an AC standpoint. The only difference is that in the AC equivalent circuit, the load resistances are effectively in parallel with the plate and cathode resistors.)

However, if the plate and cathode loads are different, the rule about equal voltages falls apart. This is why a cathodyne is not appropriate for directly driving stages that draw grid current during part or all of the cycle, and is one reason why DogZilla sports voltage followers before the finals.

This is a good opportunity to actually step through the procedure of designing a cathodyne phase splitter. Others may have different approaches, and by no means do I proscribe the following method as "the one true way." However, it works for me, and therefore might be useful for others also. It's rather intermediate between the rigorous textbook approach, and the more common "seat of the pants" approach used by most hobbyists. If you want to skip this mini-tutorial, click

First we get the plate curves for the triode we intend to use -- in this case the time-proven workhorse 6SN7. (The popular 6FQ7/6CG7 will be very similar). We'll also keep in mind the maximum power dissipation, given as 5 watts for either section, or 7.5 watts total for both. To be safe, let's say a maximum of 3.75 watts. The other important design spec is the required output voltage swing. For the 807, in Class AB

In the case of DogZilla, we have quite an array of possible supply voltages to choose from; we could even get fancy and use split (positive and negative) supplies. However, let's start simple and postulate that we'll be running from the +420 volt supply. That gives us the first point on our load-line, so we place a point at 420 volts on the X (plate voltage) axis. This represents the plate-to-cathode voltage if the tube were completely cut-off (zero plate current).

Next we do a bit of playing around. Most cathodyne circuits, in order to provide the maximum output swing, will be biased with the plate-to-cathode voltage at a little over 1/2 the supply voltage. In our case, this would be on the order of 210-250 volts. Let's say 250 volts as a starting point. Given our maximum power dissipation limit of 3.75 watts, this would mean a quiescent plate current of (3.75/250) = 15 mA. We can draw a tentative line through this point (250V, 15 mA) from the V

Note that the distance between adjacent grid lines is not constant along the load line. They get scrunched together more and more at the cutoff end (right side of the graph). This represents non-linearity, which is the fundamental cause of harmonic distortion. If you're so inclined, draw a few more load lines (or just use a straight-edge to eyeball them). You'll find that linearity gets worse with very shallow load lines. While such load lines will be very kind to your tubes, they also won't necessarily give you acceptable sound. However, there will generally be quite a wide range of load-line slopes over which linearity does not change significantly.

The thick load-line in the graph represents the value I settled on for DogZilla. If you went to the trouble of plotting the linearity, you'd find that the linearity curve is somewhat

Having settled on a load-line, let's determine the plate and cathode resistor values. The point where the load-line intercepts the Y axis represents the current that would flow if the tube were short-circuited. In the graph above, this would be 9.6 mA. If we divide this into the supply voltage, it will give our total resistance, or (420/9.6) = about 44k. Given that the plate and cathode resistors must be equal, it follows that both will be 22k in value.

Now, let's determine our maximum range of excursion along the load-line. The left-hand side is easy, since we have to remain in Class A

Next, let's calculate our plate and cathode voltages at the quiescent point. 4.4 mA through 22k represents a voltage drop of 96.8 volts. So the cathode will be at 96.8 volts, and the plate at (420-96.8)=323.2 volts. As a cross-check, the plate-to-cathode voltage will be the difference between the two voltages, or 226.4, agreeing nicely with that point on the graph. Now, let's look at our actual plate and cathode voltages at the two extremes, to see if we have enough headroom. We'll also use grid voltage = -9 volts as a reference, to get an idea of linearity. At grid-cathode volts = 0, plate voltage will be 75 volts. That means that the balance of the 420 volts will be evenly divided between the plate and cathode resistors, or (420-75)/2 = 172.5 volts, or (172.5-96.8)=75.7 volts above quiescent. At the other extreme of grid voltage swing (V

You may have noticed that the excursions are not equal in both directions. Each limit is about 10% off from the median value. Also, the midpoint of the grid voltage swing (-10 volts) is close to, but not identical with the midpoint of plate voltage swing (227.5 volts). This is due to non-linearity. Since we really don't /need/ a full 138 volts of swing, let's see if we can improve our linearity by shifting our operating point somewhat to the left. So let's try placing our quiescent grid voltage at -7 volts, corresponding to a plate voltage of 200V and current of 5.2 mA. The excursion on each output would therefore be (200-75)/2=62.5 volts leftward, and (300-200)/2=50 volts rightward, or a total of 112.5 volts total excursion, still quite reasonable. However, contrary to expectations, the deviation from the mean is now 12%! So we haven't gained anything, and might as well stay with our original figures. (It bears noting at this point that this non-linearity will be reduced by the mu of the tube in actual practise, due to local negative feedback on the cathode resistor.)

All that's left now is to define the grid bias. We've found that under quiescent conditions, the cathode will be sitting at 96.8 volts. The grid will be at -9 volts relative to the cathode, or 87.8 volts. This can be used to calculate the divider network on the grid. If you calculate it out, you'll find that with the resistor values given, the actual value comes out to 89 volts, certainly close enough. Note that the top resistor is broken into two sections, with capacitor

As a final note, it bears pointing out that the voltage divider grid bias scheme (as shown here) is an excellent alternative to the usual approach of referencing the grid to a tapped cathode resistor. The present approach is especially useful if the supply voltage is regulated, and will be extremely stable with changes in the tube's characteristics as it ages. The tapped cathode resistor approach will generally be more useful if the supply voltage can be expected to vary, as it will tend to offset such changes; however, it's more prone to drift in operating point with tube aging.

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