The Valve Wizard

The paraphase inverter is probably the simplest (in theory) and oldest of all phase inverter circuits, and requires only one extra triode in addition to the pre-amp. The name 'paraphase' can refer to several different types of phase inverter, but this is the original circuit it referred to, and its name has stuck.

The principle of the paraphase is quite elementary: A pre-amp is designed, and the output signal is fed to the grid of one power valve, and at the same time to an inverting stage (the paraphase). The inverted signal is then fed to the other power valve.

We do not require any amplification from the paraphase, since we do not want the signal reaching the second power valve to be larger than the signal at the first, as this would cause a gross imbalance. Ideally then, we would like the paraphase to have unity gain. However, because we want the paraphase to deliver a signal which is out of phase with its input, we must take the signal off its anode. This inevitably means that it will have some gain no matter what we do. Instead, we must reduce the size of its input signal to compensate. This is very simply done, by using the first power valve's grid-leak resistor as a potential divider, and taking a small sample of the signal from there.
The paraphase could be a triode or a pentode, but it is usual to use the same valve as that which is last in the pre-amp, so that the output resistance to each power valve is the same.

The following example uses half an ECC83 (12AX7) with an HT of 300V, and the power valves have grid-leak resistors (Rg1 and Rg2) of 220k.

Setting up the paraphase is quite simple. A conservative load is chosen;- usually the same as that used on the preceding stage, to keep output impedances the same (assuming you're using the same valve type for both). In this case we will assume the previous stage used an 82k anode resistor, so we will use the same for the paraphase. Draw a load line:

Biasing: Bias the triode at the optimum point which provides the maximum possible clean headroom. We should ignore the very bunched-up grid lines where amplification is very non-linear, so in this case it looks like Vgk = -1.25V would suit, at a quiescent anode current of 1.4mA. Use Ohm's law to find the cathode resistor:
1.25 / 0.0014 = 893 ohms.
The nearest standard is 1k, although 820R is fairly close too. The cathode could be left un-bypassed to limit gain even further, but this would cause the anode output impedance to rise. Therefore the cathode should be fully bypassed (unless the preceding stage has a completely unbypassed cathode, in which case we would do the same for the paraphase). The cathode bypass capacitor can be found in the usual way, and to a close aproximation the half-boost frequency is:
f = 1 / (2 * pi * Rk * Ck)
So for a half-boost of 20Hz:
Ck = 1 / (2 * pi * f * Rk)
Ck = 1 / (2 * pi * 20 * 820)
= 9.7uF
So a 10uF capacitor would do.

Attenuating the input signal: For the best balance, a potentiometer can be used in place of the power valve's grid-leak resistor (Rg1), and the signal on each power valve grid viewed on an oscilloscope. The potentiometer can then be adjusted until the two signals are of the same magnitude and the stage is in balance. The potentiometer can either be left in place, or replaced by two fixed resistors of the correct values
If an oscilloscope is not available, the design of the potential divider can be found mathematically. Although in practice this may not result in absolutely perfect balance due to variations in valves, resistor tolerances etc., it should produce acceptable results for a guitar amp. The method is again quite simple: Calculate the gain of the paraphase in exactly the same way as for a normal common cathode gain stage by using:
Av = (mu * Ra) / (Ra + ra)

We should take into account the loading effect of the following 220k grid-leak resistor which appears in parallel with Ra, making 60k instead of 82k.:
Av = (100 * 60000) / (60000 + 50000)
= 55
(The value for ra was taken from the graph, but you could approximate it using the data sheet value.) If your design has an un-bypassed cathode you will need to use the appropriate formula of:
Av = (mu * Ra)/(Ra + ra) + (mu + 1) * Rk)

Since the gain of our paraphase is 55, we now know that we need to attenuate the signal from the pre-amp by a factor of 55 before it reaches the grid of the paraphase, so we can now work out the values for our potential divider. The value of the lower resistor must be 55 times smaller than the total resistance formed by the two resistors combined. We want the total value to be 220k, or at least close.
Lower resistor = 220000 / 55 = 4000 ohms
Upper resistor = 220000 - 4000 = 196k ohms

With a potentiometer these values can be dialled in exactly, but if using fixed resistors we will need to use common values. Rather than resort to a complex combination of series / parallel resistors, we can work backwards using standard values and see what comes close. 4000 ohms is close to 3.3k:
Total grid-leak = 3300 * 55 = 181500 ohms
Upper resistor = 181500 - 3300 = 178200 ohms
So we could probably get away with the nearest standard of 180k, in combination with 3.3k. This will actually result in an attenuation factor of 55.5 which is pretty good. You may need to play with resistor and gain values several times to find an acceptable arrangement.

Output coupling capacitors: Once these resistors have all been chosen, C1 and C2 can be chosen to provide a suitable bass roll-off when feeding the power valves. For a roll-off of 70Hz:
C = 1 / (2 * pi * f * Rg)
C = 1 / 2 * pi * 70 * 220000)
= 10nF.

The other half of the ECC83 would normally be the preceding pre-amp stage.
The two triodes could also share a common cathode resistor. No bypass capacitor would be requied since the two valves amplify out of phase, and this will help to compensate for any imbalances. It should also be noted that the examples above assumed the power valves were cathode-biased. If they are fixed-biased, the paraphase will require a coupling cap at its grid to block the negative bias voltage, and a separate grid-leak resistor. Such a circuit is shown below, and it starts to look a lot like a floating paraphase.

One problem with the paraphase is that inevitably, the signal being delivered to the second power valve will always suffer at least some distortion in the paraphase, particularly since we will probably push the preceding stage to maximum swing which may overdrive the paraphase somewhat, and so there will always be some small level of imbalance. Of course, this is not necessarily a drawback in a guitar amp- many players feel the paraphase lends a 'grubbiness' to an amp, which can be quite satisfying. Classic amps using the paraphase include the early Fender Bassman and most Supro amps- which Jimmy Page used on most recordings with Led Zeppelin.