**Designing Long-Tail Pairs - The Load Line Approach**

**General**

The Marshall/Fender phase inverter is commonly known as a "long-tail pair", or "Schmitt" type phase inverter, or phase splitter (actually, the original Schmitt inverter was a differential pair with a large "tail" resistor; the "standard" guitar amplifier phase inverter is a self-biased version of this circuit that works better with positive-only power supplies and ground-referenced inputs). This paper details the design of a long-tail pair using the information obtained from the plate curves for the chosen tube.

**The design process**

Following is a schematic diagram of a typical phase inverter found in some guitar amplifiers:

For this design example, we will choose a seldom-used, but good-sounding tube, the 6SL7GT octal high-mu dual triode.

- The plate resistors

The 6SL7 has an internal plate resistance, Rp, of around 44K. Using the general rule of thumb for the "optimum" triode plate load, two times Rp, an 88K load would be good, so we'll choose 82K as the nearest standard value, and use this for the two plate resistors. Note that in most cases, the out-of-phase output plate resistor will have to be made around 10% smaller than the in-phase output to achieve perfect balance between the two outputs. However, if the second input is to be used, as a second channel input, for example, the resistors should be made equal, as it will only be balanced for one input, and the other input will be unbalanced even more.In order to plot a load line on the characteristic curves, the maximum current and maximum voltage is first determined, and a straight line is drawn to connect the two points. Since the plate voltage is at minimum when the plate current is at maximum, the a plate voltage of zero volts would mark the x-axis value of one of the end points. If the plate voltage is zero, this means that all the plate supply voltage is dropped across the load resistor, so the maximum current can be calculated by dividing the plate voltage by the load resistor. In this example, this point would be 300V/82K = 3.66mA. The intersection of the plate voltage of 0V and the plate current of 3.66mA is then one end point of the load line. The other end point is simply the intersection of the plate supply voltage and the zero current line, because the plate voltage is at maximum when the plate current is zero, as no voltage is dropped across the load resistor. A straight line is then drawn between these two points. It will have a slope equal to the negative reciprocal of the load resistance, and will intersect the x axis at the plate supply voltage as shown in the picture below.

In a normal common-cathode stage, assuming a 300V plate supply, the 6SL7 looks like it would be okay with a bias point of around -2V, which would result in a quiescent plate current of 1.1mA and a quiescent plate voltage of around 210V (the intersection of the 82K load line and the -2V grid curve on the plate curves shown below). This would require a cathode resistance of 2V/1.1mA = 1.82K, so 1.8K would be the nearest standard value. This would give a voltage swing of around +86V/-76V, or 162V p-p, at the point of clipping, as shown on the plate curves below, where the load line intersects the Vg = 0V curve and the Vg = -4V curve. This means the tube has a voltage gain of around 162V/4V = 40.5. The asymmetrical output voltage swing indicates the presence of some even order harmonic distortion. It could be minimized by careful adjustment of the load line resistance and the quiescent operating point if desired. Note that the bias point is midway between the two cursors, at a point on the load line corresponding to a grid voltage of -2V and a plate current of 1.1mA. Note also that a curve corresponding to the 1W plate dissipation limit has been added to the graph. This is the upper limit of the operating area of the tube. The load line should not cross into this area, or the tube may be damaged due to overheating the plate element.

- The cathode resistor

The preceding bias point is good for a common cathode stage using the 6SL7, however, since this will be used as a self-biased phase inverter, the plate-to-cathode voltage will be lower by around 80V or so, due to the required voltage drop across the "tail" resistor. Why 80V? Well, this is an educated guess, based on the fact that we want to drop around 25-30% of the total available plate voltage across the resistor, in order to make it a fairly large value, which is good for balance, as it better approximates a constant current source. We don't want to drop more than that, or the amount of headroom will be compromised.

This is the equivalent of running with a plate voltage of 220V instead of the 300V supply we started with. With a 220V supply and an 82K load, the 6SL7 would be best biased around -1.25V and 0.85mA, with a quiescent plate voltage of around 150V. This would give a voltage swing of around +52/-47V max, or 99Vp-p, with an 82K load. In order to achieve this bias point, a cathode resistance of 1.25V/0.86mA = 1.45K would be required. Half this value (because the resistor is shared between the two tubes) would be around 725 ohms, so a good standard value would be 680 ohms. This bias point will then result in a quiescent plate voltage of around 80V + 150V = 230V. (The 80V is the cathode voltage, and the 150V is the quiescent plate voltage relative to the cathode as determined from the plate curves). This new load line is shown below, with the cursors indicating the maximum and minimum undistorted plate voltage swings. The quiescent bias point is midway between the two cursors, at the point on the load line corresponding to a grid voltage of -1.25V and a plate current of 0.85mA. Note that the two end points of the new load line are (220V, 0mA) and (0V,2.7mA). The 2.7mA figure was arrived at by dividing the new plate-to-cathode voltage of 220V by the load resistor value of 82K, for a result of 2.7mA.

- The "tail" resistor

Now, the last thing to determine is the value of the "tail" resistor. Since we decided to raise the cathode by around 80V, and the total cathode current is 2*0.86mA = 1.72mA, the required tail resistor is 80V/1.72mA = 47K. This is a relatively large value in comparison to the bias resistor value of 680 ohms, so it should approximate a constant current source well enough for our needs. As mentioned earlier, if this resistor is too large, the headroom, i.e. available output voltage swing, will be too low. The reason for this can be seen from the above plate curves, by reducing the plate voltage and redrawing the load line. As the available plate-to-cathode voltage is lowered, the maximum available output voltage swing is reduced.

- The grid resistors and input coupling capacitors

A 0.1uF cap should be used on the second input, the grounded grid side for best low frequency balance, and the "standard" 1Meg resistors can be used for the grid bias resistors. Since the effective input impedance will be around 2Meg, the input coupling cap should be no more than around 0.005uF, for a -3dB point of 16Hz. Too high a cap can lead to unwanted "blocking" distortion, but you can probably go up to 0.022uF without too much trouble, depending on the low frequency bandwidth of the circuitry in front of the phase inverter.

- The completed design

Following is a schematic of the completed phase inverter design. Note that due to the better balance provided by the large "tail" resistor, which acts more like a constant current source, the in-phase plate load resistor is made the same size as the out-of-phase load resistor. It is not necessary to make it larger, as is the case in the "standard" Fender/Marshall type inverter, which uses a much smaller "tail" resistor.

Copyright © 2000-2014 Randall Aiken. May not be reproduced in any form without written approval from Aiken Amplification.

*Revised 02/18/14*