Purpose

In this installment we will discuss how to create a composite load line. This is "necessary" to establish the load line for a push pull amplifier, regardless of whether it is going to be Class A or Class AB. We will continue to use the same "invented" tube for this discussion. I will also use Class A, and the same "operating points" we used in the single ended examples. The data points used are consistent with the other 2 parts of this series, you should choose the operating point for your particular case as described earlier.

Building a Composite Load Line

This is again an iterative process, perhaps moreso than the single ended case as it's a bit more involved. Here's a step by step procedure to create a set of "composite curves" and load line.

1. Establish your intended operating point from the single ended curves. For this example, we will choose 300 volts and 65 mA quiescent current (-20 volt bias). This is the same operating point chosen for the single ended case. Note: For Class AB, you may choose a lower idling current point, which will allow a lower impedance loadline (and higher power output) without exceeding maximum allowable power dissipation. For instance, choosing a -25 or -30 volt quiescent point would allow you more "room" to increase tube current. To stay within Class A operation, at least some current must flow in each tube at all bias points, otherwise, you will be operating in Class AB.

2. Since the plate voltage increases on one plate and decreases on the other plate, we must have a way of representing this. The usual method is to take another set of the same tube curves, turn them around (so the maximum current is "down" and highest plate voltage is to the "left", and position them so that the quiescent voltages line up vertically, and the zero plate current lines touch each other as shown below: 3. Now, for the chosen bias point, create a new line (labeled "composite" curve) as follows (we are going to look at ONLY the -20 volt bias lines, one on the "upper" part of the graph, one on the "lower" part of the graph:

1. At the quiescent point (300V), subtract the currents (65-65=0 mA). This is the magenta points on the curves. Draw the "dot" at 300V, 0 mA on the composite curves.
2. Step to the next convenient voltage, (the dark magenta ones) note the currents in the upper and lower parts of the graph, subtract the two, and plot this point on the composites.
3. The next points are the ones shown in blue. Continue until you have the entire curve "dotted" for the -20 volt lines. The new "plate curve" for -20 volts is shown in the graph above.

4. You have now established one "line" of the composite curves. This is the line of varying the plate voltage symmetrically about the quiescent point while maintaining a constant grid voltage. (classic plate resistance line).

5.  Now consider another set of lines for your graph. This will be the -25 volt line for the "upper" set and the -15 volt line from the lower set. (We have put a 5 volt signal into the push pull stage). Again, for each plate voltage, subtract the two currents (for instance, 310V -25V bias upper and 290 volt -15V lower) and plot the dot. Continue until you have the -25/-15V line filled in. Note that if your quiescent grid voltage was, for instance, -30 volts, the second line you would add would use the -35/-25V lines instead of -25/-15V lins.

6.  Next do the same for the -30/-10, the -35/-5 and -40/0 volt lines. Then do the same for the -15/-25, the -10/-30, the -5/-35 and the 0/-40 volt lines. This completes the composite curves. The last 4 curves should be mirror images of the previous 4 you filled in. Noticing this saves you some time.

7. The last thing to do is fill in the maximum power dissipation curve. There are now going to be 2 of them, one for the one tube, one for the other. These will be symmetrical about the quiescent point. It is usually necessary to graph only half of the curve. The following is for the "upper" tube. Here's how to do it:

1. Get the current at the quiescent point. (In our case that's -20 volts, 300V, 65 mA.) Each tube will be drawing this current. Now find the current allowable via the power formula: I=P/E. In our case this is 20 watts at 300 volts or 66.7 mA. Since the "opposite" tube is pulling 65 mA, subtracting this from our 66.7 ma gives 1.7 mA, which is the most (lets call it unbalanced) current we can pull. Draw a dot at 300V, 1.7 mA.
2. Move to the next "convenient" grid line. Since, in our case, the gain is about 10, this will correspond to moving "left" by 50 volts to 250 volts. At this point the tube we have 250 volts on the upper tube, and the power formula states we could draw 20/250=80 mA. However, the "lower" tube has 350 volts on it, but is biased at -25 volts. Looking this point up on the original single ended load line indicates 57.5 mA. Subtract the 2 to obtain 22.5 mA. Draw a point at 250 volts, 22.5 mA.
3. Repeat at 200 volts. The formula produces 100 mA. From this we subtract the "lower tubes" contribution to the current at -30 volts(grid) and 400 volts(plate) which is 45.2 mA. Subtracting this from 100 mA produces 54.8 mA. Draw a dot at 200 volts and 54.8 mA.
4. Repeat at 150 volts (20/150=133 mA). The lower tube is biased at 450V and -35V grid (40 mA) so the maximum current is 133-40=93 mA. Draw a dot at 150V, 93 mA.
5. Repeat at 100 volts. The formula produces 200 mA, the lower tube biases at 500V and -40V grid (32.5 mA) so the maximum current is 167.5 mA.
6. Now connect these with a nice smooth curve.
7. Rotate for the lower tubes power condition... -1.7mA at 300V, -22.5mA at 350V (250V on the lower tube), 54.8 mA at 400V, 93 mA at 450V and 167.5 mA at 500V and draw that condition.

You should now have a graph that looks like this... By the way, if you have access to a spread sheet, this makes manipulating these things a LOT easier. The data for all these examples is in an Excel spread sheet. Feel free to load it if you like: Spreadsheet Data (platcurv.zip) Hint: If you replace the data in the spreadsheet with the data for the tube of your choice, most of the work is done for you, as there are embedded graphs for single ended and composite curves in the spreadsheet. There are a couple of items of interest to note in the curve above. You can definitely see the effect of transitioning from Class A to Class AB region; this is where the slope more-or-less abruptly changes in the curves above. The other thing to note is how much more linear the composite curves are than the SE case.

Push Pull Load Line

The same rules we used in the single ended case apply, with one exception: The impedance you plot is the impedance AS SEEN BY A SINGLE TUBE. Thus it is 1/4 the plate to plate load impedance. Lets use a 8800 ohm plate to plate load. In this case, each tube sees 2200 ohm load, and one "point" on the loadline is the 300 volt "0" mA quiescent value. Another convenient point is at "0" volts (namely, 300 volts drop across the 2200 ohm load). By ohms law, this is i = e/r = 300/2200 = 136 mA. As in the SE case, we need to obtain the same 5 currents and 2 voltages to give us the power output and distortion prediction. This is shown below: Remember the formulas we used:

Po=(Ve-Va)*(Ve-Va)/(8*RL)

HD2=75*(Ia+Ie-(2*Ic))/(Ia+Ib-Id-Ie)

HD3=50*(Ia-(2*Ib)+(2*Id)-Ie)/(Ia+Ib-Id-Ie)

HD4=25*(Ia-(4*Ib)+(6*Ic)-(4*Id)+Ie)/(Ia+Ib-Id-Ie)

These work, with the RL value equal to a single tubes load (2200 ohms in our example).

The values from the graph are: Va=104V, Ve=496V, Ia=89, Ib=44, Ic=0, Id=-44, Ie=-89.

 Power Out HD2(%) HD3(%) HD4(%) 8.7 watts -0- 0.38 -0-

Notice in the push pull configuration, even order distortion "cancels". Also, of more importance is the fact that the ODD order is reduced over the single ended case, even though the power output is more than double the SE case (clearly we could "parallel" 2 tubes SE and use half the impedance to get twice the power at the same distortion which would be 8.4 watts, and 0.7% third order distortion).

For those with the spreadsheet, it is left as an exercise for the student to see what happens if you don't feed the push pull stage with *exactly* the same signal level. For instance, allow the top tube to vary 0 to -20 volts to -40 volts bias, and have the bottom tube vary -35 to -20 to -5 instead of -40 to -20 to 0. Hint: you will get back some of the even order distortion products, but retain the other advantages of push pull.

The Effect of Plate Resistance on the Composite Curves

You might have noticed that the plate resistance, demonstrated on the composite curves seems to be much more constant than the single ended case. In Class A, this is definitely true. Using the tangent slope method we get: In this case, the plate resistance is seen to vary from 430 to 550 ohms. In the SE case, RL/RP ratio varied from 2:1 to 7:1 over the operating range. In the case of the push pull arrangement, this variation is cut to 4:1 to 5:1 over the operating range. The effect of speaker load variation will be cut down as well.

I hope this little tutorial has de-mystified load line creation and manipulation. We have gone thru how to create both a single ended and push pull loadline, shown the effects of non-ideal plate resistance, and provided some formulas to allow you to predict power output and distortion from your design.

Steve