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I highly recommend giving the article cited in post #15 a read (the one on the Valve Wizard site). Just ignore the math and stuff you don't understand; I'm sure there is a lot in it that will help you out.
I'm no expert, and learning all the time, but in a very small nutshell here is my take on it:
The operating point plotted in the middle of the load line is the idle point. Say you choose a B+, and plate and cathode resistors that put it at the intersection of the -1V grid line. An input signal applied to the grid will be added to the -1V and move the operating point left and right along the load line to the intersections of adjacent grid lines. The output signal is the corresponding plate voltage swing seen on the x-axis, the AC part of which is passed along to the next stage through the coupling cap.
In this case, if the signal is more than 2V peak-to-peak the positive part above 1V won't be seen at the output because if the grid exceeds zero volts (i.e., becomes more positive than the cathode), it will begin to conduct and effectively ground out the excess input signal. As a result, the output signal will be clipped on its negative swing, somewhat softly though, resulting in sweet-sounding even-order harmonics.
If the idle operating point is placed further to the right, say at -3V, the positive side of the input signal will come through just fine, but now the negative side will be first compressed (the plate signal for a given grid signal is reduced, because the grid lines get closer together) and then clipped, harder this time, when the tube ceases to conduct. This produces a squared-off waveform that has more odd-order and higher harmonics.
So, where you place the idle points on the load lines determines at what signal level distortion of the output will begin, and what kind of distortion it will be.
MPM
I'm no expert, and learning all the time, but in a very small nutshell here is my take on it:
The operating point plotted in the middle of the load line is the idle point. Say you choose a B+, and plate and cathode resistors that put it at the intersection of the -1V grid line. An input signal applied to the grid will be added to the -1V and move the operating point left and right along the load line to the intersections of adjacent grid lines. The output signal is the corresponding plate voltage swing seen on the x-axis, the AC part of which is passed along to the next stage through the coupling cap.
In this case, if the signal is more than 2V peak-to-peak the positive part above 1V won't be seen at the output because if the grid exceeds zero volts (i.e., becomes more positive than the cathode), it will begin to conduct and effectively ground out the excess input signal. As a result, the output signal will be clipped on its negative swing, somewhat softly though, resulting in sweet-sounding even-order harmonics.
If the idle operating point is placed further to the right, say at -3V, the positive side of the input signal will come through just fine, but now the negative side will be first compressed (the plate signal for a given grid signal is reduced, because the grid lines get closer together) and then clipped, harder this time, when the tube ceases to conduct. This produces a squared-off waveform that has more odd-order and higher harmonics.
So, where you place the idle points on the load lines determines at what signal level distortion of the output will begin, and what kind of distortion it will be.
MPM
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