Now, lets assume that this particular triode cuts plate current flow off when the grid reaches -24
volts. This point is reached at time b when Ein is -18 and the bias is -6 (-18 and -6 = -24). Plate current
remains cut off for as long as the grid is at -24 volts or greater.
With zero current flowing in the plate circuit, there is no voltage drop across RL. The entire plate-
supply voltage, Ebb (300 volts), appears as plate voltage between the cathode and the plate. This is shown
at time b in the output signal waveform. Between time b and time c, the grid voltage is greater than -24
volts. The plate current remains cutoff, and the plate voltage remains at +300. The output waveform
between time b and time c cannot follow the input because the plate voltage cannot increase above +300
volts. The output waveform is "flattopped." This condition is known as AMPLITUDE DISTORTION.
When the grid voltage becomes less negative than -24 volts, after time c, the tube starts conducting,
and the circuit again produces an output.
Between time c and time d, the circuit continues to operate without distortion. At time e, however,
the output waveform is again distorted and remains distorted until time f. Lets see what happened.
Remember that every cathode is able to emit just so many electrons. When that maximum number is
being emitted, the tube is said to be at SATURATION or PLATE SATURATION. Saturation is reached
in a triode when the voltages on the grid and plate combine to draw all the electrons from the space
Now, as our grid becomes less negative (between time c and time d), and actually becomes positive
(between time d and time e), the plate current increases, the voltage across RL increases, and the plate
Apparently when the grid voltage reached +12 volts at time e, the plate current reached saturation.
Maximum plate current (at saturation) results in maximum voltage across RL and minimum plate voltage.
Any grid voltage higher than +12 volts cannot cause further changes in the output. Therefore, between
time e and time f, the plate voltage remains at +100 volts and the waveform is distorted. This is also
This has been an explanation of one cycle of an input signal that overdrives the tube. You should
notice that, using the same circuit, a 50-volt peak-to-peak input signal caused a vastly different output
from that caused by the 6-volt peak-to-peak input signal. The 6-volt peak-to-peak signal did not overdrive
the tube. When the input signal was increased to 50-volts peak-to-peak, the tube was forced into cutoff
when the grid was driven to -24 volts, and into saturation when the grid was driven to +12 volts (the grid
voltage plus the signal voltage.) During these periods, the tube could not respond to the input signal. In
other words, the output was distorted. A method commonly used to partially overcome distortion is to
vary the bias voltage on the grid. The point at which the tube goes into cutoff or saturation can then be
For this reason tube biasing is of great importance in most tube circuits.