THE OPERATION OF VACUUM TUBES 41 



the portion of the cycle \n which plate current flows will be determined 

 by the number of degrees of the cycle in which the exciting wave is 

 above the cut-off point. Thus in this case, K will depend on the 

 amplitude of the exciting wave compared to the difference between the 

 bias and the cut-off point. When the amplitude becomes large com- 

 pared to this difference, the period of plate current flow approaches 

 180° but when the peak value of the exciting wave just reaches the 

 cut-off point, the period of flow is zero. In this case K can vary from 

 almost 0.465 to zero depending on the amplitude of the exciting wave, 

 or if the exciting voltage is sufficient to produce a saturation effect in 

 the plate current wave, K might exceed 0.465. This is the case of the 

 Class C amplifier. 



If the grid bias is more positive than the cut-off point (i.e., if the 

 tube is biased "above cut-off"), plate current flows during 360° of the 

 cycle until the amplitude of the exciting voltage is sufficient to reach 

 the cut-off point. The period of flow continues to decrease until it 

 approaches 180° as the amplitude of the exciting voltage becomes large 

 compared to the difference between the bias and cut-off potentials. 

 Here K is always greater than 0.465; see Fig. 7. This is the case of 

 the Class B amplifier. 



Fig. 10 outlines the calculation of dynamic characteristics of both 

 types from the static characteristics of Fig. 9. In these calculations, 

 the cut-off point is assumed to be constant at — 300 volts and the 

 value of K is obtained from Fig. 7 after determining the portion of the 

 cycle in which the exciting voltage is above the cut-off point. For 

 Ec = — 300 of course it is always 180°. For Ec = — 350, and a peak 

 exciting voltage of 100 volts for example, the peak eg will be — 250 

 which will give 120° for the period of flow and thus K = 0.345 from 

 Fig. 7. Then on the static characteristic. Fig. 9, taking eg = — 250, 

 and by interpolating between the K = 0.30 and K = 0.35 curves to 

 K = 0.345 we find Ip = 0.112. Multiplying this by K we get /i 

 = 0.0386. Changing to r.m.s. value, squaring and multiplying by 

 2000 ohms, we obtain 1.49 watts for the output power. It is then 

 determined that this would represent a current of 0.0995 ampere in 

 the 150-ohm dummy antenna which was used in the experimental work, 

 and which was so coupled to the output tank circuit that the impedance 

 into which the tube was working was of the value indicated. 



In the calculations of Fig. 10, the cut-off point was assumed con- 

 stant at — Eb/fJL. However, it actually varies as — ep/jx and will 

 vary sinusoidally if ep is sinusoidal. No error is introduced by this 

 fact in the case of bias at cut-off (E,.. = — 300). However, in the 

 other two cases an error is introduced. In the case for bias below 



