THE OPERATION OF VACUUM TUBES 



39 



a curve for which the saturation is so pronounced that the grid current 

 drawn has caused a depression. 



In actual tubes the static characteristics do not follow the three- 

 halves power law exactly, but for the purposes of this paper it is suffi- 

 ciently accurate to assume that they do in the portion where grid 

 current is not appreciable. 



With the information available from Figs. 6 and 7, and the static 

 characteristic of a tube, we should be able to plot the dynamic char- 

 acteristic for any value of output impedance. For an example, let 

 Fig. 9 represent the static characteristics of a tube (Western Electric 



3.00 

 2.75 

 2.50 

 2.25 



2.00 



in 



S 1.75 



UJ 

 D- 



I I. SO 



z 



1.25 

 a. 



1.00 

 0.75 

 0.50 

 0.25 



-300 -200 -100 100 200 300 400 500 



eg IN VOLTS 



Fig. 9 — Static characteristic with lines of constant K for 2000 ohms impedance. 

 (Western Electric No. 251--4 Tube.) 



No. 251-A). Then for some value of output impedance, say 2000 

 ohms, let us plot curves of constant K on the static characteristic 

 assuming £& = 3000 volts. Points on these curves are found from 



/. = 



KR, 



which is another form of (11). For example, to find where the K — 0.5 

 curve crosses the £p = 2000 curve, we have Cpm ^^ Ep = 2000, £& — Cpm 



= 1000, thus 



I. = 



1000 

 0.5 X 2000 



1.0 



(Fig. 9). 



