REFLEX OSCILLATORS 673 



APPENDIX IX 



Losses in Grids 



Although the general problem of resonator loss calculation is not treated 

 in this paper, grids seem peculiar to vacuum tubes and losses in grids will 

 be discussed briefly. 



Assume that we have a pair of grids of some mesh or network material, 

 parallel, circular, and of a diameter compared with the wavelength small 

 enough so that variation of voltage over the grids may be neglected. The 

 capacitance inside of a radius r is 



C = iirr Id 



(il) 

 e = 8.85 X 10 " farads/cm. 



Here d is the separation between the grids. For unit r.m.s. voltage across 

 the grids, the power dissipated in the part of both grids lying in the range 

 dr at r is 



dP = l{i^C)\R dr/lirr) 



(12) 

 = CO 6 irRr dr I d . 



Here R is the surface resistivity. The total power is equal to V G, where G 

 is the conductance measured at the edge of the grids, and is 



2 2 T, r-D/2 



G = P = "^^^^ [ r" dr 



CO e ttR 



'd^ Jo ■ " (i3) 



o:'e\RD'/64d\ 



It is interesting to express co in terms of X, the wavelength, c the velocity of 

 light, and then to put in numerical values 



G = (l/16)AVi?Z)Vx'<^' (i4) 



G = 1.39 X m~'RDy\^d\ (15) 



Now for the copper, the surface resistivity is 



Re = .045/ VX (i6) 



where X is measured in cm. Suppose the grid material is non-magnetic and 

 has N times the low frequency resistivity of copper. Then for it, the sur- 

 face resistivity will be A'^^ times that given by (16). Suppose that the diam- 

 eter of the grid wires is 2r and the distance center to center is o. If current 

 flowed on one half of the wire surface only, the surface resistivity parallel 

 to the wires would be 



R = N^Rcia/irr). (i7) 



