SEMICONDUCTOR DIODE GATES 1143 



(Go + G)V - GVi = 7o (1) 



- GVo +ig-\- 2G)V, - GV2 = A + gEt (2) 



- GV, + (Go + G)V2 = (3) 

 Their solutions are : 



h(g + 0+ pr^) + ih + gE,)G 



° g6o + gG + 2GoG ^^' 



_ hG + {h + gg>)((?o + GO ,. 



^' gG, + gG + 2G,0 ^°' 



The corresponding equations for the disabled gate are obtained by 

 interchanging g and G in the above. 



ENABLED GATE 



Considering the enabled gate first, it should be evident from the figure 

 that there are four requirements if the gate is to be properly enabled: 



Fi > Vo (7) 



Fi > V, (8) 



Vi<E, (9) 



Fi > (10) 



Vi is a positive voltage, so equation (6) insures that (8) will always 

 be satisfied. 

 Putting the values of Vi and Vo in (7) gives 



(/. + .^.)>(^ + ^J/o (11) 



In the case that g is much smaller than Go (which is normally true) gEb 

 is effectively the current from a constant current control generator and 

 the above inequality has a simple interpretation : 



In determining whether (7) is satisfied, and the input diode held con- 

 ducting, the above inequality compares the total current which the 

 bias and control generators put into midpoint (Vi) of the gate with the 



