CHEMICAL INTERACTIONS AMONG DEFECTS IN Ge AND Si 625 



^. = 5-i 5, = -P- B, (D18) 



KIT qJS AlO 



The requirement that No contain no Fourier terms of wavelength shorter 

 than (D15) is obviously the condition that No never pass from its 

 maximum to its minimum value in a distance shorter than D(15). Since 

 we are assuming that Nd may at places be of order Na , and at others, 

 of order zero, this amounts to the condition that No does not fluctuate 

 over ranges comparable with A^^ in distances shorter than (D15). 

 The use of (Dll), (D17), and (D18) in (DIO) yields 



kT 



u 



qNaJo 



NAiyo — I) + '^ (As sin sx + B^ cos sx) 



«=o J , 



(D19) 



^ kT (70 - 1) ,kT_ND 

 q (to) qyo Na 



which by reference to the definition (D8) for 70 proves to be identical 

 with (D7), the no-space- charge condition. 



Equation (D19) is only true when No does not fluctuate through 

 ranges of order, Na , in distances smaller than //\/7o . This distance de- 

 pends on 7o and thus on the point where V = Vo , whose neighborhood 

 is being explored. Thus, we may say that there will be no space charge 

 at all points whose Vo is such as to fix 70 at a value such that 



70 > .-2- (D20) 



Amin 



where Xmin is the minimum wavelength which needs to be considered in 

 the Fourier expansion of Nd . In terms of the definition of 70 this means 



Vo <—in^ ■ (D21) 



q (?■ 



Thus, at all points where Fo is less than the right side of (D21) the no 

 space charge approximation will hold. (D21) shows, that in the limit 

 when Xmin goes toward zero, i.e. when the infinite series must be used 

 for N n , the right side of (D21) will approach — 00 and Fo will satisfy 

 (D21) hardly anywhere. Thus space charge will exist almost eveiy where. 

 I In most diffusion problems the extremes of potential will occur in re- 

 Igions where there is no space charge. Thus in one extreme N d may equal 

 jO.9 N A and in the other it may equal zero. If there is no space charge in 

 (these extremes we may write for them 



NA-Nn = V = Na exp i-qV/kT) (D22) 



in which (D3) has been used. Setting N d equal to zero in one extreme 



