ZERO TEMPERATURE COEFFICIENT QUARTZ CRYSTALS 371 



d, = 0.8892 cef (BT) + 0.9328 cef (AT) - 0.8221 cf, . 



cu = 0.6282 cef (BT) - 0.4016 CefCAT) - 0.2264 d^ ^ 



Inserting the values from Table I and equation (9) the three elastic con- 

 stants become 



ct, = 58.58 X 10'°[1 - 171 X 10-«(r - 50) 



- 212 X 10-«(r - 50)2 _ 65 X 10-i2(r - 50)3 + . . .] 



cfe = 40.26 X W[l + 168 X 10-«(r - 50) 



(12) 



- 5 X lO-^r - 50)2 _ 167 X 10-i2(r - 50)2 + . . .] 



cl^ = 18.20 X 10'°[1 + 90 X 10-6(r - 50) 



- 270 X 10-9(r - 50)2 _ 630 x \Q-^\T - 50)^ + • • •] 



To determine the frequency and temperature coefficients for any angle, 

 one substitutes the values of the elastic constants and the temperature ex- 

 pansion coefficients in the frequency equation (4), which results in the 

 expression 



P = 10io[(3.802cos^^ + 5.526 sin^0 - 3.426 sin (9 cos (9) + 10-%T - 50) 



[668 cos^ e - 828 sin^ - 336 sin 6 cos 6- 13.5 sin^ d cos^ (9-46 



sin' e cos 0] + 10-9(r - 50)2[395 cos' ^ - 1160 sin' + 354 (13) 



sin ^ cos ^ - 12 sin' cos' ^ - 24 sin' cos 0] + 10r'^(T - 50y 



[310 cos' - 884 sin' + 1130 sin cos 0] H ] 



III. Properties of AT and BT Cut Crystals Having Zero 

 Temperature Coefficients at High Temperatures 

 The process for obtaining high frequencies cuts of the AT and BT type 

 that will have zero temperature coefficients at a high temperature — for 

 example 200°C — is to substitute for T in equation (13) the value 



T = 200° + AT (14) 



Inserting this value in (13) and collecting the results in powers of AT, we 

 find 



f X 10-10 = [3.913 cos' + 5.373 sin' - 3.465 sin<9 cos^ - 0.0023 



sin'^ cos' (9 - 0.0075 sin' l9 cos 0] + (AT) X 10-« 



[807 cos' - 1236 sin' - 154 sin ^ cos ^ - 17 sin' 



cos' 0-53 sin' cos 0] + (AT^ X 10-9[534 cos' (15) 



- 1558 sin' + 862 sin cos - 12 sin' cos' (9-24 



sin' cos 0] + (ATy X 10-i2[310 cos' - 884 sin' 



+ 1130 sin cos 0] 



