نوع فایل power point قابل ویرایش 19 اسلاید قسمتی از اسلایدها Input Stage: DC Analysis - 1 Assuming that Q10 and Q11 are matched, we can write the equation from the Widlar current source: Using trial and error, we can solve for IC10, and we get: IC10=19A Input Stage: DC Analysis -2 From symmetry we see that IC1=IC2=I, and if the npn is large, then IE3=IE4=I Analysis continues: Second Stage: DC Analysis Neglecting the base current of Q23, IC17 is equal to the current supplied by Q13b IC13b=0.75IREF where bP >> 1 Thus: IC13b=550uA=IC17 Output Stage: DC Analysis Q13a delivers a current of 0.25IREF, so we can say: IC23=IE23=0.25IREF=180mA Assuming VBE18 = 0.6V, then IR10=15mA, IE18=180-15=165mA and IC18=IE18=165mA IC19=IE19=IB18+IR10=15.8mA فهرست مطالب و اسلایدها Biasing Currents Input Stage Second Stage Output Stage Short Circuit Protection ...
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