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Unformatted text preview: = 5 k Ω , or − 10 V ≤ V tions are from the final ans , RF = 30 k Ω Vi ≤ +10 V m last slide o swer will ma , VB = 3 V of lecture 21 atch. V, and VSAT 1. Use the no T = 12 V. S oninverting ketch the g configuratio graph on Problem op amps b) Try to Lecture 2 Problem V IN <2V. Solution inverting high wh The thre referenc this case 6: Obtain a are ideal and do it yourse 21). Here VR 7: Design a n: We know g configura en input is eshold is set e voltage, w e. an expression d power sup elf. Use the e REF = 2V. N a comparator it is a nonation as outp high. t by the which is 2V n for and plo plies are at L equation for Notice that h r that has anput is V in ot V OUT vs V L + and Lvo noninvertin here = R1 / n output that IN for the fo olts. ng Schmitt T / R2. Find V is high when llowing circ Trigger (give V HYH and V H n V IN >2V an cuits. Assum en on last sli HYL . nd low when e the ide of n...
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This note was uploaded on 11/08/2011 for the course EE 230 taught by Professor Mina during the Winter '08 term at Iowa State.
 Winter '08
 mina

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