Amplifier Assignment

# Amplifier Assignment - H = V OH V IH = 0.85V, NM L = V IL V...

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The following pertain to the 7 questions on amplifiers. 1. +/- 10V supply, linear VTC through (0,0); output saturates at +7V and –9V, 50V/V gain (= slope!) for largest output signal, output centered halfway between saturation points - -1V ouput & -1V = 50 x Input Input (operating point) = -20mV output magnitude = output range/2 = 8V 2. no-load voltage gain: A Vo = 100V/V loaded gain: v out = A Vi ± v in (R L / (R L + R out )) = A V ± v in A V = 70V/V = 100 V/V (1k/ (1k + R out )) 1k + R out = 10/7 x 1k R out = 3/7 x 1k = 428.6 ² 3. 1667 V/V = A Vo (R in / (R in + 10k)) R in || R in = R in /2 is the new input impedance 909 V/V = A Vo (R in / (R in + 2 x 10k)) divide the two : 1667/909 = (R in + 20k)/ (R in + 10k) R in = 1.99k ²

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4. Want to minimize loading (voltage dividers) – match loads A v = (10 V/V) 2 x (1/2) 3 = 100/8 = 12.5V/V 5. V OL = 0 + 0.5 = 0.5V V OH = 5 – 1.5 = 3.5V Width of transition region = 3V / (10V/V) = 0.3V V IL = 2.5 – 0.15V = 2.35V V IH = 2.5 + 0.15 = 2.65V NM

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Unformatted text preview: H = V OH V IH = 0.85V, NM L = V IL V OL = 1.85V Double the width of the transition region: V IL = 2.2V, V IH = 2.8V NM H = 0.7V, NM L = 1.7V 6. Transconductance amplifier & & + & & + = L OUT OUT S IN IN MS M R R R R R R G G & large R IN , large R OUT needed to maximize G M Transresistance amplifier & & + & & + = L OUT L S IN S MO M R R R R R R R R & small R IN , small R OUT needed to maximize R M 7. R OUT = v x /i x 3 equations: & x in v v = ( ) ( ) IN S Rout x in R R i i v -= OUT in VO x Rout R v A v i-= Solve : ( ) OUT VO x Rout R A v i-= 1 from 1 and 3 Into 2 : OUT VO OUT IN S x x R A R R R i v-+ = 1 1 1 Since R IN is large, R S ||R IN ~ R S R OUT is small & the term A VO /R OUT dominates in the denominator R OUT = v x /i x -R OUT /A VO...
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## This document was uploaded on 02/09/2012.

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Amplifier Assignment - H = V OH V IH = 0.85V, NM L = V IL V...

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