Lecture 9

# Lecture 9 - • V-sources b short I-sources b open R H R F...

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Lecture 9 Superposition

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Linear vs nonlinear • Function F(x) is linear if: – F(a·x)=a·F(x) – F(x 1 +x 2 )=F(x 1 )+F(x 2 ) • Resistors are linear: – V=IR – (a·I)R=a·V – (I 1 +I 2 )R=I 1 R+I 2 R • Diodes are not linear: – I o exp(a·V/V T ) a· I o exp(V/V T )
Linear vs nonlinear • Function F(x) is linear if: – F(a·x)=a·F(x) – F(x 1 +x 2 )=F(x 1 )+F(x 2 ) • Resistors are linear: – V=IR – (a·I)R=a·V – (I 1 +I 2 )R=I 1 R+I 2 R • Diodes are not linear: – I o exp(a·V/V T ) I o exp(V/V T ) • This allows for “superposition” – Analyze circuit’s response to each input by itself – Add up responses to get response to combined inputs

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Super position • Example: portable heart-monitor: 3 sources • Set all independent sources b 0 but one, find output, do this for each input, add up resulting outputs

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Unformatted text preview: • V-sources b short, I-sources b open R H R F V H (t) gmV1 R L Vb V1 Vout I int +-R H R F gm R L V H R H R F gm R L I int R H R F gm R L V b I int =0, V b =0 VH=0, V b =0 I int =0, V H =0 Super position: example + Vout-R R/2 R/4 R V0 V1 V2 +-+-+-R V0 +-8 7 7 7 || 4 || 2 V R R R V V R R R R out = + = = 4 R V2 +-2 2 4 4 4 2 4 || 2 || V R R R V V R R R R out = + = = 2 R V1 +-4 1 6 2 6 1 6 || 4 || V R R R V V R R R R out = + = = Super position: example 8 4 1 2 2 V V V V out + + = + Vout-R R/2 R/4 R V0 V1 V2 +-+-+-Add up outputs from superposition 7/8 1 1 1 3/4 1 1 5/8 1 1 1/2 1 3/8 1 1 1/4 1 1/8 1 Vout V0 V1 V2 Digital-to-Analog conversion...
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## This note was uploaded on 11/21/2011 for the course ECE 2100 taught by Professor Kelley/seyler during the Spring '05 term at Cornell.

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Lecture 9 - • V-sources b short I-sources b open R H R F...

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