# sol2 - ECSE 330 Introduction to Electronics Solutions to...

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Unformatted text preview: ECSE 330 Introduction to Electronics Solutions to suggested problems Problem 2.16 v o i 2 i 3 i o i 1 1K Ω 10K Ω 2K Ω-0.5 V v o v i =- 10 K Ω 1 K Ω =- 10 V/V = ⇒ v o =- 10 V/V × v i =- 10 V/V × . 5 V =- 5 V i o = v o 2 K Ω =- 5 V 2 K Ω =- 2 . 5 mA i 1 = i 2 = v-- v o 10 K Ω = 0 + 5 V 10 K Ω = 0 . 5 mA i 3 = i o- i 2 =- 2 . 5 mA- . 5 mA =- 3 mA The additional current comes from the opamp. Problem 2.20 v o v i R 2 1 R 1 (a) A v = v o v i =- R 2 R 1 =- 100 V/V Since R 1 = R in = ⇒ R 1 = 1 K Ω A v = v o v i =- R 2 R 1 = ⇒ R 2 =- A v × R 1 =- 100 V/V × 1 K Ω = 100 K Ω (b) A vo = 2000 V/V = ⇒ v- =- v o 2000 KCL at v- = ⇒ v i- v- R 1 = v-- v o R 2 = ⇒ v i- (- v o 2000 ) 1 K Ω =- v o 2000- v o 100 K Ω = ⇒ v o v i =- 95 . 19 V/V (c) Let R = R x || R 1 . To restore the closed loop gain to its nominal value, we need v o v i =- 100 V/V KCL at v- = ⇒ v i- v- R = v-- v o R 2 where v- =- v o 2000 and v i =- v o 100 = ⇒ R = 949 . 5Ω since R = R x || R...
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sol2 - ECSE 330 Introduction to Electronics Solutions to...

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