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Published byPenelope Vines Modified over 2 years ago

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1 Practical OP-AMP Circuits

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2 i i Integrator

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3 Differentiator i i

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4 Voltage Subtractor or Difference Amplifier Applying KCL at node ‘a’, (V 1 – V 3 )/R 1 = (V 3 – V 0 )/R 2 (1) Applying KCL at node ‘b’, (V 2 – V 3 )/R 1 = V 3 /R 2 (2) Rearranging (1), we get, (1/R 1 + 1/R 2 )V 3 – V 1 /R 1 = V 0 /R 2 (3) Rearranging (2), we get, (1/R 1 + 1/R 2 )V 3 – V 2 /R 1 = 0 (4) (3) – (4), we get, (V 2 – V 1 )/R 1 = V o /R 2 or, V o = R 2 (V 2 – V 1 ) /R 1

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5 Ex.1) For the amplifier circuit find v o /v 1

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6 Solution : i o = i – i 1 = v i /R 1 – v/R = v i /R 1 – (- iR)/R = v i /R 1 + i = v i /R 1 + v i /R 1 = 2v i /R 1 v o = v – i o R 2 = - iR – 2v i R 2 /R 1 = - v i R/R 1 – 2v i R 2 /R 1 or, v o = - v i (R + 2R 2 )/R 1 Hence, v o /v i = -(R + 2R 2 )/R 1

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7 Ex.2) Find V o in the following circuit.

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8 V 2 = V 1 = 4/3 V Or, Vo – 4/3 = 2(4/3 +3) Or, Vo = 4/3 + 8/3 + 6 = 10 V Solution

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Lecture 251 More On Op Amps. Lecture 252 Review The ideal op amp model leads to the following conditions: i + = i - = 0 v + = v - The op amp will set.

Lecture 251 More On Op Amps. Lecture 252 Review The ideal op amp model leads to the following conditions: i + = i - = 0 v + = v - The op amp will set.

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