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chapter 06 - Chapter 6 The Operational Amplifier Exercises...

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Chapter 6: The Operational Amplifier Exercises Ex. 6.4-1 1 2 2 1 0 0 1 s s o o s v v v R R v R v R + + = = + Ex. 6.4-2 a) 2 1 2 0 3 4 0 0 a s a a R v v R R v v v R R = + + + = 4 2 3 1 2 1 1 o o a s v v 4 3 R R R v R v R R R = + = + +  b) 2 2 2 1 1 2 2 When then 1 and 1 o s 4 3 R R v R R R R R R v >> = + + ± ± R 6-1
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Ex. 6.5-1 2 1 2 1 2 0 0 o o s o s v v v R R v R v R R + + = = + Ex. 6.6–1 1 1 1 3 3 0 0 1 when 100 k and 25 k then 100 10 1 5 25 10 f in out in out in f f out in R v v v v v R R R R R v v + + = = + = = = + = 6-2
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Ex. 6.7-1 3 3 2 1 3 2 3 3 3 2 2 1 2 1 3 3 2 10 10 10 10 1 1 10 10 10 10 10 10 10 10 2 1 10 10 10 10 R R v v R R R v v R × × = + + + × + × × ×  = − + + + × × 1 3 v We require ( ) 3 1 1 4 5 v v = 2 v , so 1 3 1 3 4 2 1 10 10 10 k 10 10 R R = + = × = × and 2 3 2 2 2 3 2 1 10 10 5 2.5 k 5 10 10 R R R R R = + × = = + × 6-3
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Ex. 6.7-2 As in Ex 6.7-1 2 1 3 2 3 3 2 2 1 10 10 10 10 R R v v R = − + + + × × 1 v We require ( ) 3 1 4 6 5 v v = 2 v , so 1 3 1 3 6 2 1 20 10 20 k 10 10 R R = + = × = × and 2 3 2 2 2 3 2 4 4 40 10 5 40 k 5 10 10 R R R R R = + × = = + × Ex. 6.8-1 3 1 1 3 3 1 Analysis of the circuit in Section 6.7 showed that output offset voltage = 6 (50 10 ) For a A741 op amp, 1 mV and 80 nA so 3 output offset voltage 6 (50 10 ) 6 (10 ) (50.10 os b os b os b v i v i v i µ + × = + × + 9 )(80 10 ) 10 mV × = Ex. 6.8-2 2 2 2 1 1 1 1 o in os R R v v v R R = + + + b R i ( ) ( ) ( ) 2 1 1 3 3 9 3 When 10 k , 2 k , 5 mV and 500 nA then output offset voltage 6 5 10 10 10 500.10 35 10 35 mV os b R R v i = = × + × × = 6-4
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Ex. 6.8-3 ( ) 3 1 Analysis of this circuit in Section 6.7 showed that output offset voltage = 6 50 10 For a typical OPA1O1AM, 0.1 mV and 0.012 nA so os b os b v i v i + × = = ( ) 3 3 9 3 6 3 output offset voltage 6 0.1 10 50 10 0.012 10 0.6 10 0.6 10 0.6 10 0.6 mV × + × × × + × × = ± Ex. 6.8-4 Writing node equations 0 0 0 s o a b i s o o b v v v v v R R R R R i v A v R R v v i s R R + + = + + + = After some algebra ( ) ( ) ( ) ( ) 0 0 0 i s i f o v s f i s a f i s i R R R AR R v A v a R R R R R R R R R AR R + + = = + + + + + + For the given values, 2.00006 V/V. v A = − 6-5
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Problems Section 6-4: The Ideal Operational Amplifier P6.4-1 (checked using LNAP 8/16/02) P6.4-2 Apply KVL to loop 1: 1 1 1 12 3000 0 2000 0 12 2.4 mA 5000 i i i + + + = = = The currents into the inputs of an ideal op amp are zero so ( ) o 1 2 1 2 2.4 mA 2.4 mA 1000 0 2.4 V a i i i i v i = = = = = + = − Apply Ohm’s law to the 4 k resistor ( ) ( ) ( ) o 3 4000 2.4 2.4 10 4000 12 V o a v v i = = − × = − (checked using LNAP 8/16/02) 6-6
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P6.4-3 The voltages at the input nodes of an ideal op amp are equal so 2 V a v = − .
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