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

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Chapter 6: The Operational Amplifier Exercises Ex. 6.4-1 12 2 1 0 0 1 sso o s vvv RR vR + += =+ Ex. 6.4-2 a) 2 0 34 00 as aa R vv = + + 42 31 2 11 oo 4 3 R vR v R  = +  +  b) 22 21 2 When then 1 and 1 o s 4 3 R Rv R R v >> = + + ±± R 6-1
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Ex. 6.5-1 21 2 12 00 o os o s vv v RR vR vRR + += = + Ex. 6.6–1 11 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 vv v R v v  ++= ⇒ =+   =Ω = 6-2
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Ex. 6.7-1 33 21 32 3 2 2 10 10 10 10 11 10 10 10 10 10 10 10 10 10 10 10 10 RR vv R R    ×× =− + + +    ×+ × × ×    + + × 1 3 v We require () 31 1 4 5 2 v , so 1 3 1 3 421 1 01 0 1 0 k 10 10 R R =+ ⇒ = ×= × and 2 3 22 2 3 2 1 10 10 5 2.5 k 51 0 1 0 R R R =⇒ + × = = 6-3
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Ex. 6.7-2 As in Ex 6.7-1 21 32 33 2 10 10 10 10 RR vv R  =− + +  ×  1 v We require () 31 4 6 5 2 v , so 1 3 1 3 621 2 01 0 2 0 k 10 10 R R =+ ⇒ = ×= × and 2 3 22 2 3 2 4 44 0 1 0 5 4 0 k 51 0 1 0 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 vi µ ≤≤ × ≤ + 9 )(80 10 ) 10 mV Ex. 6.8-2 11 1 oi n o s v + + + b R i ( ) 1 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 RR v i −− =Ω=Ω ≤× × = 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 vi == ( ) 33 9 36 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 so ab i s o o b vvvv v RR R R R i vA v vv is ++ = +   +  + = After some algebra ( ) 0 00 i f o v s f is a f i RR R A R R v A v a R RRR RRRRR A R 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: 11 1 12 3000 0 2000 0 12 2.4 mA 5000 ii i ++ + = ⇒= = The currents into the inputs of an ideal op amp are zero so () o1 21 2 2 . 4 m A 2 . 4 m A 1000 0 2.4 V a vi = = =− =− =+ = Apply Ohm’s law to the 4 k resistor ( ) o 3 4000 2.4 2.4 10 4000 12 V oa vvi =− × (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 = − . Apply KCL at node a : ( ) ( ) 21 2 2 03 8000 4000 o o v v −− += = 0 V Apply Ohm’s law to the 8 k resistor 2 3.5 mA 8000 o o v i == (checked using LNAP 8/16/02) P6.4-4 The voltages at the input nodes of an ideal op amp are equal so . 5 V v = Apply KCL at the inverting input node of the op amp: 3 5 0.1 10 0 = 0 = 4 V a 10000 a v v  × −⇒   Apply Ohm’s law to the 20 k resistor 1 mA 20000 5 a v i (checked using LNAP 8/16/02) P6.4-5 The voltages at the input nodes of an ideal op amp are equal so . Apply KCL at node a : 0 V a v = 3 01 2 0 210 0 3000 4000 15 V o o v v −−− ⇒= = Apply KCL at the output node of the op amp: 0 7.5 mA 6000 3000 oo vv ii ++= = (checked using LNAP 8/16/02) 6-7
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