0132130084_ism11

0132130084_ism11 - 407 CHAPTER 11 Exercises E11.1 (a) A...

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Unformatted text preview: 407 CHAPTER 11 Exercises E11.1 (a) A noninverting amplifier has positive gain. Thus ) 2000 sin( . 5 ) ( 50 ) ( ) ( t t v t v A t v i i v o π = = = (b) An inverting amplifier has negative gain. Thus ) 2000 sin( . 5 ) ( 50 ) ( ) ( t t v t v A t v i i v o π − = − = = E11.2 375 75 25 75 500 oc = + = + = = L o L v i o v R R R A V V A 300 75 25 75 500 2000 500 2000 oc = + + = + + = = L o L v i s i s o vs R R R A R R R V V A 4 10 75 2000 375 = × = = = L i v i o i R R A I I A 6 10 75 . 3 × = = i v A A G E11.3 Recall that to maximize the power delivered to a load from a source with fixed internal resistance, we make the load resistance equal to the internal (or Thévenin) resistance. Thus we make . 25 Ω = = o L R R Repeating the calculations of Exercise 11.2 with the new value of R L , we have 250 25 25 25 500 oc = + = + = = L o L v i o v R R R A V V A 4 10 2 25 2000 250 × = × = = = L i v i o i R R A I I A 6 10 5 × = = i v A A G 408 E11.4 By inspection, . 30 and 1000 3 1 Ω = = Ω = = o o i i R R R R 3 oc 3 2 3 2 oc 2 1 2 1 oc 1 3 oc v i o i v i o i v i o v A R R R A R R R A V V A + + = = 5357 30 3000 200 3000 20 2000 100 2000 10 1 3 oc = + + = = i o v V V A E11.5 Switching the order of the amplifiers of Exercise 11.4 to 3-2-1, we have Ω Ω 100 and 3000 1 3 = = = = o o i i R R R R 1 oc 1 2 1 2 oc 2 3 2 3 oc 3 1 oc v i o i v i o i v i o v A R R R A R R R A V V A + + = = 4348 10 1000 200 1000 20 2000 300 2000 30 3 1 oc = + + = = i o v V V A E11.6 W 22.5 A) (1.5 V) 15 ( = × = s P W 5 . 20 5 . 2 5 . 5 . 22 = − + = − + = o i s d P P P P % 11 . 11 % 100 = × = s o P P η E11.7 The input resistance and output resistance are the same for all of the amplifier models. Only the circuit configuration and the gain parameter are different. Thus we have 20 and k 1 Ω = Ω = o i R R and we need to find the open-circuit voltage gain. The current amplifier with an open-circuit load is: 409 4 1000 20 200 sc sc oc oc = × = = = = i o i i i o i i i o v R R A i R R i A v v A E11.8 For a transconductance-amplifier model, we need to find the short- circuit transconductance gain. The current-amplifier model with a short- circuit load is: S 2 . 500 100 sc sc sc sc = = = = = i i i i i i i o m R A i R i A v i G The impedances are the same for all of the amplifier models, so we have . 50 and 500 Ω = Ω = o i R R E11.9 For a transresistance-amplifier model, we need to find the open-circuit transresistance gain. The transconductance-amplifier model with an open-circuit load is: Ω = × × = = = = k 500 10 10 05 . / 6 sc sc oc oc i o m i i o i m i o m R R G R v R v G i v R The impedances are the same for all of the amplifier models, so we have . 10 and M 1 Ω = Ω = o i R R E11.10 The amplifier has . k 1 and k 1 Ω = Ω = o i R R (a) We have Ω < 10 s R which is much less than i R , and we also have Ω > k 100 L R which is much larger than o R . Therefore for this source and load, the amplifier is approximately an ideal voltage amplifier. 410 (b) We have...
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0132130084_ism11 - 407 CHAPTER 11 Exercises E11.1 (a) A...

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