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Unformatted text preview: 1 CHAPTER 11 Exercises E11.1 (a) A noninverting amplifier has positive gain. Thus ) 2000 sin( . 5 ) ( 50 ) ( ) ( t v A i o = = = (b) An inverting amplifier has negative gain. Thus ) 2000 sin( . 5 ) ( 50 ) ( ) ( = = = E11.2 375 75 25 75 500 oc = + = + = = L R V 300 75 25 75 500 2000 500 2000 oc = + + = + + = = s vs 4 10 75 2000 375 = = = = I 6 10 75 . 3 = = 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 Thvenin) resistance. Thus we make . 25 = = Repeating the calculations of Exercise 11.2 with the new value of , we have 250 25 25 25 500 oc = + = + = = 4 10 2 25 2000 250 = = = = 6 10 5 = = 2 E11.4 By inspection, . 30 and 1000 3 1 = = = = o i R 3 oc 3 2 3 2 oc 2 1 2 1 oc 1 3 oc v A V + + = = 5357 30 3000 200 3000 20 2000 100 2000 10 1 3 oc = + + = = E11.5 Switching the order of the amplifiers of Exercise 11.4 to 321, we have 100 and 3000 1 3 = = = = 1 oc 1 2 1 2 oc 2 3 2 3 oc 3 1 oc + + = = 4348 10 1000 200 1000 20 2000 300 2000 30 3 1 oc = + + = = E11.6 W 22.5 A) (1.5 V) 15 ( = = s P W 5 . 20 5 . 2 5 . 5 . 22 = + = + = d % 11 . 11 % 100 = = 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 = = and we need to find the opencircuit voltage gain. The current amplifier with an opencircuit load is: 3 4 1000 20 200 sc sc oc oc = = = = = i o v R A E11.8 For a transconductanceamplifier model, we need to find the shortcircuit transconductance gain. The currentamplifier model with a shortcircuit load is: S 2 . 500 100 sc sc sc sc = = = = = m G The impedances are the same for all of the amplifier models, so we have . 50 and 500 = = E11.9 For a transresistanceamplifier model, we need to find the opencircuit transresistance gain. The transconductanceamplifier model with an opencircuit load is: = = = = = k 500 10 10 05 . / 6 sc sc oc oc The impedances are the same for all of the amplifier models, so we have . 10 and M 1 = = E11.10 The amplifier has . k 1 and k 1 = = (a) We have &lt; 10 s which is much less than , and we also have &gt; k 100 L which is much larger than . Therefore for this source and load, the amplifier is approximately an ideal voltage amplifier. 4 (b) We have &gt; k 100 s R which is much greater than i , and we also have &lt; 10 L which is much smaller than o . Therefore for this source and load, the amplifier is approximately an ideal current amplifier. (c) We have &lt; 10 which is much less than , and we also have &lt; 10 which is much smaller than . Therefore for this source and load, the amplifier is approximately an ideal transconductance amplifier....
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 Spring '07
 Haris
 Amplifier, Input impedance, Ri, Impedance matching, Electrical parameters, Output impedance

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