Microelectronic Circuit Design Third Edition - Part III Solutions to Exercises doc

1 Microelectronic Circuit Design Third Edition - Part III Solutions to Exercises CHAPTER 10 Page 509 € V o = 2P o R L = 2 20W ( ) 16Ω ( ) = 25.3 V A v = V o V i = 25.3V 0.005V = 5.06x10 3 I o = V o R L = 25.3V 16Ω = 1.58 A I i = V i R S + R in = 0.005V 10kΩ + 20kΩ = 0.167 µ A A i = I o I i = 1.58 A 0.167 µ A = 9.48x10 6 A P = P o P S = 25.3V 1.58A ( ) 0.005V 0.167 µ A ( ) = 4.79x10 10 Checking : A p = 5.06x10 3 ( ) 9.48x10 6 ( ) = 4.80x10 10 Page 510 € i ( ) A vdB = 20 log 5060 ( ) = 74.1 dB A idB = 20 log 9.48x10 6 ( ) = 140 dB A PdB = 10 log 4.80x10 10 ( ) = 107 dB ii ( ) A vdB = 20 log 4x10 4 ( ) = 92.0 dB A idB = 20 log 2.75x10 8 ( ) = 169 dB A PdB = 10 log 1.10x10 13 ( ) = 130 dB Page 511 € i ( ) The constant slope region spanning a maximum input range is between 0.4 ≤ v I ≤ 0.65, and the bias voltage V I should be centered in this range : V I = 0.4 + 0.65 2 V = 0.525 V . v i ≤ 0.65 − 0.525 = 0.125 V and v i ≤ 0.525 − 0.40 = 0.125 V. For v I = 0.8V, the slope is 0. A v = 0. (ii) v O = V O + v o For v i = 0, v I = V I = 0.6, V O = 14V and A v = +40. V o = A v V i = 40 0.01V ( ) = 4V v O = 14.0 + 4.00sin1000 π t ( ) volts V O = 14 V 2 Page 519 € g 11 = 1 20kΩ + 76 50kΩ ( ) = 0.262 µ S g 21 = 0.262 µ S 76 ( ) 50kΩ ( ) = 0.995 g 22 = 1 50kΩ + 1 20kΩ + 75 20kΩ = 3.82 mS g 12 = − 1 g 22 20kΩ ( ) = − 1 3.82mS 20kΩ ( ) = -0.0131 R in = 1 g 11 = 3.82 MΩ A = g 21 = 0.995 R out = 1 g 22 = 262 Ω Page 520 € i ( ) P = A v A i = A v 2 R S + R in R L ii ( ) V o = 2 100W ( ) 8Ω ( ) = 40 V 40 = 0.001 50kΩ 5kΩ + 50kΩ       A 8Ω 0.5Ω + 8Ω       → A = 46,800 P = I o 2 R L 2 = 0.5Ω 2 40V 8Ω       2 = 6.25 W A i = 40V 8Ω       5kΩ + 50kΩ 0.001V       = 2.75 x 10 8 iii ( ) 40 = 0.001 5kΩ 5kΩ + 5kΩ       A 8Ω 8Ω + 8Ω       → A = 160,000 P = I o 2 R L 2 = 8Ω 2 40V 8Ω       2 = 100 W! A i = 40V 8Ω       5kΩ + 5kΩ 0.001V       = 5.00 x 10 7 Page 521 € A v s ( ) = 300s s + 5000 ( ) s +100 ( ) Zeros at s = 0 and s = ∞; Poles at s = -5000 and s = −100. Page 523 € A v s ( ) = − 2 π x 10 6 s + 5000 π = −400 1+ s 5000 π → A mid = −400 f H = 5000 π 2 π = 2.50 kHz BW = f H − f L = 2.50 kHz − 0 = 2.50 kHz GBW = 400 ( ) 2.50kHz ( ) = 1.00 MHz 3 Page 524 € i ( ) A v j5 ( ) = 50 5 2 − 4 5 2 − 2 ( ) 2 + 4 5 2 ( ) = 41.87 20 log 41.87 ( ) = 32.4 dB ∠A v j5 ( ) = ∠ 5 2 − 4 ( ) − tan −1 −2 5 ( ) 5 2 − 2         = 0 − −23.5 o ( ) = 23.5 o A v j1 ( ) = 50 1 2 − 4 1 2 − 2 ( ) 2 + 4 1 2 ( ) = 67.08 20 log 41.87 ( ) = 36.5 dB ∠A v j1 ( ) = ∠ 1 2 − 4 ( ) − tan −1 −2 1 ( ) 1 2 − 2         = 180 o − −63.43 o ( ) = 243 o = −117 o Page 524 € ii ( ) A v j ω ( ) = 20 1+ j 0.1 ω 1− ω 2 A v j0.95 ( ) = 20 1 2 + 0.1 ( ) 2 0.95 2 ( ) 1− 0.95 2 ( ) 2 = 14.3 ∠A v j0.95 ( ) = ∠20 − tan −1 0.1 0.95 ( ) 1− 0.95 2         = 0 − 44.3 o ( ) = −44.3 o A v j1 ( ) = 20 1 2 + 0.1 ( ) 2 1 2 ( ) 1−1 2 ( ) 2 = 0 ∠A v j1 ( ) = ∠20 − tan −1 0.1 1 ( ) 1−1 2         = 0 − 90 o ( ) = −90.0 o A v j1.1 ( ) = 20 1 2 + 0.1 ( ) 2 1.1 2 ( ) 1−1.1 2 ( ) 2 = 17.7 ∠A v j1.1 ( ) = ∠20 − tan −1 0.1 1.1 ( ) 1−1.1 2         = 0 − −27.6 o ( ) = 27.6 o Page 526 € f H = 1 2 π 1 1kΩ 100kΩ ( ) 200 pF ( ) = 804 kHz Page 527 € A v s ( ) = 250 1+ 250 π s A o = 250 f L = 250 π 2 π = 125 Hz f H = ∞ BW = ∞ −125 = ∞ 4 Page 528 € f L = 1 2 π 1 1kΩ 100kΩ ( ) 0.1 µ F ( ) = 15.8 Hz Page 529 € i ( ) A v s ( ) = −400 1+ 100 s       1+ s 50000       A o = 400 or 52 dB f L = 100 2 π = 15.9 Hz f H = 50000 2 π = 7.96 kHz BW = 7960 −15.9 = 7.94 kHz ii ( ) ∠A v j0 ( ) = −90 − 0 − 0 = −90 o ∠A v j100 ( ) = −90 o − tan −1 100 100       − tan −1 100 50000       = −90 − 45 − 0.57 = −136 o ∠A v j50000 ( ) = −90 o − tan −1 50000 100       − tan −1 50000 50000       = −90 − 89.9 − 45 = −225 o ∠A v j∞ ( ) = −90 − 90 − 90 = −270 o Page 531 € The numerator coefficient should be 6 x10 6 . A v s ( ) = 30 2x10 5 s s 2 + 2x10 5 s +10 14 A o = 30 f o = 1 2 π 10 14 = 1.59 MHz Q = 10 7 2x10 5 = 50 BW = 1.59 MHz 50 = 31.8 kHz Page 533 € The transfer fucntion should be A v s ( ) = 6.4x10 12 π 2 s s + 200 π ( ) s + 80000 π ( ) 2 . A v s ( ) = 1000 1+ 200 π s       1+ s 80000 π       2 A o = 1000 or 60 dB f L = 200 π 2 π = 100 Hz f H = 0.644 80000 π 2 π       = 25.8 kHz BW = 25800 −100 = 25.7 kHz 5 CHAPTER 11 Page 545 € v id = 10V 100 = 0.100V =100 mV v id = 10V 10 4 = 0.001 V = 1.00 mV v id = 10V 10 6 = 1.00x10 −5 V = 10.0 µ V Page 547 € A v = − 360kΩ 68kΩ = −5.29 v O = −5.29 0.5V ( ) = −2.65 V i S = 0.5V 68kΩ = 7.35 µ A i O = −i 2 = −i S = −7.35 µ A Page 549 € I S = 2V 4.7kΩ = 426 µ A I 2 = I S = 426 µ A A v = − 24kΩ 4.7kΩ = −5.11 V O = −5.11 2V ( ) = −10.2 V Page 551 € A v = 1+ 36kΩ 2kΩ = +19.0 v O = 19.0 −0.2V ( ) = − 3.80 V i O = −3.80V 36kΩ + 2kΩ = −100 µ A Page 552 € i ( ) A v = 1+ 39kΩ 1kΩ = +40.0 A vdB = 20log 40.0 ( ) = 32.0 dB R in = 100kΩ ∞ =100kΩ v O = 40.0 0.25V ( ) = 10.0 V i O = 10.0V 39kΩ + 1kΩ = 250 µ A ii ( ) A v = 10 54 20 = 501 1+ R 2 R 1 = 501 R 2 R 1 = 500 i O = v O R 2 + R 1 10 R 2 + R 1 ≤ 0.1 mA R 1 + R 2 ≥ 100kΩ 501R 1 ≥ 100kΩ → R 1 ≥ 200 Ω There are many possibilities. (R 1 = 200 Ω, R 2 = 100 kΩ ), but ( R 1 = 220 Ω, R 2 = 110 kΩ ) is a better solution since resistor tolerances could cause i O to exceed 0.1 mA in the first case. Page 554 € Inverting Amplifier : A v = − 30kΩ 1.5kΩ = −20.0 R in = R 1 = 1.5 kΩ v O = −20.0 0.15V ( ) = −3.00 V i O = v O R 2 = −3.00V 30kΩ = −100 µ A Non - Inverting Amplifier : A v = 1+ 30kΩ 1.5kΩ = +21.0 R in = v S i S = 0.15V 0 A = ∞ v O = 21.0 0.15V ( ) = 3.15 V i O = v O R 2 + R 1 = 3.15V 30kΩ + 1.5kΩ = 100 µ A 6 Page 555 € V o1 = 2V − 3kΩ 1kΩ       = −6V V o2 = 4V − 3kΩ 2kΩ       = −6V v O = −6sin1000 π t − 6sin 2000 π t ( ) V The summing junction is a virtual ground : R in1 = v 1 i 1 = R 1 = 1 kΩ R in2 = v 2 i 2 = R 2 = 2 kΩ I o1 = V o1 R 3 = −6V 3kΩ = −2mA I o2 = V o2 R 3 = −6V 3kΩ = −2mA i O = −2sin1000 π t − 2sin2000 π t ( ) mA Page 559 € i ( ) I 2 = 3V 10kΩ + 100kΩ = 27.3 µ A ii ( ) A v = − 100kΩ 10kΩ = −10.0 V O = −10 3V − 5V ( ) = 20.0 V I O = V O − V − 100kΩ = V O − V + 100kΩ V + = V 2 R 4 R 3 + R 4 = 5 100kΩ 10kΩ + 100kΩ = 4.545V I O = 20.0 − 4.545 100kΩ = 155 µ A I 2 = 5V 10kΩ + 100kΩ = 45.5 µ A iii ( ) A v = − 36kΩ 2kΩ = −18.0 V O = −18 8V − 8.25V ( ) = 4.50 V I O = V O − V − 36kΩ = V O − V + 36kΩ V + = V 2 R 2 R 1 + R 2 = 8.25 36kΩ 2kΩ + 36kΩ = 7.816V I O = 4.50 − 7.816 36kΩ = −92.1 µ A Page 560 € I = V A −V B 2R 1 = 5.001V − 4.999V 2kΩ = 1.00 µ A V A = V 1 + IR 2 = 5.001V +1.00 µ A 49kΩ ( ) = 5.05 V V B = V 2 − IR 2 = 4.999V −1.00 µ A 49kΩ ( ) = 4.95 V V O = − R 4 R 3       V A −V B ( ) = − 10kΩ 10kΩ       5.05 − 4.95 ( ) = −0.100 V Page 564 € i ( ) A v = − R 2 R 1 = −10 26 20 = −20.0 R 1 = R in = 10kΩ R 2 = 20R 1 = 200kΩ C = 1 2 π 3kHz ( ) 200kΩ ( ) = 265 pF Closest values : R 1 = 10kΩ R 2 = 200kΩ C = 270 pF 7 Page 564 € ii ( ) R in = R 1 = 10 kΩ ΔV = − I C ΔT C = 5V 10kΩ 1 10V       1ms ( ) = 0.05 µ F t (msec) v O 2 4 6 8 -10V Page 567 € v O = −RC dv S dt = − 20kΩ ( ) 0.02 µ F ( ) 2.50V ( ) 2000 π ( ) cos2000 π t ( ) = −6.28cos2000 π t V Page 569 € i ( ) A vA = A vB = A vC = − R 2 R 1 = − 68kΩ 2.7kΩ = −25.2 R inA = R inB = R inC = R 1 = 2.7 kΩ The op - amps are ideal : R outA = R outB = R outC = 0 ii ( ) A v = A vA A vB A vC = −25.2 ( ) 3 = −16,000 R in = R inA = 2.7 kΩ R out = R outC = 0 Page 570 € A v = −25.2 ( ) 3 2.7kΩ R out + 2.7kΩ       2 ≥ 0.99 25.2 ( ) 3 2.7kΩ R out + 2.7kΩ       2 ≥ 0.99 2.7kΩ R out + 2.7kΩ ≥ 0.9950 R out ≥13.6 Ω Page 574 € i ( ) A v 0 ( ) = +1 A v s ( ) = ω o 2 s 2 + s 2 ω o + ω o 2 A v j ω ( ) = ω o 2 j ω 2 ω o + ω o 2 − ω 2 A v j ω H ( ) = 1 2 → ω o 2 ω o 2 − ω H 2 ( ) 2 + 2 ω H 2 ω o 2 = 1 2 → 2 ω o 4 = ω o 4 + ω H 4 → ω o = ω H ii ( ) 1 2 = C 1 C 2 R 2 2R → C 1 = 2C 2 → C 2 = 1 2 2.26kΩ ( ) 20000 π ( ) = 4.98nF C 2 = 0.005 µ F C 1 = 0.01 µ F 8 Page 574 € iii ( ) To decrease the cutoff frequency from 5kHz to 2 kHz, we must increase the resistances by a factor of 5kHz 2kHz = 2.50 → R 1 = R 2 = 2.50 2.26kΩ ( ) = 5.65 kΩ iv ( ) 1 2 = C C R 1 R 2 R 1 + R 2 → R 1 2 + 2R 1 R 2 + R 2 2 = 2R 1 R 2 → R 1 2 = −R 2 2 - - can't be done! Q = R 1 R 2 R 1 + R 2 dQ dR 2 = 1 R 1 + R 2 ( ) 2 R 1 R 1 + R 2 ( ) 2 R 1 R 2 − R 1 R 2         = 0 → R 2 = R 1 → Q max = 1 2 Page 575 € S C 1 Q = C 1 Q dQ dC 1 = C 1 Q 1 2 C 1 C 2 R 1 R 2 R 1 + R 2         = C 1 Q Q 2C 1 = 0.5 S R 2 Q = R 2 Q dQ dR 2 R 1 = R 2 → Q = 1 2 C 1 C 2 → S R 2 Q = 0 Page 577 € i ( ) A v j ω o ( ) = K − ω o 2 − ω o 2 + j 3 − K ( ) ω o 2 + ω o 2 = K 3 − K A v j ω o ( ) = K 3 − K ∠90 o ii ( ) f o = 1 2 π 10kΩ 20kΩ ( ) 0.0047 µ F ( ) 0.001 µ F ( ) = 5.19 kHz Q = 10kΩ 20kΩ 4.7nF +1.0nF 4.7nF 1.0nF ( ) + 1− 2 ( ) 20kΩ 1.0nF ( ) 10kΩ 4.7nF ( )         −1 = 0.829 iii ( ) S K Q = K Q dQ dK Q = 1 3 − K dQ dK = −1 3 − K ( ) 2 −1 ( ) = Q 2 S K Q = K Q dQ dK = KQ Q = 1 3 − K → KQ = 3Q −1 S K Q = 3Q −1 =1.12 Page 578 € R th = 2kΩ 2kΩ =1kΩ f o = 1 2 π 1kΩ 82kΩ ( ) 0.02 µ F ( ) 0.02 µ F ( ) = 879 Hz Q = 1 2 82kΩ 1kΩ = 4.53 9 Page 582 € ii ( ) A BP j ω o ( ) = KQ = R 2 R 1 10 = 294kΩ R 1 → R 1 = 29.4 kΩ iii ( ) f o = 1 2 π RC = 1 2 π 40.2kΩ ( ) 2nF ( ) = 1.98 kHz BW = 1 2 π R 2 C = 1 2 π 402kΩ ( ) 2nF ( ) = 198 Hz A BP j ω o ( ) = − R 2 R 1 = − 402kΩ 20.0kΩ = −20.1 iv ( ) Blindly using the equations at the top of page 580 yields f o min = 1 2 π RC = 1 2 π 1.01 ( ) 29.4kΩ ( ) 1.02 ( ) 2.7nF ( ) = 1946 Hz f o max = 1 2 π RC = 1 2 π 0.99 ( ) 29.4kΩ ( ) 0.98 ( ) 2.7nF ( ) = 2067 Hz BW min = 1 2 π R 2 C = 1 2 π 1.01 ( ) 294kΩ ( ) 1.02 ( ) 2.7nF ( ) = 195 Hz BW max = 1 2 π R 2 C = 1 2 π 0.99 ( ) 294kΩ ( ) 0.98 ( ) 2.7nF ( ) = 207 Hz A BP min = − R 2 R 1 = − 294kΩ 1.01 ( ) 14.7kΩ 0.99 ( ) = −20.4 A BP max = − R 2 R 1 = − 294kΩ 0.99 ( ) 14.7kΩ 1.01 ( ) = −19.6 The W/C results are similar if R and C are not the same for example where ω o = 1 R A R B C A C B . Page 583 € i ( ) - a ( ) R 1 = R 2 = 5 2.26kΩ ( ) = 11.3 kΩ C 1 = 0.02 µ F 5 = 0.004 µ F C 2 = 0.01 µ F 5 = 0.002 µ F f o = 1 2 π 11.3kΩ ( ) 11.3kΩ ( ) 0.004 µ F ( ) 0.002 µ F ( ) = 4980 Hz Q = 11.3kΩ 11.3kΩ 0.004 µ F ( ) 0.002 µ F ( ) 0.004 µ F + 0.002 µ F = 0.471 b ( ) R 1 = R 2 = 0.885 2.26kΩ ( ) = 2.00 kΩ C 1 = 0.02 µ F 0.885 = 0.0226 µ F C 2 = 0.01 µ F 0.885 = 0.0113 µ F f o = 1 2 π 2.00kΩ ( ) 2.00kΩ ( ) 0.0226 µ F ( ) 0.0113 µ F ( ) = 4980 Hz Q = 2.00kΩ 2.00kΩ 0.0226 µ F ( ) 0.0113 µ F ( ) 0.0226 µ F + 0.0113 µ F = 0.471 10 Page 583 € ii ( ) f o = 1 2 π 1kΩ ( ) 82kΩ ( ) 0.02 µ F ( ) 0.02 µ F ( ) = 879 Hz Q = 82kΩ 1kΩ 0.02 µ F ( ) 0.02 µ F ( ) 0.02 µ F + 0.02 µ F = 4.53 The values of the resistors are unchanged. C 1 = C 2 = 0.02 µ F 4 = 0.005 µ F f o = 1 2 π 1kΩ ( ) 82kΩ ( ) 0.005 µ F ( ) 0.005 µ F ( ) = 3520 Hz Q = 82kΩ 1kΩ 0.005 µ F ( ) 0.005 µ F ( ) 0.005 µ F + 0.005 µ F = 4.53 Page 585 € The diode will conduct and pull the output up to v O = v S = 1.0 V. v 1 = v O + v D = 1.0 + 0.6 =1.6 V For a negative input, there is no path for current through R, so v O = 0 V . The op - amp sees a -1V input so the output will limit at the negative power supply : v O = −10 V . The diode has a 10 - V reverse bias across it, so V Z > 10 V . Page 587 € i ( ) v S = 2 V : Diode D 1 conducts, and D 2 is off. The negative input is a virtual ground. v 1 = −v D2 = −0.6 V . The current in R is 0, so v O = 0 V . v S = −2 V : Diode D 2 conducts, and D 1 is off. The negative input is a virtual ground. v O = − R 2 R 1 v S = − 68kΩ 22kΩ −2V ( ) = +6.18 V v 1 = v O + v D1 = 6.78 V . v S = 15V −3.09 = −4.85 V v 1 = v O + v D1 = 6.78 V . When v O = 15 V , v D2 = -15.6 V , so V Z = 15.6 V . ii ( ) v O = 20kΩ 20kΩ 10.2kΩ 3.24kΩ       2V π = 2.00 V Page 589 € V − = − R 1 R 1 + R 2 V EE = − 1kΩ 1kΩ + 9.1kΩ 10V = −0.990 V V + = 1kΩ 1kΩ + 9.1kΩ 10V = +0.990 V V n = 0.990V − −0.990V ( ) = 1.98 V Page 591 € T = 2 10kΩ ( ) 0.001 µ F ( ) ln 1+ 0.5 1− 0.5       = 21.97 µ s f = 1 T = 45.5 kHz [...]... At the Q - point : β F = (c) Rin = v be 8mV = = 1.6 kΩ ib 5µA IC 1.5mA = = 100 I B 15µA (b) I S = IC 1.5mA = = 1.04 fA  VBE   0.700V  exp  exp   0.025V   VT  (d) With the given applied signal, the smallest value of vCE is min v CE = 5V − 0.5mA(3.3kΩ) = 3.35 V which exceeds v BE = 0.708 V (ii ) (a) v DS = 10 − 3300iDS (b) Using the peak - to - peak voltage swings, Av = vds 2.7 - 6.7 V =... to - peak voltage swings, Av = vds 2.7 - 6.7 V = = −4.0 vgs 4.0 - 3.0 V Note that there is some distortion in this amplifier since the negative output voltage excursion is larger than the positive output change (c) v min = 2.7V with vGS − VTN = 4 −1 = 3V , so the transistor has entered the triode region DS (d) Choose two points on the i - v characteristics For example, 2 2 Kn (3.5 − VTN ) and 1.0mA... t+ t = 15V → t = 6.00 ms 10kΩ(100 pF ) 100 pF Page 636 € Values taken from op - amp specification sheets (www.jaegerblalock.com or www.analog.com) Page 637 € Values taken from op - amp specification sheets (www.jaegerblalock.com or www.analog.com) € 14 Page 638 −1  1 1  REQ R1 + R2 ≥  −  = 20kΩ  4kΩ 5kΩ  Including 5% tolerances, R1 + R2 ≥ 21kΩ A v = 10 → R2 = 9R1 20V = RL ( R2 + R1 ) ≥ = 4kΩ 5mA... CHAPTER 12 Page 612 1 (i ) Aideal = β = 100 A 105 Av = = = 99.90 1+ Aβ 1+ 105 (0.01) vo 9.99V = = 99.9 µV A 105 (ii ) Values taken from OP - 27 specification sheet (www.jaegerblalock.com or www.analog.com) vo = 99.9(0.1V ) = 9.99 V vid = (iii ) Values taken from OP - 27 specification sheet € Page 613 Av = −R2 Aβ R1 1+ Aβ  −R   −R Aβ  2 2   −  R1   R1 1+ Aβ  Aβ 1  FGE = = 1− =  −R  1+ Aβ...  5x10 − 5  = 2.5x10 4 or 88 dB 10   Page 630 € Values taken from op - amp specification sheets (www.jaegerblalock.com or www.analog.com) Page 631 € (i ) V ≤ 50(0.002V ) → −0.100 V ≤ V ≤ +0.100 V (ii ) Values taken from op - amp specification sheets (www.jaegerblalock.com O O or www.analog.com) Page 633 € Values taken from op - amp specification sheets (www.jaegerblalock.com or www.analog.com) Page...   6.37kΩ in AvCE = −g m RL  = -9 .56mS 9.57kΩ 220kΩ   = −83.4 CE  0.33kΩ + 6.37kΩ   RI + Rin  ( ) € 21 Page 730 (i) The gain is proportional to RL  25.9kΩ  AvCS = −6.85  = −6.50  27.3kΩ  The corrected gain agrees more closely with thevalue from SPICE CS RL = 300kΩ Rout = 300kΩ 28.4kΩ = 25.9kΩ (ii ) µ f = g mro = 0.515mS (258kΩ) = 133  RCS   1 in iii ) v gs = v i v gs ≤ 0.2(VGS −... 3.6kΩ = 69.1 Ω gm 0.0142 CD Rout = 3.6kΩ € D L 1 1 + VDS + 5 + 2.21 iii ) ro = λ = 0.015 = 14.8kΩ ( ID 0.005 (iv ) GS RS = 3600Ω AvCD = 0.981 Rin = RG = 22 MΩ AvCD = A 3000Ω = 0.959 69.1Ω + 3000Ω Page 805 (i ) Reverse the direction of the arrow on the ermitter of the transistor as well as the values of VEE and VCC CD out (ii ) R = RE (iii ) For v CB 1 1 = 13kΩ = 75.1 Ω gm 40(331µA) CB AV = 75.1Ω (13.2mS... VCE = 15 −1.96mA 3.3kΩ = 8.45 V 51.8kΩ + 81(3.3kΩ)  80  Q3 : VEQ = 80 60 + 8.45 = 1.02 kΩ ro = = 34.9 kΩ 78.4mS 1.96mA (ii) A typical op - amp gain is at least 10,000 which exceeds the amplification factor g m = 40(1.96mA) = 78.4 mS rπ = of a single transistor € Page 815 (i ) RL1 = 478Ω 12.2kΩ = 460Ω RL2 = 3.53kΩ 54.2kΩ = 3.31kΩ  79.6mS (230Ω)   1MΩ  Av = −10mS (460Ω)(−62.8mS )(3.31kΩ) ... kΩ IC 250µA (iii ) € βF I 1.5mA = βF = C = = 100 VCE I B 15µA 1+ VA ΔiC 500µA = = 100 ΔiB 5µA rπ = 18 µ f = g mro = 10mS (360kΩ) = 3600 gm = ΔiC 0.5mA = = 62.5 mS ΔvBE 8mV βo 100 = = 1.60 kΩ g m 62.5mS Page 692 Avt = −g m RL = −9.80mS (18kΩ) = −176 € βo 50 = = 5.00 kΩ g m 10mS The slope of the output characteristics is zero, so VA = ∞ and ro = ∞ β FO = βo = rπ = Page 695 Assume the Q - point remains... 0.491mS (1kΩ)  2kΩ + 892kΩ  (iii ) R4 = 13kΩ −1kΩ = 12 kΩ RiB = 10.2kΩ  104kΩ 10.2kΩ   A = −9.80mS (18kΩ)  2kΩ + 104kΩ 10.2kΩ  = −145    892kΩ  AvCS = −0.491mS (18kΩ)  = −8.82  2kΩ + 892kΩ   1.38x10−23 V  IC 245µA iv ) VT =  = = 0.421 fA (273K + 27K ) = 25.84mV I S = ( −19  VBE   0.700V   1.602x10 K  exp  exp   0.02584V   VT  18kΩ (v ) gm RL = -9 .80mS(18kΩ) = −176, AvCE . 1 Microelectronic Circuit Design Third Edition - Part III Solutions to Exercises CHAPTER 10 Page 509 € V o = 2P o R L = 2 20W (. (a) v DS = 10 − 3300i DS (b) Using the peak - to - peak voltage swings, A v = v ds v gs = 2.7 - 6.7 4.0 - 3.0 V V = −4.0. Note that there is some distortion in this amplifier since the negative. 4.98nF C 2 = 0.005 µ F C 1 = 0.01 µ F 8 Page 574 € iii ( ) To decrease the cutoff frequency from 5kHz to 2 kHz, we must increase the resistances by a factor of 5kHz 2kHz = 2.50 → R 1 = R 2 = 2.50