# ch4 - 4.1 ＋ ＋ vin v1 K = 20 mA V － － I1 Kv1 RL Vout...

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Unformatted text preview: 4.1 ＋ ＋ vin v1 K = 20 mA / V － － I1 Kv1 RL Vout － ＋ K = 20 mA / V Vout = 15 Vin Vin = V1 Vout = 15 Vin Vin = V1 Vout = −I1RL = −KRL Vin ⇒ Vout V = −KRL ⇒ out = KRL Vin Vin ⇒ RL = 15 = 750 Ω 20 mA / V Vout = −I1RL = −KRL Vin ⇒ Vout V = −KRL ⇒ out = KRL Vin Vin ⇒ RL = 15 = 750 Ω 20 mA / V ⇒ KRL = 15 RL = 750 Ω ⇒ KRL = 15 RL = 750 Ω 4.2 RS ＋ rin v1 － I1 KV1 RL ＋ vin － vout － ＋ vout =? vin v1 = rin Vin rin + RS vout =? vin rin v1 = Vin rin + RS I1 = Kv1 ⇒ Vout vout = − RL I1 I1 = Kv1 ⇒ Vout vout = − RL I1 rin vin ⇒ Vout = −KRL rin + RS V rin = −KRL v1 ⇒ Av = out = −KRL Vin rin + RS rin v ⇒ Vout = −KRL rin + RS = −KRL v1 ⇒ Av = Vout rin = −KRL Vin rin + RS 4.3 From Soludion for problem 4.2, a>0 b>0 x≥0 a>0 b>0 x≥0 rin rin A> 0 Av = −KRL a v = −KRL rin + RS rin + RS b>0 rin = a x ax a ax a = −bRL rin = a x Av = −bxRL = −bRL Av = −bxRL x≥0 a = bx a K = bx a x + RS KR a x + RS +S + RS x x rin Av = −KRL rin + RS ax rin = a x a x x = −bRL ⇒ Av = −bRL Av ⇒ Av = −bRL = −bxRL a RS a x + RS K = bx 1 + + RS x 1 + RS x x a a Av x ⇒ Av ab RL L Av = − = −bR R RS 1 + S x a Av = −ab x RL RS Av = −ab RL RS 4.4 According to Equation (4.8), we have IC = � AE qDn n2 � VBE /VT i e −1 NB WB 1 ∝ WB We can see that if WB increases by a factor of two, then IC decreases by a factor of two . 4.5 IC1 ＋ VBE1 － VT = 26 mV IC1 = IC2 , Q1 ＋ VBE2 , － , IC2 Q2 VT = 26 mV IC1 = IC2 VBE1 − VBE2 = 20 mV VBE AEq Dn ni2 VT IC = e − 1 N EWB A q Dn ni2 VT e ⇒ IC E N EWB if IC1 = IC2 VBE1 VBE VBE1 − ,BE2 = 20 mV V equation (4.8) page 13 VBE AEq Dn ni2 VT IC = e − 1 N EWB ⇒ IC AEq Dn ni2 VT e N EWB ni2 VBE1 VBE equation (4.8) page 136 AE ≡ cross section AE ≡ cross section if IC1 = IC2 ⇒ AE ⇒ ⇒ ⇒ 1 q Dn ni2 VT q Dn ni2 e = AE2 e N EWB N EWB = e e VBE1 VT VBE2 VT VBE2 VT ⇒ AE ⇒ ⇒ ⇒ 1 q Dn q e VT = AE2 e N EWB N EWB = e e VBE1 VT VBE2 VT Dn ni2 VBE2 VT AE2 AE1 AE2 AE1 AE2 AE1 AE2 AE1 AE2 AE1 AE2 AE1 =e (VBE1 −VBE2 ) VT = e20 mV 26 mV 20 =e (VBE1 −VBE2 ) VT =e 20 mV 26 mV = e 26 2.16 = 20 e 26 2.16 4.6 (a) I X = 1 mA ⇒ IQ = IQ = 0.5 mA 1 2 VBE1 IQ1 = IS1 e VT ⇒ 5 × 10 −4 = 3 × 10 −16 e VB 26 mV 5 ⇒ VB = 26 mV ln ( × 1012 ) ⇒ VB 731.7 mV 3 VB VT − VB VT − VB 26 mV (b) IY = IS3 e ⇒ IS3 = IY e = 2.5 × 10 × e −3 = 2.5 × 10−3 × 1 5 3 × 1012 ⇒ IS3 = 1.5 × 10−15 A 4.7 (a) I X = I1 + I 2 VB VB VB ⇒ I X = IS1 e VT + IS2 e VT ⇒ VB = VT ln ( IX ) IS1 + IS2 ⇒ I X = (IS1 + IS2 )e VT IS1 = 2IS2 VB = VT ln ( IX ) 3 2 IS1 VB = 26 × 10−3 ln ( 1.2 × 10−3 ) ⇒ VB 730.6 mV 3 2 × 5 × 10−16 (b) Transistors at the edge of the active mode ⇒ VC = VB applying KVL, we have: VCC = RC I X + VB ⇒ RC = ⇒ RC = VCC − VB IX 2.5 − 0.73 1.2 × 10−3 ⇒ RC 1475 Ω 4.8 (a) Same as 4.7(a), VB 730.6 mV (b) According to 4.7(b), RC = VCC − VB 1.5 − 0.73 = IX 1.2 × 10−3 ⇒ RC 642 Ω 4.9 Q1 is at the edge of the active region ⇒ VC = VB applying KVL, we have: VCC = RC IC + VC VC =VB 500 Ω IC ＋ VCC － 2V RC VCC = RC IC + VB ⇒ VCC = RC IS eVB VT + VB VB 26 mV ⇒ 500 Ω × 5 × 10−16 e + VB = 2V VB － ＋ , , Q1 , Using numerical methods or simply, trial & error: VB 760 mV 4.10 Q1 at the edge of saturation ⇒ VC = VB Hence: VCC = RC IC + VB ⇒ VCC = RC IS eVB VT + VB VT IS = 3 × 10−16 A VCC = 3 × 10−13 eVB + VB with VCC = 2 V VB 755 mV VB(V ) ＋ 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 VCC(V ) 1 2 3 4 5 , R1 10kΩ Q1 , 1kΩ RC VCC － 4.11 VBE = 1.5 V − IE (1 kΩ) ≈ 1.5 V − IC (1 kΩ) (assuming β � 1) �� IC = VT ln IS IC = 775 µA VX ≈ IC (1 kΩ) = 775 mV 4.12 Since we have only integer multiples of a unit transistor, we need to ﬁnd the largest number that divides both I1 and I2 evenly (i.e., we need to ﬁnd the largest x such that I1 /x and I2 /x are integers). This will ensure that we use the fewest transistors possible. In this case, it’s easy to see that we should pick x = 0.5 mA, meaning each transistor should have 0.5 mA ﬂowing through it. Therefore, I1 should be made up of 1 mA/0.5 mA = 2 parallel transistors, and I2 should be made up of 1.5 mA/0.5 mA = 3 parallel transistors. This is shown in the following circuit diagram. I1 I2 VB + − Now we have to pick VB so that IC = 0.5 mA for each transistor. �� IC VB = VT ln IS � � 5 × 10−4 A = (26 mV) ln 3 × 10−16 A = 732 mV 4.13 Using the same technique as in problem12, we have: n1 n2 n2 = = I1 I2 I3 ⇒ n1 n n n n n = 2 = 3 ⇒ 1= 2= 3 0.2 0.3 0.45 4 6 9 n1 = 4 n2 = 6 n = 9 3 so lets choose Hence, VB I1 = n1IS eVB VT ⇒ 0.2 × 10−3 = 4 × 3 × 10−16 e 26 mV ⇒ VB 672 mV 4.14 From KVL, VB = R1I B + VBEQ1 I 1 mA IB = C = ⇒ I B = 10−5 A B 100 I 10-3 VBEQ1 = VT ln ( C ) = 26 × 10−3 ln ( ) IS 7 × 10−16 ⇒ VBEQ1 727.7 mV Therefore, VB = R1I B + VBEQ1 10 kΩ × 10−5 A + 728 × 10−3 ⇒ VB 0.1 + 0.728 ⇒ VB 0.828 V ＋ VB － R1 IC Q1 , , 4.15 VB − VBE = IB R1 IC = β β IC = [VB − VT ln(IC /IS )] R1 IC = 786 µA 4.16 VBE1 I X = IS1exp( V ) T VBE2 ) IY = IS2 exp( VT VBE1 = VBE2 = VBE 2IS2 I I ⇒ X = S1 = IY IS2 IS2 I X = β 1I B1 IX ⇒ = 2 IY = β 2I B2 IY β 1 = β2 ⇒ I B1 I X = =2 I B2 IY R1 ＋ VB － , , IX Q1 IY Q2 , Applying KVL: VB = R1(I B1 + I B2 ) + VBE I 1 mA VBE = VBE1 = VT ln X = 26 mA ln 742 mV IS1 4 × 10−16 I 1 mA β = 100→ = 10 μ A I B1 = X I B1 = β 100 IB1 I 10 μ A = 2 I B2 = B1 = → ⇒ I B2 ⇒ I B2 = 5 μ A IB2 2 2 Hence: VB = 5 × 103 Ω × (10 μ A + 5 μA ) + 0.742 V = 0.075 + 0.742 ⇒ VB 0.817 V 4.17 First, note that VBE 1 = VBE 2 = VBE . VB = (IB 1 + IB 2 )R1 + VBE R1 (IX + IY ) + VT ln(IX /IS 1 ) = β 5 IS 2 = IS 1 3 5 ⇒ IY = IX 3 8 R1 IX + VT ln(IX /IS 1 ) VB = 3β IX = 509 µA IY = 848 µA 4.18 Since Transistor is in forward active region, No change across VBE B ⇓ No change in IB ⇓ ＋ VBE － IB VS －＋ IC C IS e VBE VT E No change in IC 4.19 gm = IC VT V IS exp BE VT ⇒ V = V ln gmVT ⇒ gm = BE T VT IS IS = 6 × 10−16 A 1 gm = 13 Ω VBE 1 −3 13 Ω × 26 × 10 = 26 mV ln 6 × 10−16 ⇒ VBE 750 mV 4.20 gm = IC VT VBE ∆IC 1 = ∆ IS e VT VT VT BE I S e VT ∆VBE V2 T V ∆gm = ⇒ ∆gm IC ∆VBE 2 VT gm ∆VBE VT ⇒ ∆gm ⇒ ∆gm 1 ∆VBE gm VT ∆gm max 0.1 ⇒ ∆VBE max = 0.1 VT gm IC = 1 mA ⇒ ∆VBE ≤ 2.6 mV 4.21 (a) VBE = 0.8 V IC = IS eVBE /VT = 18.5 mA VCE = VCC − IC RC = 1.58 V Q1 is operating in forward active. Its small-signal parameters are gm = IC /VT = 710 mS rπ = β/gm = 141 Ω ro = ∞ The small-signal model is shown below. B + rπ vπ − gm vπ C E (b) IB = 10 µA IC = βIB = 1 mA VBE = VT ln(IC /IS ) = 724 mV VCE = VCC − IC RC = 1.5 V Q1 is operating in forward active. Its small-signal parameters are gm = IC /VT = 38.5 mS rπ = β/gm = 2.6 kΩ ro = ∞ The small-signal model is shown below. B + rπ vπ − gm vπ C E (c) VCC − VBE 1+β IC = RC β β VCC − VT ln(IC /IS ) IC = 1+β RC IE = IC = 1.74 mA VBE = VT ln(IC /IS ) = 739 mV VCE = VBE = 739 mV Q1 is operating in forward active. Its small-signal parameters are gm = IC /VT = 38.5 mS rπ = β/gm = 2.6 kΩ ro = ∞ The small-signal model is shown below. B + rπ vπ − gm vπ C E 4.22 (a) IB = 10 µA IC = βIB = 1 mA VBE = VT ln(IC /IS ) = 739 mV VCE = VCC − IE (1 kΩ) 1+β (1 kΩ) = VCC − β = 0.99 V Q1 is operating in forward active. Its small-signal parameters are gm = IC /VT = 38.5 mS rπ = β/gm = 2.6 kΩ ro = ∞ The small-signal model is shown below. B + rπ vπ − gm vπ C E (b) IE = 1+β VCC − VBE = IC 1 kΩ β β VCC − VT ln(IC /IS ) IC = 1+β 1 kΩ IC = 1.26 mA VBE = VT ln(IC /IS ) = 730 mV VCE = VBE = 730 mV Q1 is operating in forward active. Its small-signal parameters are gm = IC /VT = 48.3 mS rπ = β/gm = 2.07 kΩ ro = ∞ The small-signal model is shown below. B + rπ vπ − C gm vπ E (c) IE = 1 mA β IE = 0.99 mA IC = 1+β VBE = VT ln(IC /IS ) = 724 mV VCE = VBE = 724 mV Q1 is operating in forward active. Its small-signal parameters are gm = IC /VT = 38.1 mS rπ = β/gm = 2.63 kΩ ro = ∞ The small-signal model is shown below. B + rπ vπ − gm vπ C E (d) IE = 1 mA β IE = 0.99 mA IC = 1+β VBE = VT ln(IC /IS ) = 724 mV VCE = VBE = 724 mV Q1 is operating in forward active. Its small-signal parameters are gm = IC /VT = 38.1 mS rπ = β/gm = 2.63 kΩ ro = ∞ The small-signal model is shown below. B + rπ vπ − C gm vπ E 4.23 V IC = IS exp BE nVT gm = rπ = IC = β I B V I 1 ∂ IC IS exp BE = C = ∂VBE nVT nVT nVT ∂ VBE ∂ VBE n βVT β ∂ IB = 1 β ∂ IC = gm = IC C B ＋ vπ － rπ gmvπ E 4.24 V IC = IS exp BE , VT gm = rπ = ∂VBE ∂ IC 2 IC = αI B ⇒ ∂I B ∂ IC = 1 2 α IC = V IS I exp BE = C VT VT VT 1 ∂IC 2 α IC ∂ VBE ∂VBE = ∂ IB = 2 α IC 2 α IC = gm IC VT = 2VT α IC B ＋ rπ vπ － C gmvπ E 4.25 V V IC = IS exp BE 1 + CE VA VT V 1 ∆IC = IS exp BE ∆VCE VT VA ∆IC = IC V 1 IS exp BE ∆VCE VT VA VBE is constant ⇒ ∆IC IC min V V IS exp BE 1 + CE VA VT ∆VCE < 0.05 ⇒ < 0.05 VA + VCE min = ∆VCE VA + VCE ⇒ 20∆VCE < VA + VCE min ∆VCE = 2 V ⇒ 40 < VA + 1 ⇒ VA > 39 V VCE min = 1 V 4.26 (a) V 800 mV IC = IS exp BE = 5 × 10−17 exp 1.15 mA VT 26 mV VX = VCC − RC IC = 2.5 V − 1 kΩ × 1.15 mA VX = 1.35 V Transistor is in Forward Active Region BE CE (b) IC = IS exp V 1 + V A T V V VX 800 ⇒ IC = 5 × 10−17 exp 1 + 26 5 V equation 1 VCC − VX RC equation 2 Also we know: VX = VCC − RC IC ⇒ IC = equations 1, 2 ⇒ VCC − VX 800 VX = 5 × 10−17 exp 1+ 5 RC 26 800 VX ⇒ VX + 5 × 10−14 exp 1 + = 2 .5 5 26 ⇒ 1.2306 VX 1.347 ⇒ VX 1.095 V equation 1 IC 1.406 mA Tronsistor is in Forward Active Region 4.27 ＋ RC IS = 1 × 10−17 A 2= VAkΩ5 V － VCC IS = 1 × 10−17 A Applying KVL: VA = 5 V Applying KVL: Q1 ＋ VB , VCC = RC IC + VCE V V ⇒ VCC = RC IS exp BE 1 + CE + VCE VA VT VCC－ = RC IC + VCE , V V ⇒ VCC = RC IS exp BE 1 + CE + VCE , VA VT VBE Constant ∆VCC V 1 + 1 ∆VCE VBE Constant ∆VCC = RC IS exp BE VT VA V 1 equation 1 = RC IS exp BE + 1 ∆VCE VT VA IC = I S e VT VBE VT equation V IC = IS e VT 1 + CE ⇒ ∆IC = IS eVBE VA ⇒ ∆VCE = I Se VBE × 1 ∆VCE VA × 1 VA ∆IC VCE VBE 1 + ⇒ ∆IC = IS e VA ⇒ ∆VCE = VT × 1 ∆VCE VA × 1 VA ∆IC 1 ISeVBE VT equation 2 1 VBE VT equation 2 VBE equations ⇒ ∆IC = 1, 2 I Se VBE VT × 1 + RC IS e VBE VT 1 VA × 1 VA equations ⇒ ∆IC = 1, 2 ∆VCC ISe VT × 1 + RC IS e VBE VT 1 VA 1 × VA ∆VCC 1 VT ∆VCC = ∆VCC ⇒ ∆IC = VBE V ro + RC IS exp VA + RC IS exp BE 1 VT ∆VCC = ∆VCC ⇒ ∆IC = VT VBE ro + RC VA + RC IS exp 2.31 × 10−4 VT ⇒ ∆IC = × 0.5 ⇒ ∆IC 0.021 mA 5 + 0.4613 −4 2.31 × 10 ⇒ ∆IC = × 0.5 ⇒ ∆IC 0.021 mA 5 + 0.4613 cauld also be obtained using small V IS exp BE signal model 4.28 ＋ ＋ ∆VB － , vπ － , rπ , gmvπ , ro ∆IC RC 2 kΩ ＋ － , ∆VCC We use small signal model, Assuming that the required ∆VB is small enough. Applying Superposition, 1 gmro ∆IC = ∆VCC + ∆VB ro + RC ro + RC ∆IC = 0 ⇒ ∆VB = − 1 ∆VCC gmro VT ∆VCC VA ∆VB = − 1 IC VT VA IC ∆VCC ⇒ ∆VB = − 26 × 10−3 × (3 − 2.5) 5 ⇒ ∆VB = −2.6 mV ⇒ ∆VB = − which is small enough for small signal model 4.29 ＋ ＋ － 1 mA － VCC = 2 V I1 IS = 3 × 10−17 A ＋ VB － Q1 , IC = I S e VB VT , IS = 3 × 10−17 A , VB VB 10−3 I ⇒ VB = VT ln C = 26 mV ln 3 × 10−17 IS ⇒ VB 809.6 mV (a) IC = IS e VT 10−3 V I I = IS e VT 1 + CE ⇒ VB = VT ln C = 26 mV ln 3 × 10−17 C VT IS ⇒ VB 809.6 mV 10−3 = 3 × 10−17 eVB 1 .5 1014 VB VT = 1 + ⇒e 5 3 .9 14 10 ⇒ VB = 26 mV ln 3.9 ⇒ VB 802.8 mV VT (b) IC = IS e VT 1 + VCE VB VT VT 10−3 = 3 × 10−17 eVB 1 .5 1014 VB VT = 1 + ⇒e 5 3 .9 1014 ⇒ VB = 26 mV ln 3.9 ⇒ VB 802.8 mV 4.30 V V IC = IS exp BE 1 + CE VA VT V 1 dIC I V − ro 1 = = IS exp BE C ⇒ ro A dVCE VT VA VA IC ro > 10 kΩ ⇒ VA > 10 kΩ IC ⇒ VA > 10 kΩ × 2 mA ⇒ VA > 20 V 4.31 IC = IS e VBE /VT � � VCE 1+ VA � VCE 1+ VA � IC,T otal = nIC = nIS e gm,T otal = VBE /VT ∂IC ∂VBE IS = n eVBE /VT VT IC ≈n VT = ngm IB,T otal rπ,T otal ro,T otal = n × 0.4435 S 1 = IC,T otal β � −1 � ∂IB,T otal = ∂VBE �−1 � IC,T otal ≈ βVT �−1 � nIC = βVT rπ = n 225.5 Ω (assuming β = 100) = n �−1 � ∂IC,T otal = ∂VCE �−1 � IC,T otal ≈ VA VA = nIC ro = n 693.8 Ω = n The small-signal model is shown below. B + rπ,T otal vπ − gm,T otal vπ ro,T otal C E 4.32 (a) VBE = VCE (for Q1 to operate at the edge of saturation) VT ln(IC /IS ) = VCC − IC RC IC = 885.7 µA VB = VBE = 728.5 mV � � � � (b) Let IC , VB , VBE , and VCE correspond to the values where the collector-base junction is forward biased by 200 mV. � � VBE = VCE + 200 mV � � VT ln(IC /IS ) = VCC − IC RC + 200 mV � IC = 984.4 µA � VB = 731.3 mV � Thus, VB can increase by VB − VB = 2.8 mV if we allow soft saturation. 4.33 ＋ RC 1 kΩ － VCC IS = 7 × 10−16 A, ⇓ ro = ∞ VA = ∞ IS = 7 × 10−16 A, ⇓ Q1 ro = ∞ , Applying KVL, VCC = RC IC + VCE ⇒ RC IS e VBE = VCC RC IS e VA = ∞ , Applying KVL, VCC = RC IC + VCE VCE = VBE − 0.2 V VBE RC IC + VBE − 0.2 V = VCC VCE = VBE − 0.2 V VBE VT ⇒ RC IS e VT + VBE − 0.2 V = VCC CC RC IC = VCC − 0.2 V V VCC = VBE + VBE RC IS e T + VCC − 0.2 = VCC V + VBE − 0.2 V = VCC + VCC − 0.2 = VCC = 0 .2 V 26 mV ⇒ RC IS eVCC VT = 0 .2 V −16 VCC VT ⇒ 1 kΩ × 7 × 10 eVCC 26 mV = 0.2 V ⇒ VCC 686 mV ⇒ RC IS e VCC VT ⇒ 1 kΩ × 7 × 10−16 eVCC ⇒ VCC 686 mV = 0.2 V 4.34 VBE = VCC − IB RB VT ln(IC /IS ) = VCC − IC RB /β IC = 1.67 mA VBC = VCC − IB RB − (VCC − IC RC ) < 200 mV IC RC − IB RB < 200 mV 200 mV + IB RB RC < IC 200 mV + IC RB /β = IC RC < 1.12 kΩ 4.35 IS = 5 × 10−16 A, soft saturation VA = ∞ ⇒ ro = ∞ ＋ ＋ VB － RE , , 1 kΩ Q1 － VCC 2.5 V , ⇒ VBC = 200 mV ⇒ VB = VC + 0.2 V ⇒ VB = 2.7 V Applying I IC ⇒ VB = VBE + RE I E E KVL VBE VB = VBE + RE IC ⇒ VBE + 1 kΩ × IS e VT = 2.7 V ⇒ VBE + 5 × 10 IC = IS eVBE VT −13 e VBE VT = 2.7 V ⇒ VBE 754 mV ⇒ IC 2 mA = 5 × 10−16 e 0.026 0.754 4.36 β = 100, VA = ∞ ⇒ ro = ∞ ＋ RC VCC － VBC = 0.2 V ⇒ RP IC = 0.2 V ⇒ IC = 0 .2 V RP IB 2.5 V RP 1 kΩ IC Q1 , , VBE = VCC − RC (I B + IC ) β = 100 VBE = VCC − β+1 β + 1 RC × 0 .2 RC IC ⇒ VBE = VCC − β β RP V V IC = IS exp BE ⇒ IS = IC exp − BE VT VT ⇒ IS = 0.2 β + 1 RC VCC 0 .2 − exp β RP RP VT VT 0.2 RC VCC 0 .2 − exp RP VT RP VT β = 100 IS ⇒ IS 4.06 × 10−16 A 4.37 IS1 = 3IS2 = 6 × 10−16 A V 300 −11 I1 = IS1 exp EB1 = 6 × 10−16 exp ⇒ I1 6.155 × 10 A VT 26 V 820 I 2 = IS2 exp EB2 = 2 × 10−16 exp ⇒ I 2 10 mA 26 VT I X = I1 + I 2 ⇒ I X 10 mA 4.38 IS = 2 × 10−17 A Applying KCL, β = 100 ＋ 1.7 V － 50 kΩ RB Q1 IC , VCC － 2V ＋ VCC = VEB + RB I B + 1.7 V ⇒ 2 V = VEB + RB ⇒ 0.3 V = VEB IC + 1. 7 V β 50 kΩ + IC 100 VEB , ⇒ 0.3 V = VEB + 500 × IS e VT ⇒ 0.3 V = VEB + 10−14 eVEB VEB 26 mV ⇒ VEB 0.3 V 300 IC = IS e VT ⇒ IC = 2 × 10−17 e 26 ⇒ IC 2.05 × 10−12 A 4.39 IC = 3 mA, β = 100, RB = 23 kΩ 23 kΩ VCC Q1 IC － 1.5 V ＋ Applying KVL, VCC = VEB + RB I B ⇒ VCC = VEB + RB IC β , RB R ⇒ −IC B + VCC = VEB β IC = IS eVEB ⇒ I S = IC e VT − VEB VT , ⇒ I S = IC e 1 RB IC −VCC VT β ⇒ IS 8.85 × 10−17 A 4.40 At the edge of active ⇒ VBC = 0 IC = VB − VBC VB = RC RC 1 .2 V ⇒ IC 0.6 mA 2 kΩ VB － 1.2 V ＋ IC RC Q1 VCC ＋ － 2V ⇒ IC = V V IC = IS exp EB ⇒ IS = IC exp − EB VT VT 800 ⇒ IS = 0.6 × 10−3 exp − 26 ⇒ IS 2.6 × 10−17 A 2 kΩ 4.41 VEB = VEC (for Q1 to operate at the edge of saturation) VCC − IB RB = VCC − IC RC IC RB /β = IC RC RB /β = RC β = RB /RC = 100 4.42 IS = 3 × 10−17 A Applying KVL, 1V 1 kΩ RE Q1 ＋ － , , VCC － 2.5 V ＋ VCC = RE I E + VEB + 1 V −14 I E IC 26 mV VCC = RE IC + VEB + 1 V + VEB + 1 V , ⇒ 2.5 = 1 kΩ × 3 × 10−17 eVEB ⇒ VEB + 3 × 10 e VEB 26 mV = 1.5 V ⇒ VEB 800.5 mV VEB VT 800.5 26 IC = I S e = 3 × 10−17 e ⇒ IC 0.705 mA 4.43 IS = S = 10−17 −A , A ,= 100, VA A = ∞⇒ roro== ∞ ∞ I 3 × 3 × 10 17 β = 100, V = ∞ ⇒ (a) V= 2.52.5 .71=70.8 .8 V VEB = − 1− . = 0 V EB V V 800 IC =C S exp EB EB=3 × 10−1717 exp 800 ⇒ IIC 0..692 mA I I = IS exp = 3 × 10− exp ⇒ C 0 692 26 26 VT V T VEC = VCC CCRCRC I= 2.5 .51 kΩΩ 0.0.692 mA ⇒ VEC 1..808 V VEC = V− − IC C = 2 − − 1 k × × 692 mA ⇒ VEC 1 808 I 0.692 mA gm gm C IC = 0.692 mA ⇒gm m 26.6mS = == ⇒ g 26.6 mS VT VT 26 26 mA mA 100 100 ⇒ r 3.76 kΩ β r =π = = = r ⇒ rπ 3.76 kΩ −3 gm gm26.6 ×6 × 10−3 26. 10 , － vπ ＋ B E , rπ gmvπ ro = ∞ C 1 kΩ , I I I VEB EB = ln lnC IC V V = = Vln βBB (b) V= VT VT I ⇒ ⇒ EBEB VT T ln I S IS I SS E － vπ ＋ open , , B rπ gmvπ ro = ∞ C 500 Ω , 100 × 20 × 10−6−6 × ⇒ VEB EB = 26 mVln ln 100 × 2017 10 ⇒ V= 26 mV × × 3 × 10− −17 3 × 10 ⇒ VEB EB 827mV ⇒ V 827.6 .6 mV IC =C =B ⇒ ⇒C IC 2 mA I I β I B I = = 2 mA VEC = VCC CCRCRC I= 2.5 .50.5.kΩΩ × 2 mA ⇒ VEC = 1..5 V VEC = V− − IC C = 2 − − 0 5 k × 2 mA ⇒ VEC = 1 5 V I 2 mA = == = gm gm C IC = 2 mA⇒ ⇒gm m 77 mS r rπ = β ⇒ rr π 1..3 kΩ ⇒ 1 3 g 77 mS VT VT 26 26 mV ggm mV m 4.43 Continued...... (c) Applying KVL, VCC = VEB + (IC + I B ) × 2 kΩ VEB + 2 kΩ × IC VEB VT ⇒ VEB + 2 kΩ × IS e = VCC = 2.5 V ⇒ VEB 805 mV ⇒ VEB + 6 × 10−14 eVEB IC 26 mV VCC − VEB 2.5 − 0.805 = ⇒ IC 847.5 μ A R 2 kΩ IC 0.8475 × 10−3 = ⇒ gm 32.6 mS VT 0.026 100 β = ⇒ r π 3068 Ω gm 32.6 × 10−3 gm = rπ = E － vπ ＋ B rπ gmvπ , ro = ∞ C 2 kΩ , 4.44 (a) IB = 2 µA IC = βIB = 200 µA VEB = VT ln(IC /IS ) = 768 mV VEC = VCC − IE (2 kΩ) 1+β IC (2 kΩ) = VCC − β = 2.1 V Q1 is operating in forward active. Its small-signal parameters are gm = IC /VT = 7.69 mS rπ = β/gm = 13 kΩ ro = ∞ The small-signal model is shown below. B + rπ vπ − gm vπ C E (b) IE = VCC − VT ln(IC /IS ) 1+β IC = β 5 kΩ IC = 340 µA VEB = 782 mV VEC = VEB = 782 mV Q1 is operating in forward active. Its small-signal parameters are gm = IC /VT = 13.1 mS rπ = β/gm = 7.64 kΩ ro = ∞ The small-signal model is shown below. VCC − VEB 5 kΩ B + rπ vπ − C gm vπ E (c) IE = 1+β IC = 0.5 mA β IC = 495 µA VEB = 971 mV VEC = VEB = 971 mV Q1 is operating in forward active. Its small-signal parameters are gm = IC /VT = 19.0 mS rπ = β/gm = 5.25 kΩ ro = ∞ The small-signal model is shown below. B + rπ vπ − gm vπ C E 4.45 IS = 5 × 10−17 A IS = 5 × 10−17 A VEB = 0.8 V ＋ 1.7 V － X RC 800 IC = 5 × 10−17 e 26 800 = 5 × 10−17 e 26 ⇒ ＋ Q1 VCC － 2.5 V (a) VA = ∞ VA = ∞ ⇒ ⇒ ro = ∞ ro = ∞ VEB VT 500 Ω , , , IC = I S e IC = IS eVEB VT ⇒ ⇒ IC ⇒ IC 1.15 mA IC 1.15 mA VX = RC IC = 0.5 kΩ × 1.15 mA ⇒ VX 0.58 V VX = RC IC = 0.5 kΩ × 1.15 mA ⇒ VX 0.58 V (b) VA = 6 V VA = 6 V V VEB IC = IS e VT 1 + EC VEC VA VT IC = IS e 1 + , VA VEB , VEC = VCC − RC IC V = V − R I EC CC CC V − RC IC ⇒ IC = IS eVEB VT 1 + CC VCC − RC IC VA ⇒ IC = IS eVEB VT 1 + VA VCC IS RC VT VEBe IC 1 ⇒ IC = I S e V + I − R V V e ⇒ IC = IS eVEB VT 1 + CC A S C A VT IC − VA VA VEB VT 800 V 2 .5 17 IS eVEB VT 1 + CC 5 × 10−800e 26 1 + VEB VT VCC VA =5 × 10−17 e 26 1 +2.5 6 I ⇒ IC S=e 1 + −17 IS R VA 6 5 × 10 0.5 kΩ 800 26 × 1 + C eVEB VT = ⇒ IC = e 1+ IS RCVA EB VT V 5 × 10−17 × 0.5 kΩ 800 26 6 e 1+ e 1+ VA 6 VEB VX = RC IC = 500 × 1.44 × 10−13 ⇒ VX = 0.745 V ⇒ IC = 1.49 mA VX = RC IC = 500 × 1.44 × 10−13 ⇒ VX = 0.745 V ⇒ IC = 1.49 mA 4.46 ro = 60 kΩ, ro = VA IC ⇒ IC = 2 mA 60 × 103 Ω = VA 2 × 10−3 A ⇒ VA = 120 V 4.47 ro = 60 kΩ , ro = VA ⇒ IC IC = 1 mA VA = ro IC ⇒ VA ∝ IC ⇒ VA = 60 kΩ × 1 mA ⇒ VA = 60 V VA is half the value in problem 46 as VA is proportional to IC. 4.48 VA = 5 V ＋ ＋ 1.7 V － X 3 kΩ , Q1 VCC － (a) At the edge of active mode 2.5 V ⇒ VX = VB = 1.7 V V 1 .7 V IC = X = RC 3 kΩ ⇒ IC 0.567 mA , , IC = I S e VEB VT VEC 1 + VA ⇒ I e VT IS = C V 1 + EC VA − VEB 0.567 × 10−3 × e IS = 2 .5 − 1 .7 1+ 5 − 800 26 ⇒ IS 2.118 × 10−17 A (b) VA = ∞ VEB VT − VEB VT IC = I S e ⇒ I S = IC e − 800 26 IS = 0.567 × 10−3 e ⇒ IS 2.457 × 10−17 A IS increases 4.49 The direction of current ﬂow in the large-signal model (Fig. 4.40) indicates the direction of positive current ﬂow when the transistor is properly biased. The direction of current ﬂow in the small-signal model (Fig. 4.43) indicates the direction of positive change in current ﬂow when the base-emitter voltage vbe increases. For example, when vbe increases, the current ﬂowing into the collector increases, which is why ic is shown ﬂowing into the collector in Fig. 4.43. Similar reasoning can be applied to the direction of ﬂow of ib and ie in Fig. 4.43. 4.50 IS = 6 × 10−16 A, VA = 5 V, I1 = 2 mA IS = 6 × 10−16 A, VA = 5 V, I1 = 2 mA ＋ Q1 (a) IC = IVEBe S ＋ VB － X I1 , , VCC － VEB VT 2.5 V IC = I S e VT VEC 1 V + EC VA 1 + VA , IC ⇒ VEB = VT ln I C VEC ⇒ VEB = VT ln 1 IS V + EC VA IS 1 + VA VEC = VCC − VX IC VB = VCC − VT ln VECV VCCV VX V = =− − IC VCC − V EB CC BV = V X B CC − VT ln 1 IS V + − V VEB = VCC − VB X IS 1 + CC VA VA 2 × 10−3 ⇒ VB = 2.5 − 0.026ln ⇒ VB 1.757 V 2 × 10−3 ⇒ VB = 2.5 − 0.026 ln 6 × 10−16 1 + 2.5 −1 ⇒ VB 1.757 V 6 × 10−16 1 + 2.5 − 15 5 (b) I = I eVEB VT 1 + VEC ⇒ 1 + VEC = IC e −VEB C S IC = I S e VEB VT VEC V VEC V IC I−VEB A eS = 1 + A⇒ 1 + VA VA IS VT VT VEC = VCC − VX I VX = VCC − I C e −VEB VT − 1 VA VECV VCCV VX V = =− − −V C IS EB VT EB CC BV = V e − 1 X CC − VA VEB = VCC − VB IS dV V I ∆VXdV X ∆VEB ⇒ ∆VX A C e −VEB VT ∆VEB VA VIC I−VEB VT X EB dV∆V T eS ∆VX ⇒ ∆VX ∆VEB EB dVEB VT IS ∆VEB = −∆VB V I ∆VX V − A C e −VEB VT ∆VB ∆VEB = −∆VB I −V A VTC IS EB VT e ∆VX − ∆VB VT IS 5 2 × 10−3 2.5 − 1.757 ⇒ ∆VX − × 10−3 −16 exp2.− − 1.757 × 0.1 × 10−3 ⇒ ∆VX −24.9 mV 50.026 × 6 × 10 2 5 0.026 −3 exp − ⇒ ∆VX − × × 0.1 × 10 ⇒ ∆VX −24.9 mV 0.026 0.026 6 × 10−16 4.50 Continued...... (c) E － vπ ＋ B rπ gmvπ ro X open circunt , , ro = VA 5V = ⇒ ro 2.5 kΩ IC 2 mA IC 2 mA = ⇒ gm 76.9 mS VT 0.026 V gm = rπ = 100 β = ⇒ r π = 1.3 kΩ 2 gm 26 4.51 = 100, RB = 360 kΩ VA = ∞ ⇒ ro = ∞ 360 kΩ RB IB IE ＋ VCC － Q1 IC RC , , RB VA = ∞ β = 100,= 360 kΩ ⇒ ro = ∞ (a) given: VC = VB + 0.2 V , given: VC = VB + 0.2 V ⇒ RC IC = RB I B + 0.2 V ⇒ RC IC = RB I B + 0.2 V I 0.2 V ⇒ RC IC = RB C + 0.2 V ⇒ IC = ⇒ IC = 0.5 mA β IC 0.R V− RB 2 C ⇒ RC IC = RB + 0 .2 V ⇒ I C = ⇒ IC = 0.5 mA Rβ RC − B IC = ISVe EB IC = I S e + VT + VEB VT ⇒ IS = IC e −VEB ⇒ IS = IC e −VEB VT VT ⇒ IS = IC VCC− RB I BT e − ⇒ I S = IC e V − RB I B − CC V VT 0 .2 1 0 .2 V VCC − RB × exp − ⇒ IS = V 0.2 − RB T B .R 02 V − R RC exp − 1 V − R × β C ⇒ IS = B β R VT CC RB β RC − B RC − ⇒ IS 10−15 A = 1 fA ⇒ IS 10−15 A = 1 fA (b) gm = IC IC VT gm = VT ⇒ gm = 0 .2 V ⇒ gm 19.23 mS V RB 0 VT.2RC − ⇒ gm = ⇒ gm 19.23 mS R β VT RC − B 4.52 IS = 5 × 10−16 A, β = 100, VA = ∞ ⇒ ro = ∞ IS = 5 × 10−16 A, = 100, VA = ∞ ⇒ ro = ∞ 1 EB (a) VEB = 0 ⇒ Q1 is offC IC 0 B (b) I B = 0 V = 0 ⇒ Q is off I 0 I =0 ⇒ Q1 is off ⇒ Q1 is off V + 1 kΩ × I CC C (c) Applying KVL: VCC EBVEB + 1 kΩ × IC VEB VEB ⇒ VEB + 1 kΩ × IS e VT T CC ⇒ V ⇒ VEB + 1 kΩ × IS e V VCC VEB mV VEB + 5 × 10−13 e−13 26EB 26 mV .5 V 2 ⇒ VEB + 5 × 10 eV 2 .5 V 0.751 0.751 0.026 ⇒ e 16 0.026 − Applying KVL: V ⇒ VEB 751 mV I = 5 × 10 ⇒ VEB 751 mV C IC = 5 × 10 −16 e IC 1.8 mA ⇒ IC 1.8 mA with this current, transistor is saturated. Note VB < VC Always with this current, transistor is saturated. Note VB < VC Always BC (d) VBC = 0 ⇒ Transistor is at the edge of saturation C . IC ln ⇒ EB = V = (e) I0.5 mA mAV⇒ VEBT=lnT ln IC26 mV mV ln 0−5 mA V I = 26 10 C 0. 5 5 × 16 V = 0 ⇒ Transistor is at the edge of saturation I 0.5 mA S I 5 × 10−16 S ⇒ VEB 718 mV ⇒ VEB 718 mV Vcollector = 500 Ω × IC ⇒ VC = 0.25 V Vcollector = 500 Ω × IC ⇒ VC = 0.25 V As VB = 0, VC = 0.25 V ⇒ Transistor is soft saturated As VB = 0, VC = 0.25 V ⇒ Transistor is soft saturated 4.53 (a) VCB 2 < 200 mV IC 2 RC < 200 mV IC 2 < 400 µA VEB 2 = VE 2 = VT ln(IC 2 /IS 2 ) < 741 mV β2 IE 2 RC 1 + β2 β2 1 + β 1 IC 1 RC 1 + β 2 β1 IC 1 VBE 1 Vin < 200 mV < 200 mV < 396 µA = VT ln(IC 1 /IS 1 ) < 712 mV = VBE 1 + VEB 2 < 1.453 V (b) IC 1 = 396 µA IC 2 = 400 µA gm1 = 15.2 mS rπ1 = 6.56 kΩ ro1 = ∞ gm2 = 15.4 mS rπ2 = 3.25 kΩ ro2 = ∞ The small-signal model is shown below. + vin − rπ1 B1 + vπ1 − E1 /E2 − vπ2 + B2 C2 RC C1 gm1 vπ1 rπ2 gm2 vπ2 vout 4.54 −16 IS1 IS1 IS3I= 2 = 10−10 A, A, 1= 100, β 2= 50, 50, A VA = ∞ = 3 = 2 S 5 × 5 × 16 V =∞ β 1= 100, 2= ＋ ＋ － Vin Q2 , Vout RC , Q1 2.5V － VCC (a) V= 2 = f VB2 B Q2 Q2 Base-Collector Base-Collector V VC 2 .= 0.2 V =0 2V Forward biased 200 mV C 2 Forward biased by by 200 mV V 0 .2 V ⇒ max = VC C 2 max0.2 V ⇒ ⇒ 2 IC 2 max .= 0.4 mA I ⇒ IC 2 IC 2 max = 2 max = = = 0 4 mA RC 500 Ω Ω C max 500 RC 500 Ω 1 +β2+ 1 precisely As As shown: 1 I 1 C 2 IC 2 Note: IC1 I= 1 E2I= 2 = IC 2 IC 2precisely shown: IC C I Note: C I = E 2 β 22 I IC1 I IC1 V 1 VEB VT 1 1 IC1 I= 1 S1 IS1EBe VT 1 ⇒ ⇒ VEB1 T VT lnC ⇒ ⇒ V− Vin Vin T VT lnC VEB1 = V= ln V CC − = V= ln CI = e IS1 CC 1 1 IS IS IS1 4− 4 ×−10 4 I × ⇒ = V= VCC − ln I 1 C1 ⇒ = 2 = − 0 026 ln 4 10 ⇒ Vin Vin CC − VT VT lnC ⇒ Vin Vin .52.5 .− 0.026ln −16 −16 5× 1 IS IS1 5 × 10 10 ⇒ 1 787 V This is is minimum acceptable ⇒ Vin Vin . 1.787 V Thisminimum acceptable Vin Vin E1 － , I IC 0.40.4 mA mA (b) 1g= 1 C1 1 gm m = 26 mV VT VT 26 mV ＋ － rπ1 B1 Vin vπ1 ＋ gm1vπ1 C1 － rπ2 B2 , vπ2 ＋ E2 gm2vπ2 I 0.4 mA gm I = C 20.4 mA gm2 = 2 C 2 = = 26 mV VT VT 26 mV gm m g = gm 15 15.4 ⇒ ⇒ 1g= 1 m2 2 .4 mSmS , C2 500 Ω , Q1 in active mode RC Vout 100 β2 β 1 100 r 1 rπ1 =1 = = 0⇒ ⇒ rπ1 .= 6.Ω kΩ 2 rπ 2 =2 = 3.= 3.kΩ kΩ = r = 6 5 k5 r = 25 25 gm10.4 .4 1 gm2 gm1 gm2 26 26 3 0.4 ×−10−3 I 4 V = V= ln I 2 C 2 26 26 mV 0. × 10 −16 ⇒EB2EB2 741 mV ⇒EC1EC1 . 1.759 V V 741 mV ⇒ V V 1 759 V VEB2EB2 T VT lnC = = mV ln ln 16 5 35 10−10 ⇒ V 2 IS IS2 × 3× 4.55 (a) VBC 2 < 200 mV VBE 2 − (VCC − IC 2 RC ) < 200 mV VT ln(IC 2 /IS 2 ) + IC 2 RC − VCC < 200 mV IC 2 < 3.80 mA VBE 2 < 799.7 mV 1 + β1 IC 1 = IB 2 = IC 2 /β2 IE 1 = β1 IC 1 < 75.3 µA VBE 1 < 669.2 mV Vin = VBE 1 + VBE 2 < 1.469 V (b) IC 1 = 75.3 µA IC 2 = 3.80 mA gm1 = 2.90 mS rπ1 = 34.5 kΩ ro1 = ∞ gm2 = 146.2 mS rπ2 = 342 Ω ro2 = ∞ The small-signal model is shown below. + vin − rπ1 B1 + vπ1 − E1 /B2 C2 + rπ2 vπ2 − E2 gm2 vπ2 RC vout C1 gm1 vπ1 4.56 IS1 = 2IS2 = 6 × 10−17 A , 1= 80, 2= 100 IS1 = 2IS2 = 6 × 10−17 A , β1= 80, β 2= 100 ＋ 1= 80, (a) IC 2 = 2 mA IS1 = 2IS2 = 6 × 10−17 A , ＋ IC 2 = 2 mA Q1 2= 100 RC 500 Ω Vout VCC － 2.5 V Q2 － 2 × 10−3 Vin I C 827.6 VEB2 = VT Iln C 2 = 26 mVln × 10−3 I172= 2 mA mV 2 10− VEB2 = VT ln C 2 IS2 mV ln 3 ×17 827.6 mV = 26 , 3 ×10− IS2 , 2 × 10−3 IC 2 −3 2 × 10VEB2 = VT ln I = 26 mV ln 3 × 10−17 827.6 mV S2 2 × 10−3 IC1 100 689.9 mV 17 VBE1 = VT Iln = 26 mVln 6 × 10− VBE1 = VT ln C1 IS1 mV ln 100 17 689.9 mV = 26 IS1 2 × 10−3 6 ×10− I VBE1 = VT ln C1 = 26 mV ln 100 17 689.9 mV − IS1 6 × 10 Vin = VCC − RC IC 2 − VEB2 + VBE1 = 2.5 − 0.5 kΩ × 2 mA − 0.8276 +0.6899 V =V − R I −V +V = 2.5 − 0.5 kΩ × 2 mA − 0.8276 + 0.6899 in CC C C2 EB2 BE1 , ⇒ Vin 1.362 V ⇒ Vin 1.362 V Vin = VCC − RC IC 2 − VEB2 + VBE1 = 2.5 − 0.5 kΩ × 2 mA − 0.8276 + 0.6899 ⇒ Vin 1.362 V IC 2 2 mA ⇒ gm2 76.9 mS (b) gm2 = I 2= mA = ⇒ gm2 = C 2 VT 26 mV gm2 76.9 mS VT 26 mV B1 C1 IC 2 2 mA = ⇒ gm2 76.9 mS gm2 = ＋ I 2 mA 100 gm1 = C1 = IC1 V 2 mA26 mV ⇒ gm1 769 μS VT 26 mV 100 ＋ =T ⇒ gm1 769 S gm1 = Vn1 VT 26 mV gm1Vn1 － Vin － IC1 2 mA 100 = ⇒ gm1 769 S gm1 = 80 β rπ1 = 1 80 = VT 26 mV 1g RC 1 r1= =1 m 1 1300 gm1 E1 1300 80 ＋ r1= 1 = 1 gm1 ⇒ rπ1 = 104 kΩ rn2 Vn2 1300 ⇒ r 1 = 104 kΩ － r 100 β ⇒ r = ∞ 104 kΩ ∞ rπ2 = 2 100 = ⇒ rπ 2 = 1300 Ω VA 1 = ⇒ ro = 2g 2 ⇒ r 2 = 1300 Ω 26 = =2 VA = ∞ ⇒ ro = ∞ m gm2 2 26 100 r2= 2 = ⇒r gm2 2 26 B2 500Ω E2 gm1Vn1 C2 2 2 = 1300 Ω VA = ∞ ⇒ ro = ∞ ...
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