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Ch11s

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  • 1. Chapter 11Problem Solutions11.1(a) −0.7 − ( −3) = 0.1 ⇒ RE = 23 K RE3 − 1.5 = 0.05 ⇒ RC = 30 K RC(b) vCE 2 = 6 − iC 2 ( RC + 2 RE ) = 6 − iC 2 ( 76 )(c) vcm ( max ) ⇒ vCB 2 = 0 ⇒ vCE 2 = 0.7 VSo 0.7 = 6 − iC 2 ( 76 ) ⇒ iC 2 = 69.74 μ A( v ( max ) − 0.7 ) − ( −3) = 2 CM ( 0.06974 ) ⇒ vCM ( max ) = 0.908 V 23vCM ( min ) ⇒ VS = −3 V ⇒ vCM ( min ) = −2.3 V11.2 Ad = 180, C M RRdB = 85 dB Ad 180C M RR = 17, 783 = = ⇒ Acm = 0.01012 Acm AcmAssume the common-mode gain is negative.v0 = Ad vd + Acm vcm = 180vd − 0.01012vcmv0 = 180 ( 2sin ω t ) mV − ( 0.01012 )( 2sin ω t ) Vv0 = 0.36sin ω t − 0.02024sin ω tIdeal Output: v0 = 0.360sin ω t ( V )Actual Output: v0 = 0.340sin ω t ( V )11.3a.
  • 2. 10 − 2 ( 0.7 )I1 = ⇒ I1 = 1.01 mA 8.5 I1 1.01IC 2 = = ⇒ I C 2 ≅ 1.01 mA 2 2 1+ 1+ β (1 + β ) (100 )(101) ⎛ 100 ⎞ ⎛ 1.01 ⎞IC 4 = ⎜ ⎟⎜ ⎟ ⇒ I C 4 ≅ 0.50 mA ⎝ 101 ⎠ ⎝ 2 ⎠VCE 2 = ( 0 − 0.7 ) − ( −5 ) ⇒ VCE 2 = 4.3 VVCE 4 = ⎡5 − ( 0.5 )( 2 ) ⎤ − ( −0.7 ) ⇒ VCE 4 = 4.7 V ⎣ ⎦b.For VCE 4 = 2.5 V ⇒ VC 4 = −0.7 + 2.5 = 1.8 V 5 − 1.8IC 4 = ⇒ I C 4 = 1.6 mA 2 ⎛ 1+ β ⎞ ⎛ 101 ⎞IC 2 + ⎜ ⎟ ( 2IC 4 ) = ⎜ ⎟ ( 2 )(1.6 ) ⇒ I C 2 = 3.23 mA ⎝ β ⎠ ⎝ 100 ⎠I1 ≈ I C 2 = 3.23 mA 10 − 2 ( 0.7 )R1 = ⇒ R1 = 2.66 kΩ 3.2311.4a. Neglecting base currents 30 − 0.7I1 = I 3 = 400 μ A ⇒ R1 = ⇒ R1 = 73.25 kΩ 0.4VCE1 = 10 V ⇒ VC1 = 9.3 V 15 − 9.3RC = ⇒ RC = 28.5 kΩ 0.2b. (100 )( 0.026 )rπ = = 13 kΩ 0.2 50r0 ( Q3 ) = = 125 kΩ 0.4We have β RC (100 )( 28.5) Ad = = ⇒ Ad = 62 2 ( rπ + RB ) 2 (13 + 10 ) ⎧ ⎫ ⎪ ⎪ β RC ⎪ 1 ⎪ Acm = − ⎨ ⎬ rπ + RB ⎪ 2r0 (1 + β ) ⎪ 1+ ⎪ ⎩ rπ + RB ⎪ ⎭ ⎧ ⎫ (100 )( 28.5 ) ⎪ ⎪ 1 ⎪ ⎪ =− ⎨ ⎬ ⇒ Acm = −0.113 13 + 10 ⎪ 2 (125 )(101) ⎪ 1+ ⎪ ⎩ ⎪ 13 + 10 ⎭ ⎛ 62 ⎞C M RRdB = 20 log10 ⎜ ⎟ ⇒ C M RRdB = 54.8 dB ⎝ 0.113 ⎠c.
  • 3. Rid = 2 ( rπ + RB ) = 2 (13 + 10 ) ⇒ Rid = 46 kΩ 1Ricm = ⎡ rπ + RB + 2 (1 + β ) r0 ⎤ 2⎣ ⎦ 1 = ⎡13 + 10 + 2 (101)(125 ) ⎤ ⇒ Ricm = 12.6 MΩ 2⎣ ⎦11.5 IQ ( 0.5)(a) vCM ( max ) ⇒ VCB = 0 so that vCM ( max ) = 5 − ( RC ) = 5 − (8) 2 2vCM ( max ) = 3 V(b) Vd ⎛ I CQ ⎞ Vd ⎛ 0.25 ⎞ ⎛ 0.018 ⎞ΔI = g m ⋅ =⎜ ⎟⋅ =⎜ ⎟⎜ ⎟ = 0.08654 mA 2 ⎝ VT ⎠ 2 ⎝ 0.026 ⎠ ⎝ 2 ⎠ ΔVC 2 = ΔI ⋅ RC = ( 0.08654 ) ( 8 ) = 0.692 V(c) ⎛ 0.25 ⎞ ⎛ 0.010 ⎞ΔI = ⎜ ⎟⎜ ⎟ = 0.04808 mA ⎝ 0.026 ⎠ ⎝ 2 ⎠ΔVC 2 = ( 0.04808 )( 8 ) = 0.385 V11.6P = ( I1 + I C 4 ) (V + − V − )I1 ≅ I C 4 so 1.2 = 2 I1 ( 6 ) ⇒ I1 = I C 4 = 0.1 mA 3 − 0.7 − ( −3)R1 = ⇒ R1 = 53 k Ω 0.1 3 −1For vCM = +1V ⇒ VC1 = VC 2 = 1 V ⇒ RC = ⇒ RC = 40 k Ω 0.05One-sided output 1 0.05Ad = g m RC where g m = = 1.923 mA / V 2 0.026Then 1Ad = (1.923)( 40 ) ⇒ Ad = 38.5 211.7a. IE0 = 0.7 + ( 2 ) + I E (85) − 5 2 5 − 0.7IE = ⇒ I E = 0.050 mA 85 + 1 ⎛ β ⎞ ⎛ I E ⎞ ⎛ 100 ⎞⎛ 0.050 ⎞I C1 = I C 2 = ⎜ ⎟⎜ ⎟ = ⎜ ⎟⎜ ⎟ ⎝ 1 + β ⎠ ⎝ 2 ⎠ ⎝ 101 ⎠⎝ 2 ⎠Or I C1 = I C 2 = 0.0248 mAVCE1 = VCE 2 = ⎡5 − I C1 (100 ) ⎤ − ( −0.7 ) ⎣ ⎦So VCE1 = VCE 2 = 3.22 Vb. vcm ( max ) for VCB = 0 and VC = 5 − I C1 (100 ) = 2.52 VSo vcm ( max ) = 2.52 Vvcm ( min ) for Q1 and Q2 at the edge of cutoff ⇒ vcm ( min ) = −4.3 V(c) Differential-mode half circuits
  • 4. vd ⎛V ⎞− ′ = Vπ + ⎜ π + g mVπ ⎟ .RE 2 ⎝ rπ ⎠ ⎡ (1 + β ) ⎤ = Vπ ⎢1 + ′ RE ⎥ ⎣ rπ ⎦Then − ( vd / 2 ) Vπ = ⎡ (1 + β ) ⎤ ⎢1 + ′ RE ⎥ ⎣ rπ ⎦ 1 β RCvo = − g mVπ RC ⇒ Ad = ⋅ 2 rπ + (1 + β ) RE ′ β VT (100 )( 0.026 ) ′rπ = = = 105 k Ω RE = 2 k Ω I CQ 0.0248Then 1 (100 )(100 ) Ad = ⋅ ⇒ Ad = 16.3 2 105 + (101)( 2 )11.8a. For v1 = v2 = 0 and neglecting base currents −0.7 − ( −10 )RE = ⇒ RE = 62 kΩ 0.15b. v02 β RCAd = = vd 2 ( rπ + RB ) β VT (100 )( 0.026 ) rπ = = = 34.7 kΩ I CQ 0.075 (100 )( 50 )Ad = ⇒ Ad = 71.0 2 ( 34.7 + 0.5 ) ⎡ ⎤ ⎢ ⎥ β RC ⎢ 1 ⎥Acm = − rπ + RB ⎢ 2 RE (1 + β ) ⎥ ⎢1 + ⎥ ⎢ ⎣ rπ + RB ⎥ ⎦ ⎡ ⎤ (100 )( 50 ) ⎢ 1 ⎥ =− ⎢ ⎥ ⇒ Acm = −0.398 34.7 + 0.5 ⎢ 2 ( 62 )(101) ⎥ ⎢1 + 34.7 + 0.5 ⎥ ⎣ ⎦ 71.0C M RRdB = 20 log10 ⇒ C M RRdB = 45.0 dB 0.398c.Rid = 2 ( rπ + RB )Rid = 2 ( 34.7 + 0.5 ) ⇒ Rid = 70.4 kΩCommon-mode input resistance 1Ricm = ⎡ rπ + RB + 2 (1 + β ) RE ⎤ 2⎣ ⎦ 1 = ⎡34.7 + 0.5 + 2 (101)( 62 ) ⎤ ⇒ Ricm = 6.28 MΩ 2⎣ ⎦11.9
  • 5. (a) v1 = v2 = 1 V ⇒ VE = 1.6 9 − 1.6IE = ⇒ 18.97 μ A 390 IE = 9.49 μ A I C1 = I C 2 = 9.39 μ A 2 vC1 = vC 2 = ( 9.39 )( 0.51) − 9 = −4.21 V(b) 9.39 gm = ⇒ 0.361 mA/V 0.026 ΔI = g m d = ( 0.361× 10−3 ) ( 0.005 ) = 1.805 μ A V 2 ΔvC = (1.805 × 10−6 )( 510 × 103 ) = 0.921 V ⇒ vC 2 = −4.21 + 0.921 ⇒ −3.29 VvC1 = −4.21 − 0.921 ⇒ −5.13 V11.10(a) v1 = v2 = 0 I E1 = I E 2 ≅ 6 μ A β = 60 I C1 = I C 2 = 5.90 μ A vC1 = vC 2 = ( 5.90 )( 0.360 ) − 3 = −0.875 VVEC1 = VEC 2 = +0.6 − ( −0.875 ) = 1.475 V(b)(i) 5.90 gm = ⇒ 0.227 mA/V 0.026 Ad = g m RC = ( 0.227 )( 360 ) = 81.7 Acm = 0(ii) g R ( 60 )( 0.026 ) Ad = m C = 40.8 rπ = 2 0.0059 = 264 K − ( 0.227 )( 360 )Acm = = −0.0442 2 ( 61)( 4000 ) 1+ 26411.11
  • 6. For v1 = v2 = 0.20 VI C1 = I C 2 = 0.1 mAvC1 = vC 2 = ( 0.1)( 30 ) − 10 = −7 V 0.1gm = = 3.846 mA/V 0.026 vΔI = g m d = ( 3.846 )( 0.008 ) ⇒ 30.77 μ A 2ΔvC = ΔI ⋅ RC = ( 30.77 × 10−6 )( 30 × 103 ) = 0.923 Vv2 ↑⇒ I C 2 ↓⇒ vC 2 ↓⇒ vC1 = −7 + 0.923 = −6.077 VvC 2 = −7 − 0.923 = −7.923 V11.12RC = 50 KFor v1 = v2 = 0 −0.7 − ( −10 ) IE = 75 = 0.124 mAI C1 = I C 2 = 0.0615 mA 0.0615gm = = 2.365 mA/V 0.026 (120 )( 0.026 ) rπ = = 50.7 K 0.0615Differential Input v Vv1 = d v2 = − d 2 2Half-circuit. V ⎛ ΔR ⎞ΔI = + g m d ⇒ ΔvC1 = −ΔI ⎜ RC + ⎟ 2 ⎝ 2 ⎠ ⎛ ΔR ⎞ ΔvC 2 = +ΔI ⎜ RC − ⎟ ⎝ 2 ⎠ ⎛ ΔR ⎞ ⎛ ΔR ⎞ vo = ΔvC1 − ΔvC 2 = −ΔI ⎜ RC + ⎟ − ΔI ⎜ RC − ⎟ ⎝ 2 ⎠ ⎝ 2 ⎠ = −2ΔIRC ⎛ V ⎞ = −2 ⎜ g m d ⎟ RC ⎝ 2⎠Ad = − g m RC = − ( 2.365 )( 50 ) = −118.25Common-mode input.
  • 7. ⎛V ⎞ vcm = Vπ + ⎜ π + g mVπ ⎟ ( 2 RE ) ⎝ rπ ⎠ vcm Vπ = ⎛ β⎞ 1 + ⎜ 1 + ⎟ ( 2 RE ) ⎝ rπ ⎠ g m vcm β vcm ΔI = g mVπ = = ⎛1+ β ⎞ rπ + (1 + β )( 2 RE ) 1+ ⎜ ⎟ ( 2 RE ) ⎝ rπ ⎠ ⎛ ΔR ⎞ − β ⎜ RC + ⎟ ⋅ vcm ΔvC1 = −ΔIR1 = ⎝ 2 ⎠ rπ + (1 + β )( 2 RE ) ⎛ ΔR ⎞ − β ⎜ RC − ⎟ vcm ⎝ 2 ⎠ ΔvC 2 = −ΔIR2 = rπ + (1 + β )( 2 RE ) ⎛ ΔR ⎞ ⎛ ΔR ⎞ − β ⎜ RC + ⎟ vcm + β ⎜ RC − ⎟ vcm vo = ΔvC1 − ΔvC 2 = ⎝ 2 ⎠ ⎝ 2 ⎠ [ ] [ ] ⎛ ΔR ⎞ −2 β ⎜ ⎟ vcm = ⎝ 2 ⎠ rπ + (1 + β )( 2 RE ) − βΔR − (120 )( 0.5 ) Acm = = rπ + (1 + β )( 2 RE ) 50.7 + (121)( 2 )( 75 ) = −0.0032966 118.25C M RR = = 35,870.5 0.0032966C M R R ∫ = 91.1 dB dB11.13 v1 = v2 = 0 −0.7 − ( −10 ) IE = 75 = 0.124 mA I C1 = I C 2 = 0.0615 mA 0.0615 gm = = 2.365 mA/V 0.026Δg m = 0.01 gm g m1 = 2.377 mA/V g m 2 = 2.353 mA/V (120 )( 0.026 ) rπ = = 50.7 K 0.0615
  • 8. Vd ΔI = g m 2 VΔvC1 = − g m1 d Rc 2 VdΔvC 2 = + gm2 Rc 2 Vd V vo = ΔvC1 − ΔvC 2 = − g m1 RC − g m 2 d RC 2 2 Vd =− RC ( g m1 + g m 2 ) 2 R −50 Ad = − C ( g m1 + g m 2 ) = ( 2.377 + 2.353) ⇒ Ad = −118.25 2 2Common-Mode − g m1 RC vcm − g m 2 RC vcmΔvC1 = ΔvC 2 = ⎛ 1+ β ⎞ ⎛ 1+ β ⎞ 1+ ⎜ ⎟ ( 2 RE ) 1+ ⎜ ⎟ ( 2 RE ) ⎝ rπ ⎠ ⎝ rπ ⎠vo − ( g m1 − g m 2 ) RC − ( 2.377 − 2.353) ( 50 ) = Acm = =vcm ⎛ 1+ β ⎞ ⎛ 121 ⎞ 1+ ⎜ ⎟ ( 2 RE ) 1+ ⎜ ⎟ ( 2 )( 75 ) ⎝ rπ ⎠ ⎝ 50.7 ⎠ −1.2 = ⇒ Acm = −0.003343 358.99 C M R R ∫ = 91 dB dB11.14(a) v1 = v2 = 0 vE = +0.7 V 5 − 0.7 IE = = 4.3 mA 1I C1 = I C 2 = 2.132 mA vC1 = vC 2 = ( 2.132 )(1) − 5 = −2.87 V(b) v1 = 0.5, v2 = 0 Q2 on Q1 off ⎛ 120 ⎞I C1 = 0, I C 2 = 4.3 ⎜ ⎟ mA = 4.264 mA ⎝ 121 ⎠vC1 = −5 V vC 2 = ( 4.264 ) (1) − 5 vC 2 = −0.736 V 2.132(c) vE ≈ 0.7 V gm == 82.0 mA/V 0.026 v V (82.0 )ΔI = g m d ΔvC = ΔI ⋅ RC = g m d ⋅ RC = ⋅ Vd (1) = 41.0Vd 2 2 2Vd = 0.015 ⇒ Δvc = 0.615 VvC 2 ↓ vC1 ↑vC1 = −2.87 + 0.615 = −2.255 VvC 2 = −2.87 − 0.615 = −3.485 V11.15
  • 9. (a) IC 1gm = = = 38.46 mA/V VT 0.026 vo 1Ad = = = 100 vd 0.01Ad = g m RC100 = 38.46 RCRc = 2.6 K(b)With v1 = v2 = 0vC1 = vC 2 = 10 − (1)( 2.6 ) = 7.4 V ⇒ vcm ( max ) = 7.4 V11.16a.i. ( v01 − v02 ) = 0ii. I C1 = I C 2 = 1 mAv01 − v02 = ⎡V + − I C1 RC1 ⎤ − ⎡V + − I C 2 RC 2 ⎤ ⎣ ⎦ ⎣ ⎦ = I C ( RC 2 − RC1 ) = (1)( 7.9 − 8 ) ⇒ v01 − v02 = −0.1 Vb. ⎛v ⎞I 0 = ( I S 1 + I S 2 ) exp ⎜ BE ⎟ ⎝ VT ⎠ ⎛v ⎞ 2 × 10−3So exp ⎜ BE ⎟ = −13 −13 ⎝ VT ⎠ 10 + 1.1× 10 = 9.524 × 109 ⎛v ⎞ ⎟ = (10 )( 9.524 × 10 ) ⇒ I C1 = 0.952 mA −13 9I C1 = I S 1 exp ⎜ BE ⎝ VT ⎠I C 2 = (1.1× 10−13 )( 9.524 × 109 ) ⇒ I C 2 = 1.048 mAi. v01 − v02 = I C 2 RC 2 − I C1 RC1 ⇒ v01 − v02 = (1.048 − 0.952 )( 8 ) ⇒ v01 − v02 = 0.768 Vii. v01 − v02 = (1.048 )( 7.9 ) − ( 0.952 )( 8 ) v01 − v02 = 8.279 − 7.616 ⇒ v01 − v02 = 0.663 V11.17From Equation (11.12(b)) IQ iC 2 = 1 + evd / VT 10.90 = 1 + evd / VT 1So evd / VT = − 1 = 0.111 0.90vd = VT ln ( 0.111) = ( 0.026 ) ln ( 0.111) ⇒ vd = −0.0571 V11.18From Example 11.2, we have
  • 10. vd ( max ) 10.5 + − 4 ( 0.026 ) 1 + e − vd ( max ) / 0.026 = 0.02 v ( max ) 0.5 + d 4 ( 0.026 ) ⎡ v ( max ) ⎤ 1 0.98 ⎢ 0.5 + d ⎥= ⎢ ⎣ 4 ( 0.026 ) ⎥ 1 + e ⎦ − vd ( max ) / 0.026 10.490 + 9.423vd ( max ) = − vd ( max ) / 0.026 1+ eBy trial and errorvd ( max ) = 23.7 mV11.19a.For I1 = 1 mA, VBE3 = 0.7 V 20 − 0.7R1 = ⇒ R1 = 19.3 kΩ 1 V ⎛ I ⎞ 0.026 ⎛ 1 ⎞R2 = T ⋅ ln ⎜ 1 ⎟ = ⎜I ⎟ ⋅ ln ⎜ ⎟ ⇒ R2 = 0.599 kΩ IQ ⎝ Q⎠ 0.1 ⎝ 0.1 ⎠b. (180 )( 0.026 )rπ 4 = = 46.8 kΩ 0.1 0.1 gm = = 3.846 mA/V 0.026 100 r04 = ⇒ 1 MΩ 0.1From Chapter 10R0 = r04 ⎡1 + g m ( RE rπ 4 ) ⎤ ⎣ ⎦RE rπ 4 = 0.599 46.8 = 0.591R0 = (1) ⎡1 + ( 3.846 )( 0.591) ⎤ = 3.27 MΩ ⎣ ⎦ 100r01 = ⇒ 2 MΩ 0.05 ⎡ ⎛ r ⎞⎤Ricm ≅ ⎡(1 + β ) R0 ⎤ ⎢(1 + β ) ⎜ 01 ⎟⎥ ⎣ ⎦ ⎣ ⎝ 2 ⎠⎦ = ⎣(181)( 3.27 ) ⎦ ⎣(181)(1) ⎤ ⎡ ⎤ ⎡ ⎦ = 592 181 ⇒ Ricm = 139 MΩ(c) From Eq. (11.32(b)) − g m RCAcm = 2 (1 + β ) Ro 1+ rπ + RB 0.05 gm = = 1.923 mA / V 0.026 (180 )( 0.026 ) rπ = = 93.6 k Ω 0.05 RB = 0Then − (1.923)( 50 ) Acm = ⇒ Acm = −0.00760 2 (181)( 3270 ) 1+ 93.6
  • 11. 11.19For vCM = 3.5 V and a maximum peak-to-peak swing in the output voltage of 2 V, we need thequiescent collector voltage to beVC = 3.5 + 1 = 4.5 VAssume the bias is ±10 V , and I Q = 0.5 mA.Then I C = 0.25 mA 10 − 4.5Now RC = ⇒ RC = 22 k Ω 0.25 (100 )( 0.026 )In this case, rπ = = 10.4 k Ω 0.25Then (100 )( 22 ) Ad = = 101 So gain specification is met. 2 (10.4 + 0.5 )For CMRRdB = 80 dB ⇒ 1 ⎡ (1 + β ) I Q Ro ⎤ 1 ⎡ (101)( 0.5 ) Ro ⎤CMRR = 104 = ⎢1 + ⎥ = ⎢1 + ⎥ ⇒ Ro = 1.03 M Ω 2⎣ VT β ⎦ 2 ⎢ ( 0.026 )(100 ) ⎥ ⎣ ⎦Need to use a Modified Widlar current source.Ro = ro ⎡1 + g m ( RE1 rπ ) ⎤ ⎣ ⎦ 100If VA = 100V , then ro = = 200 k Ω 0.5 (100 )( 0.026 )rπ = = 5.2 k Ω 0.5 0.5gm = = 19.23 mA / V 0.026Then 1030 = 200 ⎡1 + (19.23)( RE1 rπ ) ⎤ ⇒ RE1 rπ = 0.216 k Ω = RE1 5.2 ⇒ RE1 = 225 Ω ⎣ ⎦Also let RE 2 = 225 Ω and I REF ≅ 0.5 mA11.20 −0.7 − ( −10 )(a) RE = ⇒ RE = 37.2 k Ω 0.25(b)
  • 12. Vπ 1 V V ⎛1+ β ⎞ Ve + g mVπ 1 + π 2 + g mVπ 2 = e or (1) ⎜ ⎟ (Vπ 1 + Vπ 2 ) = rπ rπ RE ⎝ rπ ⎠ REVπ 1 V1 − Ve ⎛ r ⎞ = ⇒ Vπ 1 = ⎜ π ⎟ (V1 − Ve ) rπ RB + rπ ⎝ rπ + RB ⎠ Vπ 2 = V2 − VeThen ⎛1+ β ⎞ ⎡ rπ ⎤ V(1) ⎜ ⎟⎢ (V1 − Ve ) + (V2 − Ve )⎥ = e ⎠ ⎣ rπ + RB ⎝ rπ ⎦ REFrom this, we find rπ + RB V1 + ⋅ V2 rπVe = ⎡ rπ + RB r + RB ⎤ ⎢ +1+ π ⎥ ⎣ RE (1 + β ) rπ ⎦NowVo = − g mVπ 2 RC = − g m RC (V2 − Ve )We have (120 )( 0.026 ) 0.125rπ = ≅ 25 k Ω, gm = = 4.81 mA / V 0.125 0.026(i) Vd VSet V1 = and V2 = − d 2 2Then ⎛ ⎛ 25 + 0.5 ⎞ ⎞ Vd Vd ⎜ 1 − ⎜ 25 ⎟ ⎟ 2 ( −0.02 ) ⎝ ⎝ ⎠⎠Ve = = 2 ⎡ 25 + 0.5 25 + 0.5 ⎤ 2.026 ⎢ +1+ ⎥ ⎣ ( 37.2 )(121) 25 ⎦SoVe = −0.00494VdNow ⎛ V ⎞ VVo = − ( 4.81)( 50 ) ⎜ − d − ( −0.00494 )Vd ⎟ ⇒ Ad = o = 119 ⎝ 2 ⎠ Vd(ii)Set V1 = V2 = Vcm Then ⎛ 25 + 0.5 ⎞ Vcm ⎜ 1 + ⎟ ⎝ 25 ⎠ V ( −2.02 )Ve = = cm ⎡ 25 + 0.5 25 + 0.5 ⎤ 2.02567 ⎢ +1+ ⎥ ⎣ ( 37.2 )(121) 25 ⎦Ve = Vcm ( 0.9972 )ThenVo = − ( 4.81)( 50 ) ⎡Vcm − Vcm ( 0.9972 ) ⎤ ⎣ ⎦ Voor Acm = = −0.673 Vcm11.21From Equation (11.18)
  • 13. v0 = vC 2 − vC1 = g m RC vd I CQgm = VTFor I Q = 2 mA, I CQ = 1 mA 1Then g m = = 38.46 mA/V 0.026Now 2 = ( 38.46 ) RC ( 0.015 )So RC = 3.47 kΩNow VC = V + − I C RC = 10 − (1)( 3.47 ) = 6.53 VFor VCB = 0 ⇒ vcm ( max ) = 6.53 V11.22The small-signal equivalent circuit isA KVL equation: v1 = Vπ 1 − Vπ 2 + v2 v1 − v2 = Vπ 1 − Vπ 2A KCL equationVπ 1 V + g mVπ 1 + π 2 + g mVπ 2 = 0 rπ rπ ⎛1 ⎞(Vπ 1 + Vπ 2 ) ⎜ + g m ⎟ = 0 ⇒ Vπ 1 = −Vπ 2 ⎝ rπ ⎠ 1 1Then v1 − v2 = 2Vπ 1 ⇒ Vπ 1 = ( v1 − v2 ) and Vπ 2 = − ( v1 − v2 ) 2 2At the v01 node:v01 v01 − v02 + + g mVπ 1 = 0RC RL ⎛ 1 1 ⎞ ⎛ 1 ⎞ 1v01 ⎜ + ⎟ − v02 ⎜ ⎟ = g m ( v2 − v1 ) (1) ⎝ RC RL ⎠ ⎝ RL ⎠ 2At the v02 node:v02 v02 − v01 + + g mVπ 2 = 0RC RL ⎛ 1 1 ⎞ ⎛ 1 ⎞ 1v02 ⎜ + ⎟ − v01 ⎜ ⎟ = g m ( v1 − v2 ) (2) ⎝ RC RL ⎠ ⎝ RL ⎠ 2From (1):
  • 14. ⎛ R ⎞ 1v02 = v01 ⎜ 1 + L ⎟ − g m RL ( v2 − v1 ) ⎝ RC ⎠ 2Substituting into (2) ⎛ R ⎞⎛ 1 1 ⎞ 1 ⎛ 1 1 ⎞ ⎛ 1 ⎞ 1v01 ⎜1 + L ⎟ ⎜ + ⎟ − g m RL ( v2 − v1/ ) ⎜ + ⎟ − v01 ⎜ ⎟ = g m ( v1 − v2 ) ⎝ RC ⎠ ⎝ RC RL ⎠ 2 ⎝ RC RL ⎠ ⎝ RL ⎠ 2 ⎛ 1 RL 1 ⎞ 1 ⎡ ⎛ RL ⎞⎤v01 ⎜ + 2 + ⎟ = g m ( v1 − v2 ) ⎢1 − ⎜ + 1⎟ ⎥ ⎝ RC RC RC ⎠ 2 ⎣ ⎝ RC ⎠⎦v01 ⎛ RL ⎞ 1 ⎛ RL ⎞ ⎜2+ ⎟ = − gm ⎜ ⎟ ( v1 − v2 )RC ⎝ RC ⎠ 2 ⎝ RC ⎠For v1 − v2 = vd 1 − g m RL v01Av1 = = 2 vd ⎛ RL ⎞ ⎜2+ ⎟ ⎝ RC ⎠ 1 g m RL v02From symmetry: Av 2 = = 2 vd ⎛ RL ⎞ ⎜2+ ⎟ ⎝ RC ⎠ v02 − v01 g m RLThen Av = = vd ⎛ RL ⎞ ⎜2+ ⎟ ⎝ RC ⎠11.23The small-signal equivalent circuit isKVL equation: v1 = Vπ 1 − Vπ 2 + v2 or v1 − v2 = Vπ 1 − Vπ 2KCL equation:
  • 15. Vπ 1 V + g mVπ 1 + g mVπ 2 + π 2 = 0 rπ rπ ⎛1 ⎞(Vπ 1 + Vπ 2 ) ⎜ + g m ⎟ = 0 ⇒ Vπ 1 = −Vπ 2 ⎝ rπ ⎠ 1Then v1 − v2 = −2Vπ 2 or Vπ 2 = − ( v1 − v2 ) 2Now v0 = − g mVπ 2 ( RC RL ) 1 = g m ( RC RL )( v1 − v2 ) 2 v0 1For v1 − v2 ≡ vd ⇒ Ad = = g m ( RC RL ) vd 211.23a. 10 − 7RD = ⇒ RD = 6 kΩ 0.5I Q = I D1 + I D 2 ⇒ I Q = 1 mAb.10 = I D ( 6 ) + VDS − VGS IDand VGS = + VTN Kn 0.5For I D = 0.5 mA, VGS = + 2 = 3.12 V 0.4and VDS = 10.12Load line is actually nonlinear.c. Maximum common-mode voltage when M 1 and M 2 reach the transition point, orVDS ( sat ) = VGS − VTN = 3.12 = 2 = 1.12VThenvcm = v02 − vDS ( sat ) + VGS = 7 − 1.12 + 3.12Or vcm ( max ) = 9 VMinimum common-mode voltage, voltage across I Q becomes zero.So vcm ( min ) = −10 + 3.12⇒ vcm ( min ) = −6.88 V11.24
  • 16. We have VC 2 = − g mVπ 2 RC = − g m (Vb 2 − Ve ) RCandVC1 = − g mVπ 1 RC = − g m (Vb1 − Ve ) RCThenV0 = VC 2 − VC1 = − g m (Vb 2 − Ve ) RC − ⎡ − g m (Vb1 − Ve ) RC ⎤ ⎣ ⎦ = g m RC (Vb1 − Vb 2 ) V0Differential gain Ad = = g m RC Vb1 − Vb 2Common-mode gain Acm = 011.25(a) vcm = 3 V ⇒ VC1 = VC 2 = 3 V 10 − 3 Then RC = ⇒ RC = 70 k Ω 0.1(b)CMRRdB = 75 dB ⇒ CMRR = 5623 Now 1 ⎡ (1 + β ) I Q Ro ⎤CMRR = ⎢1 + ⎥ 2⎣ β VT ⎦ 1 ⎡ (151)( 0.2 ) Ro ⎤5623 = ⎢1 + ⎥ ⇒ Ro = 1.45 M Ω 2 ⎢ (150 )( 0.026 ) ⎥ ⎣ ⎦Use a Widlar current source.Ro = ro [1 + g m RE ] ′Let VA of current source transistor be 100 V. 100 0.2Then ro = = 500 k Ω, g m = = 7.69 mA / V 0.2 0.026 (150 )( 0.026 )rπ = = 19.5 k Ω 0.2So 1450 = 500 ⎡1 + ( 7.69 ) RE ⎤ ⇒ RE = 0.247 k Ω ⎣ ′⎦ ′ ′Now RE = RE rπ ⇒ 0.247 = RE 19.5 ⇒ RE = 250Ω ⎛I ⎞Then I Q RE = VT ln ⎜ REF ⎟ ⎜ I ⎟ ⎝ Q ⎠ ⎛ I REF ⎞( 0.2 )( 0.250 ) = ( 0.026 ) ln ⎜ ⎜ ⎟ ⇒ I REF = 1.37 mA ⎟ ⎝ ( 0.2 ) ⎠ 10 − 0.7 − ( −10 )Then R1 = ⇒ R1 = 14.1 k Ω 1.3711.26At terminal A. R (1 + δ ) ⋅ R R (1 + δ ) RRTHA = RA R = = ≅ = 5 kΩ R (1 + δ ) + R 2+δ 2Variation in RTH is not significant ⎛ RA ⎞ + R (1 + δ )( 5 ) 5 (1 + δ )VTHA = ⎜ ⎟V = = ⎝ RA + R ⎠ R (1 + δ ) + R 2+δ
  • 17. At terminal B. RRTHB = R R = = 5 kΩ 2 ⎛ R ⎞ +VTHB = ⎜ ⎟ V = 2.5 V ⎝R+R⎠From Eq. (11.27) − β RC (V2 − V1 )VO = where V2 = VTHB and V1 = VTHA 2 ( rπ + RB ) (120 )( 0.026 )RB = 5 k Ω, rπ = = 12.5 k Ω 0.25 − (120 )( 3)(V2 − V1 )So VO = = −10.3 (V2 − V1 ) 2 (12.5 + 5 )We can find V2 − V1 = VTHB − VTHA ⎡ 5 (1 + δ ) ⎤VTHB − VTHA = 2.5 − ⎢ ⎥ ⎣ 2+δ ⎦ 2.5 ( 2 + δ ) − 5 (1 + δ ) 2.5δ − 5δ= = 2+δ 2+δ −2.5δ≅ = −1.25δ 2Then VO = − (10.3)( −1.25 ) δ = 12.9δSo for −0.01 ≤ δ ≤ 0.01We have −0.129 ≤ VO 2 ≤ 0.129 V11.27a.Rid = 2rπ (180 )( 0.026 ) rπ = = 23.4 kΩ 0.2So Rid = 46.8 kΩb. Assuming rμ → ∞, thenRicm ≅ ⎡(1 + β ) R0 ⎤ ⎣ ⎦Ricm = ⎡(181)(1) ⎤ ⎣ ⎦ = 181 ⇒ Ricm = 181 MΩ11.28(a) 10 − 0.7 − ( −10 ) I1 = = 0.5 ⇒ R1 = 38.6 K R1 0.026 ⎛ 0.5 ⎞R2 = ln ⎜ ⎟ ⇒ R2 = 236 Ω 0.14 ⎝ 0.14 ⎠(b)
  • 18. Ricm ≈ (1 + β ) Ro 0.14Ro = ro 4 (1 + g m 4 RE ) g m 4 = ′ = 5.385 mA/V 0.026 (180 )( 0.026 ) rπ 4 = = 33.4 K 0.14 ′ RE = 33.4 0.236 = 0.234 K 100 ro 4 = = 714 K 0.14 Ro = 714 ⎡1 + ( 5.385 )( 0.234 ) ⎤ ⎣ ⎦ = 1614 KRicm = (181)(1614 ) ≈ 292 MΩ(c) − g m1 RC 0.07Acm = g m1 = = 2.692 mA/V 2 (1 + β ) Ro 0.026 1+ rπ 1 (180 )( 0.026 ) rπ 1 = = 66.86 K 0.07 − ( 2.692 )( 40 )Acm = 2 (181)(1614 ) 1+ 66.86Acm = −0.012311.29 Ad 1 = g m1 ( R1 rπ 3 ) I Q1 / 2g m1 = = 19.23I Q1 VT β VT 2 (100 )( 0.026 ) 5.2rπ 3 = = = IQ2 / 2 IQ 2 IQ 2 g m 3 R2 IQ 2 / 2Ad 2 = , g m3 = = 19.23I Q 2 2 VT (19.23) I Q 2Then 30 = ⋅ R2 ⇒ I Q 2 R2 = 3.12 V 2Maximum vo 2 − vo1 = ±18 mV for linearityvo3 ( max ) = ( ±18 )( 30 ) mV ⇒ ±0.54 Vso I Q 2 R2 = 3.12 V is OK.From Ad 1 :
  • 19. ⎛ ⎛ 5.2 ⎞ ⎞ ⎜ R1 ⎜ ⎜I ⎟ ⎟⎟ ⎜ ⎝ Q2 ⎠ ⎟20 = 19.23I Q1 ( R1 rπ 3 ) = 19.23I Q1 ⎜ ⎟ ⎜ R + ⎛ 5.2 ⎞ ⎟ ⎜ 1 ⎜ IQ 2 ⎟ ⎟ ⎜ ⎟ ⎝ ⎝ ⎠⎠ 19.23I Q1 R1 ( 5.2 )20 = I Q 2 R1 + 5.2 I Q1Let ⋅ R1 = 5V ⇒ I Q1 R1 = 10 V 2 19.23 (10 )( 5.2 )Then 20 = ⇒ I Q 2 R1 = 44.8 V I Q 2 R1 + 5.2 10Now I Q1 R1 = 10 ⇒ R1 = I Q1 ⎛ 10 ⎞ ⎛ IQ2 ⎞So I Q 2 ⎜ ⎜I ⎟ ⎟ = 44.8 ⇒ ⎜ ⎜I ⎟ ⎟ = 4.48 ⎝ Q1 ⎠ ⎝ Q1 ⎠Let I Q1 = 100 μ A, I Q 2 = 448 μ AThenI Q 2 R2 = 3.12 ⇒ R2 = 6.96 k ΩI Q1 R1 = 10 ⇒ R1 = 100 k Ω11.30a. 20 − VGS 3 = 0.25 (VGS 3 − 2 ) 2I1 = 5020 − VGS 3 = 12.5 (VGS 3 − 4VGS 3 + 4 ) 2 212.5VGS 3 − 49VGS 3 + 30 = 0 ( 49 ) − 4 (12.5 )( 30 ) 2 49 ±VGS 3 = ⇒ VGS 3 = 3.16 V 2 (12.5 ) 20 − 3.16I1 = ⇒ I1 = I Q = 0.337 mA 50 IQI D1 = ⇒ I D1 = 0.168 mA 20.168 = 0.25 (VGS 1 − 2 ) ⇒ VGS1 = 2.82 V 2VDS 4 = −2.82 − ( −10 ) ⇒ VDS 4 = 7.18 VVD1 = 10 − ( 0.168 )( 24 ) = 5.97 VVDS1 = 5.97 − ( −2.82 ) ⇒ VDS 1 = 8.79 V(b)(c)
  • 20. Max vCM ⇒ VDS 1 = VDS 2 = VDS ( sat ) = VGS1 − VTN 2.82 − 2 = 0.82 VNow VD1 = 10 − ( 0.168 )( 24 ) = 5.97 VVS ( max ) = 5.97 − VDS1 ( sat ) = 5.97 − 0.82VS ( max ) = 5.15 VvCM ( max ) = VS ( max ) + VGS1 = 5.15 + 2.82vCM ( max ) = 7.97 VvCM ( min ) = V − + VDS 4 ( sat ) + VGS 1VDS 4 ( sat ) = VGS 4 − VTN = 3.16 − 2 = 1.16 VThen vCM ( min ) = −10 + 1.16 + 2.82 ⇒ vCM ( min ) = −6.02 V11.31a.I D1 = I D 2 = 120 μ A = 100 ( VGS1 − 1.2 ) ⇒ VGS 1 = VGS 2 = 2.30 V 2For v1 = v2 = −5.4 V and VDS1 = VDS 2 = 12 V ⇒ −5.4 − 2.30 + 12 = 4.3 V = VD 10 − 4.3RD = ⇒ RD = 47.5 kΩ 0.12I Q = I D1 + I D 2 ⇒ I Q = I1 = 240 μ AI1 = 240 = 200 (VGS 3 − 1.2 ) ⇒ VGS 3 = 2.30 V 2 20 − 2.3R1 = ⇒ R1 = 73.75 kΩ 0.24b. 1 1r04 = = = 416.7 kΩ λ IQ ( 0.01)( 0.24 ) 1 5.4ΔI Q = ⋅ ΔVDS = ⇒ ΔI Q ≅ 13 μ A r04 416.711.32(a) I Q = 160 μ A k′ ⎛ W ⎞ I D = n ⎜ ⎟ (VGS − VTN ) 2 2⎝L⎠ 80 80 = ( 4 )(VOS − 0.5 ) 2 2 80 = 160 (Vo5 − 0.5 ) 2 80VGS = + 0.5 = 1.207 V 160 5−2 RD = = 37.5 K VDS = 2 − ( −1.207 ) = 3.21 V 0.08(c)VDS ( sat ) = VGS − VTN = 1.207 − 0.5 = 0.707 VThen VS = VO 2 − VDS ( sat ) = 2 − 0.707 = +1.29 VAnd v1 = v2 = vcm = VGS + VS = 1.207 + 1.29 vcm = 2.50 V(b)
  • 21. 11.33 vD = 5 − ( 0.2 )( 8 ) = 3.4 V IDVGS = + VTN Kn 0.2 = + 0.8 = 1.694 V 0.25 VDS ( sat ) = VGS − VTN = 1.694 − 0.8 = 0.894 VVS = VD − VDS ( sat ) = 3.4 − 0.894 = 2.506vCM = VS + VGS = 2.506 + 1.694 ⇒ vCM = 4.2 V(b) VdΔvD = ΔI D ⋅ RD ΔI D = g m ⋅ gm = 2 Kn I D 2 =2 ( 0.25)( 0.2 ) = 0.4472 mA/VΔI D = ( 0.4472 )( 0.05 ) ⇒ 22.36 μ AΔvD = ( 22.36 × 10−6 )( 8 × 103 ) = 0.179 V vD 2 = 3.4 + ΔvD vD 2 = 3.4 + 0.179 ⇒ vD 2 = 3.58 V(c) vd = −50 mVΔI D = − ( 0.4472 )( 0.025 ) ⇒ −11.18 μ AΔvD = − (11.18 × 10−6 )( 8 × 103 ) = −0.0894 VvD 2 = 3.4 − 0.0894 ⇒ vD 2 = 3.31 V11.34a.I D1 = I D 2 = 0.5 mAv01 − v02 = ⎡V + − I D1 RD1 ⎤ − ⎡V + − I D 2 RD 2 ⎤ ⎣ ⎦ ⎣ ⎦v01 − v02 = I D 2 RD 2 − I D1 RD1 = I D ( RD 2 − RD1 )i. RD1 − RD 2 = 6 kΩ, v01 − v02 = 0ii. RD1 = 6 kΩ, RD 2 = 5.9 kΩv01 − v02 = ( 0.5 )( 5.9 − 6 ) ⇒ v01 − v02 = −0.05 Vb.
  • 22. K n1 = 0.4 mA / V 2 , K n 2 = 0.44 mA / V 2VGS1 = VGS 2I Q = ( K n1 + K n 2 )(VGS − VTN ) 21 = ( 0.4 + 0.44 )(VGS − VTN ) ⇒ (VGS − VTN ) = 1.19 2 2 I D1 = ( 0.4 )(1.19 ) = 0.476 mA I D 2 = ( 0.44 )(1.19 ) = 0.524 mAi. RD1 = RD 2 = 6 kΩv01 − v02 = ( 0.524 − 0.476 )( 6 ) ⇒ v01 − v02 = 0.288 Vii. RD1 = 6 kΩ, RD 2 = 5.9 kΩv01 − v02 = ( 0.524 )( 5.9 ) − ( 0.476 )( 6 ) = 3.0916 − 2.856 ⇒ v01 − v02 = 0.236 V11.35(a) From Equation (11.69)iD 2 1 Kn ⎛ K ⎞ 2 = − ⋅ vd 1 − ⎜ n ⎟ vd IQ 2 2IQ ⎜ 2IQ ⎟ ⎝ ⎠ 0.1 ⎡ 0.1 ⎤ 20.90 = 0.50 − ⋅ vd 1 − ⎢ ⎥ vd 2 ( 0.25 ) ⎢ 2 ( 0.25 ) ⎥ ⎣ ⎦+0.40 = − ( 0.4472 ) vd 1 − ( 0.2 ) vd 20.8945 = −vd 1 − ( 0.2 ) vd 2Square both sides0.80 = vd (1 − [ 0.2] vd ) 2 2( 0.2 ) ( vd2 ) 2 2 − vd + 0.80 = 0 2 1 ± 1 − 4 ( 0.2 )( 0.80 )vd = = 4V 2 or 1V 2 2 ( 0.2 )Then vd = ± 2 V or ± 1 V IQ 0.25But vd max = = = 1.58 kn 0.1So vd = ±1V, ⇒ vd = −1Vb. From part (a), vd ,max = 1.58 V11.36
  • 23. ⎛i ⎞d ⎜ D1 ⎟ ⎜I ⎟ ⎛ K ⎞ 2 ⎝ Q⎠= Kn ⋅ 1− ⎜ n ⎜ 2I ⎟ vd + ( ⎟ ) vd vd =0 dvd 2IQ ⎝ Q ⎠ Kn = 2IQ iD1 1 KnSo linear = + ⋅ vd IQ 2 2 IQ 1 Kn ⎡1 Kn ⎛K ⎞ 2 ⎤ + ⋅ vd ( max ) − ⎢ + ⋅ vd ( max ) ⋅ 1 − ⎜ n ⎟vd ( max ) ⎥ 2 2IQ ⎢2 2 IQ ⎝ 2I n ⎠ ⎥Then ⎣ ⎦ = 0.02 1 Kn + ⋅v 2 2 I Q d ( max ) ⎡1 Kn ⎤ ⎡1 Kn ⎛K ⎞ 2 ⎤0.98 ⎢ + ⋅ vd ( max ) ⎥ = ⎢ + ⋅ vd ( max ) ⋅ 1 − ⎜ n ⎟ vd ( max ) ⎥ ⎢2 2IQ ⎥ ⎢2 2IQ ⎜ 2I ⎟ ⎥ ⎣ ⎦ ⎣ ⎝ Q ⎠ ⎦ 0.15 ⎡1 0.15 ⎛ 0.15 ⎞ 2 ⎤0.49 + 0.98 ⋅ vd ( max ) = ⎢ + ⋅ vd ( max ) ⋅ 1 − ⎜ ⎟ vd ( max ) ⎥ ⎜ 2 ( 0.2 ) ⎟ 2 ( 0.2 ) ⎢2 2 ( 0.2 ) ⎝ ⎠ ⎥ ⎣ ⎦0.49 + 0.600 vd ( max ) = 0.50 + 0.6124 vd ( max ) ⋅ 1 − ( 0.6124 ) vd ( max ) 20.600 vd ( max ) = 0.010 + 0.6124 vd ( max ) ⋅ 1 − ( 0.6124 ) vd ( max ) 2By trial and error vd ( max ) ≈ 0.429 V11.37(b)gm = 2 K p I D = 2 ( 0.05 )( 0.008696 ) = 0.0417 mA/V VdΔI = g m = ( 0.0417 )( 0.05 ) = 0.002085 mA 2ΔvD = ( 0.002085 )( 510 ) = 1.063vD 2 ↑⇒ vD 2 = 1.063 − 4.565 = −3.502 VvD1 = −1.063 − 4.565 = −5.628 V9 = I S RS + VSG + 1I S = 2I D8 = 2 K P RS (VSG + VTP ) + VSG 28 = ( 2 )( 0.05 )( 390 )(VSG − 0.8 ) + VSG 28 = 39 (VSG − 1.6VSG + 0.64 ) + VSG 2 239VSG − 61.4VSG + 16.96 = 0 61.4 ± 3769.96 − 4 ( 39 )(16.96 )VSG = 2 ( 39 ) = 1.217 V VS = 2.217I S = 0.01739 mA I D1 = I D 2 ⇒ 8.696 μ AvD1 = vD 2 = ( 8.696 )( 0.510 ) − 9 = −4.565 V(b)
  • 24. g m = 2 K P I DQ = 2 ( 0.05 )( 0.008696 ) = 0.0417 mA/V Vd ΔvD = ΔI D ⋅ RD = ( 0.0417 )( 0.05 ) = 0.002085 mA ΔI D = g m ⋅ 2ΔvD = ( 0.002085 )( 510 ) = 1.063 Vv1 ↑, I D1 ↓, vD1 ↓vD1 = −4.565 − 1.063 = −5.628 VvD 2 = −4.565 + 1.063 = −3.502 V11.38(a) v1 = v2 = 0I D = K n (VSG + VTP ) 2ID = 6 μA 6 + 0.4 = VSG 30VSG = 0.847 V VS = +0.847 V vD = I D RD − 3 = ( 6 )( 0.36 ) − 3 = −0.84 VVSD = VS − vD = 0.847 − ( −0.84 ) vSD = 1.69 V(b)(i) Ad = g m RD g m = 2 K n I D =2 ( 30 )( 6 ) = 26.83 μ A/V Ad = ( 26.83)( 0.36 ) ⇒ Ad = 9.66 Acm = 0(ii) g R ( 26.83)( 0.36 ) Ad = m D = ⇒ Ad = 4.83 2 2 − g m RD − ( 26.83)( 0.36 ) Acm = = = −0.0448 1 + 2 g m RO 1 + 2 ( 26.83)( 4 )11.39
  • 25. For v1 = v2 = −0.30 VI D1 = I D 2 = 0.1 mA IDVSG = − VTP KP 0.1 = +1 = 2 V 0.1vD1 = vD 2 = ( 0.1)( 30 ) − 10 = −7 V gm = 2 K p I D = 2 ( 0.1)( 0.1) = 0.2 mA/V ⎛V ⎞ΔI D = g m ⎜ d ⎟ = ( 0.2 )( 0.1) = 0.02 mA ⎝ 2⎠ΔvD = ( ΔI D ) RD = ( 0.02 )( 30 ) = 0.6 VvD 2 ↑⇒ vD 2 = −7 + 0.6 ⇒ vD 2 = −6.4 V vD1 = −7 − 0.6 ⇒ vD1 = −7.6 V11.40For v1 = v2 = 0 0 = VGS + 2 I D RS − 10 10 = VGS + 2 K n RS (VGS − VTN ) 2 = VGS + 2 ( 0.15 )( 75 )(VGS − 1) 2 222.5VGS − 44VGS + 12.5 = 0So VGS = 1.61 V and I D = ( 0.15 )(1.61 − 1) ⇒ 55.9 μ A 2 gm = 2 Kn I D = 2 ( 0.15 )( 0.0559 ) g m = 0.1831 mA/VUse Half-circuits – Differential gain ⎛V ⎞⎛ ΔR ⎞vD1 = − g m ⎜ d ⎟ ⎜ RD + ⎟ ⎝ 2 ⎠⎝ 2 ⎠ ⎛V ⎞⎛ ΔR ⎞vo 2 = g m ⎜ d ⎟ ⎜ RD − ⎟ ⎝ 2 ⎠⎝ 2 ⎠ vo = vD1 − vD 2 = − g mVd RD v Ad = o = − g m RD VdNow – Common-Mode Gain
  • 26. Vi = Vgs + g mVgs ( 2 RS ) = Vcm VcmVgs = 1 + g m ( 2 RS ) ⎛ ΔR ⎞ − g m ⎜ RD + D ⎟ Vcm ⎝ 2 ⎠vD1 = 1 + g m ( 2 RS ) ⎛ ΔR ⎞ − gm ⎜ RD − D ⎟ Vcm ⎝ 2 ⎠vD 2 = 1 + g m ( 2 RS )vO = vD1 − vD 2 − g m ( ΔRD ) VcmSo vo = 1 + g m ( 2 RD ) vo − g m ( ΔRD )Acm = = Vcm 1 + g m ( 2 RS )ThenAd = − ( 0.1831)( 50 ) = −9.16 − ( 0.1831)( 0.5 )Acm = = −0.003216 1 + ( 0.1831)( 2 )( 75 )C M R R ∫ = 69.1 dB bB11.41a. Ad = g m ( r02 r04 ) VA 2 150 r02 = = = 375 kΩ I C 2 0.4 VA 4 100 r04 = = = 250 kΩ I C 4 0.4 IC 2 0.4gm = = = 15.38 mA/V VT 0.026Ad = (15.38 ) ( 375 250 ) ⇒ Ad = 2307b.RL = r02 r04 = 375 250 ⇒ RL = 150 kΩ11.41From 11.40I D1 = I D 2 = 55.9 μ A g m = 0.183 mA/V
  • 27. Vd ⎛ +V ⎞Ad : ΔvD1 = − g m1 ⋅ RD ΔvD 2 = + g m 2 ⎜ d ⎟ RD 2 ⎝ 2 ⎠ V VvO = ΔvD1 − ΔvD 2 = − g m1 d RD − g m 2 d RD 2 2 −V −V ⎛ Δg ⎛ Δg ⎞ ⎞vO = d ⋅ RD ( g m 2 + g m1 ) = d ⋅ RD ⎜ g m − m + ⎜ g m − m ⎟ ⎟ 2 2 ⎝ 2 ⎝ 2 ⎠⎠Ad = − g m RD = − ( 0.183) ( 50 ) = −9.15 ⎛ Δg ⎞ ⎛ Δg ⎞ − ⎜ g m + M ⎟ RD vcm ⎜ g m − M ⎟ RD vCM ⎝ 2 ⎠ ⎝ 2 ⎠ACM : vO = ΔvD1 − ΔvD 2 = + 1 + g m ( 2 RS ) 1 + g m ( 2 RS ) vO −Δg m RDAcm = = Δg m = ( 0.01) ( 0.183) = 0.00183 vcm 1 + g m ( 2 RS ) − ( 0.00183) ( 50 )Acm = = −0.003216 1 + ( 0.183)( 2 ) ( 75 ) C M R R ∫ = 69.1 dB dB11.42(a) v1 = v2 = 05 = 2 I D RS + VSG5 = 2 K p RS (VSG + VTP ) + VSG 25 = 2 ( 0.5 )( 2 ) (VSG − 1.6VSG + 0.64 ) + VSG 2 25 = 2VSG − 2.2VSG + 1.28 22VSG − 2.2VSG − 3.72 = 0 2.2 ± 4.84 + 4 ( 2 )( 3.72 )VSG = 2 ( 2)VSG = 2.02 V 5 − 2.02vS = 2.02 V, IS = = 1.49 mA 2 I D1 = I D 2 = 0.745 mAvD1 = vD 2 = ( 0.745 (1) − 5 ) ⇒ vD1 = vD 2 = −4.26 V(b)5 = I S RS + VSG 25 = ( I D1 + I D 2 ) RS + VSG 25 = ⎡ K p (VSG1 + VTP ) + K p (VSG 2 + VTP ) ⎤ RS + VSG 2 2 2 ⎣ ⎦VSG1 = VSG 2 − 15 = ( 0.5 )( 2 ) ⎡(VSG 2 − 1.8 ) + (VSG 2 − 0.8 ) ⎤ + VSG 2 2 2 ⎣ ⎦5 = ⎡VSG 2 − 3.6VSG 2 + 3.24 + VSG 2 − 1.6VSG 2 + 0.64 ⎤ + VSG 2 ⎣ 2 2 ⎦ 25 = 2VSG 2 − 4.2VSG 2 + 3.88 22VSG 2 − 4.2VSG 2 − 1.12 = 0 4.2 ± 17.64 + 4 ( 2 ) (1.12 )VSG 2 = 2 ( 2)
  • 28. VSG 2 = 2.339 V VSG1 = 1.339 VvS = 2.339 V = 0.5 (1.339 − 0.8 ) = 0.5 ( 2.339 − 0.8 ) 2 2I D1 I D2I D1 = 0.1453 mA I D2 = 1.184 mAvD1 = ( 0.1453)(1) − 5 vD 2 = (1.184 ) (1) − 5vD1 = −4.855 V vD 2 = −3.816 V(c) VdΔI = g m gm = 2 K p I D 2vS ≈ 2.02 V = 2 ( 0.5 )( 0.745 ) g m = 1.22 mA/VΔI = (1.22 )( 0.1) = 0.122 mAΔvD = ( ΔI ) RD = ( 0.122 )(1) = 0.122 VvD 2 ↓ vD1 ↑vD1 = −4.26 + 0.122 vD 2 = −4.26 − 0.122vD1 = −4.138 V vD 2 = −4.382 V11.43 IQa. gf = ⇒ I Q = g f ( 4VT ) = ( 8 )( 4 )( 0.026 ) 4VT⇒ I Q = 0.832 mANeglecting base currents. 30 − 0.7R1 = ⇒ R1 = 35.2 kΩ 0.832 V 100b. r04 = r02 = A = = 240 kΩ I CQ 0.416
  • 29. I CQ 0.416gm = = = 16 mA / V VT 0.026Ad = g m ( r02 || r04 ) = 16 ( 240 || 240 )⇒ Ad = 1920 (180 )( 0.026 )Rid = 2rπ , rπ = = 11.25 kΩ 0.416⇒ Rid = 22.5 kΩR0 = r02 || r04 ⇒ R0 = 120 kΩc. Max. common-mode voltage whenVCB = 0 for Q1 and Q2 .Thereforevcm ( max ) = V + − VEB ( Q3 ) = 15 − 0.7vcm ( max ) = 14.3 V Min. common-mode voltage whenVCB = 0 for Q5 .Thereforevcm ( min ) = 0.7 + 0.7 + ( −15 ) = −13.6 VSo −13.6 ≤ vcm ≤ 14.3 V 1Ricm ≅ (1 + β )( 2 R0 ) 2 V 100R0 = A = = 120 kΩ I Q 0.832Ricm = (181)(120 ) ⇒ Ricm = 21.7 MΩ11.43(a) gm = 2 Kn I D =2 ( 0.4 )(1)g m = 1.265 mA/V v 1Ad = o = = 10 vd 0.1Ad = g m RD10 = (1.265 ) RDRD = 7.91 K(b) Quiescent v1 = v2 = 0vD1 = vD 2 = 10 − (1)( 7.91) = 2.09 V ID 1VGS = + VTN = + 0.8 = 2.38 V Kn 0.4VDS ( sat ) = 2.38 − 0.8 = 1.58So vcm = vD − VDS ( sat ) + VGS = 2.09 − 1.58 + 2.38vcm = 2.89 V11.44
  • 30. g m RDAd = 2For vCM = 2.5 V IQI D1 = I D 2 = = 0.25 mA 2 10 − 3Let VD1 = VD 2 = 3 V , then RD = ⇒ RD = 28 k Ω 0.25 g m ( 28 )Then 100 = ⇒ g m = 7.14 mA / V 2 k′ ⎛ W ⎞And g m = 2 n ⎜ ⎟ ID 2⎝L ⎠ ⎛ 0.080 ⎞ ⎛ W ⎞7.14 = 2 ⎜ ⎟ ⎜ ⎟ ( 0.25 ) ⇒ ⎝ 2 ⎠⎝ L ⎠⎛W ⎞ ⎛W ⎞⎜ ⎟ = ⎜ ⎟ = 1274 (Extremely large transistors to meet the gain requirement.)⎝ L ⎠1 ⎝ L ⎠ 2Need ACM = 0.10From Eq. (11.64(b)) g m RD ACM = 1 + 2 g m Ro ( 7.14 )( 28)So 0.10 = ⇒ Ro = 140 k Ω 1 + 2 ( 7.14 ) RoFor the basic 2-transistor current source 1 1Ro = ro = = = 200 k Ω λ I Q ( 0.01)( 0.5 )This current source is adequate to meet common-mode gain requirement.11.45Not in detail, Approximation looks good.a. −V − ( −5 ) and I S = 2 I D = 2 K n (VGS 1 − VTN ) 2I S = GS 1 RS5 − VGS 1 = 2 ( 0.050 )(VGS 1 − 1) 2 205 − VGS 1 = 2 (VGS1 − 2VGS1 + 1) 2 22VGS1 − 3VGS 1 − 3 = 0 ( 3) + 4 ( 2 )( 3) 2 3±VGS1 = ⇒ VGS1 = 2.186 V 2 ( 2) 5 − 2.186IS = ⇒ I S = 0.141 mA 20 II D1 = I D 2 = S ⇒ I D1 = I D 2 = 0.0704 mA 2v02 = 5 − ( 0.0704 )( 25 ) ⇒ v02 = 3.24 Vb.g m = 2 K n (VGS − VTN ) = 2 ( 0.05 )( 2.186 − 1)g m = 0.119 mA/V 1 1r0 = = = 710 kΩ λ I DQ ( 0.02 )( 0.0704 )
  • 31. Vgs1 = v1 − VS , Vgs 2 = v2 − VSv01 v −V + g mVgs1 + 01 S = 0RD r0 ⎛ 1 1⎞ Vv01 ⎜ + ⎟ + g m ( v1 − VS ) − S = 0 (1) ⎝ RD r0 ⎠ r0v02 v − VS + g mVgs 2 + 02 =0RD r0 ⎛ 1 1⎞ Vv02 ⎜ + ⎟ + g m ( v2 − VS ) − S = 0 (2) ⎝ RD r0 ⎠ r0 v − V v − VS Vg mVgs1 + 01 S + 02 + g mVgs 2 = S r0 r0 RS v01 v02 2VS Vg m ( v1 − VS ) + + − + g m ( v2 − VS ) = S r0 r0 r0 RS v01 v02 ⎧ 2 1 ⎫g m ( v1 + v2 ) + + = VS ⎨2 g m + + ⎬ (3) r0 r0 ⎩ r0 RS ⎭From (1) ⎛ 1⎞ VS ⎜ g m + ⎟ − g m v1v01 = ⎝ r0 ⎠ ⎛ 1 1⎞ ⎜ + ⎟ ⎝ RD r0 ⎠Then ⎛ 1⎞ VS ⎜ g m + ⎟ − g m v1 ⎧ 2 1 ⎫g m ( v1 + v2 ) + ⎝ r0 ⎠ v + 02 = VS ⎨2 g m + + ⎬ (3) ⎛ 1 1⎞ r0 ⎩ r0 RS ⎭ r0 ⎜ + ⎟ ⎝ RD r0 ⎠
  • 32. ⎛ 1 1⎞ ⎛ 1⎞ ⎛ 1 1⎞ ⎧ 2 1 ⎫ ⎛ 1 1⎞g m ( v1 + v2 ) r0 ⎜ + ⎟ + VS ⎜ g m + ⎟ − g m v1 + v02 ⎜ + ⎟ = VS ⎨2 g m + + ⎬ ⋅ r0 ⎜ + ⎟ ⎝ RD r0 ⎠ ⎝ r0 ⎠ ⎝ RD r0 ⎠ ⎩ r0 RS ⎭ ⎝ RD r0 ⎠ ⎛ r ⎞ ⎛ 1 1⎞ ⎧ ⎪⎛ 2 1 ⎞⎛ r0 ⎞ ⎛ 1 ⎞⎪ ⎫g m ( v1 + v2 ) ⎜ 1 + 0 ⎟ − g m v1 + v02 ⎜ + ⎟ = VS ⎨⎜ 2 g m + + ⎟ ⎜1 + ⎟ − ⎜ gm + ⎟⎬ ⎝ RD ⎠ ⎝ RD r0 ⎠ ⎪⎝ ⎩ r0 RS ⎠ ⎝ RD ⎠ ⎝ r0 ⎠ ⎪ ⎭ ⎛ r r ⎞ ⎛ 1 1⎞ ⎧ 2 1 r 2 r 1⎫g m ⎜ v1 ⋅ 0 + v2 + v2 ⋅ 0 ⎟ + v02 ⎜ + ⎟ = VS ⎨2 g m + + + 2gm ⋅ 0 + + 0 − gm − ⎬ ⎝ RD RD ⎠ ⎝ RD r0 ⎠ ⎩ r0 RS RD RD RS RD r0 ⎭ ⎛ r ⎞ ⎛ 1 1⎞ ⎧ ⎪ 1 1 ⎛ r0 ⎞ 2 ⎫ (1 + g m r0 )⎪ (4) rg m ⎜ v1 ⋅ 0 + v2 + v2 ⋅ 0 ⎟ + v02 ⎜ + ⎟ = VS ⎨2 g m + + ⎜1 + ⎟+ ⎬ ⎝ RD RD ⎠ ⎝ RD r0 ⎠ ⎪ ⎩ r0 RS ⎝ RD ⎠ RD ⎪ ⎭ ⎛ 1 1⎞ ⎛ 1⎞Then substituting into (2), v02 ⎜ + ⎟ + g m v2 = VS ⎜ g m + ⎟ ⎝ RD r0 ⎠ ⎝ r0 ⎠ ⎡ 710 710 ⎤ ⎡1 1 ⎤Substitute numbers: ( 0.119 ) ⎢ v1 + v2 + v2 ⎥ + v02 ⎢ 25 + 710 ⎥ (4) ⎣ 25 25 ⎦ ⎣ ⎦ ⎧ 1 1 ⎛ 710 ⎞ 2 ⎫ = VS ⎨0.119 + + ⎜1 + ⎟ + ⎡1 + ( 0.119 )( 710 ) ⎤ ⎬ ⎣ ⎦ ⎩ 710 20 ⎝ 25 ⎠ 25 ⎭( 0.119 ) [ 28.4v1 + 29.4v2 ] + ( 0.0414 ) v02 = VS {0.1204 + 1.470 + 6.8392} = VS ( 8.4296 )or VS = 0.4010v1 + 0.4150v2 + 0.00491v02 ⎛ 1 1 ⎞ ⎛ 1 ⎞Then v02 ⎜ + ⎟ + ( 0.119 ) v2 = VS ⎜ 0.119 + ⎟ (2) ⎝ 25 710 ⎠ ⎝ 710 ⎠v02 ( 0.0414 ) + v2 ( 0.119 ) = ( 0.1204 ) [ 0.401v1 + 0.4150v2 + 0.00491v02 ]v02 ( 0.0408 ) = ( 0.04828 ) v1 − ( 0.0690 ) v2v02 = (1.183) v1 − (1.691) v2 vdNow v1 = vcm + 2 vd v2 = vcm − 2 ⎛ v ⎞ ⎛ v ⎞So v02 = (1.183) ⎜ vcm + d ⎟ − (1.691) ⎜ vcm − d ⎟ ⎝ 2⎠ ⎝ 2⎠Or v02 = 1.437vd − 0.508vcm ⇒ Ad = 1.437, Acm = −0.508 ⎛ 1.437 ⎞C M R RdB = 20 log10 ⎜ ⎟ ⇒ C M R RdB = 9.03 dB ⎝ 0.508 ⎠11.46KVL:
  • 33. v1 = Vgs1 − Vgs 2 + v2So v1 − v2 = Vgs1 − Vgs 2KCL:g mVgs1 + g mVgs 2 = 0 ⇒ Vgs1 = −Vgs 2 1 1So Vgs1 = ( v1 − v2 ) , Vgs 2 = − ( v1 − v2 ) 2 2Nowv02 v02 − v01 + = − g mVgs 2RD RL (1) ⎛ 1 1 ⎞ v01 = v02 ⎜ + ⎟− ⎝ RD RL ⎠ RLv01 v01 − v02 + = − g mVgs1RD RL (2) ⎛ 1 1 ⎞ v02 = v01 ⎜ + ⎟− ⎝ RD RL ⎠ RL ⎛ R ⎞From (1): v01 = v02 ⎜ 1 + L ⎟ + g m RLVgs 2 ⎝ RD ⎠Substitute into (2): ⎛ R ⎞⎛ 1 1 ⎞ ⎛ 1 1 ⎞ v02− g mVgs1 = v02 ⎜1 + L ⎟ ⎜ + ⎟ + g m RL ⎜ + ⎟ Vgs 2 − ⎝ RD ⎠ ⎝ RD RL ⎠ ⎝ RD RL ⎠ RL ⎛ R ⎞⎛ 1 ⎞ ⎛ 1 R 1 ⎞− g m ⋅ ( v1 − v2 ) + g m ⎜ 1 + L ⎟ ⎜ ⎟ ( v1 − v2 ) = v02 ⎜ + L + 2 ⎟ ⎝ RD ⎠ ⎝ 2 ⎠ ⎝ RD RD RD ⎠ 1 ⋅ g m RL1 ⎛ RL ⎞ v02 ⎛ RL ⎞ v02 gm ⎜ ⎟ ( v1 − v2 ) = ⎜ 2 + ⎟ ⇒ Ad 2 = = 22 ⎝ RD ⎠ RD ⎝ RD ⎠ v1 − v2 ⎛ RL ⎞ ⎜2+ ⎟ ⎝ RD ⎠ 1 − ⋅ g m RL v01From symmetry Ad 1 = = 2 v1 − v2 ⎛ RL ⎞ ⎜2+ ⎟ ⎝ RD ⎠ v02 − v01 g m RLThen Av = = v1 − v2 ⎛ RL ⎞ ⎜2+ ⎟ ⎝ RD ⎠11.47
  • 34. v1 − v2 = Vgs1 − Vgs 2 and g mVgs1 + g mVgs 2 = 0 ⇒ Vgs1 = −Vgs 2Then v1 − v2 = −2Vgs 2 1Or Vgs 2 = − ( v1 − v2 ) 2 gmv0 = − g mVgs 2 ( RD RL ) = ( RD RL ) ( v1 − v2 ) 2 gmOr Ad = 2 ( RD RL )11.48 Kn IQFrom Equation (11.64(a)), Ad = ⋅ RD 2 2We need Ad = = 10 0.2 K n ( 0.5 )Then 10 = ⋅ RD or K n ⋅ RD = 20 2If we set RD = 20 k Ω, then K n = 1 mA / V 2For this case VD = 10 − ( 0.25 )( 20 ) = 5 V 0.25VGS = + 1 = 1.5 V 1VDS ( sat ) = VGS − VTN = 1.5 − 1 = 0.5 VThen vcm ( max ) = VD − VDS ( sat ) + VGS = 5 − 0.5 + 1.5Or vcm ( max ) = 6 V11.49Vd 1 = − g mVgs1 RD = − g m RD (V1 − Vs )Vd 2 = − g mVgs 2 RD = − g m RD (V2 − Vs )Now Vo = Vd 2 − Vd 1 = − g m RD (V2 − Vs ) − ( − g m RD (V1 − Vs ) )Vo = g m RD (V1 − V2 )Define V1 − V2 ≡ Vd VThen Ad = o = g m RD and Acm = 0 Vd11.49 Ad = g m ( r02 r04 )g m = 2 kn I DQ =2 ( 0.12 )( 0.075 ) = 0.1897 mA/V 1 1r02 = = = 889 kΩ λn I DQ ( 0.015 )( 0.075 ) 1 1r04 = = = 667 kΩ λ p I DQ ( 0.02 )( 0.075 )Ad = ( 0.1897 ) ( 889 667 ) ⇒ Ad = 72.311.50(a)
  • 35. ⎛ K′ ⎞⎛W ⎞ ⎛ 0.080 ⎞ ⎟ (10 ) = 0.40 mA / V 2K n1 = K n 2 = ⎜ n ⎟ ⎜ ⎟=⎜ ⎝ 2 ⎠⎝ L ⎠ ⎝ 2 ⎠ ID 0.1VGS1 = VGS 2 = + VTN = + 1 = 1.5 V Kn 0.4VDS1 ( sat ) = 1.5 − 1 = 0.5 VFor vCM = +3 V ⇒ VD1 = VD 2 = vCM − VGS 1 + VDS 1 ( sat )= 3 − 1.5 + 0.5 ⇒ VD1 = VD 2 = 2 V 10 − 2RD = ⇒ RD = 80 k Ω 0.1(b) 1Ad = g m RD and g m = 2 ( 0.4 )( 0.1) = 0.4 mA / V 2 1Then Ad = ( 0.4 )( 80 ) = 16 2 16C M R RdB = 45 ⇒ C M R R = 177.8 = AcmSo Acm = 0.090 g m RD Acm = 1 + 2 g m Ro ( 0.4 )(80 )0.090 = ⇒ Ro = 443 k Ω 1 + 2 ( 0.4 ) RoIf we assume λ = 0.01 V −1 for the current source transistor, then 1 1 ro = = = 500 k Ω λ I Q ( 0.01)( 0.2 )So the CMRR specification can be met by a 2-transistor current source. ⎛W ⎞ ⎛W ⎞Let ⎜ ⎟ = ⎜ ⎟ = 1 ⎝ L ⎠3 ⎝ L ⎠ 4 ⎛ 0.080 ⎞ IQ 0.2 ⎟ (1) = 0.040 mA / V and VGS 3 = 2Then K n 3 = K n 4 = ⎜ + VTN = + 1 = 3.24 V ⎝ 2 ⎠ K n3 0.04For vCM = −3 V , VD 3 = −3 − VGS1 = −3 − 1.5 = −4.5 V ⇒ VDS 3 ( min ) = −4.5 − ( −10 ) = 5.5 V > VDS 3 ( sat )So design is OK. ⎛W ⎞On reference side: For ⎜ ⎟ ≥ 1, VGS ( max ) = 3.24 V ⎝L⎠20 − VGS 3 = 20 − 3.24 = 16.76 V 16.67Then = 5.17 ⇒ We need six transistors in series. 3.24
  • 36. 20 − 3.24VGS = = 2.793 V 6 ⎛ K′ ⎞⎛W ⎞ = ⎜ n ⎟ ⎜ ⎟ (VGS − VTN ) 2I REF ⎝ 2 ⎠⎝ L ⎠ ⎛ 0.080 ⎞⎛ W ⎞ ⎛W ⎞ ⎟⎜ ⎟ ( 2.793 − 1) ⇒ ⎜ ⎟ = 1.56 for each of the 6 transistors. 20.2 = ⎜ ⎝ 2 ⎠⎝ L ⎠ ⎝L⎠11.51 1Ad = g m RD 2gm = 2 Kn I D = 2 ( 0.25 )( 0.25) = 0.50 mA / V 1Ad = ( 0.50 )( 3) = 0.75 2From Problem 11.26
  • 37. 5 (1 + δ )V1 = VA = , V2 = VB = 2.5 V and V1 − V2 = 1.25δ 2+δThenVo 2 = Ad ⋅ (V1 − V2 ) = ( 0.75 )(1.25δ ) = 0.9375δSo for −0.01 ≤ δ ≤ 0.01−9.375 ≤ Vo 2 ≤ 9.375 mV11.52From previous results v −v Ad 1 = o 2 o1 = g m1 R1 = 2 K n1 I Q1 ⋅ R1 = 20 v1 − v2 vo3 1 1and Ad 2 = = g m 3 R2 = 2 K n3 I Q 2 ⋅ R2 = 30 vo 2 − vo1 2 2 I Q1 R1 I Q 2 R2Set = 5 V and = 2.5 V 2 2Let I Q1 = I Q 2 = 0.1 mAThen R1 = 100 k Ω, R2 = 50 k Ω 2 ⎛ 0.06 ⎞ ⎛ W ⎞ ⎛ 20 ⎞ ⎛W ⎞ ⎛W ⎞Then 2 ⎜ ⎟ ⎜ ⎟ ( 0.1) = ⎜ ⎟ ⇒ ⎜ ⎟ = ⎜ ⎟ = 6.67 ⎝ 2 ⎠ ⎝ L ⎠1 ⎝ 100 ⎠ ⎝ L ⎠1 ⎝ L ⎠ 2 2 ⎛ 0.060 ⎞ ⎛ W ⎞ ⎛ 2 ( 30 ) ⎞ ⎛W ⎞ ⎛W ⎞and 2 ⎜ ⎟ ⎜ ⎟ ( 0.1) = ⎜ ⎟ ⇒ ⎜ ⎟ = ⎜ ⎟ = 240 ⎝ 2 ⎠ ⎝ L ⎠3 ⎝ 50 ⎠ ⎝ L ⎠3 ⎝ L ⎠ 411.53 2 ⎛ v ⎞a. iD1 = I DSS ⎜ 1 − GS 1 ⎟ ⎝ VP ⎠ 2 ⎛ v ⎞iD 2 = I DSS ⎜ 1 − GS 2 ⎟ ⎝ VP ⎠ ⎛ v ⎞ ⎛ v ⎞ iD1 − iD 2 = I DSS ⎜1 − GS 1 ⎟ − I DSS ⎜ 1 − GS 2 ⎟ ⎝ VP ⎠ ⎝ VP ⎠ I DSS = ( vGS 2 − vGS1 ) VP I DSS I DSS =− ⋅ vd = ⋅ vd VP ( −VP )iD1 + iD 2 = I Q ⇒ iD 2 = I Q − iD1( ) 2 I DSS 2 iD1 − I Q − iD1 = ⋅ vd ( −VP ) 2iD1 − 2 iD1 ( I Q − iD1 ) + ( I Q − iD1 ) = I DSS 2 ⋅ vd ( −VP ) 2 1⎡ ⎤Then iD1 ( I Q − iD1 ) = I ⎢ I Q − DSS 2 ⋅ vd ⎥ 2 2⎢ ⎣ ( −VP ) ⎦ ⎥Square both sides
  • 38. 2 1⎡ I ⎤i − iD1 I Q + ⎢ I Q − DSS 2 ⋅ vd ⎥ = 0 2 D1 2 4⎢ ⎣ ( −VP ) ⎥ ⎦ 2 ⎛ 1⎞⎡ I ⎤ I Q ± I − 4 ⎜ ⎟ ⎢ I Q − DSS 2 ⋅ vd ⎥ 2 2 ( −VP ) ⎥ Q ⎝ 4⎠⎢ ⎣ ⎦iD1 = 2 ⎡ 2 ⎤ 1 2 ⎢ 2 2 I Q I DSS vd ⎛ I DSS vd ⎞ ⎥ 2 2 IQiD1 = ± IQ − IQ − +⎜ ⎟ ⎢ ( −VP ) ⎜ ( −VP ) ⎟ ⎥ 2 2 2 2 ⎝ ⎠ ⎦ ⎣Use + sign 2 IQ 1 2 I Q I DSS 2 ⎛ I ⎞iD1 = + ⋅ vd − ⎜ DSS 2 ⋅ vd ⎟ 2 2 2 ( −VP )2 ⎜ ( −V ) ⎟ ⎝ P ⎠ 2 2 IQ 1 IQ 2 I DSS ⎛ I DSS ⎞ ⎛v ⎞iD1 = + vd −⎜ ⎟ ⎜ d ⎟ 2 2 ( −VP ) ⎜ I ⎟ V IQ ⎝ Q ⎠ ⎝ P⎠Or 2 2iD1 1 ⎛ 1 ⎞ 2 I DSS ⎛ I DSS ⎞ ⎛ vd ⎞ = +⎜ ⎟ ⋅ vd −⎜ ⎟ ⎜ ⎟I Q 2 ⎝ −2VP ⎜ I ⎟ V ⎠ IQ ⎝ Q ⎠ ⎝ P⎠We hadiD 2 = I Q − iD1Then 2 2iD 2 1 ⎛ 1 ⎞ 2 I DSS ⎛ I DSS ⎞ ⎛ vd ⎞ = −⎜ ⎟ ⋅ vd −⎜ ⎟ ⎜ ⎟ I Q 2 ⎝ −2VP ⎜ I ⎟ V ⎠ IQ ⎝ Q ⎠ ⎝ P⎠b.If iD1 = I Q , then 2 2 1 ⎛ 1 ⎞ 2 I DSS ⎛ I DSS ⎞ ⎛ vd ⎞1= +⎜ ⎟ ⋅ vd −⎜ ⎟ ⎜ ⎟ 2 ⎝ −2VP ⎜ I ⎟ V ⎠ IQ ⎝ Q ⎠ ⎝ P⎠ 2 2 2 I DSS ⎛ I DSS ⎞ ⎛ vd ⎞VP = vd −⎜ ⎟ ⎜ ⎟ ⎜ I ⎟ V IQ ⎝ Q ⎠ ⎝ P⎠Square both sides
  • 39. ⎡ 2I ⎛I ⎞ ⎛ vd ⎞ ⎤ 2 2 = v ⎢ DSS − ⎜ DSS ⎟ ⎜ ⎟ ⎥ 2 2VP ⎢ IQ ⎜ I ⎟ V ⎠ ⎝ P⎠ ⎥ d ⎣ ⎝ Q ⎦ 2 2⎛ I DSS ⎞ ⎛ 1 ⎞ 2 2 2 I DSS 2 ⎟ ⎜ ⎟ ( vd ) − 2⎜ ⋅ vd + VP =0⎜ I ⎟ V⎝ Q ⎠ ⎝ P⎠ IQ 2 2 2 2 I DSS ⎛ 2I ⎞ ⎛I ⎞ ⎛ 1 ⎞ ⎟ ⎜ ⎟ (VP ) 2 ± ⎜ DSS ⎟ − 4 ⎜ DSS ⎜ I ⎟ ⎜ I ⎟ V 2 IQ ⎝ Q ⎠ ⎝ Q ⎠ ⎝ P⎠vd = 2 2 ⎛ 2I ⎞ ⎛ 1 ⎞ 2 ⎜ DSS ⎟ ⎜ ⎟ ⎜ IQ ⎟ ⎝ VP ⎠ ⎝ ⎠ 2 ⎛ IQ ⎞vd = (VP ) ⎜ 2 ⎟ ⎝ I DSS ⎠ 1/ 2 ⎛ IQ ⎞Or vd = VP ⎜ ⎟ ⎝ I DSS ⎠c. For vd small, IQ 1 IQ 2 I DSS iD1 ≈ + ⋅ ⋅ vd 2 2 ( −VP ) IQ diD1 1 IQ 2 I DSSgf = = ⋅ ⋅ 2 ( −VP ) vd → 0 d vd IQ ⎛ 1 ⎞ I Q I DSSOr ⇒ g f ( max ) = ⎜ ⎟ ⎝ −VP ⎠ 211.53 Ad = g m ( ro 2 Ro )Want Ad = 400From Example 11.15, ro 2 = 1 M ΩAssuming that g m = 0.283 mA / V for the PMOS from Example 11.15, then Ro = 285 M Ω. ⎛ k ′ ⎞⎛ W ⎞So 400 = g m (1000 285000 ) ⇒ g m = 0.4014 mA / V = 2 ⎜ n ⎟ ⎜ ⎟ I DQ ⎝ 2 ⎠ ⎝ L ⎠1 ⎛ 0.080 ⎞ ⎛ W ⎞ ⎛W ⎞ ⎛W ⎞0.04028 = ⎜ ⎟ ⎜ ⎟ ( 0.1) ⇒ ⎜ ⎟ = ⎜ ⎟ = 10.1 ⎝ 2 ⎠ ⎝ L ⎠1 ⎝ L ⎠1 ⎝ L ⎠ 211.54a.I Q = I D1 + I D 2 ⇒ I Q = 1 mAv0 = 7 = 10 − ( 0.5 ) RD ⇒ RD = 6 kΩb. ⎛ 1 ⎞ I Q ⋅ I DSSg f ( max ) = ⎜ ⎟ ⎝ −VP ⎠ 2 ⎛ 1 ⎞ (1)( 2 ) g f ( max ) = ⎜ ⎟ ⇒ g f ( max ) = 0.25 mA/V ⎝ 4⎠ 2c. g R Ad = m D = g f ( max ) ⋅ RD 2 Ad = ( 0.25 )( 6 ) ⇒ Ad = 1.5
  • 40. 11.55a. −VGS − ( −5 ) 2 ⎛ V ⎞IS = = ( 2 ) I DSS ⎜ 1 − GS ⎟ RS ⎝ VP ⎠ 2 ⎛ V ⎞5 − VGS = ( 2 )( 0.8 )( 20 ) ⎜ 1 − GS ⎟ ⎜ ( −2 ) ⎟ ⎝ ⎠ ⎛ 1 2 ⎞5 − VGS = ( 2 )16 ⎜1 + VGS + VGS ⎟ ⎝ 4 ⎠ 28VGS + 33VGS + 27 = 0 −33 ± 1089 − 4 ( 8 )( 27 )VGS = 2 (8) = −1.125 V 5 − ( −1.125 ) IS = 20 = 0.306 mA I D1 = I D 2 = 0.153 mA vo 2 = 1.17 V(b)11.56Equivalent circuit and analysis is identical to that in problem 11.36. 1 ⋅ g m RL Ad 2 = 2 ⎛ RL ⎞ ⎜2+ ⎟ ⎝ RD ⎠ 1 − ⋅ g m RL Ad 1 = 2 ⎛ RL ⎞ ⎜2+ ⎟ ⎝ RD ⎠ v02 − v01 g m RL Av = = vd ⎛ RL ⎞ ⎜2+ ⎟ ⎝ RD ⎠11.57(a) Ad = g m ( ro 2 ro 4 ) 0.1gm = = 3.846 mA/V 0.026 120ro 2 = = 1200 K 0.1 80ro 4 = = 800 K 0.1Ad = ( 3.846 ) (1200 800 )Ad = 1846(b)
  • 41. For Ad = 923 = ( 3.846 ) (1200 800 RL ) 480 RL 240 = 480 RL = ⇒ RL = 480 K 480 + RL11.58(a) ⎛ 2⎞ I Q = 250 μ A I REF = I Q ⎜ 1 + ⎟ ⎝ β⎠ ⎛ 2 ⎞ = 250 ⎜1 + ⎟ = 252.8 μ A ⎝ 180 ⎠ 5 − ( 0.7 ) − ( −5 ) R1 = ⇒ R1 = 36.8 K 0.2528(b) 0.125 Ad = g m ( ro 2 ro 4 ) gm = = 4.808 mA/V 0.026 150 ro 2 = = 1200 K 0.125 100 Ad = ( 4.808 ) (1200 800 ) ro 4 = = 800 K 0.125 Ad = 2308(c) 2 (180 )( 0.026 )Rid = 2rπ = ⇒ Rid = 74.9 K 0.125 Ro = ro 2 ro 4 = 1200 800 = 480 K = Ro(d) vcm ( max ) = 5 − 0.7 = 4.3 V vcm ( min ) = 0.7 + 0.7 − 5 = −3.6 V11.59a. ⎛ IQ ⎞ ⎛ 1 ⎞I 0 = I B3 + I B 4 ≈ 2 ⎜ ⎟ ⎜ ⎟ ⎝ 2 ⎠⎝ β ⎠ I Q 0.2I0 = = ⇒ I0 = 2 μ A β 100b. V 100 r02 = r04 = A = = 1000 kΩ I CQ 0.1 I CQ 0.1gm = = = 3.846 mA/V VT 0.026Ad = g m ( r02 r04 ) = ( 3.846 ) (1000 1000 ) ⇒ Ad = 1923c. ( Ad = g m r02 r04 RL )Ad = ( 3.846 ) (1000 1000 250 ) ⇒ Ad = 64111.60a.
  • 42. Ad = g m ( r02 r04 RL ) I CQ IQgm = = VT 2VT V 125r02 = A 2 = I CQ I CQ VA 4 80r04 = = I CQ I CQIf I Q = 2 mA, then g m = 38.46 mA/Vr02 = 125 kΩ, r04 = 80 kΩSo Ad = 38.46 ⎡125 80 200 ⎤ ⎣ ⎦Or Ad = 1508For each gain of 1000. lower the current levelFor I Q = 0.60 mA, I CQ = 0.30 mA 0.3gm = = 11.54 mA/V 0.026 125r02 = = 417 kΩ 0.3 80r04 = = 267 kΩ 0.3Ad = 11.54 ⎡ 417 267 200 ⎤ = 1036 ⎣ ⎦So I Q = 0.60 mA is adequateb.For V + = 10 V, VBE = VEB = 0.6 VFor VCB = 0, vcm ( max ) = V + − 2VEB = 10 − 2 ( 0.6 )Or vcm ( max ) = 8.8 V11.61a. From symmetry. 0.1VGS 3 = VGS 4 = VDS 3 = VDS 4 = +1 0.1Or VDS 3 = VDS 4 = 2 V 0.1VSG1 = VSG 2 = +1 = 2 V 0.1VSD1 = VSD 2 = VSG1 − (VDS 3 − 10 ) = 2 − ( 2 − 10 )Or VSD1 = VSD 2 = 10 Vb. 1 1r0 n = = ⇒ 1 MΩ λn I DQ ( 0.01)( 0.1) 1 1r0 p = = ⇒ 0.667 MΩ λP I DQ ( 0.015 )( 0.1)g m = 2 K p (VSG + VTP ) = 2 ( 0.1)( 2 − 1) = 0.2 mA / VAd = g m ( ron rop ) = ( 0.2 ) (1000 667 ) ⇒ Ad = 80(c)
  • 43. IQI D 2 = I D1 = = 0.1 mA 2 1 1 ro 4 = = = 1000 k Ω λn I D 4 ( 0.01)( 0.1) 1 1 ro 2 = = = 667 k Ω λP I D 2 ( 0.015)( 0.1) Ro = ro 2 ro 4 = 667 1000 = 400 k Ω11.62 Ad = g m ( ro 4 ro 2 ) ⎛ 0.08 ⎞gm = 2 ⎜ ⎟ ( 2.5 )( 0.05 ) ⎝ 2 ⎠ = 0.1414 mA/V 1ro 4 = = 1000 K ( 0.02 )( 0.05 ) 1ro 2 = = 1333 K ( 0.015)( 0.05 )Ad = ( 0.1414 ) (1000 1333)Ad = 80.811.63 R04 = r04 ⎡1 + g m 4 ( R rπ 4 ) ⎤ ⎣ ⎦ 80 r04 = = 800 K 0.1 0.1gm4 = = 3.846 0.026 (100 )( 0.026 ) rπ 4 = 0.1 = 26 KR rπ 4 = 1 26 = 0.963 KAssume β = 100 (100 )( 0.026 ) rπ 3 = = 26 kΩ 0.1 0.1g m3 = = 3.846 mA/V 0.026 R04 = 800 ⎡1 + ( 3.846 )( 0.963) ⎤ ⇒ 3.763 MΩ ⎣ ⎦⇒ R0 = 3.763MΩ Then Av = − g m ( r02 R0 ) 120 r02 = = 1200 kΩ 0.1 0.1 gm = = 3.846 mA/V 0.026 Av = − ( 3.846 ) ⎡1200 3763⎤ ⇒ Av = −3499 ⎣ ⎦b.For
  • 44. 80 R = 0, r04 = = 800 kΩ 0.1Av = − g m ( r02 r04 ) = − ( 3.846 ) ⎡1200 800 ⎤ ⇒ Av = −1846 ⎣ ⎦(c) For part (a), Ro = ( 3.763 1.2 ) = 0.910 M ΩFor part (b), Ro = (1.2 0.8 ) = 0.48 M Ω11.64 IE5 I +I I +II B5 = = B3 B4 = C 3 C 4 1+ β 1+ β β (1 + β )Now I C 3 + I C 4 ≈ I Q IQSo I B 5 ≈ β (1 + β ) IE6 I Q1I B6 = = 1 + β β (1 + β )For balance, we want I B 6 = I B 5So that I Q1 = I Q11.65Resistance looking into drain of M4.Vsg 4 ≅ I X R1 VX − Vsg 4I X ± g m 4Vsg 4 = r04 ⎡ R ⎤ VI X ⎢1 + g m 4 R1 + 1 ⎥ = X ⎣ r04 ⎦ r04 ⎡ R ⎤Or R0 = r04 ⎢1 + g m 4 R1 + 1 ⎥ ⎣ r04 ⎦a.
  • 45. Ad = g m 2 ( ro 2 Ro )g m 2 = 2 K n I DQ = 2 ( 0.080 )( 0.1) = 0.179 mA / V 1 1 ro 2 = = = 667 k Ω λn I DQ ( 0.015 )( 0.1)g m 4 = 2 K P I DQ = 2 ( 0.080 )( 0.1) = 0.179 mA / V 1 1 ro 4 = = = 500 k Ω λ p I DQ ( 0.02 )( 0.1) ⎡ 1 ⎤ R0 = 500 ⎢1 + ( 0.179 )(1) + = 590.5 kΩ ⎣ 500 ⎥ ⎦ Ad = ( 0.179 ) ⎡667 590.5⎤ ⇒ Ad = 56.06 ⎣ ⎦b.When R1 = 0, R0 = r04 = 500 kΩ Ad = ( 0.179 ) ⎡667 500 ⎤ ⇒ Ad = 51.15 ⎣ ⎦(c) For part (a), Ro = ro 2 Ro = 667 590.5 ⇒ Ro = 313 k ΩFor part (b), Ro = ro 2 ro 4 = 667 500 ⇒ Ro = 286 kΩ11.66Let β = 100, VA = 100 V
  • 46. VA 100ro 2 = = = 1000 k Ω I CQ 0.1Ro 4 = ro 4 [1 + g m RE ] where RE = rπ RE ′ ′Now (100 )( 0.026 )rπ = = 26 k Ω 0.1 0.1gm = = 3.846 mA / V 0.026 ′RE = 26 1 = 0.963 k ΩThen Ro 4 = 1000 ⎡1 + ( 3.846 )( 0.963) ⎤ = 4704 k Ω ⎣ ⎦Ad = g m ( ro 2 Ro 4 ) = 3.846 (1000 4704 ) ⇒ Ad = 317211.67(a) For Q2, Q4 Vx − Vπ 4 V(1) Ix = + g m 2Vπ 2 + g m 4Vπ 4 + x ro 2 ro 4 Vx − Vπ 4 V(2) g m 2Vπ 2 + = π4 ro 2 rπ 4 rπ 2(3) Vπ 4 = −Vπ 2 Vx ⎡ 1 1 ⎤From (2) = Vπ 4 ⎢ + + gm2 ⎥ ro 2 ⎢ ⎣ rπ 4 rπ 2 ro 2 ⎥ ⎦
  • 47. Now ⎛ β ⎞ ⎛ IQ ⎞ ⎛ 120 ⎞IC 4 = ⎜ ⎟⎜ ⎟=⎜ ⎟ ( 0.5 ) = 0.496 mA ⎝ 1+ β ⎠⎝ 2 ⎠ ⎝ 121 ⎠ ⎛ IQ ⎞ ⎛ 1 ⎞ ⎛ β ⎞ ⎛ 120 ⎞IC 2 = ⎜ ⎟⎜ ⎟⎜ ⎟ = ( 0.5 ) ⎜ ⎟ ⇒ I C 2 = 0.0041 mA ⎜ (121)2 ⎟ ⎝ 2 ⎠⎝ 1+ β ⎠⎝ 1+ β ⎠ ⎝ ⎠So (120 )( 0.026 )rπ 2 = = 761 k Ω 0.0041 0.0041gm2 = = 0.158 mA/V 0.026 100ro 2 = ⇒ 24.4 M Ω 0.0041 (120 ) ( 0.026 )rπ 4 = = 6.29 k Ω 0.496 0.496gm4 = = 19.08 mA / V 0.026 100ro 4 = = 202 k Ω 0.496NowVx ⎡ 1 1 ⎤ Vx = Vπ 4 ⎢ + + 0.158⎥ ⇒ which yields Vπ 4 =ro 2 ⎢ 6.29 761 24400 ⎣ ⎥ ⎦ ( 0.318) ro 2From (1), V V ⎛ 1 ⎞I x = x + x + Vπ 4 ⎜ g m 4 − g m 2 − ⎟ ro 2 ro 4 ⎝ ro 2 ⎠ ⎡ ⎛ 1 ⎞⎤Ix ⎢ 1 ⎜ 19.08 − 0.158 − ⎟ 1 24400 ⎠ ⎥ +⎝ V =⎢ + ⎥ which yields Ro 2 = x = 135 k ΩVx ⎢ 24400 202 ( 0.318)( 24400 ) ⎥ Ix ⎢ ⎥ ⎣ ⎦ 80Now ro 6 = = 160 k Ω 0.5Then Ro = Ro 2 ro 6 = 135 160 ⇒ Ro = 73.2 k Ω(b) ΔiAd = g m Ro where g m = c c vd / 2
  • 48. vdΔi = g m1Vπ 1 + g m 3Vπ 3 and Vπ 1 + Vπ 3 = 2 ⎛V ⎞Also ⎜ π 1 + g m1Vπ 1 ⎟ rπ 3 = Vπ 3 ⎝ rπ 1 ⎠ ⎛1+ β ⎞So Vπ 1 ⎜ ⎟ rπ 3 = Vπ 3 ⎝ rπ 1 ⎠ ⎛ 121 ⎞Or Vπ 1 ⎜ ⎟ ( 6.29 ) = Vπ 3 ≅ Vπ 1 ⎝ 761 ⎠ v vThen 2Vπ 1 = d ⇒ Vπ 1 = d 2 4 ⎛v ⎞ ⎛v ⎞So Δi = ( g m1 + g m 3 ) Vπ 1 = ( 0.158 + 19.08 ) ⎜ d ⎟ = 9.62 ⎜ d ⎟ ⎝ 4⎠ ⎝ 2⎠ ΔiSo g m = c = 9.62 ⇒ Ad = ( 9.62 )( 73.2 ) ⇒ Ad = 704 vd / 2Now Rid = 2 Ri where Ri = rπ 1 + (1 + β ) rπ 3Ri = 761 + (121)( 6.29 ) = 1522 k ΩThen Rid = 3.044 M Ω11.69(a) Ad = 100 = g m ( ro 2 ro 4 )Let I Q = 0.5 mA 1 1ro 2 = = = 200 k Ω λn I D ( 0.02 )( 0.25 ) 1 1ro 4 = = = 160 k Ω λP I D ( 0.025 )( 0.25 )Then 100 = g m ( 200 160 ) ⇒ g m = 1.125 mA / V ⎛ K′ ⎞⎛W ⎞gm = 2 ⎜ n ⎟ ⎜ ⎟ ID ⎝ 2 ⎠⎝ L ⎠ ⎛ 0.080 ⎞ ⎛ W ⎞ ⎛W ⎞1.125 = 2 ⎜ ⎟ ⎜ ⎟ ( 0.25 ) ⇒ ⎜ ⎟ = 31.6 ⎝ 2 ⎠⎝ L ⎠ ⎝ L ⎠n ⎛W ⎞ ⎛W ⎞Now ⎜ ⎟ somewhat arbitrary. Let ⎜ ⎟ = 31.6 ⎝ L ⎠P ⎝ L ⎠P11.70
  • 49. Ad = g m ( ro 2 ro 4 )P = ( I Q + I REF ) (V + − V − )Let I Q = I REFThen 0.5 = 2 I Q ( 3 − ( −3) ) ⇒ I Q = I REF = 0.0417 mA 1 1ro 2 = = = 3205 k Ω λn I D ( 0.015 )( 0.0208 ) 1 1ro 4 = = = 2404 k Ω λP I D ( 0.02 )( 0.0208 )ThenAd = 80 = g m ( 3205 2404 ) ⇒ g m = 0.0582 mA/V ⎛ k ′ ⎞⎛ W ⎞gm = 2 ⎜ n ⎟ ⎜ ⎟ I D ⎝ 2 ⎠ ⎝ L ⎠n ⎛ 0.080 ⎞⎛ W ⎞ ⎛W ⎞0.0582 = 2 ⎜ ⎟⎜ ⎟ ( 0.0208 ) ⇒ ⎜ ⎟ = 1.02 ⎝ 2 ⎠⎝ L ⎠ n ⎝ L ⎠n11.71 Ad = g m ( ro 2 Ro ) ≈ g m ro 2 1ro 2 = λn I D 1 = = 666.7 K ( 0.015)( 0.1)Ad = 400 = g m ( 666.7 ) g m = 0.60 mA/V ⎛ k′ ⎞⎛ W ⎞= 2 ⎜ n ⎟⎜ ⎟ ID ⎝ 2 ⎠⎝ L ⎠ ⎛ 0.08 ⎞ ⎛ W ⎞0.60 = 2 ⎜ ⎟ ⎜ ⎟ ( 0.1) ⎝ 2 ⎠⎝ L ⎠ ⎛W ⎞0.090 = 0.004 ⎜ ⎟ ⎝L⎠⎛W ⎞ ⎛W ⎞⎜ ⎟ = ⎜ ⎟ = 22.5⎝ L ⎠1 ⎝ L ⎠ 211.72
  • 50. Ad = g m ( Ro 4 Ro 6 )whereRo 4 = ro 4 + ro 2 [1 + g m 4 ro 4 ]Ro 6 = ro 6 + ro8 [1 + g m 6 ro 6 ]We have 1ro 2 = ro 4 = = 1667 k Ω ( 0.015 )( 0.040 ) 1ro 6 = ro8 = = 1250 k Ω ( 0.02 )( 0.040 ) ⎛ 0.060 ⎞gm4 = 2 ⎜ ⎟ (15 )( 0.040 ) = 0.268 mA/V ⎝ 2 ⎠ ⎛ 0.025 ⎞gm6 = 2 ⎜ ⎟ (10 )( 0.040 ) = 0.141 mA/V ⎝ 2 ⎠ThenRo 4 = 1667 + 1667 ⎡1 + ( 0.268 )(1667 ) ⎤ ⇒ 748 M Ω ⎣ ⎦Ro 6 = 1250 + 1250 ⎡1 + ( 0.141)(1250 ) ⎤ ⇒ 222.8 M Ω ⎣ ⎦(a)Ro = Ro 4 Ro 6 = 748 222.8 ⇒ Ro = 172 M Ω(b) Ad = g m 4 ( Ro 4 Ro 6 ) = ( 0.268 )(172000 ) ⇒ Ad = 4609611.73 Ad = g m ( ro 2 ro 4 ) 1ro 2 = ro 4 = λ ID 1 = = 500 K ( 0.02 )( 0.1)gm = 2 Kn I D = 2 ( 0.5)( 0.1) = 0.4472 mA/V Ad = ( 0.4472 ) ( 500 500 ) ⇒ Ad = 112 Ro = ro 2 ro 4 = 500 500 ⇒ Ro = 250 K11.74(a)I DP = K p (VSG + VTP ) 2 0.4 + 1 = VSG 3 = 1.894 V 0.5I DN = K n (VGS − VTN ) 2 0.4 + 1 = VGS 1 = 1.894 V 0.5VDS1 ( sat ) = VGS1 − VTN = 1.894 − 1 = 0.894 VV + = VSG 3 + VDS1 ( sat ) − VGS 1 + vCMV + = 1.894 + 0.894 − 1.894 + 4 ⇒ V + = 4.89 V = −V −(b)
  • 51. Ad = g m ( ro 2 ro 4 ) 1 1ro 2 = ro 4 = = = 166.7 K λ ID ( 0.015 )( 0.4 )gm = 2 Kn I D = 2 ( 0.5 )( 0.4 ) = 0.8944 mA/VAd = ( 0.8944 ) (166.7 166.7 ) ⇒ Ad = 74.511.75(a) For vcm = +2V ⇒ V + = 2.7 V If I Q is a 2-transistor current source,V − = vcm − 0.7 − 0.7V − = −3.4 V ⇒ V + = −V − = 3.4 V(b) 100 Ad = g m ( ro 2 ro 4 ) ro 2 = = 1000 K 0.1 60 ro 4 = = 600 K 0.1 0.1 gm = = 3.846 mA/V 0.026 Ad = ( 3.846 ) (1000 600 ) ⇒ Ad = 144211.76(a) V + = −V − = 3.4 V(b) 75ro 2 = = 1250 K 0.06 40ro 4 = = 666.7 K 0.06 0.06gm = = 2.308 mA/V 0.026Ad = ( 2.308 ) (1250 666.7 )Ad = 100411.77g m1 = 2 K n I Bias1 = 2 ( 0.2 )( 0.25 ) = 0.447 mA/V I CQ 0.75gm2 = = = 28.85 mA/V VT 0.026 β VT (120 )( 0.026 )rπ 2 = = = 4.16 kΩ I CQ 0.75
  • 52. i0 = g m1Vgs1 + g m 2Vπ 2Vπ 2 = g m1Vgs1rπ 2 and vi = Vgs1 + Vπ 2i0 = Vgs1 ( g m1 + g m 2 ⋅ g m1rπ 2 ) vivi = Vgs1 + g m1Vgs1rπ 2 and Vgs1 = 1 + g m1rπ 2 g m1 (1 + β )i0 = vi ⋅ 1 + g m1rπ 2 i0 g m1 (1 + β ) ( 0.447 )(121)gm = C = = vi 1 + g m1rπ 2 1 + ( 0.447 )( 4.16 )⇒ g m = 18.9 mA/V C11.78 1 1r0 ( M 2 ) = = = 500 kΩ λn I DQ ( 0.01)( 0.2 ) VA 80 r0 ( Q2 ) = = = 400 kΩ I CQ 0.2g m ( M 2 ) = 2 K n I DQ = 2 ( 0.2 )( 0.2 ) = 0.4 mA/VAd = g m ( M 2 ) ⎡ r0 ( M 2 ) r0 ( Q2 ) ⎤ ⎣ ⎦ = 0.4 ⎡500 400 ⎤ ⇒ Ad = 88.9 ⎣ ⎦If the IQ current source is ideal, Acm = 0 and C M RRdB = ∞11.79a.b. Assume RL is capacitively coupled. Then
  • 53. I CQ + I DQ = I Q VBE 0.7I DQ = = = 0.0875 mA R1 8 I CQ = 0.9 − 0.0875 = 0.8125 mA g m1 = 2 K P I DQ = 2 (1)( 0.0875 ) ⇒ g m1 = 0.592 mA/V I CQ 0.8125gm2 = = ⇒ g m 2 = 31.25 mA/V VT 0.026 β VT (100 )( 0.026 ) rπ 2 = = ⇒ rπ 2 = 3.2 kΩ I CQ 0.8125c.V0 = ( − g m1Vsg − g m 2Vπ 2 ) RLVi + Vsg = V0 ⇒ Vsg = V0 − ViVπ 2 = ( g m1Vsg ) ( R1 rπ 2 )V0 = − ⎡ g m1Vsg + g m 2 g m1Vsg ( R1 rπ 2 ) ⎤ RL ⎣ ⎦V0 = − (V0 − Vi ) ⎣ g m1 + g m 2 g m1 ( R1 rπ 2 ) ⎦ RL ⎡ ⎤ ⎡ g m1 + g m 2 g m1 ( R1 rπ 2 ) ⎦ RL ⎤ = ⎣ V0Av = Vi 1 + ⎡ g m1 + g m 2 g m1 ( R1 rπ 2 ) ⎤ RL ⎣ ⎦We findg m1 + g m 2 g m1 ( R1 rπ 2 ) = 0.592 + ( 31.25 )( 0.592 ) ( 8 3.2 ) = 42.88 ( 42.88 )( RL )Then Av = 1 + ( 42.88 )( RL )11.80a. Assume RL is capacitively coupled. 0.7I DQ = = 0.0875 mA 8I CQ = 1.2 − 0.0875 = 1.11 mAg m1 = 2 K p I DQ = 2 (1)( 0.0875 ) ⇒ g m1 = 0.592 mA/V I CQ 1.11gm2 = = ⇒ g m 2 = 42.7 mA/V VT 0.026 β VT (100 )( 0.026 )rπ 2 = = ⇒ rπ 2 = 2.34 kΩ I CQ 1.11b.
  • 54. Vsg = VXI X = g m 2Vπ 2 + g m1Vsg(g V m1 sg )(R1 rπ 2 ) = Vπ 2I X = VX ⎡ g m1 + g m 2 g m1 ( R1 rπ 2 ) ⎤ ⎣ ⎦ VX 1R0 = = IX g m1 + g m 2 g m1 ( R1 rπ 2 ) 1 = ⇒ R0 = 21.6 Ω 0.592 + ( 0.592 )( 42.7 ) ( 8 2.34 )11.81(a) Vo − ( −Vπ )(1) g m 2Vπ + =0 ro 2 Vo − ( −Vπ ) −Vπ −Vπ ⎛ 1 1⎞(2) g m 2Vπ + = g m1Vi + + or 0 = g m1Vi − Vπ ⎜ + ⎟ ro 2 ro1 rπ ⎝ ro1 rπ ⎠ g m1ViThen Vπ = ⎛ 1 1⎞ ⎜ + ⎟ ⎝ ro1 rπ ⎠From (1)
  • 55. ⎛ 1 ⎞ Vo⎜ g m 2 + ⎟ Vπ + =0⎝ ro 2 ⎠ ro 2 ⎛ 1 ⎞ ⎜ gm2 + ⎟ ⎛ 1 ⎞Vo = −ro 2 ⎜ g m 2 + ⎟ Vπ = −ro 2 g m1Vi ⎝ ro 2 ⎠ ⎝ ro 2 ⎠ ⎛ 1 1⎞ ⎜ + ⎟ ⎝ ro1 rπ ⎠ ⎛ 1 ⎞ − g m1ro 2 ⎜ g m 2 + ⎟ VAv = o = ⎝ ro 2 ⎠ Vi ⎛ 1 1⎞ ⎜ + ⎟ ⎝ ro1 rπ ⎠Nowg m1 = 2 K n I Q = 2 ( 0.25)( 0.025 ) = 0.158 mA / V IQ 0.025gm2 = = = 0.9615 mA / V VT 0.026 1 1ro1 = = = 2000 k Ω λ IQ ( 0.02 )( 0.025) VA 50ro 2 = = = 2000 k Ω I Q 0.025 β VT (100 )( 0.026 )rπ = = = 104 k Ω IQ 0.025Then ⎛ 1 ⎞ − ( 0.158 )( 2000 ) ⎜ 0.9615 + ⎟Av = ⎝ 2000 ⎠ ⇒ Av = −30039 ⎛ 1 1 ⎞ ⎜ + ⎟ ⎝ 2000 104 ⎠To find Ro; set Vi = 0 ⇒ g m1Vi = 0
  • 56. Vx − ( −Vπ )I x = g m 2Vπ + ro 2Vπ = − I x ( ro1 rπ )Then ⎛ 1 ⎞ VI x = ⎜ g m 2 + ⎟ ( − I x ) ( ro1 rπ ) + x ⎝ ro 2 ⎠ ro 2Combining terms, Vx ⎡ ⎛ 1 ⎞⎤ Ro = = ro 2 ⎢1 + ( ro1 rπ ) ⎜ g m 2 + ⎟ ⎥ Ix ⎣ ⎝ ro 2 ⎠ ⎦ ⎡ ⎛ 1 ⎞⎤ = 2000 ⎢1 + ( 2000 104 ) ⎜ 0.9615 + ⎟ ⇒ Ro = 192.2 M Ω ⎣ ⎝ 2000 ⎠ ⎥ ⎦(b) Vo − ( −Vgs 3 )(1) g m 3Vgs 3 + =0 ro3 Vo − ( −Vgs 3 ) −Vgs 3 − ( −Vπ 2 ) ⎛ 1 ⎞ Vgs 3(2) g m 3Vgs 3 + = g m 2Vπ 2 + or 0 = Vπ 2 ⎜ g m 2 + ⎟ − ro3 ro 2 ⎝ ro 2 ⎠ ro 2 Vπ 2 −Vgs 3 − ( −Vπ 2 ) ( −Vπ 2 )(3) + g m 2Vπ 2 + = g m1Vi + rπ 2 ro 2 ro1 Vgs 3From (2), Vπ 2 = ⎛ 1 ⎞ ro 2 ⎜ g m 2 + ⎟ ⎝ ro 2 ⎠Then ⎛ 1 1 1⎞ Vgs 3(3) Vπ 2 ⎜ + gm2 + + ⎟ = g m1Vi + ⎝ rπ 2 ro 2 ro1 ⎠ ro 2or
  • 57. Vgs 3 ⎡ 1 1 1⎤ Vgs 3 ⎢ + gm2 + + ⎥ = g m1Vi + ⎛ 1 ⎞ rro 2 ⎜ g m 2 + ⎟ ⎣ π 2 ro 2 ro1 ⎦ ro 2 ⎝ ro 2 ⎠ Vgs 3 ⎡ 1 1 1 ⎤ Vgs 3 ⎛ 1 ⎞⎣ ⎢104 + 0.9615 + 2000 + 2000 ⎥ = 0.9615Vi + 20002000 ⎜ 0.9615 + ⎦ ⎟ ⎝ 2000 ⎠Then Vgs 3 = 1.83 × 105 Vi ⎛ 1 ⎞ −V ⎛ 1 ⎞ ⎟ (1.83 ×10 ) Vi 5From (1), ⎜ g m 3 + ⎟ Vgs 3 = o or Vo = −2000 ⎜ 0.158 + ⎝ ro 3 ⎠ ro3 ⎝ 2000 ⎠ VAv = o = −5.80 × 107 ViTo find Ro Vx − ( −Vgs 3 )(1) I x = g m 3Vgs 3 + ro3 Vx − ( −Vgs 3 ) −Vgs 3 − ( −Vπ 2 )(2) g m 3Vgs 3 + = g m 2Vπ 2 + ro 3 ro 2(3) Vπ 2 = − I x ( ro1 rπ 2 ) ⎛ 1 ⎞ VFrom (1) I x = Vgs 3 ⎜ g m 3 + ⎟ + x ⎝ ro 3 ⎠ ro3 ⎛ 1 ⎞ VxI x = Vgs 3 ⎜ 0.158 + ⎟+ ⎝ 2000 ⎠ 2000 V Ix − xSo Vgs 3 = 2000 0.1585
  • 58. From (2), ⎡ 1 1 ⎤ V ⎛ 1 ⎞Vgs 3 ⎢ g m 3 + + ⎥ + x = Vπ 2 ⎜ g m 2 + ⎟ ⎣ ro 3 ro 2 ⎦ ro 3 ⎝ ro 2 ⎠ ⎡ 1 1 ⎤ Vx ⎛ 1 ⎞Vgs 3 ⎢ 0.158 + + ⎥ + 2000 = Vπ 2 ⎜ 0.9615 + 2000 ⎟ ⎣ 2000 2000 ⎦ ⎝ ⎠ ⎡ I − Vx / 2000 ⎤ VxThen ⎢ x ⎥ ( 0.159 ) + 2000 = − I x ( 2000 104 ) ( 0.962 ) ⎣ 0.1585 ⎦ VWe find Ro = x = 6.09 × 1010 Ω Ix11.82Assume emitter of Q1 is capacitively coupled to signal ground. ⎛ 80 ⎞I CQ = 0.2 ⎜ ⎟ = 0.1975 mA ⎝ 81 ⎠ 0.2I DQ = = 0.00247 mA 81 (80 )( 0.026 )rπ = = 10.5 k Ω 0.1975 0.1975g m ( Q1 ) = = 7.60 mA / V 0.026gm ( M1 ) = 2 K n I D = 2 ( 0.2 )( 0.00247 )g m ( M 1 ) = 0.0445 mA / V VπVi = Vgs + Vπ and Vπ = g m ( M 1 ) Vgs rπ or Vgs = g m ( M 1 ) rπ ⎛ 1 ⎞ ViThen Vi = Vπ ⎜ 1 + ⎜ g (M )r ⎟ or Vπ = ⎟ ⎝ 1 π ⎠ ⎛ 1 ⎞ ⎜1 + m ⎜ g (M ) r ⎟ ⎟ ⎝ m 1 π ⎠ V − g m ( Q1 ) RCVo = − g m ( Q1 ) Vπ RC ⇒ Av = o = Vi ⎛ 1 ⎞ ⎜1 + ⎜ g (M )r ⎟ ⎟ ⎝ m 1 π ⎠ − ( 7.60 )( 20 )Then Av = ⇒ Av = −48.4 ⎛ 1 ⎞ ⎜1 + ⎜ ( 0.0445 )(10.5 ) ⎟ ⎟ ⎝ ⎠
  • 59. 11.83Using the results from Chapter 4 for the emitter-follower: ⎡ rπ 9 + r07 R011 ⎤ ⎢ rπ 8 + ⎥ 1+ β R0 = R4 || ⎢ ⎥ ⎢ 1+ β ⎥ ⎢ ⎥ ⎣ ⎦ β VT (100 )( 0.026 ) rπ 8 = = = 2.6 kΩ IC8 1 IC 8 1 IC 9 ≈ = = 0.01 mA β 100 (100 )( 0.026 ) rπ 9 = = 260 kΩ 0.01 V 100 r07 = A = = 500 kΩ I Q 0.2 VA 100 r011 = = = 500 kΩ I Q 0.2 0.2R011 = r011 [1 + g m RE ] , g m = ′ = 7.69 0.026 (100 ) ( 0.026 ) rπ 11 = = 13 kΩ 0.2 ′ RE = 0.2 13 = 0.197 kΩR011 = 500 ⎡1 + ( 7.69 )( 0.197 ) ⎤ = 1257 kΩ ⎣ ⎦Then ⎡ 260 + 500 1257 ⎤ ⎢ 2.6 + ⎥R0 = 5 || ⎢ 101 ⎥ ⎢ 101 ⎥ ⎢ ⎣ ⎥ ⎦ = 5 0.0863 ⇒ R0 = 0.0848 K ⇒ 84.8 Ω11.84 Ri = rπ 1 + (1 + β ) rπ 2 (100 )( 0.026 )rπ 2 = = 5.2 kΩ 0.5 (100 )( 0.026 ) (100 ) ( 0.026 ) 2rπ 1 = = = 520 kΩ ( 0.5 /100 ) 0.5 Ri = 520 + (101)( 5.2 ) ⇒ Ri ≅ 1.05 MΩ rπ 3 + 50 (100 )( 0.026 ) R0 = 5 , rπ 3 = = 2.6 kΩ 101 1 2.6 + 50 R0 = 5 = 5 0.521 ⇒ R0 = 0.472 kΩ 101
  • 60. ⎛V ⎞V0 = − ⎜ π 3 + g m 3Vπ 3 ⎟ ( 5 ) ⎝ rπ 3 ⎠ ⎛1+ β ⎞V0 = −Vπ 3 ⎜ ⎟ ( 5) (1) ⎝ rπ 3 ⎠Vπ 3 (V − V ) = g m 2Vπ 2 + 0 π 3rπ 3 50 ⎛ 1 1 ⎞ Vg m 2Vπ 2 = Vπ 3 ⎜ + ⎟− 0 (2) ⎝ rπ 3 50 ⎠ 50 ⎛V ⎞Vπ 2 = ⎜ π 1 + g m1Vπ 1 ⎟ rπ 2 ⎝ rπ 1 ⎠ (3) ⎛ 1+ β ⎞ = Vπ 1 ⎜ ⎟ rπ 2 ⎝ rπ 1 ⎠andVin = Vπ 1 + Vπ 2 (4) 0.5gm2 = = 19.23 mA/V 0.026Then ⎛ 101 ⎞V0 = −Vπ 3 ⎜ ⎟ ( 5 ) ⇒ Vπ 3 = −V0 ( 0.005149 ) (1) ⎝ 2.6 ⎠And ⎛ 1 1 ⎞ V19.23Vπ 2 = −V0 ( 0.005149 ) ⎜ + ⎟− 0 ⎝ 2.6 50 ⎠ 50 (2) = −V0 ( 0.02208 )Or Vπ 2 = −V0 ( 0.001148 )AndVπ 1 = Vin − Vπ 2 = Vin + V0 ( 0.001148 ) (4)So ⎛ 101 ⎞−V0 ( 0.001148 ) = ⎡Vin + V0 ( 0.001148 ) ⎤ ⎜ ⎣ ⎦ 520 ⎟ ( 5.2 ) (3) ⎝ ⎠ V0−V0 ( 0.001148 ) − V0 ( 0.001159 ) = Vin (1.01) ⇒ Av = = −438 Vin11.85
  • 61. 5I2 = = 1 mA 5 1VGS 2 = + 0.8 = 2.21 V 0.5 2.21 − ( −5 )I1 = = 0.206 mA 35V0 = ( g m 2Vgs 2 ) ( R2 r02 )Vgs 2 = ( g m1Vsg1 ) ( r01 R1 ) − V0 and Vsg1 = −VinSo Vgs 2 = − ( g m1Vin ) ( r01 R1 ) − V0ThenV0 = g m 2 ( R2 r02 ) ⎡ − ( g m1Vin ) ( r01 R1 ) − V0 ⎤ ⎣ ⎦ V0 − g m 2 ( R2 r02 ) g m1 ( r01 R1 )Av = = Vin 1 + g m 2 ( R2 r02 )gm2 = 2 Kn2 I D 2 = 2 ( 0.5 )(1) = 1.414 mA / Vg m1 = 2 K p1 I D1 = 2 ( 0.2 )( 0.206 ) = 0.406 mA / V 1 1r01 = = = 485 kΩ λ1 I D1 ( 0.01)( 0.206 ) 1 1r02 = = = 100 kΩ λ2 I D 2 ( 0.01)(1)R2 r02 = 5 100 = 4.76 kΩR1 r01 = 35 485 = 32.6 kΩ − (1.414 )( 4.76 )( 0.406 )( 32.6 )Then Av = 1 + (1.414 )( 4.76 )So ⇒ Av = −11.5Output Resistance—From the results for a source follower in Chapter 6. 1 1R0 = R2 r02 = 5 100 gm2 1.414 = 0.707 4.76So R0 = 0.616 kΩ11.86a.
  • 62. 5R2 = ⇒ R2 = 10 kΩ 0.5 I D2 0.5VSG 2 = − VTP 2 = + 1 = 2.41 V K p2 0.25 5 − ( −2.41)R1 = ⇒ R1 = 74.1 kΩ 0.1b.V0 = − ( g m 2Vsg 2 ) ( r02 R2 )Vsg 2 = V0 − ⎡ − ( g m1Vgs1 ) ( r01 R1 ) ⎤ and Vgs1 = Vin ⎣ ⎦ V0 − ( g m 2 ) ( r02 R2 ) ( g m1 ) ( r01 R1 )Av = = Vin 1 + ( g m 2 ) ( r02 R2 )g m1 = 2 K n1 I D1 = 2 ( 0.1)( 0.1) = 0.2 mA / Vgm2 = 2 K p 2 I D 2 = 2 ( 0.25)( 0.5) = 0.707 mA / V 1 1r01 = = = 1000 kΩ λ1 I D1 ( 0.01)( 0.1) 1 1r02 = = = 200 kΩ λ2 I D 2 ( 0.01)( 0.5 )r02 R2 = 200 10 = 9.52 kΩr01 R1 = 1000 74.1 = 69.0 kΩ − ( 0.707 )( 9.52 )( 0.2 )( 69 )Then Av = 1 + ( 0.707 )( 9.52 )So ⇒ Av = −12.0 1 1R0 = R2 r02 = 10 200 gm2 0.707 = 1.414 9.52Or R0 = 1.23 kΩ11.87a.I C 2 = 0.25 mA 5−2R= ⇒ R = 12 kΩ 0.25 v − VBE ( on ) 2 − 0.7I C 3 = 02 ⇒ RE1 = ⇒ RE1 = 2.6 kΩ RE1 0.5 5 − v03 5 − 3RC = = ⇒ RC = 4 kΩ IC 3 0.5 ⎡ v03 − VBE ( on ) ⎤ − ( −5 )IC 4 = ⎣ ⎦ RE 2 3 − 0.7 + 5RE 2 = ⇒ RE 2 = 2.43 kΩ 3
  • 63. b. Input resistance to base of Q3, Ri 3 = rπ 3 + (1 + β ) RE1 (100 )( 0.026 ) rπ 3 = = 5.2 kΩ 0.5 Ri 3 = 5.2 + (101)( 2.6 ) = 267.8 kΩ v 1 Ad 1 = 02 = g m 2 ( R Ri 3 ) vd 2 0.25 gm2 = = 9.62 mA/V 0.026 1 Ad 1 = ( 9.62 ) (12 267.8 ) ⇒ Ad 1 = 55.2 2 v − β ( RC Ri 4 )Now 03 = v02 rπ 3 + (1 + β ) RE1where Ri 4 = rπ 4 + (1 + β ) RE 2 v0 (1 + β ) RE 2and = v03 rπ 4 + (1 + β ) RE 2 (100 )( 0.026 )rπ 4 = = 0.867 kΩ 3v0 (101)( 2.43) = = 0.9965v03 0.867 + (101)( 2.43)Ri 4 = 0.867 + (101)( 2.43) = 246.3 kΩrπ 3 = 5.2 kΩ v03 − (100 ) ( 4 246.3)So = = −1.47 v02 5.2 + (101)( 2.6 ) v0So Ad = = ( 55.2 )( 0.9965 )( −1.47 ) ⇒ Ad = −80.9 vdc. Using Equation (11.32b) − g m 2 ( R Ri 3 )Acm1 = 2 (1 + β ) R0 1+ rπ 2 (100 )( 0.026 )rπ 2 = = 10.4 kΩ 0.25 − ( 9.62 ) (12 267.8 )Acm1 = = −0.0569 = Acm1 2 (101)(100 ) 1+ 10.4 ⎛ v0 ⎞⎛ v03 ⎞Then Acm = ⎜ ⎟⎜ ⎟ ⋅ Acm1 ⎝ v03 ⎠⎝ v02 ⎠= ( 0.9965 )( −1.47 )( −0.0569 ) ⇒ Acm = 0.08335 ⎛ 80.9 ⎞C M RRdB = 20 log10 ⎜ ⎟ ⇒ C M RRdB = 59.7 dB ⎝ 0.08335 ⎠11.88a. 10 − v01 10 − 2RC1 = = ⇒ RC1 = 80 kΩ I C1 0.1 10 − v04 10 − 6RC 2 = = ⇒ RC 2 = 20 kΩ IC 4 0.2
  • 64. b. v01 − v02 Ad 1 = = − g m1 ( RC1 rπ 3 ) vd 0.1 g m1 = = 3.846 mA/V 0.026 (180 )( 0.026 ) rπ 3 = = 23.4 kΩ 0.2 Ad 1 = − ( 3.846 ) ( 80 23.4 ) ⇒ Ad 1 = −69.6 v04 1 Ad 2 = = g m 4 RC 2 v01 − v02 2 0.2 gm4 = = 7.692 mA/V 0.026 1 Ad 2 = ( 7.692 )( 20 ) = 76.9 2Then Ad = ( 76.9 )( −69.6 ) ⇒ Ad = −535211.89a. Neglect the effect of r0 in determining the differential-mode gain. v02 1Ad 1 = = g m 2 ( RC Ri 3 ) where Ri 3 = rπ 3 + (1 + β ) RE vd 2 − β RC 2A2 = rπ 3 + (1 + β ) RE 12 − 0.7 − ( −12 ) 23.3I1 = = = 1.94 mA ≈ I C 5 R1 12 1 ⋅ (1.94 )gm2 = 2 = 37.3 mA/V 0.026 ( 200 )( 0.026 )rπ 3 = IC 3 1v02 = 12 − (1.94 )(8) = 4.24 V 2 4.24 − 0.7IC 3 = = 1.07 mA 3.3 ( 200 )( 0.026 )rπ 3 = = 4.86 kΩ 1.07Ri 3 = 4.86 + ( 201)( 3.3) = 668 kΩ 1Ad 1 = ( 37.3) ⎣8 668⎦ = 147.4 ⎡ ⎤ 2ThenAd = Ad 1 ⋅ A2 = (147.4 )( −1.197 ) ⇒ Ad = −176 VA 80R0 = r05 = = = 41.2 kΩ I C 5 1.94 − g m 2 ( RC Ri 3 )Acm1 = 2 (1 + β ) R0 1+ rπ 2 ( 200 )( 0.026 )rπ 2 = = 5.36 kΩ 1 ⋅ (1.94 ) 2
  • 65. − ( 37.3) ( 8 668 )Acm1 = = −0.09539 2 ( 201)( 41.2 ) 1+ 5.36A2 = −1.197Acm = ( −0.09539 )( −1.197 ) ⇒ Acm = 0.114b. vd = v1 − v2 = 2.015sin ω t − 1.985sin ω t vd = 0.03sin ω t ( V ) v +vvcm = 1 2 = 2.0sin ω t 2 v03 = Ad vd + Acm vcm = ( −176 )( 0.03) + ( 0.114 )( 2 )Or v03 = −5.052sin ω tIdeal, Acm = 0Sov03 = Ad vd = ( −176 )( 0.03)v03 = −5.28sin ω tc.Rid = 2rπ 2 = 2 ( 5.36 ) ⇒ Rid = 10.72 kΩ2 Ricm ≅ 2 (1 + β ) R0 (1 + β ) r0 VA 80r0 = = = 82.5 kΩ IC 2 1 ⋅ (1.94 ) 22 Ricm = ⎡ 2 ( 201)( 41.2 ) ⎤ ⎡( 201)( 82.5 ) ⎤ ⎣ ⎦ ⎣ ⎦ = 16.6 MΩ 16.6 MΩSo ⇒ Ricm = 4.15 MΩ11.90a. 24 − VGS 4 = kn (VGS 4 − VTh ) 2I1 = R124 − VGS 4 = ( 55 )( 0.2 )(VGS 4 − 2 ) 224 − VGS 4 = 11 (VGS 4 − 4VGS 4 + 4 ) 2 211VGS 4 − 43VGS 4 + 20 = 0 ( 43) − 4 (11)( 20 ) 2 43 ±VGS 4 = = 3.37 V 2 (11) 24 − 3.37I1 = = 0.375 mA = I Q 55 ⎛ 0.375 ⎞v02 = 12 − ⎜ ⎟ ( 40 ) = 4.5 V ⎝ 2 ⎠v02 − VGS 3 = I D 3 = kn (VGS 3 − VTh ) 2 R5
  • 66. 4.5 − VGS 3 = ( 0.2 ) ( 6 ) (VGS 3 − 4VGS 3 + 4 ) 2 21.2VGS 3 − 3.8VGS 3 + 0.3 = 0 ( 3.8) − 4 (1.2 ) ( 0.3) 2 3.8 ±VGS 3 = = 3.09 V 2 (1.2 ) 4.5 − 3.09I D3 = = 0.235 mA 6 ( 0.2 ) ⎛ 0.375 ⎞ gm2 = 2 Kn I D2 = 2 ⎜ ⎟ ⎝ 2 ⎠ = 0.387 mA/V 1 1 Ad 1 = g m 2 RD = ( 0.387 )( 40 ) ⇒ Ad 1 = 7.74 2 2 − g m 3 RD 2 A2 = 1 + g m 3 R5 g m3 = 2 K n I D3 = 2 ( 0.2 )( 0.235 ) = 0.434 mA/V − ( 0.434 ) ( 4 ) A2 = = −0.482 1 + ( 0.434 )( 6 )So Ad = Ad 1 ⋅ A2 = ( 7.74 ) ( −0.482 ) ⇒ Ad = −3.73 1 1 R0 = r05 = = = 133 kΩ λ I Q ( 0.02 )( 0.375 ) − g m 2 RD − ( 0.387 ) ( 40 ) Acm1 = = 1 + 2 g m 2 R0 1 + 2 ( 0.387 ) (133) = −0.149 Acm = ( −0.149 )( −0.482 ) ⇒ Acm = 0.0718b. vd = v1 − v2 = 0.3sin ω t v +vvcm = 1 2 = 2sin ω t 2 v03 = Ad vd + Acm vcm = ( −3.73)( 0.3) + ( 0.0718 )( 2 ) ⇒ v03 = −0.975sin ω t ( V )Ideal, Acm = 0v03 = Ad vd = ( −3.73)( 0.3)Or ⇒ v03 = −1.12sin ω t ( V )11.91The low-frequency, one-sided differential gain is
  • 67. v02 1 ⎛ r ⎞Av 2 = = g m RC ⎜ π ⎟ vd 2 ⎝ rπ + RB ⎠ 1 ⋅ β RC = 2 rπ + RB (100 )( 0.026 ) rπ = = 5.2 kΩ 0.5 1 ⋅ (100 )(10 )Av 2 = 2 ⇒ Av 2 = 87.7 5.2 + 0.5CM = Cμ (1 + g m RC ) 0.5 gm = = 19.23 mA/V 0.026CM = 2 ⎡1 + (19.23)(10 ) ⎤ ⇒ CM = 387 pF ⎣ ⎦ 1 fH = 2π ⎡ rπ RB ⎤ ( Cπ + CM ) ⎣ ⎦ 1 = So ⇒ f H = 883 kHz 2π ⎣5.2 0.5⎦ × 10 × ( 8 + 387 ) × 10−12 ⎡ ⎤ 311.92 1 1a. From Equation (11.117), f Z = = 2π R0 C0 2π ( 5 × 106 )( 0.8 × 10−12 )Or f Z = 39.8 kHzb. From Problem 11.69, f H = 883 kHz. From Equation (11.116(b)), the low-frequencycommon- mode gain is − g m RC Acm = ⎡⎛ RB ⎞ 2 (1 + β ) R0 ⎤ ⎢⎜ 1 + ⎟+ ⎥ ⎣⎝ rπ ⎠ rπ ⎦ rπ = 5.2 kΩ, g m = 19.23 mA/VSo − (19.23)(10 )Acm = ⎡⎛ 0.5 ⎞ 2 (101) ( 5 × 106 ) ⎤ ⎢⎜ 1 + ⎟+ ⎥ ⎢⎝ 5.2 ⎠ 5.2 × 103 ⎥ ⎣ ⎦ −4 = −9.9 × 10 ⎛ 87.7 ⎞C M RRdB = 20 log10 ⎜ −4 ⎟ = 98.9 dB ⎝ 9.9 × 10 ⎠
  • 68. 11.93 gma. From Equation (7.72), fT = 2π ( Cπ + Cμ ) 1gm = = 38.46 mA/V 0.026 38.46 × 10−3Then 800 × 106 = 2π ( Cπ + Cμ )Or Cπ + Cμ = 7.65 × 10−12 F = 7.65 pFAnd Cπ = 6.65 pFCM = Cμ (1 + g m RC ) = 1 ⎡1 + ( 38.46 )(10 ) ⎤ ⎣ ⎦ = 386 pF 1fH = 2π ⎡ rπ RB ⎤ ( Cπ + CM ) ⎣ ⎦ (120 )( 0.026 ) rπ = = 3.12 kΩ 1 1 fH = 2π ⎡3.12 1⎤ × 10 × ( 6.65 + 386 ) × 10−12 ⎣ ⎦ 3Or f H = 535 kHz 1 1b. From Equation (11.140), f Z = = 2π R0 C0 2π (10 × 106 )(10−12 )Or f Z = 15.9 kHz11.94The differential-mode half circuit is: ⎛v ⎞ ⎛1⎞ g m ⎜ d ⎟ RC ⎜ ⎟ β RCv02 = ⎝ 2⎠ or Av = ⎝ 2⎠ ⎛1+ β ⎞ rπ + (1 + β ) RE 1+ ⎜ ⎟ RE ⎝ rπ ⎠ (100 )( 0.026 ) rπ = = 5.2 kΩ 0.5 ⎛1⎞ ⎜ ⎟ (100 )(10 ) 500 Av = ⎝ ⎠ 2 = 5.2 + (101) RE 5.2 + (101) RE
  • 69. a. For RE = 0.1 kΩ : Av = 32.7b. For RE = 0.25 kΩ : Av = 16.4