Chapter15

Microelectronic Circuit Design

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CHAPTER 15 15.1 a () I C = α F I E = 1 2 β F F + 1 12 V BE R EE = 1 2 100 101 12 0.7 2.7 x 10 5 = 20.7 µ A | V C = 12 3.3 x 10 5 I C = 5.17 V V CE = V C −− 0.7 V = 5.87 V Q Point = 20.7 A , 5.87 V b A dd =− g m R C 40 20.7 A 330 k 273 R id = 2 r π = 2 o V T I C = 2 100 0.025 V 20.7 A = 243 k R od = 2 R C = 660 k c A cc o R C r + o + 1 2 R EE 100 330 k 122 k Ω+ 2 101 270 k 0.604 A dd g m R C 2 137 A cd = A cc CMRR = 137 0.604 = 227 or 47.1 dB (very low) R ic = r + o + 1 2 R EE 2 = 122 k 2 101 270 k 2 = 27.3 M 15.2 a I E = 1 2 1.5 0.7 75 x 10 3 V = 5.33 A I C = F I E = 60 61 I E = 5.25 A V CE = 1.5 10 5 I C 0.7 = 1.68 V | Q-Pt : 5.25 A , 1.68 V b g m = 40 I C = 0.210 mS r = 60 g m = 286 k A dd g m R C 0.210 mS 100 k 21.0 A cc o R C r + o + 1 2 R EE 60 100 k 286 k 61150 k 0.636 For differential output : CMRR = 21.0 0 =∞ For single-ended output : CMRR = 21.0 2 0.636 = 16.5, a paltry 24.4 dB! R id = 2 r = 572 k R ic = r + o + 1 2 R EE 2 = 286 + 2 k Ω= 4.72 M R od = 2 R C = 200 k R oc = R C 2 = 50 k 15.3 ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 15-1
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*Problem 15.3 VCC 2 0 DC 12 VEE 1 0 DC -12 VIC 8 0 DC 0 VID1 4 8 AC 0.5 VID2 6 8 AC -0.5 RC1 2 3 330K RC2 2 7 330K Q1 3 4 5 NBJT Q2 7 6 5 NBJT REE 5 1 270K .MODEL NBJT NPN BF=100 VA=60 IS=1FA .OP .AC LIN 1 1KHZ 1KHZ .PRINT AC IM(VID1) IP(VID1) VM(3,7) VP(3,7) .TF V(7) VIC .END Results: A dd = VM (3,7) =− 241 | R id = 1 IM ( VID 1) = 269 k | A cc =-0.602 | R ic = 23.2 M V(IVOUT) Time (s) Problem15.45(b)-Transient-7 -6.000 -4.000 -2.000 +0.000e+000 +2.000 +4.000 +6.000 +0.000e+000 +1.000m +2.000m +3.000m +4.000m Simulation results from B 2 SPICE. Problem15.3(b)-Fourier-Table FREQ mag phase norm_mag norm_phase +0.000 +49.786n +0.000 +0.00 +0.000 +1.000k +5.766 +180.000 +1.000 +0.000 +2.000k +99.572n +93.600 +17.268n -86.400 +3.000k +80.305m -180.000 +13.927m -360.000 +4.000k +99.572n +97.200 +17.268n -82.800 +5.000k +1.161m +179.993 +201.326u -7.528m +6.000k +99.572n +100.800 +17.268n -79.200 +7.000k +13.351u -179.005 +2.315u -359.005 +8.000k +99.572n +104.400 +17.268n -75.600 Using the Fourier analysis capability of SPICE, THD = 1.39% 15-2 ©Richard C. Jaeger and Travis N. Blalock - 3/10/07
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15.4 a () I E = 18 V 0.7 V 24 .7 x 10 4 = 184 µ A | I C = α F I E = 100 101 I E = 182 A V CE = 18 10 5 I C −− 0.7 = 0.92 V | Q- point : 182 A , 0.92 V Note that R C is quite large and the common- mode input range is poor. More realistic choices might be 47 k or 51 k b g m = 40 I C = 7.28 mS r π = 100 g m = 13.7 k A dd =− g m R C 7.28 mS 100 k 728 A cc β o R C r + o + 1 2 R EE 100 100 k 13.7 k Ω+ 101 94 k 1.05 For differential output : CMRR = 33.7 0 =∞ For single-ended output : CMRR = 728 2 1.05 = 346, a paltry 50.8 dB! R id = 2 r = 27.4 k R ic = r + o + 1 2 R EE 2 = 13.7 + 101 94 2 k Ω= 4.75 M R od = 2 R C = 200 k R oc = R C 2 = 50 k ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 15-3
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15.5 a () I C = α F I E = 1 2 β F F + 1 12 V BE R EE = 1 2 100 101 12 0.7 2.7 x 10 5 = 20.7 µ A V C 1 = V C 2 = 12 2.4 x 10 5 I C = 7.03 V | V CE = V C −− 0.7 V = 7.73 V Q Point = 20.7 A , 7.73 V r π = 100 0.025 V 20.7 A = 121 k A cc =− o R C r + o + 1 2 R EE = 100 240 k 121 k Ω+ 101 540 k 0.439 v ic = 5.000 + 5.000 2 = 5.00 V v ic = 5 V , v C 1 = v C 2 = 7.03 + A cc v ic = 7.03 0.439 5 = 4.84 V Note that the BJT's are just beyond the edge of saturation! b I C = F I E = 1 2 F F + 1 5 V V BE 12 V R EE = 1 2 100 101 17 V 0.7 V 2.7 x 10 5 = 29.9 A V C 1 = V C 2 = 12 2.4 x 10 5 I C = 4.82 V Part (a) has a small error of 0.02 mV c The common - mode signal voltage applied to the base-emitter junction is v be = v ic
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This note was uploaded on 03/05/2008 for the course EE 307 taught by Professor Braun during the Spring '08 term at Cal Poly.

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Chapter15 - CHAPTER 15 15.1 (a) I C = F IE = VCE = VC -...

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