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L16 Part2 Darlington_1

# L16 Part2 Darlington_1 - EE 311 Lecture 16 EE0 311 Lecture...

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1 EE 311 Lecture 16 & EE0 311 Lecture 15 Part 2 6.11.2 The Darlington Ref: Basic SS Amp Eqs. Slide 6 Ref: EE 311 L12 & EEO 311 L11 Part 2 Slide 9 R10 & R20

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2 V B1 = 0 VDC V E2 = V B1 - V BE1 - V BE2 = 0 - (V BE1 + V BE2 ) I E2 = [V E2 - (- V EE )] / R E = [V EE - (V BE1 + V BE2 )] / R E I B2 = I E2 / ( β 2 + 129 = I E1 Design for I E2 = 100 mA (h FE2 = 100) 6.11.2 Darlington Configuration
3 I B2 = I E2 / ( β 2 + 129 = 100 mA / (100 + 1) = 1 mA I E1 = I B2 = 1 mA = I C1 (h FE1 = 70) I C = I s e (VBE/VT) or V BE = V T ln(I C / I S ) V BE2 - V BE1 = V T [ln(I C2 /I S ) - ln(I C1 /I S )] = V T ln(I C2 /I C1 ) = V T ln(I C2 /I B2 ) = V T ln( β 2 ) V BE2 - V BE1 = V T ln( β 2 ) = 25 mV * ln (100) = 115 mV Assume V BE2 = 0.700 V Then V BE1 = 0.700 - 0.115 = 0.585 V Let V EE = 15 V R E = (V EE - V BE2 - V BE1 ) / I E2 = (15 - .7 - .585) / 100 mA =13.7V/100mA= 0.137 k = 0.14 k R INQ1 = r x1 + r π1 + ( β 1 + 129 R INQ2 R INQ2 = r x2 + r π2 + ( β 2 + 129 * (R E || R L ) R L = 100 = .1 k R E || R L = 0.137k || 0.1k = .058 k  =  58   r π2 = ( β 2 + 1)r e2 = 101 * (V T / I E2 ) = 101*(25mV/100mA) = 101*(0.25 ) = 25 Assume r x2 = r π2 = 25 R IN,Q2 = 25 + 25 + 101*58 = 5900 = 5.9 k R π1 = ( β 1 + 129 r e1 = ( β 1 + 129 *(V T /I E1 ) = (71)*(25 mV / 1mA) = 71*(25 ) = 1775  = 1.775 k Assume r x1 = r π1

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4 R OUT = R E || R OUT, Q2 R OUT, Q2 = (r π2 + r x2 + R OUT Q1 ) / ( β 2 + 129 Correction: β 1 β 2 R OUT, Q1 = (r π1 + r x1 + R S ) / ( β 1 + 129 Assume R S = 10 k R OUTQ1 = (1.775 + 1.775 + 10) / 71 = 13.5 / 71 = 0.19 k  = 190  R OUTQ2 = (25 + 25 + 190) / 101 = 240 / 101 = 0.0024 k  = 2.4  R OUT = R E || R OUTQ2 = 137 || 2.4 2.4 Open-Circuit Time Constants Q1 R C1 = 0 CC R b1 = R S = 10 k
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