Lesson+9 - The day the education of American youth was...

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Lesson 9 1 The day the education of American youth was transformed due to the power of the electrical engineer.
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Opamps again Ch.8-5 Lesson 9 Lesson 9 µ741 2
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V=12 V=24 S R1 R2 C L Challenge 8 Lesson 9 What is v C (t), the capacitor’s final voltage? What is v C (t)? v C (t) i(t) R 1 =10 , R 2 =2 , L=2H, C=0.25F t=0 3
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V=12 V=24 S R1 R2 C L v C (t) i(t) Lesson 9 Using elementary circuit theory KVL: Ldi(t)/dt + R 1 i(t) + v C (t) = 24 KCL: i(t) = C dv C (t)/dt + v C (t)/R 2 Differentiate i(t), substitute, and combine d 2 v C (t)/dt 2 +((1/R 2 C)+(R 1 /L) dv C (t)/dt +((R 1 +R 2 )/(R 1 R 2 )) v C (t) = 24/LC Characteristic Equation: s 2 +7s+12 = 0 (s 1 = -3, s 2 = -4), overdamped) Switch in final ( red ) position. { Other solution strategies possible } 4
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V=12 V=24 S R1 R2 C L v C (t) Lesson 9 Eigenvalues (s 1 =-3, s 2 =-4, overdamped) Assume the solution form. The input being a constant, suggests that the steady-state output sldo contains a constant term: v C (t) = K 1 e -3t + K 2 e -4t + K 3 K 3 called final value by authors. K 3 = v C (t= ) = (2/(10+2)) 24 = 4V (voltage divider) Capacitor’s final voltage i(t) 5
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Lesson 9 Derive the 1 st initial condition v C (0+) . v C (0+)= v C (0-)= (2/(10+2))12 = 2V (voltage divider) Therefore v C (0+) = 2 = K 1 + K 2 + 4 or K 1 + K 2 = -2 (1 st IC) V=12 V=24 S R1 R2 C L v C (t) R 1 =10 , R 2 =2 , L=2H, C=0.25F 6 Derive the initial conditions. t<0 i(t)
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Lesson 9 Therefore, from physics: Cdv C (t)/dt| t=0 = i C (0) = i(0) – i R2 (0) = i(0) – (v C (0)/R 2 ) or dv C (t)/dt| t=0 = 4 – 4 = 0 = -3 K 1 -4 K 2 V=12 V=24 S R1 R2 C L v C (t) i(t) i(0+) = i(0-) = 12/12 = 1A (Ohms Law and inductor’s property) Derive the 2 nd initial condition. R 1 =10 , R 2 =2 , L=2H, C=0.25F i.e., Capacitor’s voltage cannot change instantly. 7
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Solve: K 1 = -8; K 2 = 6 Finally: v(t) = 4 - 8e -3t + 6e -4t v C (t= ) = 4; v C (t=0) = 4-8+6 = 2 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 2 2.2 2.4 2.6 2.8 3 3.2 3.4 3.6 3.8 4 Step response Lesson 9 V=12 V=24 S R1 R2 C L v C (t) i(t) 8 Lesson 9
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Lesson 9 Lesson 9 Op Amps are commonly are used to implement 2 nd order RLCK circuits?
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