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Vin 2.0.4 A Basic RC Circuit R Figure 7: RC circuit — integrator. I C Vout Consider the basic RC circuit in Fig. 7. We will start by assuming that Vin is a DC voltage source (e.g. a battery) and the time variation is introduced by the closing of a switch at time t = 0. We wish to solve for Vout as a function of time. Applying Ohm’s Law across R gives Vin − Vout = IR. The same current I passes through the capacitor according to I = C(dV/dt). Substituting and rearranging gives (let V ≡ VC = Vout): dV dt 1 1 + V = RC RC Vin The homogeneous solution is V = Ae −t/RC ,whereAis a constant, and a particular solution is V = Vin. The initial condition V (0) = 0 determines A, and we find the solution V (t) =Vin � � −t/RC 1 − e This is the usual capacitor “charge up” solution. Similarly, a capacitor with a voltage Vi across it which is discharged through a resistor to ground starting at t = 0 (for example by closing a switch) can in similar fashion be found to obey V (t) =Vie −t/RC 2.0.5 The “RC Time” In both cases above, the rate of charge/discharge is determined by the product RC which has the dimensions of time. This can be measured in the lab as the time during charge-up or discharge that the voltage comes to within 1/e of its asymptotic value. So in our charge-up example, Equation 3, this would correspond to the time required for Vout to rise from zero to 63% of Vin. 2.0.6 RC Integrator From Equation 2, we see that if Vout ≪ Vin then the solution to our RC circuit becomes Vout = 1 � RC Vin(t)dt (4) 7 (2) (3)
Note that in this case Vin can be any function of time. Also note from our solution Eqn. 3 that the limit Vout ≪ Vin corresponds roughly to t ≪ RC. Within this approximation, we see clearly from Eqn. 4 why the circuit above is sometimes called an “integrator”. 2.0.7 RC Differentiator Let’s rearrange our RC circuit as shown in Fig. 8. Vin C I Figure 8: RC circuit — differentiator. R Vout Applying Kirchoff’s second Law, we have Vin = VC + VR, where we identify VR = Vout. By Ohm’s Law, VR = IR,whereI=C(dVC/dt) by Eqn. 1. Putting this together gives Vout = RC d dt (Vin − Vout) In the limit Vin ≫ Vout, we have a differentiator: Vout = RC dVin dt By a similar analysis to that of Section 2.0.6, we would see the limit of validity is the opposite of the integrator, i.e. t ≫ RC. 8
Page 1 and 2: Lecture Notes for Analog Electronic
Page 3 and 4: 1.2.1 Resistors in series I3 I1 I2
Page 5 and 6: The goal is to deduce VTH and RTH t
Page 7: Thus, we should try to keep the rat
Page 11 and 12: Thus, we have an expression of the
Page 13 and 14: The decibel scale works as follows:
Page 15 and 16: The general solution is then the re
Page 17 and 18: Vin L C Figure 12: A RLC circuit. S
Page 19 and 20: A diode is fabricated from a pn jun
Page 21 and 22: Class Notes 5 5 Transistors and Tra
Page 23 and 24: 5.1.2 Transistor Switch and Saturat
Page 25 and 26: Class Notes 6 Following our discuss
Page 27 and 28: V in C R 1 R 2 V cc R C R E V out F
Page 29 and 30: In 1 and similarly for output 2 R C
Page 31 and 32: Class Notes 8 5.8 More on Transisto
Page 33 and 34: We see that IC is not intrinsically
Page 35 and 36: Hence, variations in β are attenua
Page 37 and 38: 2. The inputs draw no current. ( Th
Page 39 and 40: Perhaps the best way to beat these
Page 41 and 42: 6.6.1 Gain IN C - + R OUT Figure 34
Page 43 and 44: Therefore, the efect of the closed
Page 45 and 46: 7 Comparator Circuits 7.1 Simple Co
Page 47 and 48: V h Vl +5 v in V out Figure 40: Exa
Page 49 and 50: 8 Radio Basics Class Notes 11 In th
Page 51 and 52: of interest are actually in a narro
Page 53 and 54: ≈ cos ωct − Ap sin ωct cos ω

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