9/24/11 1 Lecture 9 – Sep 26 Chem 4101 – Fall 2011 High-pass, loss-pass filters 1- Capacitors and Capacitance 2- AC currents 3- Low-pass and High-pass filters -Section 2B Suggested exercises: 2-13, 2-14, 2-15, 2-18, 2-19 Lecture 9 – Sep 26 Chem 4101 – Fall 2011 Periodic fluctuations (Alternating currents) Figure 2-5, Section 2B Power Signal (analysis) Control Period (s): tp Frequency (1/tp), (Hertz, Hz): f Angular frequency (2πf), (Radians, Rad): ω Maximum amplitude: Vo, Io, So Lecture 9 – Sep 26 Chem 4101 – Fall 2011 Capacitors and Capacitance Figure 2-8, Section 2B -2 Q= CV Q: Coulombs (charge) C: Farads (capacitance) V: Volts (potential) (eq. 2-28) i = Ii e-t/RC vR = Vi e-t/RC vc = Vi (1-e-t/RC) RC: time constant (s) When t=RC, VR=0.37Vi On your own Understand eq. 2-29, 2,30, 2-35, 2-36, 2-37 Switch (S) to 1 Switch (S) to 29/24/11 2 Lecture 9 – Sep 26 Chem 4101 – Fall 2011 Pulsed signals Figures 2-13, Section 2B-6 circuit Lecture 9 – Sep 26 Chem 4101 – Fall 2011 Alternating currents Figure 2-7, Section 2C Maximum amplitudes, Ip (A) and Vp (V) Period, tp (s) Frequency, f = 1/tp (Hz) Angular frequency, ω = 2πf (Rad) Phase, Φ (Rad) [2π = 360°; π//2 = 90°] Phase € i = IPsin(ωt + φ) = IPsin(2πft + φ) € Irms=Ip22= 0.707IpSee eq. 2-25 to 2-27 (Average amplitude) Lecture 9 – Sep 26 Chem 4101 – Fall 2011 Voltage lags the current in capacitive circuits Figure 2-9, Section 2B-4 (Equation 2-44) € Vp= IpXC € Xc=VpIp=1ωC=12πfC(Equation 2-45) (Equation 2-46) XC : Capacitive Reactance (Ω)9/24/11 3 Lecture 9 – Sep 26 Chem 4101 – Fall 2011 Lecture 9 – Sep 26 Chem 4101 – Fall 2011 Impedance (Z) of a series RC circuit Figure 2-9, Section 2B-4 Equation 2-49 Equation 2-50 Impedance (Z) Ξ Total resistance AC circuit DC circuit The vectors representing R and Xc are 90° out of phase Lecture 9 – Sep 26 Chem 4101 – Fall 2011 Example of Impedance Calculations Equations 2-50 and 2-51 For the circuit in the previous slide, calculate the requested in row 4. Confirm the following values and then write on the board: Table 1 – Cell I7, Table 2 – H7, Table 3 – Cell G7, Table 4 – Cell E7, Table 5 – Cell D79/24/11 4 Lecture 9 – Sep 26 Chem 4101 – Fall 2011 High-pass filter On your own: confirm that equation 2-53 provides the results in the graph!Gain or attenuation: 20 log[(Vp)o/(Vp)I Dimensions: Decibels Figures 2-11, 2-12, Section 2B-5 R = 10 kΩ, C = 0.11 µF ‘Voltage divider’ High-pass Filter Lecture 9 – Sep 26 Chem 4101 – Fall 2011 Low-pass filter On your own: confirm that equation 2-53 provides the results in the graph Gain or attenuation: 20 log[(Vp)o/(Vp)I Dimensions: Decibels Figures 2-11, 2-12, Section 2B-5 R = 1 MΩ, C = 1 µF ‘Voltage divider’ Low-pass
View Full Document
Unlocking...