Chapter 10 RC Circuits ISU EE C.Y. Lee Objectives Describe the relationship between current and voltage in an RC circuit Determine impedance and phase angle in a series RC circuit Analyze a series RC circuit Determine the impedance and phase angle in a parallel RC circuit Analyze a parallel RC circuit Analyze series-parallel RC circuits Determine power in RC circuits ISU EE 2 C.Y. Lee Sinusoidal Response of RC Circuits The capacitor voltage lags the source voltage Capacitance causes a phase shift between voltage and current that depends on the relative values of the resistance and the capacitive reactance ISU EE 3 C.Y. Lee Impedance and Phase Angle of Series RC Circuits The phase angle is the phase difference between the total current and the source voltage The impedance of a series RC circuit is determined by both the resistance (R) and the capacitive reactance (XC) (Z = 1/jωC= −jX ) (Z= R+Z = R−jX ) C ISU EE (Z = R) C ) (Z = −jX 4 C C (Z = R−jXC) C C.Y. Lee Impedance and Phase Angle of Series RC Circuits In the series RC circuit, the total impedance is the phasor sum of R and jXC Impedance magnitude: Z = √ R2 + X2C Phase angle: θ = tan-1(XC/R) ISU EE 5 C.Y. Lee Impedance and Phase Angle of Series RC Circuits Example: Determine the impedance and the phase angle Z = √(47)2 + (100)2 = 110 Ω θ = tan-1(100/47) = tan-1(2.13) = 64.8° ISU EE 6 C.Y. Lee Analysis of Series RC Circuits The application of Ohm’s law to series RC circuits involves the use of the quantities Z, V, and I as: V = IZ V I= Z V Z= I ISU EE 7 C.Y. Lee Analysis of Series RC Circuits Example: If the current is 0.2 mA, determine the source voltage and the phase angle XC = 1/2π(1×103)(0.01×10-6) = 15.9 kΩ Z = √(10×103)2 + (15.9×103)2 = 18.8 kΩ VS = IZ = (0.2mA)(18.8kΩ) = 3.76 V θ = tan-1(15.9k/10k) = 57.8° ISU EE 8 C.Y. Lee Relationships of I and V in a Series RC Circuit In a series circuit, the current is the same through both the resistor and the capacitor The resistor voltage is in phase with the current, and the capacitor voltage lags the current by 90° I VR VS VC ISU EE 9 C.Y. Lee KVL in a Series RC Circuit I From KVL, the sum of the voltage drops must equal the applied voltage (VS) VR VS Since VR and VC are 90° out of phase with each other, they must be added as phasor quantities VS = √V2R + V2C θ = tan-1(VC/VR) ISU EE VC VR= IR VC= I(−jXC) 10 VS= IZ = I(R−jXC) C.Y. Lee KVL in a Series RC Circuit Example: Determine the source voltage and the phase angle VS = √(10)2 + (15)2 = 18 V θ = tan-1(15/10) = tan-1(1.5) = 56.3° ISU EE 11 C.Y. Lee Variation of Impedance and Phase Angle with Frequency For a series RC circuit; as frequency increases: – R remains constant – XC decreases – Z decreases – θ decreases ISU EE 12 f Z R C.Y. Lee Variation of Impedance and Phase Angle with Frequency Example: Determine the impedance and phase angle for each of the following values of frequency: (a) 10 kHz (b) 30 kHz (a) XC = 1/2π(10×103)(0.01×10-6) = 1.59 kΩ Z = √(1.0×103)2 + (1.59×103)2 = 1.88 kΩ θ = tan-1(1.59k/1.0k) = 57.8° (b) XC = 1/2π(30×103)(0.01×10-6) = 531 kΩ Z = √(1.0×103)2 + (531)2 = 1.13 kΩ θ = tan-1(531/1.0k) = 28.0° ISU EE 13 C.Y. Lee Impedance and Phase Angle of Parallel RC Circuits Total impedance in parallel RC circuit: Z = (RXC) / (√ R2 +X2C) Phase angle between the applied V and the total I: 1 1 1 -1 = + θ = tan (R/XC) Z R − jX C 1 − jX C + R = Z R (− jX C ) Z = RX C X C + jR ⎛ RX C ⎞ ⎟⎟ V = IZ = I ⎜⎜ ⎝ X C + jR ⎠ ISU EE 14 C.Y. Lee Conductance, Susceptance and Admittance Conductance is the reciprocal of resistance: G = 1/R Capacitive susceptance is the reciprocal of capacitive reactance: BC = 1/XC Admittance is the reciprocal of impedance: Y = 1/Z ISU EE 15 C.Y. Lee Ohm’s Law Application of Ohm’s Law to parallel RC circuits using impedance can be rewritten for admittance (Y=1/Z): I V= Y I = VY I Y= V ISU EE 16 C.Y. Lee Relationships of the I and V in a Parallel RC Circuit The applied voltage, VS, appears across both the resistive and the capacitive branches Total current, Itot, divides at the junction into the two branch current, IR and IC Itot= V/Z IC= V/(−jXC) = V((XC+jR)/RXC) IR= V/R Vs, VR, VC ISU EE 17 C.Y. Lee KCL in a Parallel RC Circuit From KCL, Total current (IS) is the phasor sum of the two branch currents Since IR and IC are 90° out of phase with each other, they must be added as phasor quantities IC= V/(−jXC) Itot= V/Z = V((XC+jR)/RXC) Itot = √ I2R + I2C θ = tan-1(IC/IR) IR= V/R ISU EE 18 C.Y. Lee KCL in a Parallel RC Circuit Example: Determine the value of each current, and describe the phase relationship of each with the source voltage IR = 12/220 = 54.5 mA IC = 12/150 = 80 mA Itot = √(54.5)2 + (80)2 = 96.8 mA θ = tan-1(80/54.5) = 55.7° ISU EE 19 Vs C.Y. Lee Series-Parallel RC Circuits An approach to analyzing circuits with combinations of both series and parallel R and C elements is to: – Calculate the magnitudes of capacitive reactances (XC) – Find the impedance (Z) of the series portion and the impedance of the parallel portion and combine them to get the total impedance – … ISU EE 20 C.Y. Lee 生物組織的阻抗 一個生物組織的交流阻抗(Z)是生物組織的電阻性阻 抗(Z')與電容性阻抗(Z")的聯合表現: Z= Z'− jZ" 1 1 1 = + Z R R i + (− jX m ) Z = R e2 R i + R e R i2 + R e X m2 (R e + R i )2 + X m2 − j 細胞的等效電路 ISU EE 細胞的簡化模型 21 R e2 X m (R e + R i )2 + X m2 C.Y. Lee 生物組織的阻抗 身體右半邊的等效電路 ISU EE 22 C.Y. Lee 生物組織的阻抗 典型生物組織之Z-plot Z = Zs:通過之電流為直流時 組織的阻抗 (Z ′)2 + (− Z ′′)2 c Z∞:通過之交流電流為無 限大頻率時組織的阻抗 c fC (特徵頻率): Z"具有最大值時的頻率 各種組織器官的特性參數 ISU EE 23 C.Y. Lee 生物組織的阻抗 各種組織器官之Z-plot 脾臟 肌肉 肝臟 肺臟 ISU EE 24 C.Y. Lee RC Lag Network The RC lag network is a phase shift circuit in which the output voltage lags the input voltage ⎛ XC ⎞ φ = 90 − tan ⎜ ⎟ ⎝ R ⎠ o ISU EE 25 −1 C.Y. Lee RC Lead Network The RC lead network is a phase shift circuit in which the output voltage leads the input voltage ⎛ XC ⎞ φ = tan ⎜ ⎟ ⎝ R ⎠ −1 ISU EE 26 C.Y. Lee Frequency Selectivity of RC Circuits A low-pass circuit is realized by taking the output across the capacitor, just as in a lag network ISU EE 27 C.Y. Lee Frequency Selectivity of RC Circuits The frequency response of the low-pass RC circuit is shown below, where the measured values are plotted on a graph of Vout versus f. 10 V ISU EE 28 C.Y. Lee Frequency Selectivity of RC Circuits A high-pass circuit is implemented by taking the output across the resistor, as in a lead network ISU EE 29 C.Y. Lee Frequency Selectivity of RC Circuits The frequency response of the high-pass RC circuit is shown below, where the measured values are plotted on a graph of Vout versus f. 10 V ISU EE 30 C.Y. Lee Frequency Selectivity of RC Circuits The frequency at which the capacitive reactance equals the resistance in a low-pass or high-pass RC circuit is called the cutoff frequency: 1 fC = 2πRC ISU EE 31 C.Y. Lee Coupling an AC Signal into a DC Bias Network ISU EE 32 C.Y. Lee Summary When a sinusoidal voltage is applied to an RC circuit, the current and all the voltage drops are also sine waves Total current in an RC circuit always leads the source voltage The resistor voltage is always in phase with the current In an ideal capacitor, the voltage always lags the current by 90° ISU EE 33 C.Y. Lee Summary In an RC circuit, the impedance is determined by both the resistance and the capacitive reactance combined The circuit phase angle is the angle between the total current and the source voltage In a lag network, the output voltage lags the input voltage in phase In a lead network, the output voltage leads the input voltage A filter passes certain frequencies and rejects others ISU EE 34 C.Y. Lee