Download Chapter 1 0 - RC Circuits

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