Transcript Document

electronics fundamentals
circuits, devices, and applications
THOMAS L. FLOYD
DAVID M. BUCHLA
วงจร RC
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 10
การตอบสนองต่อสัญญาณไซน์ของวงจร RC
เมื่อมีท้งั ค่าความต้านทานและค่าการเก็บประจุอยูใ่ นวงจรอนุกรม มุมของเฟสระหว่าง
แรงดันที่ป้อนและกระแสรวมจะอยูร่ ะหว่าง 0 และ 90 ขึ้นอยูก่ บั ค่าของความ
ต้านทานและค่ารี แอคแตนซ์
VR
VC
V R leads VS
V C lags V S
R
C
VS
I
I leads V S
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 10
อิมพีแดนซ์ในวงจรอนุกรม RC
ในวงจรอนุกรม RC ค่าอิมพีแดนซ์รวมมีค่าเท่ากับผลรวมของเฟสเซอร์ของ R และ XC
R จะพล็อตตามแนวแกน x ซีกบวก.
XC พล็อตตามแนวแกน y ซีกลบ
 XC 

 R 
  tan 1 
R
R


Z เป็ นเส้นทะแยงมุม
XC
XC
Z
Z
เปลี่ยนตาแหน่งของเฟสเซอร์ให้เป็ นสามเหลี่ยม
อิมพีแดนซ์.
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 10
จงวาดภาพสามเหลี่ยมอิมพีแดนซ์ และแสดงค่าของ R = 1.2 kW
และ XC = 960 W
Z
1.2 kW 
2
+  0.96 kW 
 1.33 kW
  tan 1
0.96 kW
1.2 kW
 39
Electronics Fundamentals 8th edition
Floyd/Buchla
2
R = 1.2 kW

39o
Z = 1.33 kW
XC =
960 W
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 10
การวิเคราะห์วงจรอนุกรม RC
ใช้กฎของโอห์ม โดยใช้ Z, V, I
V  IZ
V
I
Z
V
Z
I
เนื่องจากในวงจรอนุกรม I มีค่าเท่ากันในทุก ๆ ที่ของวงจร จึงสามารถหาแรงดัน
ตกคร่ อมอุปกรณ์ต่าง ๆ ด้วยค่าอิมพีแดนซ์ของอุปกรณ์ตวั นั้นกับค่าของกระแส
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 10
จากตัวอย่างที่ผา่ นมา สมมติให้กระแสมีค่า 10 mArms จงวาดเฟสเซอร์ไดอะแกรม
ของแรงดัน
แรงดันของเฟสเซอร์ไดอะแกรมหาได้จากกฎของโอห์ม ด้วยการคูณอิมพีแดนซ์เฟส
เซอร์แต่ละตัวด้วย 10 mA
R = 1.2 kW

39o
Z = 1.33 kW
Electronics Fundamentals 8th edition
Floyd/Buchla
x 10 mA
=
XC =
960 W
VR = 12 V

39o
VS = 13.3 V
VC =
9.6 V
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 10
การเปลี่ยนแปลงของมุมเฟสเมื่อเทียบกับความถี่
เมื่อความถี่เปลี่ยน สามเหลี่ยม
อิมพีแดนซ์ของวงจรอนุกรม RC ก็จะ
เปลี่ยนไปด้วยดังภาพ เนื่องจาก XC
จะมีค่าน้อยลง ถ้าเพิ่มความถี่ f
ลักษณะแบบนี้เรี ยกว่า การตอบสนอง
ต่อความถี่ของวงจร RC
R
3
Increasing f
2
1
Z3
f3
XC 2
f2
XC 1
f1
Z2
Z1
Electronics Fundamentals 8th edition
Floyd/Buchla
XC 3
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 10
การนาไปใช้งาน
ถ้ากาหนดความถี่ให้ วงจรอนุกรม RC สามารถใช้ในการสร้างเฟสล้าหลังได้ดว้ ยการ
กาหนดปริ มาณของแรงดันอินพุต และเอาต์พตุ ที่ได้จากแรงดันตกคร่ อมตัวเก็บประจุ
วงจรนี้เรี ยกว่าวงจร low-pass filter เป็ นวงจรที่ยอมให้ความถี่ต่าผ่านไปได้เท่านั้น
V
R

Vin
C
Vout
VR
f
(phase lag)
Vout
Electronics Fundamentals 8th edition
Floyd/Buchla
Vout
f
Vin
Vin
(phase lag)
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 10
ในทานองเดียวกัน ถ้ากลับอุปกรณ์ในวงจร ก็จะได้วงจร high-pass filter วงจรที่
ยอมให้ความถี่สูงผ่านไปได้ จากความถี่ที่กาหนด (cutoff frequency)
C
V
Vout

Vin
(phase lead)
Vin
R
Vout
Vout
VC
Electronics Fundamentals 8th edition
Floyd/Buchla
Vin

(phase lead)
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 10
Summary
Applications
An application showing how the phase-shift network is
useful is the phase-shift oscillator, which uses a
combination of RC networks to produce the required 180o
phase shift for the oscillator.
Amplifier
Rf
Phase-shift network
C
C
C
R
Electronics Fundamentals 8th edition
Floyd/Buchla
R
R
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 10
Summary
Sinusoidal response of parallel RC circuits
For parallel circuits, it is useful to introduce two new
quantities (susceptance and admittance) and to review
conductance.
Conductance is the reciprocal of resistance.
Capacitive susceptance is the reciprocal
of capacitive reactance.
G
1
R
1
BC 
XC
1
Admittance is the reciprocal of impedance. Y 
Z
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 10
Summary
Sinusoidal response of parallel RC circuits
In a parallel RC circuit, the admittance phasor is the sum
of the conductance and capacitive susceptance phasors.
The magnitude can be expressed as Y  G 2 + BC 2
 BC 

G
From the diagram, the phase angle is   tan 1 
BC
Y
VS
G
BC

Electronics Fundamentals 8th edition
Floyd/Buchla
G
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 10
Summary
Sinusoidal response of parallel RC circuits
Some important points to notice are:
G is plotted along the positive x-axis.
BC is plotted along the positive y-axis.
 BC 

G
  tan 1 
Y is the diagonal
BC
Y
VS
G
BC

Electronics Fundamentals 8th edition
Floyd/Buchla
G
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 10
Summary
Sinusoidal response of parallel RC circuits
Draw the admittance phasor diagram for the circuit.
The magnitude of the conductance and susceptance are:
G
1
1

 1.0 mS
R 1.0 kW
Y  G 2 + BC 2 
BC  2 10 kHz  0.01 mF  0.628 mS
1.0 mS +  0.628 mS  1.18 mS
2
2
BC = 0.628 mS
VS
f = 10 kHz
R
1.0 kW
C
0.01 mF
Y=
1.18 mS
G = 1.0 mS
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 10
Summary
Analysis of parallel RC circuits
Ohm’s law is applied to parallel RC circuits using
Y, V, and I.
V
I
Y
I  VY Y 
I
V
Because V is the same across all components in a
parallel circuit, you can obtain the current in a given
component by simply multiplying the admittance of
the component by the voltage as illustrated in the
following example.
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 10
Summary
Analysis of parallel RC circuits
If the voltage in the previous example is 10 V, sketch the
current phasor diagram. The admittance diagram from the
previous example is shown for reference.
The current phasor diagram can be found from
Ohm’s law. Multiply each admittance phasor by 10 V.
BC = 0.628 mS
Y=
1.18 mS
G = 1.0 mS
Electronics Fundamentals 8th edition
Floyd/Buchla
x 10 V
=
IC = 6.28 mA
IS =
11.8 mA
IR = 10 mA
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 10
Summary
Phase angle of parallel RC circuits
Notice that the formula for capacitive susceptance is
the reciprocal of capacitive reactance. Thus BC and IC
are directly proportional to f: BC  2 fC
As frequency increases, BC
and IC must also increase,
so the angle between IR and
IS must increase.
Electronics Fundamentals 8th edition
Floyd/Buchla
IC
IS

IR
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 10
Summary
Equivalent series and parallel RC circuits
For every parallel RC circuit there is an equivalent
series RC circuit at a given frequency.
The equivalent resistance and capacitive
reactance are shown on the impedance triangle:
Req = Z cos 

Z
Electronics Fundamentals 8th edition
Floyd/Buchla
XC(eq) = Z sin 
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 10
Summary
Series-Parallel RC circuits
Series-parallel RC circuits are combinations of both series and
parallel elements. These circuits can be solved by methods from
series and parallel circuits.
Z1
Z2
For example, the
R1
C1
components in the
R2
C2
green box are in
series: Z1  R12  X C21
The components in
the yellow box are
R2 X C 2
in parallel: Z2 
R22  X C2 2
Electronics Fundamentals 8th edition
Floyd/Buchla
The total impedance can be found by
converting the parallel components to an
equivalent series combination, then
adding the result to R1 and XC1 to get the
total reactance.
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 10
Summary
Measuring Phase Angle
An oscilloscope is commonly used to measure phase angle in
reactive circuits. The easiest way to measure phase angle is to
set up the two signals to have the same apparent amplitude and
measure the period. An example of a Multisim simulation is
shown, but the technique is the same in lab.
Set up the oscilloscope so that
two waves appear to have the
same amplitude as shown.
Determine the period. For the
wave shown, the period is
 20 μs 
T  8.0 div 
  160 μs
 div 
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 10
Summary
Measuring Phase Angle
Next, spread the waves out using the SEC/DIV control in order
to make an accurate measurement of the time difference between
the waves. In the case illustrated, the time difference is
 5 μs 
t  4.9 div 
  24.5 μs
 div 
The phase shift is calculated from
 t 
 24.5 μs 
o
    360  
 360  55
T 
 160 μs 
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 10
Summary
The power triangle
Recall that in a series RC circuit, you could multiply
the impedance phasors by the current to obtain the
voltage phasors. The earlier example is shown for
review:
R = 1.2 kW

39o
Z = 1.33 kW
Electronics Fundamentals 8th edition
Floyd/Buchla
x 10 mA
=
XC =
960 W
VR = 12 V

39o
VS = 13.3 V
VC =
9.6 V
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 10
Summary
The power triangle
Multiplying the voltage phasors by Irms gives the power triangle
(equivalent to multiplying the impedance phasors by I2). Apparent
power is the product of the magnitude of the current and magnitude of
the voltage and is plotted along the hypotenuse of the power triangle.
The rms current in the earlier example was 10 mA.
Show the power triangle.
VR = 12 V
x 10 mA
=

39o
VS = 13.3 V
Electronics Fundamentals 8th edition
Floyd/Buchla
VC =
9.6 V
Ptrue = 120 mW

39o
Pa = 133 mVA
Pr = 96
mVAR
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 10
Summary
Power factor
The power factor is the relationship between the
apparent power in volt-amperes and true power in
watts. Volt-amperes multiplied by the power factor
equals true power.
Power factor is defined mathematically as
PF = cos 
The power factor can vary from 0 for a purely reactive
circuit to 1 for a purely resistive circuit.
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 10
Summary
Apparent power
Apparent power consists of two components; a true
power component, that does the work, and a
reactive power component, that is simply power
shuttled back and forth between source and load.
Some components such
as transformers, motors,
and generators are rated
in VA rather than watts.
Electronics Fundamentals 8th edition
Floyd/Buchla
Ptrue (W)

Pa (VA)
Pr (VAR)
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 10
Summary
Frequency Response of RC Circuits
When a signal is applied to an RC circuit, and the output is taken
across the capacitor as shown, the circuit acts as a low-pass filter.
As the frequency increases, the output amplitude decreases.
Vin
10 V dc
Vout
10
10VVrms
rms
10 V rms
0
10 V dc
100
100W
W
W
100
20
kHz
ƒƒƒ == 110
kHz
kHz
V 8.46
rms V rms
10 V dc 1.57
0.79V
rms
0
11 mm
mFF
F
Vout (V)
9.98
Plotting the response:
8.46
1.57
0.79
9
8
7
6
5
4
3
2
1
0.1
Electronics Fundamentals 8th edition
Floyd/Buchla
1
10 20
100
f (kHz)
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 10
Summary
Frequency Response of RC Circuits
Reversing the components, and taking the output across the resistor as
shown, the circuit acts as a high-pass filter.
As the frequency increases, the output amplitude also increases.
Vin
10 V dc
Vout
10V
V rms
rms
10
10 V rms
0
10 V dc
m
F
11mm
mF
F
1
F
ƒ
=
100
Hz
ƒ
=
1
kHz
ƒ = 10 kHz
9.87V rms
5.32 V rms
0.63 V rms
100 W
W
100
100
100 W
0 V dc
Vout (V)
9.87
Plotting the response:
5.32
0.63
Electronics Fundamentals 8th edition
Floyd/Buchla
10
9
8
7
6
5
4
3
2
1
0
0.01
0.1
1
10
f (kHz)
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 10
Selected Key Terms
Impedance The total opposition to sinusoidal current
expressed in ohms.
Phase angle The angle between the source voltage and the
total current in a reactive circuit.
Capacitive The ability of a capacitor to permit current;
suceptance (BC) the reciprocal of capacitive reactance. The
unit is the siemens (S).
Admittance (Y) A measure of the ability of a reactive circuit to
permit current; the reciprocal of impedance.
The unit is the siemens (S).
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 10
Selected Key Terms
Power factor The relationship between volt-amperes and
true power or watts. Volt-amperes multiplied
by the power factor equals true power.
Frequency In electric circuits, the variation of the output
response voltage (or current) over a specified range of
frequencies.
Cutoff The frequency at which the output voltage of
frequency a filter is 70.7% of the maximum output
voltage.
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 10
Quiz
1. If you know what the impedance phasor diagram looks
like in a series RC circuit, you can find the voltage phasor
diagram by
a. multiplying each phasor by the current
b. multiplying each phasor by the source voltage
c. dividing each phasor by the source voltage
d. dividing each phasor by the current
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 10
Quiz
2. A series RC circuit is driven with a sine wave. If the
output voltage is taken across the resistor, the output
will
a. be in phase with the input.
b. lead the input voltage.
c. lag the input voltage.
d. none of the above
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 10
Quiz
3. A series RC circuit is driven with a sine wave. If you
measure 7.07 V across the capacitor and 7.07 V across the
resistor, the voltage across both components is
a. 0 V
b. 5 V
c. 10 V
d. 14.1 V
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 10
Quiz
4. If you increase the frequency in a series RC circuit,
a. the total impedance will increase
b. the reactance will not change
c. the phase angle will decrease
d. none of the above
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 10
Quiz
5. Admittance is the reciprocal of
a. reactance
b. resistance
c. conductance
d. impedance
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 10
Quiz
6. Given the admittance phasor diagram of a parallel RC
circuit, you could obtain the current phasor diagram by
a. multiplying each phasor by the voltage
b. multiplying each phasor by the total current
c. dividing each phasor by the voltage
d. dividing each phasor by the total current
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 10
Quiz
7. If you increase the frequency in a parallel RC circuit,
a. the total admittance will decrease
b. the total current will not change
c. the phase angle between IR and IS will decrease
d. none of the above
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 10
Quiz
8. The magnitude of the admittance in a parallel RC circuit
will be larger if
a. the resistance is larger
b. the capacitance is larger
c. both a and b
d. none of the above
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 10
Quiz
9. The maximum power factor occurs when the phase
angle is
a. 0o
b. 30o
c. 45o
d. 90o
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 10
Quiz
10. When power is calculated from voltage and current for an
ac circuit, the voltage and current should be expressed as
a. average values
b. rms values
c. peak values
d. peak-to-peak values
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 10
Quiz
Answers:
Electronics Fundamentals 8th edition
Floyd/Buchla
1. a
6. a
2. b
7. d
3. c
8. d
4. c
9. a
5. d
10. b
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.