AU1343
Audio Electronics II
Week 3, July 29
I Reactance and Impedance: Review and Denitions
Last week we learned about reactance (XC for a capacitor and XL for an
inductor) and impedance (ZC for a capacitor and a resistor and ZL for an inductor and a resistor). We will start today's lesson by reviewing these terms and
equations and then we will practice using these equations to solve for dierent
unknowns
Abbreviations:
f = Frequency in Hertz (abbreviated Hz)
C = Capacitance of capacitor in Farads (abbreviated F)
L = Inductance of inductor in Henry (abbreviated H)
R = Resistance of resistor in Ohms (abbreviated Ω)
XC = Reactance of capacitor in Ohms (abbreviated Ω)
XL = Reactance of inductor in Ohms (abbreviated Ω)
ZC = Impedance of capacitor and resistor in series, in Ohms (abbreviated
Ω)
ZL = Impedance of inductor and resistor in series, in Ohms (abbreviated Ω)
XC =
1
2Πf c
XL = 2Πf L
√
ZC = XC2 + R2
√
ZL = XL2 + R2
It is important to see the relationship between frequency and these quantities, and to understand what this means:
• The reactance of a capacitor is inversely related to frequency
• The reactance of an inductor is directly related to frequency
• The impedance of a capacitor and resistor is inversely related to frequency
• The impedance of an inductor and resistor is directly related to frequency
II Reactance and Impedance: Practice
Students take turns coming up to the board to solve a problem:
Example 1:
1
R
f
C
Given:
Figure 1
C = 1µF
R = 1KΩ
f = 1KHz
Calculate the reactance of the capacitor and then the impedance of the series
combination of the capacitor and resistor at the given frequency
Solution:
∗ 103 = 0.159KΩ = 159Ω
√
√
√
√
= XC2 + R2 = 1592 + (1K)2 = 25281 + 1 ∗ 106 = 1025281 =
XC =
ZC
1013Ω
1
2Πf c
=
1
2Π(1∗103 )(1∗10−6 )
=
1
2Π(1∗10−3 )
=
1
2Π
Exercise 1:
Given:
For the circuit in Figure 1:
C = 2.2µF
R = 15KΩ
f = 100Hz
Calculate the reactance of the capacitor and then the impedance of the series
combination of the capacitor and resistor
Exercise 2:
Given:
For the circuit in Figure 1:
C = .047µF
R = 5KΩ
f = 1.5KHz
Calculate the reactance of the capacitor and then the impedance of the series
combination of the capacitor and resistor
Exercise 3:
For the circuit in Figure 1:
2
C = .1µF
R = 22KΩ
f = 100Hz
Calculate the reactance of the capacitor and then the impedance of the series
combination of the capacitor and resistor
Exercise 4:
For the circuit in Figure 1:
C = 100µF
R = 2Ω
f = 2KHz
Calculate the reactance of the capacitor and then the impedance of the series
combination of the capacitor and resistor
Exercise 5:
For the circuit in Figure 1:
C = 0.16µF
R = 220Ω
f = 12KHz
Calculate the reactance of the capacitor and then the impedance of the series
combination of the capacitor and resistor
Exercise 6:
For the circuit in Figure 1:
C = 1µF
R = 330Ω
f = 800Hz
Calculate the reactance of the capacitor and then the impedance of the series
combination of the capacitor and resistor
Exercise 7:
For the circuit in Figure 1:
C = 0.4µF
R = 330Ω
f = 1KHz
Calculate the reactance of the capacitor and then the impedance of the series
combination of the capacitor and resistor
II Voltage Divider: Review
Last semester we learned that the Voltage Divider is a fundamental building
block of electronics, and last week you learned that the Voltage Divider appears
in the simple resistor-capacitor (RC) and inductor-capacitor (LC) lters.
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The purpose of a voltage divider is to divide an incoming voltage into two
(or more) lesser output voltages, and that the ratio of the output voltages is
equal to the ratio of the resistors that make up the Voltage Divider:
R1
V1
R2
V2
Vin
Figure 2
If V1 is the voltage across resistor R1 , and if V2 is the voltage across resistor
R2 , then the following is easily derived from KVL and Ohm's Law:
V1
V2
=
R1
R2
2
V2 = Vin R1R+R
2
As we saw last week, we can take advantage of the fact that a capacitor's
reactance changes with frequency to create a voltage divider that divides the
voltage dierently at dierent frequencies, thereby creating a lter:
C
Vin
R
Vout
Figure 3
We use the very same voltage divider equation above, substituting the impedance
of capacitor and resistor in series in place of R1 + R2 :
Vout = Vin ZRc
Since Zc is inversely proportional to frequency, this means that Vout is directly proportional to frequency, meaning that the higher the frequency, the
higher the output. This is a High Pass Filter: low frequencies are blocked by
the capacitor while higher frequencies are passed.
A good way to see this is to calculate the value of Vout at dierent frequencies,
and to plot these calculations on a graph:
Exercise 8:
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For the circuit in Figure 3:
Vin = 10Vpp
C = 0.016µF
R = 1KΩ
A) Calculate the output voltage for frequencies of 100Hz, 1KHz, 10KHz, and
100KHz
B) Plot these out on a graph. Create a logarithmic graph by using the same
distance between the four frequencies.
You can see that the output voltage is the same as the input voltage at the
highest frequency, and is nearly zero at the lowest frequency. Thus, this is a
High Pass Filter.
KVL tells us that the opposite must be true on the capacitor, and thus, we
can make a Low Pass Filter simply by swapping the resistor and capacitor:
R
Vin
C
Vout
Figure 4
And now we have:
c
Vout = Vin X
Zc
Exercise 9:
For the circuit in Figure 4:
C = 0.1µF
R = 1KΩ
Vin = 10Vp−p
A) Calculate the output voltage for frequencies of 100Hz, 1KHz, 10KHz, and
100KHz
B) Plot these out on a graph. Create a logarithmic graph by using the same
distance between the four frequencies.
You can see that the output voltage is the same as the input voltage at the
lowest frequency, and is nearly zero at the highest frequency. Thus, this is a
Low Pass Filter.
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An ideal lter would allow everything to pass above or below a certain frequency, and would block everything else. Our lter changes from passing to
blocking over some frequency range. We dene the cuto frequency to be the
frequency at which the output voltage is half of the input voltage.
If we set Vout = 12 Vin and substitute this into our equations for the voltage
dividers above, we can solve for the frequency as a function of the capacitor and
resistor.
It turns out that we get the same equation for both the High Pass Filter and
the Low Pass Filter (this makes sense due to KVL they must complete to Vin
which is xed):
fc =
1
2ΠRC
Since halving the voltage creates a change of -3dB, this is also referred to as
f−3dB
Now we have a way to predict at what frequency our lter will pass (or
block).
We can also use this same equation to help us select the appropriate resistor
or capacitor for a particular lter.
Exercise 9:
What is the cuto frequency of a low pass lter with the following components:
C = 0.1µF
R = 1KΩ
Exercise 10:
What is the cuto frequency of a high pass lter with the following components:
C = 0.1µF
R = 1KΩ
Exercise 11:
What is the cuto frequency of a high pass lter with the following components:
C = 0.018µF
R = 15KΩ
Exercise 12:
Design a high pass lter with the following properties:
fc = 500Hz
R = 15KΩ
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Exercise 13:
Design a low pass lter with the following properties:
fc = 500Hz
R = 15KΩ
Exercise 14:
Discuss whether a high pass lter should be designed with a larger or smaller
resistor. What about a low pass lter?
Exercise 15:
Design a low pass lter with the following properties:
fc = 500Hz
R = 1.5KΩ
Homework:
1) Read about lters at http://www.electronics-tutorials.ws/lter/lter_1.html.
You are not responsible for the details that are beyond what we've learned in
class, but the more you can understand, the more comfortable you will feel with
lters.
2) Play with the design tool at http://sim.okawa-denshi.jp/en/Fkeisan.htm.
The rst two (RC LPF and RC HPF) are the two that we've learned about so
far. We will also discuss active lters, primarily the Sallen-Key active lter. See
how they respond to changing resistors, which we will use in our project.
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