EEE2042S / EEE2049W
Lecture 2: Electronic Fundamentals
Outline
❑ Why Analogue Electronics?
❑ Resistors
❑ Capacitors
❑ Inductors
❑ Passive Filters
❑ Oscillators
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Filters
❑ There are many different types of filters ranging from active to passive to digital
filters. Here we will only discuss passive filters
❑ Passive filters are made up of passive components such as resistors, capacitors
and inductors.
❑ There are essentially three categories of filters:
▪ Low pass filter (LPF)
▪ High pass filter (HPF)
▪ Band pass filter (BPF)
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Low Pass Filters
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Passive Low Pass Filters
❑ A simple passive RC Low Pass Filter or LPF, can be easily made by connecting
together in series a single Resistor with a single Capacitor.
❑ This circuit acts as a frequency dependent voltage divider where the capacitor
acts in place of a resistor.
❑ This type of filter is known generally as a “first-order filter” because it has only
one reactive component, the capacitor, in the circuit.
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Passive Low Pass Filters
❑
If we treat this circuit as a voltage divider, we see that Vout is given by:
𝑉𝑜𝑢𝑡 = 𝑉𝑖𝑛 ×
𝑅2
𝑅1 + 𝑅2
❑ We also know that the capacitive reactance of a capacitor is given by:
𝑋𝐶 =
1
in Ohms
2𝜋𝑓𝐶
❑ In an AC circuit, the opposition to current flow is called impedance, Z, and for a series circuit consisting
of a single resistor and capacitor like the one above, the impedance is calculated as:
𝑍=
𝑅2 + 𝑋𝐶2
❑ This gives us a voltage divider output of:
𝑉𝑜𝑢𝑡 = 𝑉𝑖𝑛 ×
𝑋𝐶
𝑅2 + 𝑋𝐶2
= 𝑉𝑖𝑛
𝑋𝐶
𝑍
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Passive Low Pass Filters – Example
❑ A LPF circuit consisting of a resistor of 4.7kΩ in series with a capacitor of 47nF
is connected across a 10v sinusoidal supply. Calculate the output voltage (Vout)
at a frequency of 100Hz and at 10kHz.
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Passive Low Pass Filters – Example
A LPF circuit consisting of a resistor of 4.7kΩ in series with a capacitor of 47nF is connected across a 10v
sinusoidal supply. Calculate the output voltage (Vout) at a frequency of 100Hz and at 10kHz.
Vout at 100Hz:
𝑋𝐶 =
1
1
=
= 33863Ω
2𝜋𝑓𝐶 2𝜋 × 100 × 47 × 10−9
∴ 𝑉𝑜𝑢𝑡 = 𝑉𝑖𝑛 ×
𝑋𝐶
= 10 ×
𝑅 2 + 𝑋𝐶2
33863
47002 + 338632
= 9.9v
Vout at 10kHz:
𝑋𝐶 =
1
1
=
= 338.6Ω
2𝜋𝑓𝐶 2𝜋 × 10 000 × 47 × 10−9
∴ 𝑉𝑜𝑢𝑡 = 𝑉𝑖𝑛 ×
𝑋𝐶
𝑅2 + 𝑋𝐶2
= 10 ×
338.6
47002 + 338.62
= 0.178v
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Passive Low Pass Filters – Frequency Response
❑ We can see from the example that as the frequency increases, the voltage at the
output drops . If we plot the output voltage with respect to input frequency, we
get what is known as the Frequency Response Curve or Bode Plot.
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Passive Low Pass Filters – Cut-off Frequency
❑ The so-called cut-off frequency is defined as being the frequency where the capacitive
reactance and resistance are equal. When this occurs, the output signal is attenuated by
3dB.
❑ The cut-off frequency is therefore calculated by rearranging the impedance formula for a
capacitor:
𝑋𝐶 =
1
, where X C = R
2𝜋𝑓𝐶
∴ 𝑓𝑐 =
1
2𝜋𝑅𝐶
❑ The phase shift of the signal at this cutoff frequency is determined by:
𝜔 = − arctan(2𝜋𝑓𝑅𝐶)
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High Pass Filters
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Passive High Pass Filters
❑ Whereas the low pass filter only allowed signals to pass below its cut-off
frequency point, ƒc, the passive high pass filter circuit as its name implies, only
passes signals above the selected cut-off point, ƒc eliminating any low frequency
signals from the waveform.
❑ In this circuit arrangement, the reactance of the capacitor is very high at low
frequencies so the capacitor acts like an open circuit and blocks any input
signals at Vin until the cut-off frequency point (ƒc) is reached.
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Passive High Pass Filters – Frequency Response
❑ In the case of a HPF, as the frequency increases, the voltage at the output
increases towards the input voltage. We can plot the Frequency Response as:
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Passive High Pass Filters – Cut-off Frequency
❑ The cut-off frequency of this circuit is determined in the same manner as for a
LPF:
𝑋𝐶 =
1
, where XC = R
2𝜋𝑓𝐶
1
∴ 𝑓𝑐 =
2𝜋𝑅𝐶
❑ However the phase shift is now different as the output signal phase now leads
the input signal:
𝜔 = arctan
1
2𝜋𝑓𝑅𝐶
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Passive High Pass Filters – Example
❑ Calculate the cut-off frequency for a simple passive HPF consisting of an 82pF
capacitor connected in series with a 240kΩ resistor.
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Passive High Pass Filters – Example
❑ Calculate the cut-off frequency for a simple passive HPF consisting of an 82pF
capacitor connected in series with a 240kΩ resistor.
𝑓𝑐 =
1
1
=
= 8kHz
2𝜋𝑅𝐶 2𝜋 × 240 000 × 82 × 10−12
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Band Pass Filters
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Passive Band Pass Filters
❑ Band Pass Filters can be used to isolate or filter out certain frequencies that lie
within a particular band or range of frequencies.
❑ By connecting or “cascading” together a single Low Pass Filter circuit with a
High Pass Filter circuit, we can produce another type of passive RC filter that
passes a selected range or “band” of frequencies that can be either narrow or
wide while attenuating all those outside of this range.
❑ This new type of passive filter arrangement produces a frequency selective filter
known commonly as a Band Pass Filter or BPF for short.
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Passive Band Pass Filters – Frequency Response
❑ In a BPF, the pass band lies either side of the LPF and HPF cut-off frequencies. As such,
the BPF cut-off frequencies are calculated in the same way as with a LPF and a HPF.
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Passive Band Pass Filters – Example
❑ A second-order band pass filter is to be constructed using RC components that
will only allow a range of frequencies to pass above 1kHz and below 30kHz.
Assuming that both the resistors have values of 10kΩ, calculate the values of
the two capacitors required.
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Passive Band Pass Filters – Example
HPF stage:
𝐶1 =
1
1
=
= 15.9nF
2𝜋𝑓𝐿 𝑅 2𝜋 × 1 000 × 10 000
𝐶2 =
1
1
=
= 530pF
2𝜋𝑓𝐻 𝑅 2𝜋 × 30 000 × 10 000
LPF stage:
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Passive Band Stop Filters
❑ You also get what are known as band stop filters that create a stop band instead
of a pass band as shown in the figure.
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Passive Band Stop Filters
❑ This is easily achieved by using a high pass filter and low pass filter that do not
have overlapping pass bands.
❑ It must be noted that all band pass or stop filters are inherently second-order
filters as they contain two reactive components.
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Questions?
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