Facilitator Feedback to Unit 2 Tutorial Problems

Unit 2 Tutorial: Basic Radio Frequency Circuits and Transmission Lines

2.1 LO4: Analyse a basic radio frequency (RF) Circuit
[Knowledge area: Engineering Science; GA1 partially assessed]
2.1.1 The small RF amplifier below has the following data:
VCC = 20 V, VBE = 0.7 V, Ct = 100 pF, R1 = 12 kΩ, R2 = 68 kΩ, RE = 1.2 kΩ,
RL = 75 Ω, R3 = 10 kΩ, LP = 6.8 μH, ac collector resistor rc = 250 kΩ, unloaded
quality factor Qu = 100, primary to secondary turns ratio = 5.
Determine the RF amplifier’s voltage gain in decibels at resonant frequency and
amplifier’s bandwidth.

Cc 8 T1
Q1 Ct R3 4

R1 R2 6 RL
Vin Vout
1

RE CE 5

VCC
Rdn

Cdn 100

VCC

R2

Q1

R1 RE

𝑅1 12000
𝑉𝐵 = 𝑉𝐶𝐶 = × 20 = 3 𝑉
𝑅1 +𝑅2 12000+68000

𝑉𝐸 = 𝑉𝐵 − 𝑉𝐵𝐸 = 3 − 0.7 = 2.3 𝑉

𝑉𝐸 2.3
𝐼𝐸 = = = 1.92 𝑚𝐴
𝑅𝐸 1200
𝑉𝑇 26 𝑚𝑉 26 𝑚𝑉
𝑟𝑒 = = = = 13.56 𝛺
𝐼𝐸 𝐼𝐸 1.92 𝑚𝐴

1 1 1 1
𝑓𝑂 = = = = = 6.098 𝑀𝐻𝑧
2𝜋√𝐿𝐶 2𝜋√𝐿𝐶 2𝜋√𝐿𝑃×𝐶𝑡 2𝜋√6.8𝜇𝐻×100𝑝𝐹

𝑍𝐷 = 𝑄𝑢 𝑋𝐿 = 100 × 2𝜋 × 6.098𝑀𝐻𝑧 × 6.8𝜇𝐻 = 26.05 𝑘Ω

𝑅𝐿 (𝑟𝑒𝑓) = 𝑛2 𝑅𝐿 = 52 × 75 = 1.875 𝑘Ω
𝑅𝐶 (𝑒𝑓𝑓) = 𝑟𝑐 ‖𝑅3 ‖𝑍𝐷 ‖𝑅𝑟 = 250‖10‖26.05‖1.875 =1480 Ω
 𝑅𝑐 (𝑒𝑓𝑓)
𝐴𝑣 (𝑎𝑡 𝑓𝑂 )𝑖𝑛 𝑑𝐵 = 20𝑙𝑜𝑔 ( )
𝑛𝑟𝑒

1480
= 20𝑙𝑜𝑔 ( )
5 × 13.56
= 26.78 𝑑𝐵
𝑅𝐶 (𝑒𝑓𝑓) 1480
𝑄𝐿 = = = 5.68
𝑋𝐿 2𝜋×6.098×106 ×6.8×10−6

𝑓𝑜 6.098 × 106
𝐵𝑊 = = = 1.07 𝑀𝐻𝑧
𝑄𝐿 5.68

2.1.2 Signals at 99.9 MHz from the local oscillator and 89.2 MHz from the RF amplifier
are applied to the input of a mixer. Determine the output frequencies if the mixer
is a multiplier. [10.7 MHz, 189.1 MHz]
𝑓𝑜𝑢𝑡 (𝑚𝑖𝑛) = 𝑓𝐿𝑂 − 𝑓𝑅𝐹 = 99.9 − 89.2 = 10.7 𝑀𝐻𝑧

𝑓𝑜𝑢𝑡 (𝑚𝑎𝑥) = 𝑓𝐿𝑂 + 𝑓𝑅𝐹 = 99.9 + 89.2 = 189.1 𝑀𝐻𝑧

2.1.3 It is desired to generate a 5.1 MHz signal locked in phase to a given 3.4 MHz
signal. Determine the smallest integer values of R and N in the following
diagram to achieve the objective.

𝑁 𝑓𝑜𝑢𝑡 5.1
= = = 1.5 When R = 1, N = 1.5R = 1 x 1.5 = 1.5;
𝑅 𝑓𝑜𝑠𝑐 3.4

When R = 2, N = 1.5R = 1.5 x 2 = 3 [R = 2, N = 3]

2.1.4 A Colpitts oscillator shown below has the following parameters:
C2 = 220 pF, C3 = 24 pF, L1= 51 µH
VCC

L1

R1

C4
1

C1
Q1

R2

L2
1 2

C2 C3

a) Determine the resonant frequency of the oscillator. [4.79 MHz]

𝐶2 𝐶3 220×24
𝐶𝑇 = = = 21.64 𝑝𝐹
𝐶2 +𝐶3 220+24

Assuming that Q>10, then

1 1
𝑓0 = = = 4.79 𝑀𝐻𝑧
2𝜋√𝐿1 𝐶𝑇 2𝜋√51 × 10−6 × 21.64 × 10−12
b) Derive the expression for the feedback attenuation β.
1
𝑉𝑓 𝐼𝑋𝐶2 2𝜋𝑓𝑜 𝐶2 𝐶3
𝛽= = = =
𝑉𝑜𝑢𝑡 𝐼𝑋𝐶3 1 𝐶2
2𝜋𝑓𝑜 𝐶3

c) Determine the required gain for oscillation. [9.17]

For start of oscillation 𝐴𝑣 𝛽 > 1

1 𝐶2 220
𝐴𝑣 > > > > 9.17
𝛽 𝐶3 24
2.2 LO5: Design a basic RF circuit [Knowledge area: Design and Synthesis;
GA3 partially assessed]

2.2.1 An RF amplifier should be designed based on an ideal transformer and the

following schematic.
Data:
PO = 5 mW, RL = 75 Ω, β = 100, VCC = 9 V, C (stray) = 5 pF, CCB = 2 pF,
VBE = 0.7 V, rc = 10 kΩ, η = 50%.
Provide a partial design solution which only has values for biasing resistors:
R1, R2, and RE.

8 T1
C1
Q1 Ct 4

R1 R2 6 RL
Vin Vout
1

RE CE 5

VCC
Rdn

Cdn 100

2×5×10−3
𝐼𝐸 = 𝐼𝐶 = = 1.23 𝑚𝐴
0.9×9

As a Rule of Thumb let VE = 0.1 × 9 = 0.9 V

𝑉𝐸 0.9
𝑅𝐸 = = = 731.7 Ω
𝐼𝐸 1.23 × 10−3
1.23 × 10−3
𝐿𝑒𝑡 𝐼1 = 10𝐼𝐵 = 10 = 123 𝜇𝐴
100
𝑉𝐸 + 𝑉𝐵𝐸 0.9 + 0.7
𝑅1 = = = 13000 Ω
𝐼1 123𝜇𝐴
1.23 × 10−3
𝐿𝑒𝑡 𝐼2 = 11𝐼𝐵 = 11 = 135 𝜇𝐴
100
𝑉𝐶𝐶 − (𝑉𝐸 + 𝑉𝐵𝐸 ) 9 − 1.6
𝑅2 = = = 54815 Ω
𝐼2 135 𝜇𝐴
2.2.2 Use Table 1 to design a 4th order low-pass filter for source and load impedance
of 75 Ω with a cut-off frequency of 3.5 MHz. Draw the filter circuit using π
configuration. State the E24 value for each calculated component value.
[470 pF, 6.2 µH, 1 nF, 2.7 µH]

𝐶𝑛 (𝑡𝑎𝑏𝑙𝑒) 0.7654
𝐶1 = = = 464𝑝𝐹(470𝑝𝐹)
𝜔𝑅𝐿 2𝜋×3.5×106 ×75

𝑅𝐿 𝐿𝑛 (𝑡𝑎𝑏𝑙𝑒) 75 × 1.8478
𝐿2 = = = 6.3𝜇𝐻(6.2𝜇𝐻)
𝜔 2𝜋 × 3.5 × 106

𝐶𝑛 (𝑡𝑎𝑏𝑙𝑒) 1.8478
𝐶3 = = = 1.1𝑛𝐹(1𝑛𝐹)
𝜔𝑅𝐿 2𝜋 × 3.5 × 106 × 75

𝑅𝐿 𝐿𝑛 (𝑡𝑎𝑏𝑙𝑒) 75 × 0.7654
𝐿4 = = = 2.6𝜇𝐻(2.7𝜇𝐻)
𝜔 2𝜋 × 3.5 × 106

2.2.3 Use Table 1 to design a fourth order high-pass filter for 52 Ω source and load
impedances, with a cut-off frequency of 5 MHz. Draw the filter circuit using π
configuration. State the E24 value for each calculated component value.
[2.2 𝜇𝐻, 1.2 𝑛𝐹, 9.1 𝜇𝐻, 4.3 𝑛𝐹]

Rs C2 C4

2 2
Vs RL
L1 L3

1 1

𝑅𝐿 52
𝐿1 = = = 2.2𝜇𝐻(2.2 𝜇𝐻)
𝜔𝑅𝐿 𝐶1 (𝑡𝑎𝑏𝑙𝑒) 2𝜋×5×106 ×0.7654

1 1
𝐶2 = = = 1.7 𝑛𝐹(1.8 𝑛𝐹)
𝜔𝑅𝐿 𝐿2 (𝑡𝑎𝑏𝑙𝑒) 2𝜋 × 5 × 106 × 1.8478
𝑅𝐿 52
𝐿3 = = = 9.0 𝜇𝐻(9.1 𝜇𝐻)
𝜔𝑅𝐿 𝐶3 (𝑡𝑎𝑏𝑙𝑒) 2𝜋×5×106 ×1.8478

1 1
𝐶4 = = = 4.2 𝑛𝐹(4.3 𝑛𝐹)
𝜔𝑅𝐿 𝐿4 (𝑡𝑎𝑏𝑙𝑒) 2𝜋 × 5 × 106 × 0.7654
2.3 LO6: Choose a suitable transmission line for a required application
[Knowledge area: Engineering Science; GA1 partially assessed]

2.3.1 A certain microwave transmitter operating at 2.2 GHz must deliver most of its
output power to the transmit antenna. The transmitter power is 120 W. The
distance between the transmitter and antenna is 20 m. The following
transmission lines are available for this application:

Coaxial cable with attenuation of 8 dB/100 m

Elliptical waveguide with attenuation of 2.68 dB/100 m

Choose a suitable transmission line that meets the requirement.

Power delivered to the antenna using the coaxial transmission line:

7.09𝑑𝐵
𝐿𝑐𝑜𝑎𝑥𝑖𝑎𝑙 (𝑑𝐵) = × 32𝑚 = 2.27𝑑𝐵
100𝑚
𝑃𝑑𝑒𝑙𝑖𝑣𝑒𝑟𝑒𝑑 = 𝑃𝑡 − 𝐿𝑐𝑎𝑏𝑙𝑒
= 10𝑙𝑜𝑔160 − 2.27
= 19.77𝑑𝐵
= 101.869
= 94.84𝑊
Power delivered to the antenna using the waveguide transmission line:
2.68𝑑𝐵
𝐿𝑤𝑎𝑣𝑒𝑔𝑢𝑖𝑑𝑒 (𝑑𝐵) = × 32𝑚 = 0.86𝑑𝐵
100𝑚
𝑃𝑑𝑒𝑙𝑖𝑣𝑒𝑟𝑒𝑑 = 𝑃𝑡 − 𝐿𝑤𝑎𝑣𝑒𝑔𝑢𝑖𝑑𝑒
= 10𝑙𝑜𝑔160 − 0.86
= 21.18𝑑𝐵
= 102.118
= 131.22𝑊
Comparison table
Transmission line Power delivered
Coaxial cable 94.84 W
Waveguide 131.22 W

The above results indicate that the waveguide is a suitable transmission line
as it delivers 131.22W whereas the coaxial cable delivers 94.84W.
2.3.2 Contrast between fibre-optic and coaxial transmission lines under the
following factors.
Construction, attenuation, bandwidth, cost, weight, cable diameter, tapping,
safety, data transmission rate, external magnetic field, cable installation and
fragility.

Construction: Fibre cable is constructed from glass or plastic while coaxial

cable is constructed from plastic, metal foil and metal wire.
Attenuation: Fibre-optic cable has less signal attenuation over a given
distance than coaxial cable.
Bandwidth: Fibre-optic line has a wider bandwidth than a coaxial line.
Cost: Optical fibre cable is highly expensive while coaxial cable is less
expensive.
Weight: Fibre-optic cable is lighter than coaxial cable.
Cable diameter: Practical fibre-optic cable is much smaller in diameter than
coaxial cable.
Tapping: Fibre-optic line cannot be as easily tapped as it the case with a
coaxial line, and it does not radiate signals that can be picked up for
eavesdropping purposes.
Safety: Fibre-optic line provides greater safety for there are no conductors.
Data transmission rate: Fibre optic line provides higher data rates than coaxial
line.
External magnetic field: Fibre cable is not affected by external magnetic while
coaxial cable is affected.
Cable installation: Installation of coaxial cable is easier than fibre optic cable.
Fragility: Glass fibre is much more fragile and easily broken than a coaxial
cable.
2.3.3 Coaxial cable and fibre optic cable are considered for connecting two towns
that are 45 kilometres apart. This application requires a transmission line with
less signal attenuation, wide bandwidth, not affected by interference and safe.
Choose a suitable transmission line for this required application.
Comparison table for transmission line selection

Choice factor Coaxial Fibre optic

Signal Lossy, increasing with Provides less signal attenuation

attenuation frequency over a given distance than coaxial
cable.
Bandwidth Wide bandwidth Provides wider bandwidth than
coaxial cable
Interference Affected by Immune to interference
electromagnetic and radio
frequencies
Safety Not safe as there are Provide greater safety for there are
conductors no conductors.

The above comparison table indicates that the fibre optic is suitable for this application.