6.1 Describe the term linear amplifier with reference to amplifiers - NSC Electrical Technology Electronics - Question 6 - 2022 - Paper 1

The operating frequency range of the amplifier.

The saturation point occurs in region A of the DC load line.

Class AB amplifiers would be represented at Q3 along the DC load line.

Changing temperatures can affect the quiescent voltages and currents at Q1.

Class B amplifiers conduct during half of the input signal cycle, effectively amplifying only one half of the waveform. This results in a more efficient operation as they do not consume power when not active, but they can introduce distortion at the waveform crossover point.

The purpose of using multi-stages in amplifier circuits is to increase the overall voltage gain.

The total voltage gain in decibels (dB) can be calculated using the formula:

$A_{v} = 20 \log_{10}(A_{1} \times A_{2})$

Substituting the given values:

$A_{v} = 20 \log_{10}(10 \times 15) = 20 \log_{10}(150) = 20 \times 2.1761 \approx 43.52 \text{dB}$

The output waveform should show a sinusoidal shape, consistent in amplitude with the input waveform but may be inverted depending on amplifier configuration.

Capacitor C2 maintains a fixed DC operating point on the base, ensuring it is unaffected by the applied AC input signal.

One advantage is improved impedance matching, resulting in maximum voltage and current gain.

Using an output transformer allows for impedance matching, which can help to minimize power losses in the circuit.

The amplifier is a radio frequency amplifier.

The radio frequency amplifier is designed to amplify a single frequency while suppressing other frequencies, ensuring signal clarity.

Capacitor C3 is used to improve the selectivity of the circuit by being able to resonate at the required frequency and suppress the unwanted frequencies.

The power loss can be calculated using the formula:

$P_{loss} = 10 \log_{10}\left( \frac{P_{initial}}{P_{final}} \right)$

Substituting the values:

Initial Power (20 mW) and Final Power (100 mW):

$P_{loss} = 10 \log_{10}\left( \frac{20}{100} \right) = 10 \log_{10}(0.2) \approx -7 \text{dB}$

The circuit diagram represents a Colpitts oscillator.

The amplifier circuit serves to overcome losses and establish a gain of 1 (unity gain) while maintaining feedback to sustain oscillation.

The tank circuit determines the oscillation frequency of the oscillator and creates a 180° phase shift that is fed to the input of the transistor.

The RF choke suppresses all harmonic frequencies, while capacitor C3 passes the oscillatory frequency signal back to the tank circuit, causing positive feedback at the operational frequency.

R5 serves as the potential divider with R4.

The RC-phase shift oscillator utilizes multiple sets of RC combinations to create a phase shift of 180° to the output waveform from the amplifier while the additional 180° phase shift is created through the RC network, effectively yielding a total phase shift of 360°.

The formula for the oscillation frequency is given by:

$f_{0} = \frac{1}{2 \pi R C}$

Substituting the values:

$f_{0} = \frac{1}{2 \pi \times 10 \times 10^3 \times 0.01 \times 10^{-6}} \approx 6.50 kHz$