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

One key consideration when determining the amplifier class is the operating frequency range, which defines the frequency over which the amplifier will function effectively.

The saturation point of the biased transistor occurs in Region A.

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

One condition that affects the quiescent voltages and quiescent currents at Q1 is changing temperatures.

Class B amplifiers operate by conducting current during one half of the input signal cycle (positive or negative), which leads to higher efficiency. Each transistor in a complementary pair alternates conduction, allowing for amplification without crossover distortion seen in Class A amplifiers.

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

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

$A_{v} = 20 \times \text{log}(A_{V1} \times A_{V2})$ $A_{v} = 20 \times \text{log}(10 \times 15) = 20 \times \text{log}(150) \approx 43.52 \text{ dB}$

The function of C2 is to maintain a fixed DC operating point on the base, which will be unaffected by the applied AC input signal.

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

It is preferable to connect the load via an output transformer to reduce loading effects on the circuit and to improve power transfer.

The amplifier is a radio frequency amplifier.

Radio frequency amplifiers are constructed in such a way that they are able to amplify a single frequency while suppressing other frequencies.

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 as:

$P_{loss} = 10 \times \text{log}\frac{P1}{P2} = 10 \times \text{log}\frac{20 \times 10^{-3}}{100 \times 10^{-3}} = 10 \times \text{log}(0.2) \approx -7.96 \text{ dB}$

The circuit is identified as a Colpitts oscillator.

The function of the amplifier circuit is to overcome the losses and establish a gain of 1 (unity gain). Positive feedback to the tank circuit ensures that oscillation is maintained.

The RF choke suppresses all harmonic frequencies while C3 passes the oscillatory frequency signal back to the tank circuit, causing positive feedback only on the resonant frequency, thus acting as a filter.

It serves as the potential divider with Rb.

The RC-phase shift oscillator uses three sets of RC combinations to create a phase shift of 180° to the output waveform from the amplifier and attenuates it. Together with the 180° phase shift from the base voltage and the inverting feature of the amplifier, the output signal with a total shift of 360° is achieved and amplified before supplying it to the RC-network. The cycle repeats itself.

The oscillation frequency can be calculated as:

$f_{0} = \frac{1}{2\text{π}RC} = \frac{1}{2\text{π} \times 10 \times 10^{3} \times 0.1 \times 10^{-6}} \approx 6.50 \text{ kHz}$