Mathematical Analysis of Linear Audio Amplifier

Linear Amplifier

A linear amplifier is an electronic device that amplifies an input signal without introducing any significant distortion to the waveform of the output signal. In other words, the output signal is a scaled-up version of the input signal, and the amplifier maintains the same shape and characteristics of the input signal.

A linear amplifier can be used to amplify a variety of signals, including audio, video, and radio frequency (RF) signals. It is commonly used in communication systems, audio equipment, and scientific instruments.

The amplification process in a linear amplifier is achieved through the use of active components such as transistors, operational amplifiers (op-amps), or vacuum tubes. These components are designed to operate in their linear region, which means that the output voltage or current is directly proportional to the input voltage or current.

Linear amplifiers are important because they provide a way to increase the power of a signal without introducing distortion or nonlinear effects that can degrade the quality of the signal. This makes them particularly useful in applications where signal fidelity is critical, such as in high-fidelity audio systems, instrumentation, and communication system.

Linear Audio Amplifier

Here a linear audio amplifier such as used in FM receiver or AM receiver is analyzed and its important parameters and specification described with example calculation. Consider the following two port circuit diagram of linear audio amplifier.

linear audio amplifier
In the above circuit diagram, \(v_i\) is the input signal and \(v_o\) is the output signal. The load \(R_L\) of 8Ohm represents the speaker. The resistor \(R_I\) is the input resistor, the resistance \(R_{in}\) represents the resistance of the audio amplifier. 

The input signal \(v_i\) can be written as a sum of individual sine wave.

$$v_i(t) = \sum_{j=1}^{\infty} V_j sin(w_j t +\phi_j)$$

where \(V_j\) represents the amplitude of jth component signal, \(w_j\) is the frequency of the signal component in radian and \(\phi_j\) is the phase.

The Fourier series decomposition of a periodic signal represents it as a sum of harmonic sinusoids of different frequencies, where each sinusoid has a specific amplitude and phase angle. The complex Fourier coefficients \(V_j\) describe the amplitude and \(\phi_j\) phase angle of each harmonic component in the signal.

The Fourier series representation of a signal is useful in analyzing and understanding the frequency content of a periodic signal, as well as in designing filters and other signal processing techniques to manipulate the signal in the frequency domain.

Since the amplifier is linear, we can apply superposition principle and treat each individual signal component separately and we can write the input signal as,

$$v_i(t) = V_i sin(w_i t)$$

where we have taken the phase as zero as this signal is taken as the reference signal.

The output signal \(v_o\) is as follows,

$$v_o(t) = V_o sin(w_i t + \theta)$$

Voltage and Current at the Load

Let us calculate the voltage and current required at the output if the power at the load is 100W.

Let us assume that the input signal amplitude is \(V_i\)=1mV and that the output power delivered at the load by the amplifier is 100W. The using the following power relation we can calculate the amplitude of the output signal at the load.

$$P_o = \frac{V_{rms}^2}{R_L}$$

Since $$V_{rms}=\frac{V_o}{\sqrt{2}}$$

we have,

$$P_o = \frac{(\frac{V_o}{\sqrt{2}})^2}{R_L}$$

And solving for \(V_o\) we have,

\(V_o=\sqrt{2 P_o R_L}=40V\) 

The output current equation,

  $$i_o(t) = I_o sin(w_i t + \theta)$$

and the current amplitude required is,

\(I_o = \frac{V_o}{R_L}=\frac{40V}{8\Omega}=5A\) 

So in order to have 100W power at the load of 8Ohm we need a voltage of 40V and current of 5A. And since the load is purely resistive, there is no phase difference between the current and the voltage.

Voltage Gain, Current Gain and Power Gain

In order to specify the amplifier we also need the current gain, voltage gain and the power gain of the amplifier.

The phasor representation of the input and output signal can be written as,

\(v_i=V_i \angle 0\)

and, \(v_o=V_o \angle \theta\)

The voltage gain can be expressed as the ratio of these phaser output to input voltage,

$$A_v = \frac{v_o}{v_i} = \frac{V_o \angle \theta}{V_i \angle 0}$$ 

The magnitude and phase representation of the voltage gain are therefore,

\(|A_v| = \frac{V_o}{V_i}\) and \(\angle A_v = \theta \)

For the audio amplifier circuit example above, the magnitude of the voltage gain is,

\(|A_v| = \frac{40V}{1mV} = 40000\) 

To achieve this voltage gain several stages of amplifier is needed. The phase of the voltage gain depends upon the phase shift of the signal passing through the audio amplifier.

The current gain can be expressed as the ratio of these phaser output to input current,

$$A_i = \frac{i_o}{i_i} = \frac{I_o \angle \theta}{I_i \angle 0}$$ 

The magnitude and phase representation of the voltage gain are therefore,

\(|A_i| = \frac{I_o}{I_i}\) and \(\angle A_i = \theta \)

For the audio amplifier circuit example above, the magnitude of the voltage gain is,

\(|A_i| = \frac{5A}{I_i}\)

The magnitude of the input current \(I_i\) can be calculated as follows,

\(I_i = \frac{V_i}{R_I+R_{in}} = \frac{1mV}{5k\Omega+50k\Omega}=1.82 \times 10^{-8} A\)

where \(R_{in}\) is the input resistance of the amplifier

Hence the current gain of the audio amplifier is,

 \(|A_i| = \frac{5A}{1.82 \times 10^{-8} A} = 2.75 \times 10^{8}\)

The power gain of the amplifier is,

For the above audio amplifier circuit the power gain is,

Decibel Unit

The amplifier gain are usually expressed in decibel unit because gain are usually large numbers and it is easier to add and divide the gain when expressed in decibel units. The following are the voltage gain, current gain and power gain expressed in decibel units.

Voltage gain in decibel

\(A_{v(dB)} = 20 log(|A_v|)\)

For the above amplifier circuit the voltage gain in dB is,

\(A_{v(dB)} = 20 log(40000) = 86.02 dB\)
Current gain in decibel is,
\(A_{i(dB)} = 20 log(|A_i|)\)
For the above amplifier circuit the current gain in dB is therefore,
 \(A_{i(dB)} = 20 log(2.75 \times 10^{8}) =170.34 dB\)
Power gain in decibel is,
\(A_{p(dB)} = 10 log(A_p) \)
\(A_{p(dB)} = 10 log(1.10 \times 10^13) = 130.04 dB\)

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