Chapter 12

Amplifier of Hi-Fi channel

Take 15 minutes to prepare this exercise.

Then, if you lack ideas to begin, look at the given clue and start searching for the solution.

A detailed solution is then proposed to you.

If you have more questions, feel free to ask them on the forum.

A stereo amplifier can be modeled by the following circuit diagram in which the input resistance will be considered infinite :

Amplifier of Hi-Fi channel

It performs the following two tests :

• Test n°1 :

, , RMS , .

Is measured with a digital oscilloscope effective (RMS) output value equals to .

• Test n°2 :

, , RMS , .

We then measure the effective (RMS) output value equals to .

Moreover, we find that during each test, the two output signals keep, regardless of the frequency, the same effective (RMS) value and are in phase with .

Question

Determine the gain and the complex output impedance .

Solution

We denote, in complex notation and noting and effective (RMS) values of the input and output voltages and  the phase shift of with respect to :

And :

The voltage divider rule gives :

By denoting :

Where the real and imaginary parts and a priori depend on the frequency, the effective value of the output voltage can be written :

The phase shift between and is zero being regardless of the frequency, and deduce and :

As the effective (RMS) value does not depend on the frequency, the output of the stereo impedance is finally real and equivalent to a single resistor of constant value , independent of frequency.

Tests with two resistance  values then lead to the system of two following equations :

And :

With in .

and

Question

The amplifier being supplied with a voltage :

What should be the load resistor so that it provides the maximum average power at the constant amplitude input voltage ?

Solution

The average electrical power received by the load resistance is equal to :

It will be extremal, at given , and (amplifier characteristics) when :

But :

Therefore :

if .

The power is then effectively maximum and is equal to :