Q.The microphone convert audio frequency sound into electrical signal of 5mV and is connected to CB amplifier having input impedance of 25$\Omega$ and output impedance of 500k$\Omega$ with load resistance 5k$\Omega$. Find the output voltage.

Every amplifier is characterised by a small set of parameters that let us compare designs and predict performance. Three of them are gains and two are impedances:

• Current gain $A_i = i_o / i_i$ — ratio of output to input current.
• Voltage gain $A_v = v_o / v_i$ — ratio of output to input voltage.
• Power gain $A_p = p_o / p_i = A_v \times A_i$ — the product of the voltage and current gains.
• Input impedance $Z_i = v_i / i_i$ — the impedance the source sees.
• Output impedance $Z_o = v_o / i_o$ — the impedance the load sees looking back.

The block model of an amplifier makes the amplification intuitive: the input signal drives a current through a low input impedance, so $v_i = i_{in} \times Z_{in}$ is small, while a nearly equal current flows through a high output impedance, so $v_o = i_o \times Z_o$ is much larger. Because $Z_o \gg Z_{in}$ with the two currents almost equal, a small input appears greatly enlarged at the output — this impedance transformation is the essence of voltage amplification. A microphone signal of a few millivolts fed through such a stage, for instance, can emerge amplified a couple of hundred times.