Question

In a class B push-pull amplifier, the AC power to be delivered in $1.125$ watts, ${{V}_{CC}}=9volts$ . The DC current under signal condition is given by:
(A) $0$
(B) $40mA$
(C) $80mA$
(D) $250mA$

Answer 1

Hint In the given question, we have been given some information regarding a class B push-pull amplifier and we are asked to find out the DC current. For doing that, we need to first establish a relation between the AC power output, the maximum current and the voltage across the amplifier. Using this relation, we will find the maximum current; and then we will use the relation between the DC current and the maximum current to find our required answer. Let’s proceed to a detailed solution of the given question.
Formula Used: ${{\left( {{P}_{O}} \right)}_{ac}}=\dfrac{{{\left( {{I}_{C}} \right)}_{\max }}\times {{V}_{CC}}}{2}$

Complete Step by Step Solution
We know the relation between the AC power output, the maximum current and the voltage across the amplifier is given as
${{\left( {{P}_{O}} \right)}_{ac}}=\dfrac{{{\left( {{I}_{C}} \right)}_{\max }}\times {{V}_{CC}}}{2}$ where ${{\left( {{P}_{O}} \right)}_{ac}}$ is the AC power to be delivered, ${{V}_{CC}}$ is the output voltage and ${{\left( {{I}_{C}} \right)}_{\max }}$ is the maximum current through the amplifier
Substituting the values, ${P_O} = 1.125W$ and ${V_{CC}} = 9volts$ in the above equation, we get

& 1.125=\dfrac{{{\left( {{I}_{C}} \right)}_{\max }}\times 9volts}{2} \\\ & \Rightarrow {{\left( {{I}_{C}} \right)}_{\max }}=\dfrac{2.25}{9}=0.25A \\\ \end{aligned}$$ We have obtained the value of the current in amperes and we must convert into mill amperes since the options are given in mill amperes. Since we know that $$1A=1000mA$$ , we can say that the current in the circuit is $$0.25A\times 1000=250mA$$ **Therefore, we can say that option (D) is the correct answer to the question.** **Note** Sometimes, the students may be asked to find the value of the DC current in a half cycle or a half-sine loop. In that case, we will find the maximum current through the amplifier and then divide the obtained value by pi to find the required DC current. The push-pull configuration of the amplifier is introduced to compensate for the increased distortion and the non-utilization of the input power.