Question

If a l V rms signal source with an internal resistance of 50 Ω is connected to the amplifier in Figure what is the actual rms signal applied to the amplifier input? Assume Bac 200 +15 V RI 1.0 kΩ 01 Di C3 02 75 Ω 1.0 kΩ

Answer 1

As per given that Draw the circuit as given in figure 1 Consider the following values The load resistance is, R_L = 75 Ohm The resistance in the voltage divider circuit are R_1 = 1 k Ohm and R_2 = 1 k Ohm The actual ac beta beta = 200 The voltage of the signal source is V_in = (1) (V_rms) = squareroot 2 sin omega t The internal resistance of the signal source R_int = 50 Ohm \r\n Consider with internal resistance in the source, the Q- point won't charge Calculate the drop across internal resistance V_50 Ohm = (3.4 times 10^-5 Ohm) (50 Ohm) = 1.7 mV Calculate the voltage of the signal source. V_in = (1) (V_rms) = squareroot 2 sin omega_t v Calculate the actual voltage V_actual = (squareroot 2 sin omega_t - 1.7 times 10^-3) V As this is a non-linear equation. V_actual = (1 - 1.7 times 10^-3) V = 0.9983 V Therefore, the actual rms voltage V_actual is 0.9983 V Write the expression for average power dissipation for maximum output of a class amplifier