Problem 53-5    Source:  Problem developed by the site author.

In the Figure 53-5.1 we have V = 4 sin (200 t + 30°). Determine the value of V_o.

Solution of the Problem 53-5

Initially we must calculate the equivalent circuit Thévenin for the point V_x, as show in the Figure 53-5.1. As we know, the inverter input of OP 1 does not circulate current. Therefore, we can calculate a voltage divider for the input circuit. Let us calculate the parallel of the resistance of 6 Ω and the capacitor, whose reactance is -j3:

Now we can apply a voltage divider and calculate the value of V_x. Note that we must add the resistor value of 10 Ω with Z_i1 for we find the denominator value. So, we get Z_i1 + 10 = 11.2 - j2.4 = 11.45 ∠-12.1°. Then:

To determine the impedance of Thévenin, we shorted the voltage source V. Therefore, the resistor of 10 ohms results in parallel with capacitor - j3 and the resistor of 6 ohms. Calculating this parallel we find:

So, we get the circuit shown in Figure 53-5.2.

We clearly have an operational amplifier in the inverter configuration, and we know that the gain of this circuit, adapted to AC, is given by:

Where   Z_i = Z_th  and  Z_f it is necessary find your value. For that, we must resolve the parallel of the resistor of 8 Ω and the capacitor, whose reactance is -j4. This parallel results in 1.6 - j3.2.To find Z_f, we must add the resistor value of 3 Ω. So, the result is:

As we know the values ​​of Z_f and Z_i = Z_th, we can calculate the value of A_v, or:

Repare no sinal negativo de A_v. To eliminate this sign, we can add or subtract 180° from the value found. We normally want the resulting angle to be less than 180°. So, let's subtract. Soon:

As we know the values ​​of V_th = 3.5 ∠ 42.1°, we can calculate the value of V_o, or:

Substituting for the values ​​already found, we obtain:

It is also possible to write the final result in trigonometric form. See below.