Problem 65-6    Source:   Problem elaborated by the author of the site.

Let be the circuit shown in the Figura 65-06.1. Assume that V_i = 18 sin 1000 t and V_D = 0.7 volts on the conduction. Calculate the voltage V_o and sketch a graph with the circuit transfer characteristic.

Solution of the Problem 65-6   

For the positive part of the sine, this circuit will respond in two different ways, depending on the input voltage. So let's look at each situation separately.

When the input voltage V_i begins to grow in the positive peak direction, note that while V_i ≤ +2.7 , there will be no current flow through the two circuits containing the D₁ and D₂ diodes. So the output voltage is exactly the same as the input voltage, that is:

When the input voltage increases its value and V_i > +2.7, the circuit containing the D₁ diode will be conducting, but the circuit containing D₂ will still be cut off and we can disregard it. Then we can use a circuit solving technique by replacing the diode conduction voltage with a battery voltage of 0.7 V. Thus, representing the diode D₁ as a constant voltage source of 0.7 V, we can add its value to the battery of 2 volts, resulting in a single voltage source of 2.7 volts. Thus, we obtain the equivalent circuit shown on the left, in the Figura 65-06.2. Performing a transformation of sources in the circuit, in the end we obtain the circuit represented in the figure on the right.

To reach the final circuit we add the two current sources and calculate the parallel of the two resistors of 100 Ω each. We found a source of value (V _i + 2.7) / 100 and an equivalent resistance of 50 Ω. Multiplying the value of the current source by the equivalent resistance, we find the value of the voltage source. This circuit is exactly the Thévenin equivalent of the circuit shown in Figure 65-06.1 for the positive sine cycle and when the input voltage, V_i > +2.7. See the Figure 65-06.3. Since the circuit does not have a load connected to the output, then the calculated voltage appears fully on the output. Like this:

Now analyzing the circuit for the negative part of the sinusoid, clearly the diode D₁ will be in the cutoff zone at whole negative cycle of the sinusoid. However, the circuit containing D₂ will be in conduction after the input voltage reaches a certain value. Thus, when the input voltage V_i begins to grow in value towards the negative peak, note that while V_i ≥ -4.7 , there will be no current flow through the two circuits containing the D₁ and D₂ diodes . In this case, the output voltage will be exactly the same as the input voltage. So we can write:

And when V_i ≤ -4.7 the D₂ diode will be in the zone conducting. Using the same methodology as above, we will replace the D₂ diode and 4.0 volt battery with a single battery 4.7 volts as shown in the Figura 65-06.4 on the left. Pay close attention to the current sources. They are pointing down as we are analyzing the negative cycle of the sinusoid.

To reach the final circuit we add the two current sources and calculate the parallel of the two resistors. We find a source of value (V _i + 9.4) / 200 and an equivalent resistance of 66.67 Ω. Multiplying the value of the current source by the equivalent resistance, we find the value of the voltage source. This circuit is exactly the Thévenin equivalent of the circuit shown in Figure 65-06.1 for the sinusoid negative cycle and when the input voltage, V_i < -4.7. See the Figure 65-06.5. Since the circuit does not have a load connected to the output, then the calculated voltage appears fully on the output. Like this:

Note that in the above expression we use the V_i module and pass the negative sign in front of the expression. You can also type as V_o = - (V _i / 3) - 3.13 . The result is the same.

In the Figura 65-06.6 we show the transfer characteristic of the circuit. Note that in this problem, a compression occurs in both the positive sinusoid cycle and the negative cycle. In the graph, for a real diode, the slope change of the lines should be smoothed. As we are using the model of an ideal diode is worth the above representation.