Problem 65-3    Source:   Problem elaborated by the author of the site.

Given the circuit shown in the Figure 65-03.1, calculate the output voltage V_o, assuming that the input voltage is given by V_i = 10 sin 1000 t. Sketch the waveform of V_o. Consider the voltage on the conduction diode as V_D = 0.7 V.

Solution of the Problem 65-3   

When the input voltage V_i initiates the growth of its value towards the positive peak, we already know in advance that the diode D₂ will remain cut across the positive portion of the input voltage. On the other hand, the D₁ diode will not conduct until the input voltage reaches the value of 0.7 V. From that moment, the current that circulates through the diode and the two resistors is given by:

And as the output voltage V_o is given by V_o = 100 I, so we easily conclude that:

Therefore, the correct answer for the positive part of the sinusoid is:

Now let's look at what happens to the negative part of the sinusoid. Note that D₁ will always be cut-off. So the circuit will work for the negative part of the sinusoid as if it didn't exist. So the circuit comes down to the voltage source V_i , the two resistors, the 3V battery and the D diode ₂ . Due to the presence of the 3V battery, it is clear that the diode will not conducting while V_i ≥ - 3.7 V. As soon as the input voltage reaches - 3.7 V and continue until it reaches the peak value, that is, - 10 V , the diode D₂ is conducting. Then we can calculate the value of I by making the mesh:

Algebraically working this equation and knowing that V_D = 0.7 V, we get:

However, we know from the circuit that V_o = - 100 I. So replacing we have:

Therefore, the correct answer for the negative part of the sinusoid is:

In the Figure 65-03.2 we see the graph of the input voltage and the output voltage. Note that next to t = 0 there is a jump in output voltage from zero to 0.7 V. This is due to the need for a voltage of 0. V over the diode to enter the conduction zone. For the negative part of the input voltage, the diode only conducts when the input voltage reaches the value of - 3,7 V.

Note that there is a range in the input voltage value that generates a null output voltage and is perfectly visible in the graph above. So we can write that: