Problem 64-10    Source:  Question 2 of the Electronic Circuits test, Ufrgs, 2014.

In the circuit shown in Figure 64-10.1, assuming that V_D = 0.65 volts, V _Z = 3.6 volts, determine the relation V_o = f (V_i), indicating the points where the zeners conduct / do not conduct, when V_i ≤ 0.

Solution of the Problem 64-10   

From the circuit, with the polaridad of V_i, we see that when D₃ conducts like a zener diode, D₁ will conduct like an ordinary diode. Thus, considering the null currents in the circuit, the voltage on D₃ will appear on the output, ie, V_o = - 3.6 volts. Note that D₁ will conduct simultaneously with D₃, when V_i = - 3.6 - 0.65 = - 4.25 volts. Therefore, it remains for us to discover when D₂ will conduct.

As the voltage across D₁ is constant, we can replace the diode D₁ with a voltage source of 0.65 volts in series with V_i, and of opposite polarity, which results in a single source with voltage equal to V_i - V_D / 2. Now doing the equivalent Thévenin for the point a and the ground, we find the circuit shown in the Figure 64-10.2, where we change the topology a bit in order to make the diagram more intelligible. Note the new voltage value of the source and the resistor in series with the source.

It is easy to see that D₂ will start to conduct when V_ba = 3.6 volts, considering I₃ = 0. In this case we have:

In this condition we can calculate the value of V_i. Making the mesh highlighted in blue we have:

Making numerical substitutions by V_D = 0.65, V_Z = V_ba = 3.6 and 50 x 0.036 = 1.8 we can perform the calculation and find for V_i the value of:

Note that for this value of V_i we continue with V_o = - 3.6 volts because we assume that I₃ = 0.

Therefore, the conduction points of the diodes are:

Determining the Equation    ⇒    V_o = f(V_i)

From this value of V_i , the output voltage module increases as a function of the voltage drop across the resistor where I₃ circulates. We must determine which equation governs this increase in the magnitude of V_o. Analyzing the circuit, we can write the equation that determines the voltage ratio between points a - b, or:

So we can find a relationship between I₁ and I₃, because:

Making the equations of a mesh that passes through D₂ we have:

Replacing I₁ by the value found previously and algebraically rearranging, we have:

On the other hand, from the circuit we have:

Isolating the value of I₃ as a function of V_i and substituting in the above equation, we finally find the desired relationship, or:

Isolating the value of I₃ as a function of V_i and substituting in the above equation, we finally find the desired relationship, or:

In Figure 64-10.3 we can see the graph of the circuit's response as a function of V_i.