Problem 64-5    Source:   Problem elaborated by the author's of the site.

Look at the circuit shown in Figure 64-05.1 . Assume that V_i varies randomly between 16 V and 22 V, and the load for the zener is formed by some resistors that can be activated by switches. Additionally, there is a variable load of 0 to 80 mA. Determine the values of R_S and the power dissipated by the zener and also in R_S .

Solution of the Problem 64-5   

Note that the input voltage is variable and so is the load. Therefore, CASE 4 studied in item Theory applies. Initially we must find the value of the minimum load and maximum load. The minimum load occurs when the two keys, S₁ and S₂ are open and the current source I₁ = 0. In that case, we only have the 120 Ω resistor as a load. Then the minimum current is:

Maximum load occurs when the two keys, S₁ and S₂ are closed and the current source provides I₁ = 80 mA. So, in that case, the three resistors are in parallel which results in an equivalent resistor of R_eq = 80 Ω. Then the maximum load current is:

Therefore, we can calculate the variation in current in the load, or

Therefore, this is the current variation that the zener must absorb when we go from maximum load to minimum load. We must add to this value the minimum current required by the zener to maintain the voltage constant at 10 V when the load varies. Using a zener of 2 W power, the maximum current that the zener can handle is

Therefore, I_Zmin will be:

Note that the maximum current through the zener is

This value is lower than the maximum current supported by the zener. In principle, this zener satisfies the needs of the circuit. With these values, it is possible to calculate R_Smin and R_Smax using the equations eq. 64-14 and eq. 64-15, already studied. Let's repeat them here.

Remembering that in this problem, V_i(min) = 16 volts and V_i(max) = 22 volts . Soon:

And in turn:

Note that R_Smax < R_Smin making the project unfeasible. Therefore, a zener diode with 2 W of power is not suitable for the project. Therefore, we will use a 5 W diode. Now we must recalculate the values. For I_Zmax we have

Therefore, I_Zmin will be:

The other variables remain the same. Therefore, let's calculate the new values of R_Smin and R_Smax.

And in turn:

Now, notice that R_Smax > R_Smin making the project viable. Therefore, a zener diode with a power of 2 W which, at first, seemed viable, turned out to be unfeasible after the calculations. Therefore, we adopted a zener with a power of 5 W, making the project viable. To find a definitive value for R_S, we can calculate the arithmetic mean between the two values found. Doing the calculations, we obtain:

Thus, the 5 watt zener used in the project fully meets the needs of the circuit. To find the power that the zener will dissipate, we must analyze the worst situations for him. In this case, it will be when the input voltage is maximum, that is, V_imax = 22 V, and when the load is minimum, that is, I_Lmin = 0.08 A. Thus, we can calculate the current, I''_Z, over the zener by making I''_Z = I_RS - I_Lmin. Then:

Carrying out the calculation, we find:

Therefore, the maximum power dissipated by the zener is

To calculate the power dissipated by the resistor R_S, we must calculate the maximum current that will flow through it. Carrying out the calculation we will find the following value: I_RS = ( V_imax - V_Z ) / R_S = ( 22 - 10 )/ 22 = 0.545 A. Soon