Problem 64-2    Source:  Problem elaborated by the author of the site.

Be the circuit shown in the Figure 64-02.1. Assume that Vi = 20 volts, V_Z = 10 volts and the current in the load ranges from 30 to 90 mA. Verify that zener 1N758 that can dissipate up to 0.5 W of power can be used in the design.

Solution of the Problem 64-2   

Note that the input voltage is constant but the load is variable. Then there will be variation in the current flowing through the load and also through the zener. Therefore, we apply the case 2 studied in the Zener Diode Circuit Design item (Here!!). Since zener 1N758 must be used and the zener power is 0.5 watt, we can calculate I_Zmax and I_Zmin, or:

Calculating I_Zmin, or:

With these two values in mind, we can calculate R_Smax and R_Smin using the two equations studied. See below:

And in turn:

As is well known, R_Smin > R_Smax making the project unfeasible. Therefore, we conclude that we can not use zener 1N758 in this project.

To work around this problem, use a zener that allows 1 watt to dissipate, such as the 1N4740 model. So by recalculating the values to see if you can run the project.

Calculating I_Zmin, or:

Therefore, the new values of R_Smax and R_Smin, will be:

And in turn:

Now R_Smax > R_Smin. The commercial value of resistor between these values is 82 ohms and 100 ohms. The value of 100 ohms is discarded because the component would be working at the acceptable limit. Therefore, the best choice for series resistor lies in the value of R_Smax = 82 ohms. So:

We must determine the power that the resistor R_S will dissipate in the circuit. The current passing through R_S is given by:

Note that the current through R_S is constant. It does not depend on variations in load because these variations are absorbed by the zener. Then the power dissipated is:

Therefore, a resistor of 2 W, or rather one of 5 watts may be used. Thus, a greater gap is achieved.

To conclude, calculate the power dissipated by zener. Note that to calculate the power dissipated by the zener, we must use the largest current that flows through it. Thus, the worst situation is when the load consumes the least current. So, I_Zmax = I_S - I_Lmin = 0.09195 A = 91.95 mA. Then:

Since the power is less than 1 watt then this zener can be used. In addition, a better solution is to use a zener diode that supports 5 watts.