Question 1.13: Objective: Consider a simple constant-voltage reference circ...
Objective: Consider a simple constant-voltage reference circuit and design the value of resistance required to limit the current in this circuit.
Consider the circuit shown in Figure 1.45. Assume that the Zener diode breakdown voltage is V_{Z} = 5.6 V and the Zener resistance is r_{z} = 0. The current in the diode is to be limited to 3 mA.
As before, we can determine the current from the voltage difference across R divided by the resistance. That is,
I = \frac{V_{P S} − V_{Z}}{R}
The resistance is then
R = \frac{V_{P S} − V_{Z}}{I} = \frac{10 − 5.6}{3} = 1.47 k \Omega
The power dissipated in the Zener diode is
P_{Z} = I_{Z}V_{Z} = (3)(5.6) = 16.8 mW
The Zener diode must be able to dissipate 16.8 mW of power without being
damaged.
Comment: The resistance external to the Zener diode limits the current when the diode is operating in the breakdown region. In the circuit shown in the figure,the output voltage will remain constant at 5.6 V, even though the power supply voltage and the resistance may change over a limited range. Hence, this circuit provides a constant output voltage. We will see further applications of the Zener diode in the next chapter.