Final answer:
Using Ohm's law and power equations with the diode specifications, the maximum voltage Vs that can be applied without exceeding the zener diode's maximum power rating of 40 mW is 16.3 V.
Step-by-step explanation:
To find the maximum voltage Vs that can be applied to a zener diode without exceeding its maximum power rating of 40 mW, we must use the given zener diode characteristics and apply Ohm's law and power equations.
First, we must find the maximum current that the zener diode can safely conduct without exceeding its power rating of 40 mW.
The maximum power (P) rating of the diode is given by:
P = VZ × IZ,max
Where VZ is the zener voltage and IZ,max is the maximum zener current. Substituting the known values:
40 mW = 3 V × IZ,max
IZ,max = 40 mW / 3 V = 13.3 mA
Next, we calculate the total voltage drop across the series resistor (RS) and the zener diode when IZ,max is flowing through the circuit:
VS = VR + VZ
VR = IZ,max x RS = 13.3 mA × 1 kΩ
VR = 13.3 V
Now, we add the zener voltage to the voltage drop across the series resistor:
VS = VR + 3 V = 13.3 V + 3 V
VS = 16.3 V
Thus, the maximum voltage that can be applied without exceeding the zener's maximum power rating is 16.3 V.