A Zener diode having a breakdown voltage of 12 V and a maximum power dissipation rating of 0.6 W is connected in series with a 100 resistance. The maximum unregulated DC supply voltage that can be applied across this combination without damaging the diode is ______ V.

Step-by-step solution

Given Zener breakdown voltage V_Z = 12 V and maximum power dissipation P_ max = 0.6 W. The maximum safe current that can flow through the Zener diode is: I_ max = P_ max V_Z = 0.6 12 = 0.05 A. The voltage drop across the series resistor R = 100 when maximum current flows is: V_R = I_ max R = 0.05 100 = 5 V. The maximum safe supply voltage V_S is the sum of the voltage drop across the resistor and the Zener voltage: V_S = V_Z + V_R = 12 + 5 = 17 V. Answer: 17